CSU Math ELM Test Study Guide. Ch 1. ELM Test  Numbers and Data: Basic Arithmetic Calculations... Pg.9


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1 CSU Math ELM Test Study Guide Ch 1. ELM Test  Numbers and Data: Basic Arithmetic Calculations... Pg.9 Lesson 1  What are the Different Types of Numbers? Lesson 2  Arithmetic with Whole NumbersTake Quiz Lesson 3  Arithmetic Calculations with Signed Numbers Lesson 4  What Is The Order of Operations in Math?  Definition & Examples Lesson 5  How to Find the Prime Factorization of a Number Lesson 6  How to Find the Greatest Common Factor Lesson 7  How to Find the Least Common Multiple Ch 2. ELM Test  Numbers and Data: Rational Numbers... Pg.31 Lesson 1  How to Build and Reduce Fractions Lesson 2  How to Find Least Common Denominators Lesson 3  Comparing and Ordering Fractions Lesson 4  Changing Between Improper Fraction and Mixed Number Form Lesson 5  How to Add and Subtract Like Fractions and Mixed Numbers Lesson 6  How to Add and Subtract Unlike Fractions and Mixed Numbers Lesson 7  Multiplying Fractions and Mixed Numbers Lesson 8  Dividing Fractions and Mixed Numbers Lesson 9  Practice with Fraction and Mixed Number Arithmetic Lesson 10  Estimation Problems using Fractions Lesson 11  Solving Problems using Fractions and Mixed Numbers Lesson 12  How to Solve Complex Fractions Lesson 13  Calculations with Ratios and Proportions Ch 3. ELM Test  Numbers and Data: Decimals and Percents... Pg.65 Lesson 1  What is a Decimal Place Value? Lesson 2  Comparing and Ordering Decimals
2 Lesson 3  Arithmetic with Decimal Numbers Lesson 4  Solving Problems Using Decimal Numbers Lesson 5  What is a Percent?  Definition & Examples Lesson 6  Changing Between Decimals and Percents Lesson 7  Solve Problems Using Percents Lesson 8  Changing Between Decimals and Fractions Ch 4. ELM Test  Numbers and Data: Irrational Numbers... Pg.90 Lesson 1  Evaluating Square Roots of Perfect Squares Lesson 2  Estimating Square Roots Lesson 3  Simplifying Square Roots When not a Perfect Square Lesson 4  Simplifying Expressions Containing Square Roots Ch 5. ELM Test  Numbers and Data: Data & Statistics... Pg.102 Lesson 1  Statistical Analysis with Categorical Data Lesson 2  Understanding Bar Graphs and Pie Charts Lesson 3  Summarizing Categorical Data using Tables Lesson 4  How to Calculate Percent Increase with Relative & Cumulative Frequency Tables Lesson 5  What is a TwoWay Table? Lesson 6  Make Estimates and Predictions from Categorical Data Lesson 7  What is Quantitative Data?  Definition & Examples Lesson 8  What is a Histogram in Math?  Definition & Examples Lesson 9  What are Center, Shape, and Spread? Lesson 10  How to Calculate Mean, Median, Mode & Rang Lesson 11  Describing the Relationship between Two Quantitative Variables Lesson 12  Reading and Interpreting Line Graphs Lesson 13  Making Estimates and Predictions using Quantitative Data Ch 6. ELM Test  Algebra: Basic Expressions... Pg.136 Lesson 1  What is a Variable in Algebra? Lesson 2  Expressing Relationships as Algebraic Expressions Lesson 3  Evaluating Simple Algebraic Expressions Lesson 4  The Commutative and Associative Properties and Algebraic Expressions Lesson 5  The Distributive Property and Algebraic Expressions Lesson 6  Combining Like Terms in Algebraic Expressions
3 Lesson 7  Practice Simplifying Algebraic Expressions Lesson 8  Negative Signs and Simplifying Algebraic Expressions Ch 7. ELM Test  Algebra: Exponents... Pg.158 Lesson 1  What Are the Five Main Exponent Properties? Lesson 2  How to Define a Zero and Negative Exponent Lesson 3  How to Simplify Expressions with Exponents Lesson 4  Rational Exponents Lesson 5  Simplifying Expressions with Rational Exponents Ch 8. ELM Test  Algebra: Linear Equations & Inequalities... Pg.174 Lesson 1  What is a Linear Equation? Lesson 2  How to Write a Linear Equation Lesson 3  Problem solving using Linear Equations Lesson 4  Solving Linear Equations with Literal Coefficients Lesson 5  Solving Linear Equations: Practice Problems Lesson 6  What is a System of Equations? Lesson 7  How Do I Use a System of Equations? Lesson 8  Solving a System of Equations with Two Unknowns Lesson 9  Solving Problems involving Systems of Equations Lesson 10  What is an Inequality? Lesson 11  Solving Linear Inequalities: Practice Problems Ch 9. ELM Test  Algebra: Absolute Value Equations & Inequalities... Pg.212 Lesson 1  What is an Absolute Value? Lesson 2  How to Evaluate Absolute Value Expressions Lesson 3  How to Solve an Absolute Value Equation Lesson 4  Solving Absolute Value Practice Problems Lesson 5  How to Solve and Graph an Absolute Value Inequality Lesson 6  Solving and Graphing Absolute Value Inequalities: Practice Problems Ch 10. ELM Test  Algebra: Polynomials... Pg.232 Lesson 1  What are Polynomials, Binomials, and Quadratics? Lesson 2  How to Add, Subtract and Multiply Polynomials Lesson 3  Multiplying Binomials Using FOIL and the Area Method
4 Lesson 4  Multiplying Binomials Using FOIL & the Area Method: Practice Problems Lesson 5  How to Factor Quadratic Equations: FOIL in Reverse Lesson 6  Factoring Quadratic Equations: Polynomial Problems with a Non1 Leading Coefficient Lesson 7  How to Divide Polynomials with Long Division Lesson 8  How to Use Synthetic Division to Divide Polynomials Lesson 9  Dividing Polynomials with Long and Synthetic Division: Practice Problems Lesson 10  How to Solve a Quadratic Equation by Factoring Lesson 11  How to Solve Quadratics That Are Not in Standard Form Lesson 12  How to Complete the Square Lesson 13  Completing the Square Practice Problems Lesson 14  How to Use the Quadratic Formula to Solve a Quadratic Equation Lesson 15  Using the Quadratic Formula to Solve Equations with Literal Coefficients Lesson 16  Solving Problems using the Quadratic Formula Ch 11. ELM Test  Algebra: Rational Expressions... Pg.292 Lesson 1  How to Add and Subtract Rational Expressions Lesson 2  Practice Adding and Subtracting Rational Expressions Lesson 3  How to Multiply and Divide Rational Expressions Lesson 4  Multiplying and Dividing Rational Expressions: Practice Problems Lesson 5  How to Solve a Rational Equation Lesson 6  Rational Equations: Practice Problems Lesson 7  Solving Rational Equations with Literal Coefficients Lesson 8  Solving Problems Using Rational Equations Lesson 9  Finding Constant and Average Rates Lesson 10  Solving Problems Using Rates Ch 12. ELM Test  Geometry: Perimeter, Area & Volume... Pg.329 Lesson 1  Perimeter of Triangles and Rectangles Lesson 2  Perimeter of Quadrilaterals and Irregular or Combined Shapes Lesson 3  Area of Triangles and Rectangles Lesson 4  Area of Complex Figures Lesson 5  Circles: Area and Circumference Lesson 6  Volume of Prisms and Pyramids
5 Lesson 7  Volume of Cylinders, Cones, and Spheres Lesson 8  What is Area in Math?  Definition & Formula Ch 13. ELM Test  Geometry: Properties of Objects... Pg. 353 Lesson 1  Properties of Shapes: Rectangles, Squares and Rhombuses Lesson 2  Properties of Shapes: Quadrilaterals, Parallelograms, Trapezoids, Polygons Lesson 3  How to Identify Similar Triangles Lesson 4  Applications of Similar Triangles Lesson 5  Properties of Congruent and Similar Shapes Lesson 6  Parallel, Perpendicular and Transverse Lines Lesson 7  Types of Angles: Vertical, Corresponding, Alternate Interior & Others Lesson 8  Angles and Triangles: Practice Problems Lesson 9  The Pythagorean Theorem: Practice and Application Lesson 10  Applying Scale Factors to Perimeter, Area, and Volume of Similar Figures Ch 14. ELM Test  Geometry: Graphing Basics... Pg. 382 Lesson 1  What Is a Number Line? Lesson 2  What Are the Different Parts of a Graph? Lesson 3  Plotting Points on the Coordinate Plane Ch 15. ELM Test  Geometry: Graphing Functions... Pg.390 Lesson 1  Graph Functions by Plotting Points Lesson 2  Identify Where a Function is Linear, Increasing or Decreasing, Positive or Negative Lesson 3  Linear Equations: Intercepts, Standard Form and Graphing Lesson 4  How to Find and Apply The Slope of a Line Lesson 5  How to Find and Apply the Intercepts of a Line Lesson 6  Graphing Undefined Slope, Zero Slope and More Lesson 7  Equation of a Line Using PointSlope Formula Lesson 8  How to Use The Distance Formula Lesson 9  How to Use The Midpoint Formula Lesson 10  What is a Parabola? Lesson 11  Parabolas in Standard, Intercept, and Vertex Form Lesson 12  How to Graph Cubics, Quartics, Quintics and Beyond
6 How to Prepare for the ELM Most students learn the math covered on the ELM in high school but may be rusty on some of the concepts. We offer a full collection of video lessons that will help you review or learn all of the math included on the exam. If you want a complete review, start at the beginning and take the quiz associated with each lesson to ensure you've mastered the material. You can also jump to specific chapters if you just need to review some of the material. Here is a breakdown of the material covered in this course: Basic Arithmetic Calculations Review the rules of different types of numbers, including whole numbers and signed numbers. Practice factoring and learn how to find the greatest common factor and least common multiple. Rational Numbers Review the rules of factors and practice with like, unlike, and complex fractions as well as mixed numbers. This chapter also reviews ratios and proportions. Decimals and Percents Learn basics of decimals and percents, including arithmetic and ordering. Practice converting decimals to percents or fractions. Irrational Numbers Work with square roots, including evaluating, estimating, and simplifying square roots. Learn how to simplify complex expressions that contain square roots, and complete square root practice problems. Data & Statistics Practice reading bar graphs, pie charts, and representations of data. You'll also learn how to work with mean, median, and mode and use data to make predictions.
7 Basic Expressions in Algebra Learn properties associated with algebraic expressions, including the commutative, associative, and distributive properties. This chapter also includes lessons on simplifying expressions and combining like terms. Exponents Use the five main properties of exponents to simplify expressions. Learn how to work with zero and negative exponents. Linear Equations & Inequalities Practice solving linear equations with one and two variables. Learn a simple stepbystep process for solving linear equations and inequalities. Absolute Value Equations and Inequalities Learn the rules of absolute value and applying them when solving and graphing absolute value equations. This chapter also teaches you how to solve and graph inequalities with absolute value. Polynomials Learn about polynomials, including quadratic equations. This chapter includes FOIL, synthetic division, and practice problems for the quadratic formula. Rational Expressions Practice adding, subtracting, multiplying, and dividing rational expressions and then review how to solve a rational equation. Apply rational expressions to the real world by solving problems using constant and average rates. Perimeter, Area & Volume Find the perimeter and area of basic shapes, including triangles and rectangles. Also practice finding area and circumference of circles. Learn how to find the volume of prisms, cylinders, and other 3dimensional shapes. Properties of Objects Review the properties of shapes such as circles and quadrilaterals and then dig deep into triangles, angles, and lines. Finally, use the Pythagorean theorem to find the area of triangles. Graphing Basics
8 Learn how to read a graph or number line and practice plotting points on a coordinate plane. Graphing Functions Practice graphing functions and working with slope. Learn the different forms of linear equations, including standard form and pointslope formula. Then, learn about graphing parabolas and the different forms of quadratic functions. Practice We provide practice questions along with each lesson. You can also complete ELM sample questions provided by CSU.
9 What are the Different Types of Numbers? Chapter 1 / Lesson 1 Natural Numbers This is a lesson on the different classifications of numbers, which is basically a big word for just saying that numbers are part of families and numbers have different homes depending on what kind of number they are, just like people do. I'll start by talking about myself a little bit. A lot of the time when I meet someone new, they ask me where I live. You don't give them your specific address. I don't tell them 202 Calvert Drive; that doesn't really mean much. But I do tell them that I live in Cupertino. That's probably the most specific name I use to describe where I live. The number 0 is a whole, but not natural, number On the numbers side of things, the most basic number is the number 1, and the most specific descriptor of where it lives is called the natural numbers. The natural numbers are all the numbers that you learn when you're a baby, like 1, 2, 3, 4, 5, 6 and on and on. The natural numbers are also sometimes call the counting numbers because they're the first numbers you learn how to count. Whole Numbers Back to me real fast, although I do live in Cupertino, Cupertino is part of Santa Clara County. This means that since I'm a resident of Cupertino, I'm also a resident of Santa Clara County. But the reverse isn't necessarily true, because not everyone that lives in Santa Clara County lives in Cupertino, so you have to be careful going back and forth.
10 Numbers are the exact same way. If we get a tiny bit less specific, what we come to are called the whole numbers. All of the natural numbers are part of the whole numbers, just like all the people who live in Cupertino also live in Santa Clara County. But there is one whole number that is not a natural number. That is the number 0. So when we're talking about the whole numbers, we're talking about the numbers that start with 0 and starting going up 1, 2, 3 and on and on. Integers Taking a step further back in my situation, Cupertino and Santa Clara County are both parts of California. Again, not everyone that lives in California lives in Santa Clara or, even more specifically, Cupertino. But if you live in Cupertino or Santa Clara, you're definitely living in California. Taking a step out in terms of the numbers brings us to what are called the integers. Again, not all of the integers are whole numbers and natural numbers. But all of the whole numbers and natural numbers are integers. The integers now also add in the negatives: 1, 2, 3 and on and on in the negative direction as well. Diagram showing the different types of rational numbers Rational Numbers Taking another step back with me, I think most people are aware that California is a state that is part of the U.S. So I am also a resident of the United States. Again, everyone that is a resident of California is also a resident of the U.S. But not everyone that lives in the U.S. lives in California. That trend continues with the numbers. As we take another step back, we come to what are called the rational numbers. The new additions to the club are fractions. This means we could have things like 1/2 or 1/3 or 3/4 or maybe 11/7.
11 The list becomes a little bit harder to write but, again, you could imagine that there are a lot of different kinds of numbers in here. It's still true that all of the previous numbers we mentioned are rationals, but not all of the rationals, specifically these fractions, are part of the numbers in the integers, whole numbers or natural numbers. Irrational Numbers Now that we've taken where I live all the way out to the fact that I live in America, we probably shouldn't leave out the people that live in different countries. But they're in a different group. For people that live in Japan or France, for example, you're not living in America, so you're part of a separate group. The irrational number pi has no end There's a similar thing going on with the numbers. There's a separate group of numbers that don't fit with the rest of these. Those are called the irrational numbers. Irrational numbers are all the numbers that can't be described as fractions. There are a lot of different numbers like this. One of the ones you might have heard of is called pi, which is and that's as much as I know, but it does go on forever, and there is no way you can represent pi with a fraction to make it the exact right decimal. Another is the square root of 2, which is something like 1.4, but again, the decimal goes on forever and there's no fraction that can represent it. These irrational numbers are their own group. They are not part of the rationals, or naturals or whole numbers; they are not related, but separate. Real Numbers I don't like ending the story on separatism. Regardless of what country you live in, we're all residents of the planet Earth. Numbers are the same way. They're happy and
12 they like to be together. They all live under the banner of what are called the real numbers. The irrational and the rational come together, and with them combined, they form the real numbers. Lesson Summary To review, the most specific set of numbers are the natural numbers and they are 1, 2, 3 and so on and so forth. The next step out is the whole numbers, and you simply add 0 to the group; everything else stays the same. When we come to the integers, we add the negatives: 3, 2, 1, 0, 1, 2, 3. Next come the rational numbers, and the rational numbers give us the fractions. The irrational are separate to this and they are the numbers that can't be fractions, like pi and the square root of 2. The real numbers bring the rational and the irrational together in one big set. Lesson 2 Arithmetic with Whole Number Four Operations There are four operations that can be performed when working with Whole Numbers. These four operations are Addition, Subtraction, Multiplication, and Division. We use the operations in everyday life for a multitude of tasks. Addition The most basic of these four operations is Addition. Addition is the operation that involves calculating the total amount of a represented group. To add a set of numbers together, we first need to line them up vertically by place value. Then, add each column of numbers from right to left. If a column has a sum greater than 9, we will carry the tens place to the next column. Let's take a visit to the local mall to visit my good friends Val and Keri. They own a local cookie shop where they are busy getting ready for their customers.
13 Val and Keri have several orders placed for today and need to know how many cookies they should bake. They have orders of 116 chocolate chip cookies, 524 sugar cookies, and 75 snicker doodle cookies. Val asks Keri to calculate how many cookies they must bake today. Keri knows that to start adding these numbers together she must line them up vertically by place value. So, Keri writes down 524, 116, and 75 lined up on the order form. Keri begins by adding the column on the right equals 15. She knows that since her value is greater than 9 she must carry the tens place to the next column. So, she writes down the 5 and carries the 1 to the next column. To add the next column, Keri will need to add the 1 that we carried. The sum of these numbers is 11. Again, Keri must carry the tens value since the total is greater than 9. Next, she needs to add the last column the carried 1 equals 7. Keri now knows that they need to bake 715 cookies to fill all of their customer orders. Subtraction The next operation is Subtraction. Subtraction is the operation that involves taking away value from the total amount. To subtract whole numbers, we will need to line the numbers up by place value from right to left. Next, we will subtract each place value. Occasionally, the value that we are subtracting is not large enough to take away the value. When this happens, we must borrow value from the next place value. When you borrow from the next place value, you are borrowing a set of ten. This set of ten is added to the existing value, and subtraction is continued. Let's check back in on Val and Keri to see how they are doing on the cookie order. Val and Keri have cooked all of the chocolate chip and snicker doodle cookies but still need to cook the sugar cookies. As they are preparing to start the sugar cookies, the phone rings. Val answers and it's their customer. She tells Val that she needs to reduce her order for sugar cookies. She had originally ordered 524 sugar cookies but needs to subtract 134 of them. Val now needs to subtract the customer's original order of 524 sugar cookies by 134. To do this, he must first line up the number from right to left by place value. To begin, Val starts with the values on the right. 4 minus 4 is 0.
14 Next, Val needs to subtract 2 minus 3. However, he knows that you cannot take 3 from 2 so he must borrow. So, he borrows a set of ten from the next place value. As he borrows a set from the 5, its value becomes a 4. The set of ten also increases the value of the 2 to 12. Val can now subtract, 12 minus 3 is 9 and 4 minus 1 is 3. Val now knows that he only needs to cook 390 sugar cookies to fill the customer's order. Multiplication Multiplication is another one of the four operations. Multiplication is the process of adding multiple sets of the same number. To multiply, we will take each place value and multiply it to every value in the other number. If the number being multiplied has more than one value, you must add a zero as a place holder each time you move to the next place value. This process continues until all of the places have been multiplied. Back in the cookie shop, Val and Keri have finished their daily order. As they talk, Keri wonders if they cooked this many cookies daily, how many cookies would they cook in 126 days. To do this, she will need to multiply the daily cookie order of 715 by 126 days. To start this problem, Keri needs to multiply the 6 by each of the values on top from right to left. If a value is greater than 9, the tens place will be carried to the next place and added to the multiplied value. Six times 5 is 30, so we keep the zero and carry the 3. Next, we multiply 6 times 1 which is 6 and then add the carried three to equal 9. The value 9 is written below. Then we multiply the 6 times the 7 which is 42. Since there are no additional places on top to be multiplied, we write the 42 below. Next, we must multiply each place on top by the 2. Since we moved one place on the bottom, we will need to add a zero to the next line below. As we multiply the 2 to each value on top we get As we move to the next place to multiply, we must now add two zeros to the bottom because we have moved over two place values. Next we need to multiply the 1 by each value on the top, which gives us
15 The last step to multiplying is to add these values together from right to left. After adding these values, the product is 90,090. Keri knows that if they sell 715 cookies for 126 days they will sell 90,090 cookies. Division The last of the four operations is Division. Division is the process of taking a total value and dividing it into equal parts. The first step is to set up the problem. The value that is being divided is called the dividend and the value that we are dividing into is called the divisor. To set up a division problem you will use the format dividend divided by divisor. When dividing, we take the divisor and divide it into each value of the dividend from left to right. The number of times that the value will divide into the dividend goes on top. The value is then multiplied and subtracted from the dividend. The process continues until all of the places in the dividend have been divided. If there is a value that remains at the end, this is called your remainder. The wonderful smell of cookies has now attracted customers to the cookie shop. Val and Keri are glad to see such a huge crowd. There are 12 customers that have gathered at the cookie counter. Val and Keri know that they have cooked only 715 cookies. Val wants to know how many cookies each of the 12 customers order. In this problem, 715 is the dividend and 12 is the divisor. The problem would be set up as 715 divided by 12. The first step is to take the 12 and see which value it will divide into. 12 will not divide into 7 so we must use the first two values 71. Twelve will go into 71 five times. So, we write the 5 on top. Next, we need to multiply the 5 times 12 which is 60. This value gets written below the 71 in the dividend. Now, we must subtract 71 minus 60, which is 11. We must also bring down the next value from the dividend. The new dividend is 115. Next, we must see how many times 12 will divide into 115, which is 9. This 9 gets written on top. At this point we repeat the same process as in the previous step. We need to multiply 9 times the 12 which is 108, which will be written below the 115.
16 Subtract 108 from 115, which is 7. Since we are out of values in the dividend, 7 is our remainder. Val can now see that he can sell each customer 59 cookies and he will have 7 cookies leftover. Lesson Summary There are four operations that can be performed when working with Whole Numbers. These four operations are Addition, Subtraction, Multiplication, and Division. The most basic of these four operations is Addition. Addition is the operation that involves calculating the total amount of a represented group. The next operation is Subtraction. Subtraction is the operation that involves taking away value from the total amount. Multiplication is another one of the four operations. Multiplication is the process of adding multiple sets of the same number. The last of the four operations is Division. Division is the process of taking a total value and dividing it into equal parts. Lesson 3 Arithmetic Calculations with Signed Numbers Signed numbers are often referred to as integers. Integers included both positive and negative numbers. In this lesson, you will learn how to add, subtract, multiply, and divided integers. Signed Numbers Signed numbers are also referred to as integers. Integers are the set of whole numbers and their opposites. The set of integers would include values 3, 2, 1, 0, 1, 2, 3 etc. Sam's friends decided to surprise him with a scavenger hunt for his birthday. His friends started the scavenger hunt from the big oak tree at the park. Sam knows that his journey will begin at the oak tree. So the oak tree will represent zero on a number line.
17 Adding Integers Adding integers is the process for adding both positive and negative numbers. When adding integers, if the signs of the values are the same, you will add the two values. If the signs are different, subtract the two values. You will always keep the sign of the largest value. Back at the oak tree, Sam finds a note that tells him to take 15 steps forward and 26 steps backwards. Sam has decided to just add these two values. 15 steps forward would be positive 15 and 26 steps backwards would be 26. So Sam decides to add Looking at these two values, Sam sees that they have different signs. This means that he will subtract the two values. 26 minus 15 would equal 11. Since 26 is the largest value and it's negative, the 11 would also be negative. So Sam knows that the answer to this problem is negative 11. This means that he will need to take 11 steps backward from the tree. Subtracting Integers Subtracting integers is actually the process of adding the opposite of the stated value. To work a subtracting integer problem, you must first change the problem by changing the operation to addition and the sign of the last number to its opposite. Let's check back with Sam as he continues his scavenger hunt. Sam is now 11 paces back from the tree where he started. This number would be represented by the integer 11. On the ground, Sam finds a fortune cookie. Inside the fortune cookie, the directions tell Sam to subtract 8 paces from his current location. Sam knows that he needs to subtract 11 minus 8. Sam knows that in order to start the subtraction problem that he must change the problem to adding its opposite. To do so, Sam will change the subtraction sign to addition and the positive 8 to a negative 8. Now Sam has an adding integer problem. Since the signs are the same, he will add the two values. 11 plus 8 equals 19. The sign of the 19 would be negative since the larger value, 11, was also negative. Sam knows that his next clue will be 19 paces backwards from the tree.
18 Multiplying and Dividing Integers When multiplying and dividing integers, you will need to multiply and divide as normal. The sign of your answer is determined by the sign of the values that were multiplied and divided. If the signs of the two values were the same, your answer will be positive. If the signs of the two values are different, your answer will be negative. Sam feels that the end of the scavenger hunt is getting near. As he looks up, he sees a balloon floating above him. Sam takes the balloon and pops it. Inside he finds his final clue. The clue states that he needs to take his current location and multiply by negative 3. Sam now knows that he needs to multiply negative 19 times negative times 3 equals 57. Because the signs of the 19 and 3 are the same, negative, the answer to this problem will be positive. Sam knows that the prize for the scavenger hunt can be found 57 paces forward from the tree. To reach the destination of positive 57, Sam must return to the oak tree, which represented zero. Sam now needs to go 57 paces past the oak tree in order to reach the clue. Sam sprints past the oak tree 57 paces to a set of bushes. Behind the bushes, Sam finds all of his friends hiding and waiting to wish him happy birthday. Lesson Summary Let's recap these four operations that we can perform with signed numbers. Adding integers is the process for adding both positive and negative numbers. When adding integers, if the signs of the values are the same, you will add the two values. If the signs of the values are different, subtract the two values. You will always keep the sign of the largest value. Subtracting integers is actually the process of adding the opposite of the stated value. To work a subtracting integer problem, you must first change the problem by changing the operation to addition and the sign of the last number to its opposite. When multiplying and dividing integers, you will need to multiply and divide as normal. The sign of your answer will be determined by the sign of the values that were being multiplied and divided. If the signs of the two values were the same, your answer will be positive. If the signs of the two values are different, your answer will be negative.
19 Lesson 4 What Is The Order of Operations in Math?  Definition & Examples The order of operations is the steps used to simplify any mathematical expression. In this video, learn how to solve problems using these steps and easy tricks to remember them. What Is the Order of Operations? Hello! My name is Bob and I live with my wonderful Aunt Sally. My Aunt Sally does a great job of raising me. She fixes my meals, cleans the house and tucks me in at night. However, Aunt Sally often gets the order of these events mixed up. For example, yesterday my Aunt Sally cleaned the house, tucked me into bed and then made my meals. I keep telling my Aunt Sally that order is important, but she doesn't seem to understand. Just like in math, there is a particular order that we work problems. Without this order, it is possible that we could all get different answers. The order that we use to simplify expressions in math is called the order of operations. The order of operations is the order in which we add, subtract, multiply or divide to solve a problem. Order of Operations Steps The steps we use to solve any mathematical expression are: 1. Simplify all of the parentheses. This includes all forms of grouping symbols, such as brackets and braces, in addition to parentheses. 2. Simplify all exponents. 3. Simplify all multiplication and division from left to right. When simplifying the multiplication and division, work from left to right. 4. Simplify all addition and subtraction from left to right. Again, when simplifying the addition and subtraction, work from left to right. By following this order, we can all solve the problem and get the same solution.
20 PEMDAS After explaining all of those rules to my Aunt Sally, she seems a little overwhelmed. However, I do have a shortcut to help her remember these steps. It's called PEMDAS. It stands for: P  Parenthesis E Exponents M Multiplication D  Division A  Addition S  Subtraction Remember that the steps for multiplication and division is one step. We work all of the multiplication and division in one step from left to right. Multiplication does not always come before division; they are worked in the order that they appear. This is also true for addition and subtraction. They are worked in the same step from left to right. An easy way for me to remember these steps is to remember the phrase Please Excuse My Dear Aunt Sally, where the: P  Parenthesis  Please E  Exponents  Excuse M  Multiplication  My D  Division  Dear A  Addition  Aunt S  Subtraction  Sally Example Let's show Aunt Sally how the order of operations helps us to solve problems. I want to show Aunt Sally a problem from my homework tonight. The problem is 3 +[6 ( )] 8 x 2. The problem is x x 2. Remember, to work this problem, we will follow the order of operations. Let's think PEMDAS.
21 The first step to solve this problem is to work the P (Parentheses). In this problem, they used both parentheses and brackets. We will need to start inside the parentheses and work out until we complete all of the grouping symbols. Also, when working inside the grouping symbols, we must follow the remaining order. To begin, we will need to add the and then subtract 4, which is 8. We still must now work inside the bracket, 6 times 8 is 48. The next letter in our acronym is E for Exponents. Since there are no exponents, we continue on. The next step is to simplify the M and the D (Multiplication and Division) from left to right. Since division actually comes first, we work it from left to right. We'll first divide 48 8, which is 6. There is still Multiplication in our problem, so next we will need to multiply 6 times 2, which equals 12. The only step remaining is AS (Addition and Subtraction). There is only one thing left in this problem, which is 3 plus 12, which equals 15. So, as you can see Aunt Sally, the answer to this problem would be 15. Lesson Summary The order of operations is the order in which we add, subtract, multiply or divide to solve a problem. The steps that we use to solve any mathematical expression are: 1. Simplify all of the parentheses. This includes all forms of the grouping symbols such as brackets, braces and parentheses. 2. Simplify all exponents. 3. Simplify all multiplication and division from left to right. When simplifying the multiplication and division, make sure you work from left to right. 4. Simplify all addition and subtraction from left to right. Again, when simplifying the addition and subtraction, work from left to right. A helpful way for me to remember this order is Please Excuse My Dear Aunt Sally, which stands for: P  Parentheses E Exponents M Multiplication D  Division A  Addition S  Subtraction
22 Remember that multiplication and division are included in the same step and simplified from left to right. This is also the same for addition and subtraction. Lesson 5 How to Find the Prime Factorization of a Number The prime factorization of a number involves breaking that number down to it's smallest parts. This lesson will show you two different ways to discover the prime factorization of any number. Factors of a Number When you are trying to come to a conclusion about a problem, you often say that there are many 'factors' to consider. This means that there are many parts that make up the whole problem of what you are trying to decide. If the decision is where to go for dinner, the factors involved in that decision might be price, how far away the restaurant is, and how well you will enjoy the food. Numbers also have factors, the parts that make up the whole number. The factors of a number are the numbers that, when multiplied together, make up the original number. For example, factors of 8 could be 2 and 4 because 2 * 4 is 8. And factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24, because 1 * 24 is 24, 2 * 12 is 24, 3 * 8 is 24 and so is 4 * 6. So all of these numbers are said to be factors of 24. Prime Numbers A prime number is any number that is only divisible by itself and 1. Some examples of prime numbers include 2, 5 and 17. Numbers such as 15 or 21 are not prime, because they are divisible by more than just themselves and 1. Prime Factorization To factor a number is to break that number down into smaller parts. To find the prime factorization of a number, you need to break that number down to its prime factors.
23 How to Determine the Prime Factorization of a Number There are two main ways for determining the prime factors of a number. I will demonstrate both methods, and let you decide which you like best. Both methods start out with a factor tree. A factor tree is a diagram that is used to break down a number into its factors until all the numbers left are prime. The first way you can use a factor tree to find the factorization of a number is to divide out prime numbers only. Let's factor 24 using this method. Since 24 is an even number, the first prime number that can be factored out is a 2. This leaves us with 2 * 12. Again, 12 is an even number, so we can factor out another 2, leaving us with 2 * 2 * 6. Since 6 is even, we can factor out a third two, leaving 2 * 2 * 2 * 3. All of these numbers are prime, so the factorization is complete. The other method for using a factor tree to find the prime factorization of a number is just to pull out the first factors that you see, whether they are prime or not. Looking back at our example from above, let's factor 24 again using this method.
24 The first thing you might notice is that 6 * 4 is 24, so that is one set of factors for 24. Since neither of these numbers are prime, we can continue to factor both of them. 6 can be broken down to 2 * 3, and 4 can be broken down to 2 * 2. Now all of our factors are prime, and the factorization of 24 is complete, again giving the answer of 2 * 2 * 2 * 3. Both of these methods work equally well, and can be used interchangeably. There are people who like to use certain tricks to pull out prime numbers first without having to decide what other numbers might be factors of the original number. The tricks to find some of the prime numbers are: 1. Any even number is divisible by If you add up the digits in a large number and the sum you get is divisible by 3, the number is also divisible by A number that ends with a 5 or 0 is divisible by 5. These little tricks can help you factor larger numbers where it might not be easily apparent where to start. Let's try another example. Find the prime factors of 117. The first thing I notice about this number is that if you add the digits (1+1+7), you get 9. This means that the number is divisible by 3. Since it is not even, and does not end with a 5 or 0, it is not divisible by 2 or 5, so we can start with the 3.
25 117 divided by 3 is 39, so our first two factors are 3 and is also divisible by three because = 12 39/3 equals is a prime number, so our factorization is complete, and the factors of 117 are 3 * 3 * 13. Lesson Summary The prime factors of a number are all the prime numbers that, when multiplied together, equal the original number. You can find the prime factorization of a number by using a factor tree and dividing the number into smaller parts. You can begin by finding a prime number and factoring out that number, then continuing on in that manner. Or, you can just divide the number into any two numbers, even if they aren't prime and continue on from there until every number is prime. Lesson 6 How to Find the Greatest Common Factor If the factors of a number are the different numbers that you can multiply together to get that original number, then the greatest common factor of two numbers is just the biggest one that both have in common. See some examples of what I'm talking about
26 here! Introduction The greatest common factor is usually used when simplifying fractions, and it's one of those topics that you learn pretty early on in your education but can easily forget or mistake for a different math idea, mainly the least common multiple. But before we can talk about the greatest common factor, often written as the GCF, we first have to know what a plain old regular factor is. What Is a Factor? Simply put, the factors of a number are the smaller numbers that make up that original one. Saying that slightly more mathematically sounds like this: The factors of a number are the different numbers that you can multiply together to get that original number. But a lot of math topics are best shown with examples, and this is probably one of them. Let's start by looking at the factors of 6. The factors of 6 are going to be 2 and 3, because 2 x 3 = 6. It's also true that 1 and 6 are factors, then, because 1 x 6 is also equal to 6. That gives us our full list for the factors of 6 as 1, 2, 3, and 6. How about the factors of 60? Well, I know I can always do 1 times the number, so that works for this. Also, 60 is even, so I know 2 works, and 2 x 30 = 60. If we try dividing 60 by 3, we get 20, so that means I can add 3 and 20 to the list. If we continue on up, we find that 4 and 15, 5 and 12, 6 and 10 work  but the next one that works is 10 and 6, and this is basically just a repeat of 6 and 10. So at this point we can stop, because the rest of the multiplication problems we do are just all repeats of numbers we've already got on our list. That makes the factors of 60 all the different numbers you see here: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. Quite a lot of them. The Greatest Common Factor So now we're ready to talk about what this lesson is really on: the greatest common factor, or GCF. In order to find a GCF, we need to be looking at two numbers, say 10 and 22. Then we simply ask ourselves, of the factors these two numbers have, which one's the biggest that they have in common?
27 Well, factors of 10 are 1 and 10 and 2 and 5, while the factors of 22 are 1 and 22 and 2 and 11. That makes the greatest common factor of 10 and 22 2, because it's the biggest number I see on both the lists. That's it! Examples Let's try a few more examples just to make sure you've got it. Maybe this one: find the GCF of 27 and 45. We'll start by listing out the factors of each of these numbers individually, just like we learned earlier. Looking at 27 first, 1 will always work, so we can start there. 27 is odd, so 2 is not going to work, but we can do 3 x 9. The next one that works is 9 x 3, so you've started repeating and we can stop. That makes our lists for the factors for 27 pretty short  just 1, 3, 9 and 27. Next with 45  after we count 1 and 45, we can again rule 2 out, but 3 and 15 is good, 4 doesn't work, but 5 x 9 does, and the next one is 9 x 5, so we've hit our repeating point. That makes our list of factors of 45 what you see here: 1, 3, 5, 9, 15, 45. So answering the original question, 'What is the GCF of these two numbers?', is as easy as picking out the biggest number that is on both of these lists. Looks like 9 is our winner! Last example: Find the greatest common factor between 4 and 16. We again begin by writing out all the factors of these two numbers. For 4, we get 1, 2 and 4, while for 16 we get 1, 2, 4, 8 and 16. So the greatest common factor of 4 and 16 is the biggest number that's on both lists. That's 4. Notice that 4 was one of the original numbers from the problem. That's totally okay! Some people get a little freaked out that this isn't allowed and decide to go with 2 instead because it's the next one down on the list. Don't do that! It's okay if the GCF is one of the original numbers from the problem. Lesson Review To review: the factors of a number are the different numbers that you can multiply together to get that original number. The greatest common factor of two numbers is the biggest factor that they both have in common and is often written as the GCF for short. It's okay to have the GCF of two numbers be one of the original numbers itself.
28 And finally, as a side note, the place in math where you'll use this the most often is when you're simplifying fractions. Lesson 7 How to Find the Least Common Multiple The least common multiple of two numbers is the smallest number that can be divided evenly by your two original numbers. See some examples of what I'm talking about here! Least Common Multiple The least common multiple is a math topic you usually use when you're trying to find a common denominator between two fractions, and it's one of those things that you learn pretty early on in your education, but it can easily be forgotten or mistaken for a different math idea, usually the greatest common factor. So let's start by reminding you exactly what it is. The least common multiple of two numbers, often written as the LCM, is the smallest number that can be divided evenly by those original two numbers. For example, the LCM of 5 and 6 is 30, because it is the smallest number that both 5 and 6 go into. And that's it  least common multiple. What Is a Multiple? But, in order to not forget what a least common multiple is in the future, it's probably best to understand some of the vocabulary in the name  mainly, what is a multiple? Well, the multiples of a number are just what you get when you multiply that number by 1, then 2, then 3 and so on. For example, the multiples of 2 are 2, 4, 6, 8, 10, 12 and on and on. Or the multiples of 7 are 7, 14, 21, 28, 35 and on and on and on. Notice that 2 is a multiple of 2, and 7 is a multiple of 7. That means that any number is a multiple of itself; we'll need to remember that a little bit later on. Anyways, now that we know this vocabulary, we can say that the least common multiple just means the smallest multiple that two numbers have in common.
29 How to Find the Least Common Multiple The most foolproof way to find a least common multiple is to list out all the multiples of each number and then find the first one they have in common, like I've been showing you so far. LCM of 8 and 6? Well, multiples of 8 are 8, 16, 24, and the multiples of 6: 6, 12, 18, 24  hey! Got it But this way can get kind of annoying, especially when the numbers get bigger, so there are some shortcuts as well. Besides just knowing your times tables really well and being able to do it all in your head, it's often the case that the least common multiple of two numbers is just what you get when you multiply those two numbers together. If we look at the first example we did, the LCM of 5 and 6, the answer was 30, which is exactly what 5 times 6 is. But the second example we did didn't follow this pattern. The least common multiple of 6 and 8 was 24, which is not equal to 6 times 8. Now, the reason this trick works for the first one but not the second one is the fact that the numbers in the first one do not share any factor. Factors are kinda like the opposites of multiples. If multiples are the bigger numbers that 6 go into (6, 12, 18, 24...) factors are the smaller numbers that go into 6 (1, 2, 3, 6). For example, the factors of 15 are 1, 3, 5 and 15, because those are all the numbers that you can divide 15 by and get a nice number out. Or we can say the factors of 22 are 1, 2, 11 and 22. So, back to the trick. I could just write out all the multiples of 11 and 12 and find the first one they have in common, or I could use the shortcut and just do 11 times 12 to get my answer of 132, because I know that 11 and 12 don't have any factors in common except 1. But then, when I do another problem, maybe like the least common multiple of 6 and 20, I can't use the shortcut because 6 and 20 share a factor that isn't 1. In this case, it's 2. That means 6 times 20 is not my answer, and I have to instead just list out all the numbers and find the first one they have in common, which appears to be 60. If you want to get really fancy, you actually could still use the shortcut, but it's a slightly longer shortcut, and after you do 6 times 20, you divide that by whatever factor they have in common, which in this case was 2, so you could end up with the correct answer.
30 LCM of Multiples Before we finish, I should tell you that sometimes the LCM of two numbers can be one of the original numbers itself. Remember how we said earlier that a number is a multiple of itself? Well that means that, for example, if we were doing the LCM of 3 and 9, we'd just get 9. Or if we were doing the LCM of 4 and 16, it's just 16. So if you see this happen in one of your problems, don't worry, totally okay. Lesson Summary Let's quickly review what we've learned. The multiples of a number are the numbers you get when you times that first number by 1, then by 2, and then 3 and 4 and so on. The least common multiple of any two numbers is the smallest multiple that they have in common, and it's also written as the LCM for short. To find the LCM of any two numbers, you can write out a list of their multiples until you find one that's the same. Or there is a shortcut that says as long as they don't have any factors in common, you can just multiply those two numbers together. Finally, as a side note, the place where you'll probably find yourself using the LCM the most often is when you are trying to find a common denominator between two fractions.
31 How to Build and Reduce Fractions Chapter 2 / Lesson 1 Fractions are a fundamental part of everyday life. Fractions represent a comparison of a part to the whole. To reduce fractions, we will divide by the greatest common factor, or the GCF. What Is a Fraction? A fraction is a comparison of the part to the whole. When writing a fraction, we use the form 'part over whole.' The part in a fraction is called the numerator. The whole of the fraction is called the denominator. How to Build a Fraction Now that we know fractions represent a comparison of the part to the whole, let's see how to build a fraction. To build a fraction, we need to see how many whole items would be included in our set. The number of whole items in our set would be placed on the bottom as the denominator. Next, we would need to see how many parts of our set are included. The part of the set would be placed on the top of the fraction and called the numerator. For example, Andrew coaches an area soccer team. There are 12 players on his team. At Tuesday's practice, only 8 of the players were able to attend. Andrew wants to know what fraction of his team made it to practice on Tuesday. Andrew knows that his whole team has 12 players. The number 12 would represent the whole, also known as the denominator, and would be placed on the bottom of our fraction. There were only 8 players who showed up for practice, which is the part of the team. The number 8 would represent the part, also known as the numerator, and
32 would be placed on the top of the fraction. Andrew can now see that only 8/12 of the team arrived at practice on Tuesday. How to Reduce a Fraction A fraction is considered reduced, or in simplest form, when the only value that will divide into the numerator and denominator evenly is 1. To reduce a fraction, divide both the numerator and denominator by the greatest common factor, or the GCF. Fractions can also be reduced by dividing by a common factor multiple times until the only common factor remaining is 1. Let's check back in on Andrew after his team's practice. After practice, Andrew was discussing how only 8/12 of his team showed up for his practice with the other coaches. Andrew wanted to be sure to tell them the fraction in the simplest form. So Andrew started thinking, 'what is the simplest form of the fraction 8/12?' Andrew starts by thinking about his numbers 8 and 12. He knows that both numbers are even, so they would both divide by 2. However, he wants to see if a larger value will divide into both numbers. Andrew realizes that both 8 and 12 will divide by 4, which would be the GCF. To simplify his fraction, he will divide both the numerator and denominator by 4. 8 divided by 4 is 2 and 12 divided by 4 is 3. Andrew can now see that 8/12 in simplest form is 2/3. Andrew can now tell the other coaches that only 2/3 of his team showed up for practice tonight. Lesson Summary So in review, a fraction is a comparison of the part to the whole. When writing a fraction, we use the form 'part over whole.' The part in a fraction is called the numerator. The whole of a fraction is called the denominator. A fraction is considered reduced, or in simplest form, when the only value that will divide into the numerator and denominator evenly is 1. To reduce a fraction, divide both the numerator and denominator by the greatest common factor, or the GCF. Fractions can also be reduced by dividing by a common factor multiple times until the only common factor remaining is 1.
33 How to Find Least Common Denominators Chapter 2 / Lesson 2 Finding a Least Common Denominator is an important tool when working with fractions. To find a Least common denominators, we will use multiples of our numbers. What is a Multiple? When finding the least common denominator, we will be using multiples. Multiples are the products of a given number. For example, the multiples of 2 would be 2, 4, 6, 8, 10, etc We often call the least common denominator the least common multiple, or the LCM. Finding a Least Common Denominator Using Multiples Using these multiples, we can now find a common denominator. The least common denominator is the small common multiple of both denominators. To begin, let's look at two fractions, 2/4 and 6/9. We can see that our denominators are 4 and 9. The first thing we need to do is to list out several multiples of each number. A good amount to start with is 10 of each. The first 10 multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, and 40. The first 10 multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, 81, and 90. Starting with the multiples of the smallest number, 4, we are looking to see if there is a common multiple within the multiples of 9. We can see that both 4 and 9 have the multiple of 36. Since 36 is the smallest common multiple, it is called our least common multiple, or LCM; so the LCM of 4 and 9 is 36. Example Jerry and Tom both collect baseball cards. Jerry puts his cards into pages that contain 16 cards each, and Tom puts his cards into pages that contain 12 cards each. Both Jerry and Tom are curious about how many full pages of cards that they will need to have the same amount of cards.
34 Jerry and Tom know that they must find the least common denominator to see at what number they will have the same number of cards. To begin, Jerry lists the first 10 multiples of 16, which are 16, 32, 48, 64, 80, 96, 112, 128, 144, and 160. Next, Tom lists the first 10 multiples of 12, which are 12, 24, 36, 48, 60, 72, 84, 96, 108, and 120. Tom and Jerry can see that both of their sets contain the numbers 48 and 96. Since 48 is the smallest number, it is called the least common denominator. Example 2 Jerry and Tom now know that they will have the same amount of cards when they both reach full pages containing 48 cards. As the boys finish putting away their cards into the collector's books, they suddenly realize that they are hungry. The boys decided to snack on some chips. Tom chose to eat the Doritos, which he could eat 4 at a time. Jerry chose to eat the Cheetos, which are smaller. He could eat 9 Cheetos at a time. The boys want to know how many handfuls of chips they must eat to eat the same number of chips. To begin, Tom lists the first 10 multiples of 4, which are 4, 8, 12, 16, 20, 24, 28, 32, 36, and 40. Jerry now lists the first 10 multiples of 9, which are 9, 18, 27, 36, 45, 54, 63, 72, 81, and 90. The boys can see that they must eat handfuls totaling 36 chips in order to eat the same amount. Lesson Summary So in review, to find the least common denominator, we will be using multiples. Multiples are the products of a given number. We often call the least common denominator the least common multiple or the LCM. Using these multiples, we can find a common denominator. The least common denominator is the smallest common multiple of both denominators. Comparing and Ordering Fractions Chapter 2 / Lesson 3
35 Comparing and ordering fractions is way to examine fractions that contain differentsized sets. In order to compare these fractions, you must find a common denominator and make equivalent fractions. Comparing and Ordering Fractions When comparing and ordering fractions, you must have a common denominator. After you find a common denominator, you can simply compare the numerators of the fractions. If the fraction also includes a whole number, you would compare the whole number before finding the common denominator. For example, 6 ¾ would be greater than 5 ½. We can see that because the whole number is larger in 6 ¾ than it would be in 5 ½. Finding the Common Denominator There are several ways to find a common denominator when getting ready to compare fractions. The easiest method to find a common denominator is to multiply your denominators together. After you have multiplied the denominators together, the product will become your new denominator. For example, when comparing 7/8 and 5/6, the two denominators are 8 and 6. To find a common denominator, I would multiply the 8 and 6 together, which equals 48. The common denominator of this problem would be 48. Now that we know our common denominator is 48, we must get equivalent fractions. Looking at the fraction 7/8, 8 times 6 is 48, so we must also multiply the numerator times 6. 7 times 6 equals 42. The fraction 7/8 is equivalent to 42/48. Now we must find the equivalent to 5/6. Since we multiplied the denominator 6 times 8, we must also multiply the numerator 5 times 8, which equals 40. The fraction 5/6 is equivalent to 40/48. Comparing Fractions In the previous example, we were comparing 7/8 to 5/6. The first step was to find a common denominator, which we found as 48. The next step was to find equivalent fractions for each of the fractions. Now we are comparing 42/48 to 40/48. Since we know that 42 is larger than 40, we can see that 42/48 would be greater than 40/48. The equivalent fraction for 42/48 was 7/8, which would be larger than 5/6.
36 Comparing Fractions: Example Comparing fractions can help us compare several items that may not all be the same size. By finding a common denominator, we change our fractions so that the items will be much easier to compare. Jimmy and Dan both had pizza for supper last night. Jimmy's family ordered pizza from Pizza Palace. Pizza Palace slices their pizzas into sixteen slices. Jimmy's family ate 10/16 of the pizza for supper last night. Dan's family ordered pizza from Magic Pizza, which slices their pizza into 8 slices. Dan's family ate 7/8 of their pizza for supper. Jimmy and Dan are both curious which one of their families ate more pizza for supper last night. To begin comparing the fractions of pizzas that their families ate, they must first find the common denominator. Since the denominators are 16 and 8, the easiest way to find the common denominator is to multiply the denominators together. 16 times 8 equals 128. So the common denominator for both of these fractions will be 128. Next, they must make equivalent fractions. Jimmy's family ate 10/16 of their pizza. To find the common denominator, Jimmy had to multiply his denominator by 8, which equaled 128. He must also multiply the numerator 10 times 8, which would equal 80. So Jimmy's family ate an equivalent of 80/128 slices of pizza for supper last night. Dan must now find the equivalent fraction for the amount of pizza his family has eaten. To find the common denominator, Dan multiplied his denominator 8 by 16, which equaled 128. He must now multiply the numerator 7 times 16, which is 112. Dan now knows that his family ate an equivalent of 112/128 slices of pizza. So Jimmy's family ate 80/128 of their box of pizza and Dan's family ate 112/128 of their box of pizza. Jimmy and Dan now know that Dan's family ate more pizza than Jimmy's family last night, because the numerator 112 is larger than 80. Ordering Fractions: Example Later that day at recess, Jimmy and Dan met up with their friends Sal and Rex. They decided to run laps around the track at their school. All four of them take off and run as fast as they can. Dan makes it only ½ of the track before he must stop. Sal makes it
37 ¾ of the track before he must stop. Jimmy makes it 4/6 of the track before he must stop. Rex makes it 8/10 of the track. All four of them are exhausted as they slowly walk toward the finish line. The four of them want to know in what order from greatest to least they were able to run without stopping. The boys know that their distances were ½, ¾, 4/6, and 8/10 of the track. In order to compare these four distances, the boys know that they will need a common denominator. The easiest way to get a common denominator is to multiply the denominators together. However, in this example, it will cause our denominator to be a much larger number that may be impractical to work with. Another way to get a common denominator is to think about what number all four of these denominators will divide into evenly. Since our denominators are 2, 4, 6, and 10, the lowest number that all four of these numbers will divide into is 60, so the boys decided to use 60 as their common denominator. Dan knows that he ran ½ of the track. In order to get the common denominator of 60, Dan must multiply the numerator and denominator by 30. This would mean that Dan's equivalent fraction would be 30/60. Next, Sal knows that he ran ¾ of the track without stopping. In order for him to compare his fraction, he must use the denominator of 60. Sal will need to multiply both the numerator and denominator by 15 in order to make an equivalent fraction. Sal determines that he ran 45/60 of the track. Jimmy also ran hard and ran 4/6 of the track without stopping. Jimmy uses the common denominator of 60 by multiplying his numerator and denominator by 10. This makes Jimmy's equivalent fraction as 40/60. Rex was able to run 8/10 of the track without stopping. For Rex to compare his fraction to his friends, he must also use the common denominator of 60. In order to do so, Rex must multiply his numerator and denominator by 6, which would make his equivalent fraction as 48/60. All of the boys now know their equivalent fractions in order to compare their distances. Dan ran 30/60 of the track, Sal ran 45/60 of the track, Jimmy ran 40/60 of the track and Rex ran 48/60 of the track. The boys can now compare their numerators to see in what order they ran the furthest to the least. By comparing their numerators, we can see that Rex ran the furthest, Sal ran the second furthest, Jimmy ran the third furthest, and Dan ran the least far.
38 Lesson Summary So, in review, when comparing and ordering fractions, you must have a common denominator. After you find a common denominator, you can simply compare the numerators of the fractions. The easiest method to find a common denominator is to multiply your denominators together. Once you have made equivalent fractions using your common denominator, you can just compare the numerators of your new equivalent fractions. Another way to find a common denominator is to look for a number that all of your denominators will divide into evenly. Changing Between Improper Fraction and Mixed Number Form Chapter 2 / Lesson 4 Improper fractions and mixed numbers are two different types of fractions. They both have different purposes when using fractions in a problem. In this lesson, you will learn how to convert between these two different forms. What Is an Improper Fraction? An improper fraction is a fraction where the numerator is larger than the denominator. An example of this would be the fraction 12/8. Improper fractions represent a value that is greater than the total value of one set. Often times, it's easier to leave your fraction in the improper fraction form to work with fractions. When multiplying and dividing fractions, you must use an improper fraction rather than mixed numbers. What Is a Mixed Number? Mixed numbers are fractions that contain a numerator, a denominator, and a whole number. These types of fractions contain whole sets and a fraction of the remaining set. An example of a mixed number would be 4 3/4. Mixed numbers are used to represent the final answer when working with fractions. They are also helpful when adding and subtracting fractions.
39 Changing From an Improper Fraction to a Mixed Number The challenge with these two different forms of fractions is being able to convert between them easily. When converting an improper fraction to a mixed number, we will think of the fraction bar as division. For example, using the improper fraction 12/8, we would divide the numerator 12 by the denominator 8. To convert an improper fraction to a mixed number, we will start by dividing the numerator by the denominator. Once you are finished dividing, your quotient will become your whole number. Your remainder will also become your numerator, and you will keep the same denominator. Looking at the example, 12 will divide by 8 one time, with 4 left as our remainder. This will make our mixed number 1 4/8. Improper to Mixed Example Let's check with a friend of mine, Adam, who works at a local cookie factory. Adam spends his day stacking delicious cookies into boxes that can hold 10 cookies each. As he works steadily, the machine suddenly speeds up. Adam realizes that he has run out of boxes. In a panic, he press the emergency stop button. Adam must go get enough boxes to pack the overflow of cookies. He counts the cookies and sees that there 78 cookies to fit in the boxes that only hold 10 cookies each. Adam knows that the fraction to represent this would be 78/10. Adam needs to change this improper fraction to a mixed number so that he can see how many boxes he needs. Adam begins by dividing the numerator 78 by the denominator will divide into 78 7 times. After subtracting, the remainder will be 8. Adam can now see that the quotient 7 will become his whole number, the remainder 8 will become the numerator, and the denominator will stay 10. Adam's mixed number is 7 8/10. Adam can see that he will be able to fill 7 full boxes and 8 out of 10 cookies in the next box. As he packs the last cookie, he presses the start button, and the machine starts sending more cookies down the line.
40 Changing From a Mixed Number to an Improper Fraction Occasionally when working with fractions, you will need to use an improper fraction. To change a mixed number to an improper fraction, we will multiply the whole number times the denominator. Next, we'll add that value and the numerator. This value will become your new numerator, and you will keep the same denominator. Mixed to Improper Example Let's visit the local soccer fields where James is setting up the youth soccer league. James puts 8 kids on each team and has 3 kids left over. He currently has 9 3/8 teams. James knows that this means he has 9 full teams and 3 kids out of 8 left over. James wants to know what number of kids signed up for soccer if he has 9 3/8 teams. To begin, James multiplies the whole number 9 times the denominator 8, which equals 72. Now, using that value, 72, he will add the numerator, 3, which equals 75. James know that he will keep the same denominator. So the fraction that would represent the number of players playing soccer this season is 75/8. Lesson Summary So, in review, an improper fraction is a fraction where the numerator is larger than the denominator. Mixed numbers are fractions that contain numerators, denominators, and whole numbers. To convert an improper fraction to a mixed number, we will start by dividing the numerator by the denominator. Once you are finished dividing, your quotient will become your whole number, your remainder will become your numerator, and you will keep the same denominator. To change a mixed number to an improper fraction, multiply the whole number times the denominator. Next, you'll need to add that value and the numerator. This new value will become your numerator, and you will also keep the same denominator.
41 How to Add and Subtract Like Fractions and Mixed Numbers Chapter 2 / Lesson 5 When adding and subtracting fractions, you must be sure to have a common denominator. Once you have a common denominator, you just add or subtract your numerators and then your whole numbers. Learn how in this lesson. Adding Like Fractions and Mixed Numbers To add fractions you must have a common denominator. When looking at your fraction, you need to make sure that the fractions you are adding have the same denominator. Once your denominators are the same, you can simply add the numerators. You will keep the same denominator for your answer. If your fraction contains a whole number, it would be called a mixed number. When adding mixed numbers, first add your numerators and then your whole numbers. Again, the denominator in your answer will stay the same. Adding Like Fractions Example Todd and James work on a farm collecting eggs from the hens. They both work to fill cartons that can contain up to 12 eggs each. Todd and James have a basket full of eggs that they must now put into the cartons. They both start stacking each carton with 12 eggs. As Todd grabs the last egg, he sees that he has collected 4 full cartons and 5 eggs in the next carton. His fraction would represent 4 5/12. James starts counting his eggs; he collected 6 full cartons and 2 eggs in the next carton. His fraction would represent 6 2/12. The two boys now want to know how many eggs they collected together. They will need to add 4 5/ /12. The two boys can see that both of their fractions have a denominator of 12. Since they have a common denominator, they can just add their numerators and then their whole numbers. The boys know that they will keep the same denominator. So is 7, and the whole numbers, 4 + 6, are 10. This makes their combined amount of eggs 10 and 7/12. The boys know that they collected 10 full cartons and 7 out of 12 eggs in the next carton.
42 Subtracting Like Fractions and Mixed Numbers Same as adding fractions, to subtract fractions you must have a common denominator. When looking at your fraction, you need to make sure that the fractions you are subtracting have the same denominator. Once your denominators are the same, you can simply subtract the numerators. When subtracting mixed numbers, first subtract your numerators and then subtract your whole numbers. Subtracting Fractions Example Let's look at an example of subtracting mixed numbers. Donald works at the Apple store and things are really busy with the release of the new iphones. Donald knows that the iphones come in cases that hold 6 phones per case. He can see that they have 10 full cases and only 4 phones in another case. His fraction would represent 10 4/6 cases of iphones in stock. Throughout the day, the store was full of customers buying phones. At the end of the day, Donald learns that they sold 7 2/6 cases of iphones. He now needs to know how many cases he has left to sell. Donald must subtract 10 4/67 2/6 to see how many cases the store has remaining. To begin, Donald can see that he has a common denominator of 6 in both fractions. He can now subtract his numerators and then his whole numbers. 42 is 2 and the whole numbers, 107, are 3. Donald can see that his fraction is 3 2/6. He knows that he needs to report the number in simplest form. So 2 and 6 will both divide by 2, making his fraction 3 1/3. Donald now knows that the store only has 3 1/3 cases of iphones remaining. Lesson Summary So let's look back at this skill of adding and subtracting like fractions and mixed numbers. To add and subtract fractions you must have a common denominator. When looking at your fraction, you need to make sure that the fractions you are adding or subtracting have the same denominator. Once your denominators are the same, you can simply add or subtract the numerators. When you're adding and subtracting mixed numbers, first add or subtract your numerators, and then you'll add or subtract your whole numbers.
43 How to Add and Subtract Unlike Fractions and Mixed Numbers Chapter 2 / Lesson 6 Simple fraction arithmetic gets a little more complicated when our denominators don't match. In this lesson, we'll learn how to add and subtract unlike fractions. Then we'll do the same with mixed numbers. Handling Fractions Some fractions are easy to add. Let's consider the fractions of burgers. What if you eat a 1/4 pound burger, then another 1/4 pound burger? How much burger did you eat? 1/4 + 1/4 is 2/4, so you ate 1/2 a pound of burger. We just added the numerators, or the numbers on the top. That's simple, tasty math. Adding Unlike Fractions But what if you ate a 1/4 pound burger and then a 1/2 pound burger? That's a lot of meat. To add unlike fractions, you need to find the least common denominator. The denominator is the number on the bottom. The least common denominator is the smallest shared multiple of the denominators. With 2 and 4, it's 4! So we multiply 1/2 by 2/2 to get 2/4. Then we have 1/4 + 2/4, which is 3/4. So you ate 3/4 of a pound of beef. Here's another example: After all that meat, you start watching what you eat. You're making your own salad dressing for your new, healthier lunch. The recipe calls for 3/8 of a teaspoon of salt and 1/3 of a teaspoon of sugar. 1/3 of a teaspoon? Where did you get this cookbook? Anyway, how much salt and sugar is that? 3/8 + 1/3  those are different denominators. What's the least common denominator? Sometimes it's easiest to just multiply them together. 8 * 3 is 24. That actually is the least common multiple. We multiply 3/8 by 3/3 to get 9/24. We multiply 1/3 by 8/8 to get 8/24. 9/24 + 8/24 is 17/24. So 17/24 of a teaspoon  that's your total of salt and sugar.
44 Subtracting Unlike Fractions To subtract unlike fractions, we do the same thing  find the least common denominator. Here's an example: Eating salads was ok, but you think exercise is a better way to get healthy. You decide to take up rock climbing. You're on a practice wall and you get 4/5 of the way to the top. The next day, you're very sore and you only get 1/4 of the way. How much less far did you climb? This is 4/51/4. The least common denominator is 20. So 4/5 becomes 16/20 and 1/4 becomes 5/20. 16/205/20 is 11/20. So the difference between day one and day two is 11/20 of the wall. Here's another example: After rock climbing makes you hurt all over, you try running. You run a loop around a local park and it takes 3/4 of an hour. The next time, it only takes 2/3 of an hour. How much time did you cut? That's 3/42/3. The smallest multiple? It's 12. 3/4 becomes 9/12. 2/3 becomes 8/12. 9/128/12 is 1/12. So you cut 1/12 of an hour. That's five whole minutes. Nice work! Adding Unlike Mixed Numbers Let's move on to adding unlike mixed numbers. This just requires the additional step of adding the whole numbers. With like mixed numbers, this is pretty straightforward. 2 1/ /4 is just 2 + 1, or 3, and 1/4 + 1/4, or 2/4. So it's 3 2/4. Let's try an unlike one in context. You decide to add swimming to your exercise options. The first day, you swim 7 1/2 laps. The next day, you swim 9 1/4 laps. How many total laps did you swim? First, let's find common multiples. It's 4, so 9 1/4 stays the same. 7 1/2 becomes 7 2/ is 16 and 1/4 + 2/4 is 3/4. So you swam 16 3/4 laps. That's pretty good! You're a long way from those multiple burger meals by this point. In fact, your everexpanding athletic repertoire now includes cycling. You go out for a ride and cover 7 1/10 miles. You stop for a quick snack. Then you ride another 11 2/5 miles. How far did you go? Our smallest common multiple is 10, so 7 1/10 is good. 11 2/5 becomes 11 4/ is 18. Then 1/10 + 4/10 is 5/10, or 1/2. So you went 18 1/2 miles. I think you're ready for a triathlon.
45 Subtracting Unlike Mixed Numbers Let's try subtracting unlike mixed numbers. Think of these like two separate subtraction problems, where you subtract the whole numbers, then the fractions. Let's try it in context. Let's say you're deep into triathlon training and you're all about the protein smoothies. You've been drinking a 3 5/8 cup smoothie, but you decide to cut that back to 2 1/4. How much less it that? This is 3 5/82 1/4. Let's get those least common denominators. With 8 and 4, it's just 8. So we make 2 1/4 into 2 2/8. Now let's consider the whole numbers and fractions separately. 32 is 1. And 5/82/8 is 3/8. So you've cut 1 3/8 cups of smoothie. Let's do one more. Let's jump forward in time. We saw you go from burger mania to triathlon ready, let's say you're now a world class Ironman triathlon competitor. You complete one triathlon in 14 5/6 hours. That's pretty good. But you train and train, then do another in 9 1/3 hours. That seems like a huge improvement, but how much is it? It's 14 5/69 1/3. Let's make 9 1/3 into 9 2/6, then consider the parts separately is 5. And 5/62/6 is 3/6, or 1/2. So you cut 5 1/2 hours off your time. That's amazing! Lesson Summary In summary, did you know you were an amazing Ironman triathlete? Wait, you're not? Well, okay. But you do know how to add and subtract unlike fractions and mixed numbers! When you add or subtract unlike fractions, you first need to find the least common denominator. This is the smallest common multiple. Then you just add or subtract the numerators. With mixed numbers, you still need the least common denominator. But then you handle the whole numbers and fractions separately. Multiplying Fractions and Mixed Numbers Chapter 2 / Lesson 7 Multiplying fractions is much more straightforward than adding, subtracting or dividing fractions. In this lesson, learn how it works. We'll also learn how to multiply mixed numbers.
46 Fractions of Dollars How'd you like to earn some money? Let's say I get paid $100 to wash the windows of a skyscraper. But I'm afraid of heights, so I pay my buddy Larry 1/2 of what I'd get to do it for me. Larry then convinces his brother Darryl to do it for 1/2 of what he'd earn. Darryl convinces his other brother Darryl (whom we call Darryl 2) 1/2 of what he'd earn. Then Darryl 2 comes to you and offers you 1/2 of what he'd earn. That's probably worth it, right? Well, that's 1/2 of 1/2 of 1/2 of 1/2 of $100. 1/2 * 1/2 * 1/2 * 1/2 is 1/16, and 1/16 of $100 is $6.25. Would you wash the windows of this building for $6.25? I wouldn't. But I wouldn't do it for $100, either. And what did we just do to figure that out? We multiplied fractions! Let's learn more about it without the acrophobia. Multiplying Fractions To multiply fractions, we follow three steps. Step one: multiply the numerators; those are the top numbers. Step two: multiply the denominators; those are the bottom numbers. Step three: simplify the fraction, if necessary. So if we had 1/4 * 2/5, we start with 1 * 2, which is 2. Then we do 4 * 5, which is 20. Now we have 2/20. We can simplify that to 1/10. So 1/4 * 2/5 is 1/10. This is just like what we did with the window washing example Practice Multiplying Fractions Let's look at another one in context. Let's say we're in a band and we're eating before our big gig. We get a megasupercarnivore pizza. I'm super hungry, so I take 1/2 of it, which is 4 slices. But after eating just 1 slice, which is 1/4 of what I took, I can feel my arteries clogging, so I stop. How much did I eat? I ate 1/4 of 1/2, so let's do 1/4 * 1/2. 1 * 1 is 1. 4 * 2 is 8. So, I ate 1/8 of the pizza. Okay, now I need to get over my illness, because we're going onstage soon. We're opening the gig for another group. It's a soldout show at a 3,600person club. If 5/9 of the total crowd is there early to hear us play, and 1/10 of them buy our tshirts after being blown away by our awesome sound, how many shirts do we sell?
47 So, we do 5/9 * 1/10. 5 * 1 is 5 and 9 * 10 is 90. That's 5/90. We can simplify that to 1/18. What is 1/18 of 3,600? It's 200. That's a lot of shirts! Well, it is for us. And it's enough gas money to get home. On the way, our drummer spends 2/3 of the trip telling us about his fantasy kickball league. It's pretty boring, and you sleep through a good 7/8 of the story. How much of the trip did you spend sleeping? It's 2/3 * 7/8. 2 * 7 is 14 and 3 * 8 is 24. So, 14/24, which simplifies to 7/12. That's some good shuteye. Multiplying Mixed Numbers Okay, but what do we do if the fractions get more complicated? Let's talk about how we multiply mixed numbers. This is a fourstep process. Step one: convert to improper fractions. Step two  well, really, we just follow the steps from earlier: multiply the top, multiply the bottom, then simplify. If we have 3 1/2 * 2 1/4, first we convert to improper fractions. To convert 3 1/2, multiply the whole number times the denominator, that's 3 * 2, or 6, then add that to the numerator. So, that's 7/2. With 2 1/4, we do 2 * 4, or 8, and get 9/4. 7/2 * 9/4? 7 * 9 is 63 and 2 * 4 is 8. So, it's 63/8. That simplifies to 7 7/8. Practice Multiplying Mixed Numbers How about one in context? Let's say I quit the band and start training to make the NBA. I run 3 1/4 miles on Tuesday. If I want to run 2 1/2 times as far on Wednesday, how far will I go? First, convert to improper fractions. 3 1/4 becomes 13/4. 2 1/2 becomes 5/2. So we have 13/4 * 5/2. 13 * 5 is * 2 is 8. So, 65/8. That simplifies to 8 1/8. Can I do 8 1/8 miles? I just multiplied mixed numbers, so I think so! Well, actually, there's no proven correlation between multiplying mixed numbers and distance running. I have another problem. I'm 6'3' '. That's 6 3/12 feet. Tall? Sure, but not NBA tall. Then I meet an old wizard selling magic growing cookies. Magic and cookies? I'm all in. He says they'll make me grow 2 5/6 as tall as I am now. That seems like a random amount, but he's a wizard selling magic cookies, so I go with it. What is 6 3/12 * 2 5/6? With 6 3/12, we do 6 * 12, which is 72, then add that to 3 to get 75/ /6 becomes 17/6. With 75/12 * 17/6, we do 75 * 17, which is And then 12 * 6, which we
48 know is 72. So, 1275/72. That simplifies to 17 51/72, then to 17 17/24. So I'll be over 17 feet tall. I'll definitely make the NBA at that height. Okay, let's do one more. After the magic cookies totally fail to make me even an inch taller, though they were tasty, I get a job feeding animals at the zoo. The lions get 12 2/5 pounds of meat at a time. The tigers get 1 2/3 times that. What do the tigers get? And why does the zoo write their instructions like this? I don't know, but I do know how to figure out the tigers' food. That's important, because no one wants hungry tigers. 12 2/5 converts to 62/5. 1 2/3 converts to 5/3. 62 * 5 is * 3 is 15. So, 310/15. That simplifies to 20 10/15, or 20 2/3. That's a lot of meat, but they're big cats. Lesson Summary In summary, to multiply fractions, we follow three steps: first, multiply the numerators; second, multiply the denominators; finally, simplify as needed. With mixed numbers, we start by converting them to improper fractions, then we just multiply the fractions! Dividing Fractions and Mixed Numbers Chapter 2 / Lesson 8 Dividing fractions and mixed numbers? It sounds daunting, but it's not as tricky as it sounds. In this lesson, we'll learn how to divide fractions and mixed numbers. Let's Divide Let's talk about cookies. Baking cookies is a form of chemistry  tasty, sometimes chocolate chip chemistry. But there also seems to be an awful lot of math involved. This is especially true if you're like me and you don't always have an unlimited supply of every ingredient. For example, maybe you have only 1 1/2 cups of flour left and you need to modify your recipe so that you maximize your cookie potential. You're going to need to divide fractions. Let's learn more about this critical skill.
49 Dividing Fractions To divide fractions, we follow three steps. Step one: flip the second fraction. This gives you its reciprocal. Step two: multiply the fractions. This means you multiply the numerators, then the denominators. Finally, step three: simplify as needed. So with 9/10 divided by 4/5, we flip the 4/5 to get 5/4. That's its reciprocal. Then we multiply. 9 * 5 is 45. And 10 * 4 is 40. So, 45/40. We simplify that to 9/8, or 1 1/8. Practice Dividing Fractions Let's try a few in context. Let's say you're throwing a party. You have a punch bowl that has 8/9 of a gallon of punch left. If your cups will hold one cup of punch, which is 1/16 of a gallon, how many cups can you fill? We need to divide 8/9 by 1/16. Okay, first step? Flip the second fraction. So 1/16 becomes 16/1. Then multiply 8/9 by 16/1. That's 128/9. That simplifies to 14 2/9. So you can fill 14 cups and then someone gets stuck with 2/9 of a cup, which isn't great, but 14 people get full cups. Next up, you're dishing out pizza. Unfortunately, you didn't plan this well. You only have 7/8 of a pizza on hand. You somehow only ordered one pizza and then you ate one slice while filling punch cups. There are 21 people at your party. How much of the pizza will each person get? This is 7/8 divided by 21. Remember, 21 is the same as 21/1, so to get its reciprocal, we do 1/21. Now we multiply 7/8 by 1/21. That's 7/168. That simplifies to 1/24. So everyone gets 1/24 of the pizza. I call the pepperoni slice! Let's try one more of these. After the pizza debacle, you're rationing your other snacks. You turn to your guacamole, of which you have 4/5 of a pound. You've overanalyzed this a bit and determined that people use about a tablespoon, or 1/32 of a pound, on each chip. How many chips can dip in your guacamole? This is 4/5 divided by 1/32. Let's flip 1/32 to get 32/1. Then multiply 4/5 by 32/1. That's 128/5, or 25 3/5. Hmmm 25 chips. This isn't only a good lesson in dividing fractions; it's also a good lesson in party planning.
50 Dividing Mixed Numbers Rather than focus on effective party preparation tips, let's take our fraction division to the next level and discuss dividing mixed numbers. To divide mixed numbers, we follow four steps. Step one: convert to improper fractions. So, if we have 3 1/3, we multiply the whole number times the denominator. That's 3 * 3, or 9. Then add that to the numerator. So we get 10/3. Next, we follow the steps to divide fractions. Flip the second one, then multiply, then simplify. So, with 3 1/3 divided by 5 1/2, we convert both to improper fractions. We know 3 1/3 is 10/3. With 5 1/2, 5 * 2 is 10. Add 10 and 1 and get 11/2. Then we flip that one to get 2/11. So, 10/3 * 2/ * 2 is * 3 is 33. So, 20/33. That looks like it should simplify, but it actually doesn't. So we're done! Practice Dividing Mixed Numbers Let's try some in context. We're going to leave the party behind and talk about flowers. Let's say you're building a raised flower bed out of some scrap lumber. You have a board that's 10 3/4 feet long. You want pieces that are 2 1/2 feet long for the flower bed. How many pieces can you get? We divide 10 3/4 by 2 1/2. First, convert to improper fractions. 10 3/4 becomes 43/4. 2 1/2 becomes 5/2. Next, flip 5/2 to get 2/5. Now multiply 43/4 times 2/5. That's 86/20. That simplifies to 4 3/10. You needed four sides, so you have enough wood! Ok, now let's say you're planting flowers. You have 6 1/4 square feet of space and your flowers need 1 1/8 square feet to flourish. How many flowers can you plant? So, 6 1/4 divided by 1 1/8. 6 1/4 becomes 25/4. 1 1/8 becomes 9/8. We flip 9/8 to get 8/9. And 25/4 times 8/9? 25 * 8 is * 9 is 36. So we get 200/36. That simplifies to 5 5/9. Well, you can't really plant 5/9 of a flower, but you know you have space for 5 flowers and a little spare room. After planting your flowers, you decide to make some cookies. Remember when I mentioned those earlier? I haven't stopped thinking about them and now it's time to solve that problem once and for all. Plus, all that building and planting earned you some cookies, right? You realize you're running a bit low on sugar, though. You have 2 1/5 cups of sugar. Your recipe says each batch requires 1 1/6 cups of sugar. How many batches can you make?
51 So, 2 1/5 divided by 1 1/6. 2 1/5 becomes 11/5. 1 1/6 becomes 7/6. We flip 7/6 to get 6/7. Then we multiply 11/5 by 6/7. 11 * 6 is * 7 is /35 simplifies to 1 31/35. Oh no! You can definitely make one batch. But then you only have 31/35 of what you need for a second batch. This was definitely a two batch kind of day, so it looks like you'll be asking a neighbor for 4/35 of a cup of sugar. Let's hope you have the 4/35 cups measuring cup handy. Lesson Summary In summary, there are three steps involved with dividing fractions. First, flip the second fraction to get its reciprocal. Next, multiply the fractions. Finally, simplify as needed. When you want to divide mixed numbers, you start by converting them to improper fractions, then follow the steps to divide fractions. Practice with Fraction and Mixed Number Arithmetic Chapter 2 / Lesson 9 Adding whole numbers is one thing. But adding fractions and mixed numbers? That's not so simple. Or is it? In this lesson, we'll practice arithmetic with both fractions and mixed numbers. Fraction Style Fashion styles come and go, but there are certain fashion rules that never change. And you don't have to be a fashion expert to know some of these. For example, socks and sandals? Super comfortable, but not okay. Stripes and plaid? Don't do that. Vertical and horizontal stripes? That's just weird. Fractions operate in much the same way. Depending on what you're trying to do with your fractions, you may need to colorcoordinate. But sometimes you can just pick any two fractions and you're okay. It's all about the operation you're trying to perform. Let's work on practicing arithmetic with fractions and mixed numbers. First, though, let's quickly review what we're talking about. A fraction is simply a part of a whole number. 1/2, 3/4, 25/26  these are fractions. 26/26 would just be simplified to 1, which is not a fraction. Fractions consist of two parts, a numerator (which is the top number) and a denominator (which is the bottom
52 number). A mixed number is a whole number and a fraction. Let's say you had 3/2. You could simplify that to 1 1/2. That's a mixed number. Okay, let's practice! Multiplication Let's start with multiplication. This is the easiest fraction operation. We just multiply the numerators, then multiply the denominators. It's like our closet only has matching colors. So, 1/2 * 1/4 is just 1/8. 2/3 * 3/7? That's 6/21. 5/6 * 7/8? That's 35/48. When you multiply fractions, people are jealous of how good you look no matter what you wear. Sometimes you need to do a little simplifying at the end. If you multiply 3/4 * 2/3, you get 6/12. That simplifies to 1/2. With mixed numbers, we can follow the same process, but we first have to convert to an improper fraction. This just means multiplying the whole number times the denominator, then adding it to the numerator. So 1 1/2 becomes 3/2. 3 2/3 becomes 11/3. This is like working with a more colors, though maybe they're all Earth tones, so they kind of all still go together. So what is 4 1/2 * 2 3/4? 4 1/2 becomes 9/2 and 2 3/4 becomes 11/4. 9/2 * 11/4 is 99/8. We then convert that back to a mixed number by reversing the process we did before: divide the numerator by the denominator. The answer becomes the whole number and the remainder becomes the numerator. So we have 12 3/8. What about 1 1/5 * 1 1/6? 1 1/5 becomes 6/5 and 1 1/6 becomes 7/6. 6/5 * 7/6 is 42/30, or 1 12/30. We can simplify that to 1 2/5. Addition & Subtraction If you want to add or subtract fractions, you need to abide by the fashion police. You can't just add 1/2 to 1/4. That's like going out wearing two kinds of plaid. You're not in a '90s grunge band, are you? So we need to coordinate. To do that, we need to get the same denominator on both fractions. If we want to add 1/4 and 1/4, we just add the numerators and get 2/4. With 1/2 and 1/4, we need to find the least common denominator, which is 4. We multiply 1/2 * 2/2 to get 2/4. Now we can add them to get 3/4.
53 What about 3/5 + 2/3? The least common denominator is 15. So we multiply 3/5 * 3/3 to get 9/15. We multiply 2/3 * 5/5 to get 10/15. 9/ /15 is 19/15. That can be simplified to 1 4/15. Subtraction works the same. We still can't wear that purple shirt with the neon green pants. Why do we have neon green pants, anyway? Oh, but subtraction practice. What is 5/61/3? The least common denominator is 6, so we convert 1/3 to 2/6. 5/62/6 is 3/6, or 1/2. What about 1/27/8? 8 is our least common denominator, so 1/2 becomes 4/8. 4/87/8 is 3/8. With mixed numbers, it's simplest to convert to improper fractions. So for 3 1/32 1/2, we convert 3 1/3 to 10/3 and 2 1/2 to 5/2. 6 is our least common denominator. So 10/3 becomes 20/6. 5/2 becomes 15/6. 20/615/6 is 5/6. What about 7 1/ /5? 7 1/4 becomes 29/4. 5 4/5 becomes 29/5. Our least common denominator is 20. So 29/4 becomes 145/20. 29/5 becomes 116/20. Add them together and we have 261/20, which is 13 1/20. Division Finally, there's dividing fractions. This is for fashion pioneers. When we divide fractions, we turn the second fraction upside down and then multiply the fractions together. It's like the crazy stuff you see at those fancy fashion shows. Is she wearing pants on her head? Are those sunglasses on her knees? So with (1/2) / (4/5), we turn 4/5 upside down to 5/4, which is its reciprocal. Then we do 1/2 * 5/4, which is 5/8. What about (5/6) / (3/8)? 3/8 becomes 8/3. And 5/6 * 8/3 is 40/18. We can simplify that to 20/9, or 2 2/9. Dividing mixed numbers is just like multiplying. We just need to first convert them to improper fractions. So (2 2/3) / (1 1/2)? 2 2/3 becomes 8/3. And 1 1/2 becomes 3/2. The reciprocal of 3/2 is 2/3. And 8/3 * 2/3 is 16/9. That simplifies to 1 7/9. What about (4 3/4) / (2 1/5)? 4 3/4 becomes 19/4. 2 1/5 becomes 11/5. We flip 11/5 to get 5/11. And 19/4 * 5/11 is 95/44. That converts to 2 7/44.
54 Lesson Summary In summary, multiplying fractions involves just multiplying the numerators by each other, then the denominators. Adding and subtracting requires us to find the least common denominator. Then we just add or subtract the numerators. With division, we're in high fashion territory. We flip the second fraction to its reciprocal. Then we multiply them together. When we have mixed numbers, we convert them to improper fractions. And no matter what, no socks with sandals! I'm serious about that. Estimation Problems using Fractions Chapter 2 / Lesson 10 When working with fractions, estimating can save time and effort while getting you very close to the correct answer. In this lesson, we'll learn how to use estimation to quickly solve problems involving fractions. When Close Counts Did you ever hear the saying 'Close only counts in horseshoes and hand grenades'? I immediately imagine people playing horseshoes with hand grenades, which is kind of dangerous. But the point of that saying is not entirely true. There are a few other situations in which close counts just fine, like when I iron a shirt  it's close to wrinklefree, and that's pretty good, or at least I think so. There's also baking cookies  close to done is when they're still gooey in the middle, which is actually better than being done in my opinion  or folding laundry, and especially a fitted sheet. I've never gotten better than close and my life seems to go on just fine. Estimating Fractions And then there are fractions. Fractions can be onerous. What if you have 5/322 and you need to add it to 420/863? You'd need to find the common denominator. (For the record, it's 277,886.) But did you ever have 5/322nds of something? Let's say you're sharing pizza with some math geek friends and 5/322nds is your share. Do you know what you have? Maybe a small piece of one piece of pepperoni  in other words, pretty much zero.
55 And what about 420/863rds? If a glass can hold 863 drops of water and you put in 420, what do you have? A lot of free time, apparently. But you also have a glass that's about half full  or half empty. That's up to you. My point is that 420/863 is pretty close to 1/2. What we just did is estimate the value of fractions in order to make them much simpler to work with. When you are estimating fractions, round the fraction to the nearest whole number, 1/2 or 1/4. Practice Estimating Here's what this looks like: You know 50/100 is 1/2. Think of 50/100 as $50 you're willing to spend out of the $100 you have. What if there's a dress you want to buy and it's $48? That's pretty much $50. With tax, it might even be a little more. And if it's $54? You might also think of that as being right around $50. You'd still be spending close to what you intended. Let's consider another fraction: 60/80. That's 3/4. Well, 57/80 is also pretty much 3/4. And that saves you having to find a denominator like 277,886. Then you can get through your work way faster. Practice Problems Let's try a few practice problems. Let's say we want to add 84/90 to 32/29. 84/90 is very close to 90/90, or 1. And 32/29? That's also close to 1. So if we estimate the solution, it's going to be about 2. Note that if we didn't estimate, we'd need a denominator of We'd add 2436/2610 to 2880/2610 to get 5316/2610 to get 2 96/2610, or 2.04, which is super close to just 2. It's ever so slightly more than 2, so it's like a cookie that's just slightly burned. Some people love those burnedonthebottom cookies. Here's one: 7 97/ /10. 97/99 is very close to 1, so let's make that first number 8. 6/10 is close to 1/2, so let's make that second number 2 1/ /2 is 10 1/2. What about 14/207/15? Well, 15/20 is 3/4, so we can estimate that 14/20 is nearly 3/4 as well. And 7/15? 7/14 is 1/2, so 7/15 is close to 1/2. Therefore, 14/207/15 is close to 3/41/2, or 1/4. I think this one's like a steak. If you wanted sort of medium rare, then that's what I can deliver  sort of medium rare. Sometimes it's more rare, sometimes it's more medium. But we're in the right ballpark.
56 Let's try a multiplication one. What is 11/23 * 3 5/9? This one's like parallel parking. You don't need to be right up next to the curb, but you don't want to be at a weird angle with one end of your car out in traffic. Close enough to the curb would be saying 11/23 is very close to 1/2. Sticking way out would be calling it 3/4. If you think in quarters, 11/23 is much closer to 1/2 than 3/4. And 5/9 is also close to 1/2. So 1/2 * 3 1/2 is 1 3/4. How about 8/13 * 7/25? 8/13 is close to 1/2, but it's closer to 3/4. Why? 8/12 is 3/4. So 8/13 is just a little bigger than that. 7/25 is close to 1/4. What's 7 * 4? 28. So 7/28 is 1/4. 7/30 is just a little smaller. 3/4 * 1/4 is 3/16. What about division? What is 72/150 divided by 37/75? Without estimating, those fractions are like fitted sheets. Seriously, though, I can't fold a fitted sheet. But I can estimate. And both of those fractions are close to 1/2. 1/2 divided by 1/2 is 1/2 * 2/1, or 1. With multiplication and division, the further you go from the exact numbers, the less accurate you get. Small differences are, well, multiplied. But this is still fairly accurate. Maybe this division problem is like handgrenade horseshoes. You're throwing it in the general vicinity and, after the smoke clears, you probably earned yourself some points. Lesson Summary In summary, we learned that close counts in way more than horseshoes and hand grenades. Most importantly, estimating fractions is a great way to save time and effort. The simplest way to estimate fractions is to use the closest whole number, 1/2 or 1/4. For example, 18/19 is pretty much 1, 7/10 is close to 3/4 and 9/16 is very nearly 1/2. Solving Problems using Fractions and Mixed Numbers Chapter 2 / Lesson 11 Fractions don't just live in math problems. They're all around us, helping us through our lives. In this lesson, we'll look at real scenarios where your ability to work with fractions can save the day.
57 Fractions Are Everywhere We encounter fractions every day. Think about food. When you bake, you might need a half cup of sugar or a quarter teaspoon of vanilla. When there's pizza, you may hope for more than just 1/8 of the pie. When you're sneaking cookies, maybe you have 3 1/2, because 4 would just be gluttonous. Fractions are also very common in sports. At the track, you might run 4 3/4 laps. When you watch a football game, sometimes it's the final quarter that's most exciting. On the golf course, you might celebrate making par on 5 out of 9 holes. You can get through life without knowing how to solve problems with fractions, but what kind of life would that be? Half full? Half empty? You would never know. Comparing Fractions Let's look at a few situations involving fractions and mixed numbers. Remember, a fraction is a part of a whole number, like 1/2 or 5/8. A mixed number is a whole number and a fraction, like 33 1/3. Let's say you and your friend Ginger team up for a talent competition reality show. In order to win the final round, you need to get more votes than your competitors. Your flaming sword juggling routine gets 4/9 of the votes. The family that does trapeze tricks with a tiger gets 2/7 of the votes. The guy conducting the monkey string quartet gets 1/3 of the votes. Who wins? This question is just asking us which fraction is largest. To do this, we need the least common denominator, or the smallest multiple of the denominators. If we started listing multiples, we'd find that 63 is what we want. To convert 4/9 to 63rds, we multiply it by 7/7. That'll be 28/63. With 2/7, we multiply by 9/9 to get 18/63. With 1/3, we multiply by 21/21 to get 21/63. Now we just compare numerators  28, 18, 21. You win! Addition and Subtraction Unfortunately, someone notices that = 67, and 67/63 is more than 100% of votes, thereby invalidating your victory due to voter fraud. You and Ginger try to start fresh on a cupcake baking show. Together, you need to bake 1,000 cupcakes in
58 one hour. If you make 1/2 of the cupcakes and Ginger makes 2/5, did you make enough? We need to add 1/2 and 2/5. To do this, we again need the least common denominator. Here it's 10. To convert 1/2 to 10ths, we multiply it by 5/5. So it's 5/10. With 2/5, we multiply by 2/2, so it's 4/10. We then add the numerators, so 5/10 and 4/10 gives us 9/10. So 9/10 of our cupcakes are ready. Of course, we need 10/10. And now time is up! To deal with your loss, you eat your way through many of those cupcakes. Then you and Ginger decide to go on a competitive weight loss show. One of the challenges is to run on a treadmill as far as you can go. You run 3 2/3 miles. Ginger ran 4 5/6 miles. How many more miles did Ginger run than you? We want to know what 4 5/63 2/3 is. With a mixed number, we need to first convert it to an improper fraction, or a fraction with a flair for offcolor jokes. Wait, no, it's a fraction with a larger numerator than denominator. To do this, just multiply the whole number by the denominator, then add it to the numerator. 4 5/6 becomes 29/6, and 3 2/3 becomes 11/3. Now we need a common denominator, which is 6. We multiply 11/3 by 2/2 to get 22/6. And then we just subtract numerators is 7. So Ginger ran 7/6, or 1 1/6, more miles than you. Multiplication and Division Things are getting tense between you and Ginger and, after the weight loss show, you both go on a competitive dancing show. Now you're competing headtohead. Of all the types of dances you do, Ginger is an expert at 5/6. And when you were getting along better, she taught you 2/3 of what she knows. How many dances can you do? We need to multiply 5/6 * 2/3. To multiply fractions, simply multiply the numerators, then multiply the denominators. 5 * 2 is 10. And 6 * 3 is 18. So you know 10/18, or 5/9. That's a little better than 1/2, but definitely not as much as Ginger. After Ginger beats you on the dance show with an amazing paso doble, you find out that she was behind that talent show voter fraud. And she sabotaged the cupcake competition. Plus, she cheated on the treadmill race. So you decide to take her down on a fashion design show. You've gotten so good at working with fractions that you think you have an edge here. You tell her she has 6 2/3 yards of polka dot fabric. She needs to divide that into 1/4 yard segments for an awesome leisure suit pattern you gave her. How many segments
59 can she get with that fabric? Okay, this is just 6 2/3 divided by 1/4. To divide fractions, just flip the second fraction, which gives you its reciprocal. But first, we need to convert 6 2/3 to an improper fraction. 6 * 3 is 18, and is 20. So it's 20/3. 20/3 divided by 1/4 is 20/3 * 4/1, or 80/3. If we convert that back to a mixed number, we get 26 2/3. So she'll have 26 complete 1/4 yard segments and 2/3 of a segment. That's going to be one awesome leisure suit. And, in fact, it was. She won the competition. Lesson Summary In summary, let's look at what we've learned here. First, Ginger is really not the friend you thought she was, is she? And it's not nice to seek fashion design vengeance. More important, fractions and mixed numbers are everywhere. And it only takes a few simple tricks to work with them. Mixed numbers should first be converted to improper fractions. Then, with addition and subtraction, always find the least common denominator. Multiplication is simpler  just multiply the numerators, then the denominators. With division, flip the second fraction to get its reciprocal, then multiply. And now your glass is half full! How to Solve Complex Fractions Chapter 2 / Lesson 12 Did you know fractions can nest within fractions? In this lesson, we'll learn about complex fractions and look at two different methods for solving them. Getting Complex Have you ever seen Russian nesting dolls? You know, those wooden figures that you can open up and there's a smaller figure inside. Then you open that one and there's an even smaller figure inside, and then another, and another, and so on. When you look at the outermost doll, it's just an ordinary figure. But when you open it, you realize that it has these cool layers just below the surface.
60 Complex Fractions Complex fractions are just like nesting dolls. They're like regular fractions but then, whoa, there's another fraction inside that fraction. A complex fraction is a fraction that contains fractions in the numerator, denominator or both. Here's an example: (3/4 + 2/5)/(1/21/6) Our numerator, the number on top, is 3/4. And our denominator, the number on the bottom, is 1/2. Method One When you put a nesting doll back together, you start with the innermost doll and work outwards. You solve complex fractions in much the same way. There's usually only one way to put a nesting doll back together. But there are actually two ways to solve a complex fraction. The first way involves handling the top and bottom separately. Step 1: Simplify the numerator. Here's an example. The top is 3/4 + 2/5. We need to make the denominators the same by finding the least common denominator, or the smallest shared multiple of the denominators. Here, that's 20. To make the denominator of 3/4 20, we multiply it by 5/5, which gets us 15/20. We multiply 2/5 by 4/4 to get 8/20. 15/20 + 8/20 is 23/20. Now for Step 2: Simplify the denominator. That's 1/21/6. What's the least common denominator? 6. So 1/6 is fine. We multiply 1/2 by 3/3 to get 3/6. 3/61/6 is 2/6, or 1/3. Now we have (23/20)/(1/3). And we're ready for the final step: Step 3: Divide the Fractions. (23/20)/(1/3) is really 23/20 divided by 1/3. That's all this line right here means. To divide fractions, we first flip the second one, giving us its reciprocal. So 1/3 becomes 3/1. Then we multiply them together. So 23/20 * 3/1 is 69/20. That can be simplified to 3 9/20. That's almost as pretty as a nesting doll!
61 Method Two That's a perfectly fine way to handle complex fractions. But maybe you get a little impatient dealing with the numerator, then the denominator. Maybe you just want to throw all those dolls together in one big flurry. That's kind of like the second method, which just has two steps. Step 1: Find the least common denominator for all fractions. Let's try an example with variables this time: (x + 1/x)/(1 + 4/2x). Remember that x is the same as x/1 and 1 is the same as x/x or 1/1. Our least common denominator here is 2x. Now, Step 2: Multiply all fractions by the common denominator. x * 2x is 2x^2. 1/x * 2x is 2x/x, which is 2. So our new numerator is 2x^ * 2x is 2x. 4/2x * 2x is 8x/2x, or just 4. So our new denominator is 2x + 4. Now we have (2x^2 + 2)/(2x + 4). We can pull a 2 out of both parts, which cancels out, leaving us with (x^2 + 1)/(x + 2). That's not the prettiest nesting doll you've ever seen, but it is simplified. So which method is better? It's really more of a personal preference. Sometimes, one method will be simpler than another. But most people just find that they like using one method more than another. We could have solved both examples we looked at using either method and gotten the same answers. Lesson Summary In summary, a complex fraction is just a fraction that contains fractions in the numerator, denominator or both. There are two methods of solving them. The first one treats the numerator and denominator separately. You simplify each by finding the least common denominator. Then you divide the fractions, which involves multiplying the numerator by the reciprocal of the denominator. The second method involves finding the least common denominator for all fractions. Then you multiply all terms by that common denominator and simplify as far as you can.
62 In the end, you'll have taken a bunch of separate fractions and turned them into a much simpler, single fraction, just like putting nesting dolls back together. Calculations with Ratios and Proportions Chapter 2 / Lesson 13 Even if only 99 out of 100 people need to know how to work with ratios, the odds that they'll be useful to you are very high. In this lesson, we'll practice performing calculations with ratios and proportions. Let's Compare Comparisons can get kind of a bad rap. I might say my favorite football team is better than yours. Or I could say that I'm betterlooking than you. Or maybe I tell you that my dog is smarter than yours. First of all, my dog is sweet, but I don't think he's winning any IQ contests. And aside from their possible lack of accuracy, these kinds of comparisons are kind of mean. But comparisons can serve useful purposes  math purposes. And I don't mean abstract, makeyounervous, pop quiz math purposes; I mean legitimately useful math purposes. Let's look at how this works. Definitions First, let's get some definitions on the table to give us a framework. When we're talking about comparisons, we mean ratios and proportions. A ratio is a comparison between two things. For example, a soccer team may need 5 soccer balls for every 10 players during practice. The ratio of soccer balls to players if 5:10. A proportion is a pair of ratios that are equal to each other. It's like this: a/b = c/d or a:b = c:d. For example, let's say you save $5 out of every $100 you earn, so you save at a ratio of 5:100. If you earn $500, you'll save $25. The proportion can be written as 5/100 = 25/500. Ratio Problems Okay, now let's try some math. Let's start with ratio problems.
63 Here's a bowl of fruit. What is the ratio of apples to oranges? There are 3 apples and 4 oranges, so the ratio is 3:4. Now, I like apples way more than oranges, so this ratio just won't do. I'm more of an allapplestonooranges ratio kind of guy. Let's talk about milk. You can buy milk in quarts or gallons or other sizes not relevant to this question. What's the ratio of quarts to gallons? This question is really asking how many quarts are in a gallon. If you know your milk, you know it's 4. So the ratio of quarts to gallons is 4:1. Incidentally, this ratio holds true for chocolate milk; it's just tastier that way. Also, always pay attention to the order of the ratio. 4:1 is not the same as 1:4. How about this? In a town with 400 commuters, 120 people drive to work. The rest ride their bikes. What is the ratio of drivers to cyclists? We just need to know the number of each. We know there are 120 drivers. And if there are a total of 400 commuters, then the cyclists must be , or 280. So the ratio of drivers to cyclists is 120:280. This is one bikeloving town. Proportion Problems Let's get more complicated. What if Angie's Car Batteries and Decorative Garden Gnomes sells car batteries and gnomes in a ratio of 35:2? By the time Angie has sold 280 car batteries, how many gnomes did she sell? We just need to set this up as a proportion. Again, that's two ratios that are equal to each other. Our first ratio is 35:2. Our second ratio is 280:g, where g is the number of gnomes. Just crossmultiply and we find that Angie has sold 16 gnomes. We also learned that Angie should probably stick to car batteries, as those gnomes just aren't moving very well. What about this? In the annals of garage rock band history, guitar players outnumber drummers 13:3. In Rockville, which is your typical garage rockloving town, there are 12 drummers. How many guitar players are there? We set this up as 13:3 = g:12, where 12 is our Ringos and g is our number of aspiring Eddie Van Halens. Just solve for g. We get 3g = 13*12. So 3g = 156 and g = 52. So Rockville has 52 guitarists. Some of those people are really going to need to learn bass.
64 Here's another. Little Timmy is up late watching the weather. He knows that if 12 inches of snow falls overnight, school will be canceled. If snow is falling at a rate of 2 inches per hour, how many hours will it take for 12 inches to fall? So the rate of snow to hours is 2:1. Our proportion is 2:1 = 12:h, where h is the number of hours it will take. We crossmultiply and find that it will take 6 hours for school to get canceled. Little Timmy looks at the clock, sees that it's already 4 AM, and knows that the snow better speed up  or he better get started on that big English paper due today. Let's try a spacethemed problem. If 1 in 10,000 planets in the universe could possibly support life, and there are an estimated 3 trillion planets in the universe, how many planets might support life? These numbers are big, but the math is the same. 1:10,000 = p:3,000,000,000,000, where p represents all the Tatooines, Capricas and Gallifreys out there. If we crossmultiply, we'll find that 300 million planets could support life. That's a lot. Lesson Summary In summary, there's probably life somewhere else in the universe. More pertinent, though, a ratio is simply a comparison of two things. It could be apples and oranges or drummers and guitarists. A proportion is a set of ratios that are equal to each other. With proportions, you can find out all sorts of fun things. As long as you know the ratio of two things or events, you can determine the details of things that are directly proportional.
65 What is a Decimal Place Value? Chapter 3 / Lesson 1 Decimals Before we can talk about decimal place values, we need to quickly review what a decimal is. Decimals are numbers with one visible point or dot somewhere in the number. This point is called the decimal point. We actually see them everywhere around us, especially when we go shopping. What do we see on most everything we purchase? Well, we see a price tag. What kinds of numbers do you most often see on price tags? That's right! You see decimals! You see things like 6.99, 1.99, 0.99 and so on. What Is a Place Value? All those numbers you see on price tags and every other number you see and encounter must follow the rules of place values. What exactly is a place value? Well, a place value tells you the location of each digit in a number. The number 699, for example, has a 9 in the units place, a 9 in the tens place and a 6 in the hundreds place. All numbers follow the same place value naming criteria. The number 211 has a 1 in the units place, a 1 in the tens place and a 2 in the hundreds place. The number 423 has a 3 in the units place, a 2 in the tens place and a 4 in the hundreds place. Notice how the tens place is always the second digit to the left and that the hundreds place is always the third digit to the left. Believe it or not, you already know most of the place values since you use them to count. Look at this large number broken down into its place values. The number is 123,456. Hundred Thousands Ten Thousands Thousands Hundreds Tens Units
66 1 2 3, How would you say this number? Yes, you would say one hundred twentythree thousand four hundred and fiftysix. Let's break this down into its place values. While we read from the left to the right, we will read our place values from the right to the left, in the opposite direction. We have 6 units, 5 tens, 4 hundreds, 3 thousands, 2 ten thousands and 1 hundred thousand. If you were asked how many hundreds there are in 123,456, you would answer 4 because there are 4 hundreds in that number. You can think of the number as an addition problem. 123,456 is the same as 100,000, plus 20,000, plus 3,000, plus 400, plus 50, plus 6. Just like our place values told us, we have 1 hundred thousand, 2 ten thousands, 3 thousands, 4 hundreds, 5 tens and 6 units. Where the Decimal Is Matters When you add a decimal point into the mix, the location of the decimal point now determines your place values. Your decimal point tells you where to begin counting. The decimal number 6.99 has a 6 in the units place. The units place value is always the first number to the left of the decimal. To count your place values to the right of the decimal, you start from the left and go to the right. So essentially, to count place values, you always start from the decimal and work your way out. Remember that every number has a decimal even if none is shown. The numbers 211 and 423 both have a decimal point at the end, but we don't write it or show it because there are no numbers after it. When you have a decimal, the place values will look like this. Units. Tenths Hundredths Thousandths Ten Thousandths Hundred Thousandths Notice here that you begin with tenths when you go the right of the decimal point and that you begin with the units when you go to the left. We know what tens, hundreds and thousands mean, but what about tenths, hundredths and thousandths? What do they mean? A tenth means one tenth or 1/10. In decimal form, it is 0.1. Notice the position of the 1. It's in the tenths place. 'Hundredth' means a hundredth or 1/100. In decimal form, it is 0.01.
67 Notice again the position of the 1. Where is it at? That's right, in the hundredths place. As we continue to the right, the numbers will continue to get smaller, with each step being 10 times smaller than the one before. Going to the left, each place value is bigger than the previous by 10 times. If you wanted to tell someone how many tenths a certain decimal number has, you would find the location of that place value and tell him or her the number that's there. For the number , there are 2 tenths. Like regular numbers, you can also think of decimal numbers as addition problems. The only difference would be that your last addition is a fraction of everything after the decimal. You can think of 0.7 as 0 plus 7 / 10. We are dividing by 10 because that's what the tenth place tells us to do. If we have 1.23, we can rewrite it as 1 plus 23 / 100 since the hundredths place tells us to divide by 100. What about a number like 1.234? How would you write that? Yes, you can write that as 1 plus 234 / 1000 since the last place value is the thousandths place, which is telling us to divide by a thousand. Reading Decimals When you read decimals out loud, you will use the place values. The number 0.7 would read as seven tenths. The number 1.23 would read as one point twenty three hundredths. Like regular numbers, you also read decimal numbers from the left to the right. So, when you have a decimal number, your end word will be the last place value that you reach. Lesson Summary To review, decimals are numbers with a point somewhere in the number. That point is called a decimal point. The place values are counted from the decimal point. Going to the left, they are units, tens, hundreds, thousands, ten thousands, hundred thousands and so on. Going to the right, they are tenths, hundredths, thousandths, ten thousandths, hundred thousandths and so on. Each place value is 10 times bigger than the place value to its right or 10 times smaller than the place value to its left. When you are looking at any number, going to the right means your numbers are getting smaller and going to the left means your numbers are getting bigger. When you read numbers, you are actually basing it off their place values.
68 Comparing and Ordering Decimals Chapter 3 / Lesson 2 Decimals Before we talk about comparing and ordering decimals, let's cover some definitions. A decimal number is defined as a number that has a decimal point in it. A decimal point is a point or dot used to show the beginning of digits that are smaller than 1. Decimal numbers are another way to represent fractions. For example, the decimal number 21.5 is another way of saying 21 1/2. The decimal point is placed before the 5 to separate the 21 (the whole number) from the.5 (the fraction). The point shows that the 5 is smaller than, or just a fraction of, the whole number 1. We know that 0.5 is actually half of 1. The same is true of the decimal The number after the decimal, 0.14, is smaller than 1. Using Decimals to Count We count decimals in a similar manner to regular numbers. For regular numbers, we start with 1, then 2, then 3, then 4, etc. As the numbers get bigger, the numbers also get longer. For example, 201 is bigger than 31. This is the way we count for the part of the decimal number that is to the left of the decimal point (the whole part of the number). But once we reach the number to the right of the decimal point (the fraction part of the number), the way we count changes a bit. I want you to stop for a moment and think about how we alphabetize names. Picture a filing cabinet with a drawer open. What do you see? I have a filing cabinet in my office, and the way I alphabetize is by the first letter, followed by the second letter, and so on and so forth. Each successive letter in a word helps me to get even more detailed in my filing. A folder called 'BOY' is more specific than something called 'BO,' and I can file it after 'BO' but before 'BP.' Counting decimals is similar. The first digit after the decimal point is like our first letter. The second digit after the decimal point is like our second letter.
69 Similar to the alphabet, we can count with just our first digit after the decimal point or we can get even more detailed. Just like we can say A, B, C, D, E, F, G, and so on, we can count with just the first digit after the decimal point like this: 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9 Just like we can get more specific when alphabetizing by filing things by AA, AB, AC, AD, AE, AF, AG, and so on, we can count with another digit to be more even specific, like this: 0.11, 0.12, 0.13, 0.14, 0.15, 0.16, 0.17, 0.18, 0.19 I stopped counting when the last digit reached 9 because at this point, something unique happens to decimal numbers. For regular numbers, what happens after you reach 9 or 19? Why, the digit to the left increases by 1, and your 9 changes to a 0. The 9 becomes a 10 or the 19 becomes a 20. And then you keep counting by increasing your last digit by ones again. For decimal numbers, when you reach 0.9 or 0.19, what you do is you also increase the digit to the left of the 9 by 1. Your decimal point stays put. So the 0.9 becomes a 1.0 and the 0.19 becomes a Unless the problem tells you otherwise, we don't write zeroes at the ends of decimals. So the 1.0 will be written as 1 and the 0.20 will be written as 0.2. You can compare this to alphabetizing. With alphabetizing, when you reach 'BZ,' you roll over to 'C.' After the 'C,' you can start getting specific again, with 'CA,' 'CB,' and so on. If we continue counting from the 0.9, we would get 1, then 1.1, then 1.2, and so on, until we reach 1.9. When we reach this point, we increase our digit to the left of the 9 by 1 to get 2. Then, we will continue again at 2.1 and so on. If we continue counting from the 0.19, we would get 0.2, then 0.21, then 0.22, and so on. When we reach 0.29, we would go to 0.3 and then continue to 0.31 and so on. Comparing Two Decimals To compare two decimals to each other to see which is greater, think about alphabetizing again. If you were going to file two different folders, one called 'T' and the other called 'BEACH,' which one would you put in front? You would file the folder called 'T' in the back and the one called 'BEACH' in the front because the 'T' is bigger than the 'B' and therefore goes in the back. When alphabetizing, you look at the first letter, then the second, and so on. If the first letter is bigger, then it goes behind the other folder no matter how large of a word the other folder has.
70 So, just like with alphabetizing and looking at the first letter, then the next, and so on, you do the same with decimal numbers. You look at the first digit after the decimal point, then the next, and so on. If the first digit after the decimal point is bigger, then that number is automatically bigger no matter how long the other decimal is. If the first digit to the right of the decimal point is the same, then you look at the digit after that. For example, 0.8 is bigger than 0.5 and even even though has four digits instead of just one. This is where decimal numbers differ from regular numbers. In regular numbers, 3,401 would definitely be bigger than 8. Remember this distinction. Think of alphabetizing. Just like 'AB' comes before 'AC,' so 0.34 comes before And just like 'APPLE' comes before 'APY,' comes before Ordering Decimals Now that you know about counting and comparing decimals, you can use both of these skills to put decimal numbers in order. If you were given a bunch of decimals like these, you would use what you have just learned to order them from least to greatest , 1.34, 2.1, 5.312, 0.89, 0.7 You can think about alphabetizing and compare each number. Look at the first digit to the right of the decimal point. If that first digit is the same in two different decimal numbers, then you would look at the digit after that to see which one is greater. Looking at the first two numbers, and 1.34, I see right away that 1.34 is bigger than because 1 is bigger than 0. These are the numbers before the decimal, so I use the rule of regular numbers. The third number is 2.1, and this is bigger than 1.34, so I put that in third place is next, and that one is in fourth place because the 5 is bigger than the 2. The next one is 0.89, and that one is less than 1.34, but is it bigger than 0.432? I look at the first digit to the right of the decimal point and I see a 4 in the one and an 8 in the other. Okay. The 8 is bigger, so the 0.89 goes after the and before the Now what about the 0.7? Where does that one go? Well, I think about alphabetizing and I see that 0.7 is between 0.89 and 0.432, so I will place it there. So, my final ordering from smallest to largest is 0.432, 0.7, 0.89, 1.34, 2.1, and
71 Lesson Summary So what did we learn? We recalled that decimal numbers are numbers with a decimal point. And a decimal point is just a dot that is placed before the digits that tells you about the fraction part of the number  the numbers smaller than 1. For example, 0.5 is smaller than 1. In the decimal number 3.14, the 0.14 is smaller than 1. Counting decimal numbers is similar to counting regular numbers until you get to the decimal point. When you get to the decimal point, you count like you would when alphabetizing. You go from 0.39 to 0.4 the same way you would go from 'TZ' to 'U.' When comparing and ordering decimal numbers, you start by comparing the first digit immediately after the decimal point to see which is bigger. If those two are the same, you compare the next digits. If those two are the same, then you move on and compare the next digits. This is similar to alphabetizing, such as when you are comparing 'CAT' to 'CAR.' When you order decimal numbers, you use the skills you learned from counting and comparing decimals to place all the decimals in order. Arithmetic with Decimal Numbers Chapter 3 / Lesson 3 Decimal Numbers Mr. Prof is our little professor for this video. He's about to show us how he adds, subtracts, multiplies and divides decimal numbers. He's picking up his chalk and he's getting ready. First order of business he wants to take care of is the difference between decimal numbers and regular numbers. Decimal numbers, he says, are numbers with a decimal point, while regular numbers are the numbers you normally use to count with. A decimal point is a dot used to show that a particular number has a part of it that is less than 1. For example, the number 1.5 means that we have a whole one plus half a one. The 0.5 is the half a one. All of these are examples of decimal numbers because all of them have a decimal point: 1.5, 2.01, 0.7, 0.95, 3.14,
72 Now that we've talked about what decimal numbers are, let's see how we go about adding and subtracting decimals. Watch how Mr. Prof does it. Adding and Subtracting Let's add 0.5 and 0.81 together and see what happens. Mr. Prof has just written these on his board. He's written the 0.5 on the first line and immediately under, he's written But something looks different. Instead of the 5 and the 1 lining up because they are the last digits, the 5 and the 8 are lined up. Why is that? This is because when you are working with decimals, the decimal point matters. You want to line things up according to the decimal, like the 8 and the 5 here. Now that we have our problem correctly set up and written down, our next step is to drop that decimal point down into our answer area. So we write a decimal point in line with the other two decimal points. Once we have our decimal point where we want it, we will now go ahead and add these numbers the way we know how. I don't see a number above the 1, but since it's after the decimal point, I can go ahead and add a zero there. This is exactly what Mr. Prof has done. So I add 0 plus 1 equals 1. So I put a 1 underneath the 1 in the answer field. Next, I have 5 plus 8 equals 13. Because 13 is greater than 9, I carry the 1 over and write a little 1 on top of the 0, and I write 3 underneath the 8 in my answer field. Next, I have 1 plus 0 plus 0, which equals 1. So I write 1 underneath the 0 in my answer field. So my answer, as Mr. Prof also shows, is So when adding decimal numbers together, what you have to remember is to line them up with the decimal point when you set up your problem. After that, you add the way you know how. Now, what about subtracting decimal numbers? Yes, you guessed it, we subtract in a similar way  by setting up our problem with the decimal points lined up. Let's subtract 0.5 from 0.81 to see how it works. Since I am subtracting 0.5 from 0.81, I will write 0.81 on the top line and 0.5 on the second line, remembering to line them up according to the decimal point. Mr. Prof, I see, has already done that. With everything lined up, I then write a decimal point in the answer field in line with the other decimal points. Then I go ahead and subtract the usual way. Since I am working on numbers after the decimal point, if I don't see a number, I can add a 0. So
73 I have 1 minus 0, which is 1. Then I have 8 minus 5, which is 3. Okay. So my answer is Also, for subtracting, we remember to always subtract the smaller number from the larger. If the problem is asking us to subtract a larger number from a smaller number, then our answer will be negative. For example, if we were subtracting 0.81 from 0.5, then our answer is still 0.31, but negative Mr. Prof gives us a high five for making it this far. Now let's go on to multiplying. Multiplying Multiplying decimals is similar to multiplying our regular numbers with the exception of the last step, where we place our decimal point in our answer. Up until that point, we essentially ignore the decimal points. When we set up our problem for multiplication, instead of lining it up according to the decimal, we line it up according to the last digit. So, to multiply 0.81 and 0.5 together, we line up the 1 and the 5 because they are the last digits. Once we have the problem set up, we go ahead and multiply the way we know how. I have 5 times 1 equals 5. Then I have 5 times 8, which equals is greater than 9, but since we are multiplying our last two digits together, I don't have to carry the 4, and I can write 40 down in my answer line. I'm not done yet, though. I need to figure out where my decimal point goes for my answer. To figure this out, I'm going to count how many decimal places I have from the numbers I'm multiplying together has two decimal places and 0.5 has one decimal place. So that means I have a total of three decimal places. That means I need to have three digits after my decimal point in the answer. So I look at the number in the answer  the and I count three spaces to the left starting from the right. I end up before the 4, so that is where I put my decimal. So my answer is Mr. Prof wants another high five for doing this one correctly! Dividing Dividing decimals is not the same as multiplying. While we left all the decimal points in place while multiplying, we might have to move the decimal points when dividing. This is because for our final result to make sense, we need to divide by a whole number if we are to do this by hand without a calculator. We will be dividing 0.81 by
74 0.5. Since we are dividing by 0.5, we have to move the decimal point one place to the right to change that 0.5 to a 5, a whole number. Because we moved the decimal point one place to the right in the 0.5, we also have to move the decimal point one place to the right in the If we don't do this, we would be changing the problem. So moving the decimal point one place to the right in the 0.81 gives us 8.1. So now we are dividing 8.1 by 5. We set up our division problem like we normally do, with the 8.1 inside the division bracket thing and the 5 outside. At this point, we can put our decimal point into place in our answer field. We located the decimal point between the 8 and the 1, and we write another decimal point in line with it on top in our answer field. Now that we have our decimal point in place in our answer, we can go ahead and divide the way we normally do. 8 divided by 5 is 1 with a remainder of 3. The 3 is smaller than 5, so I know 1 is the correct digit for that part of my answer. I multiply 1 by 5 to get 5 and I put 5 underneath the 8. I subtract 5 from the 8 to get 3. I bring down the 1. Now I divide 31 by 5 to get 6 with a remainder of 1. 1 is smaller than 5, so 6 is correct for that digit of my answer. I multiply 6 by 5 to get 30 and I write 30 on the bottom. I subtract 30 from 31 to get 1. I don't have any more numbers to bring down, but I have a remainder that I still need to finish dividing. Because I've already placed my decimal point in my answer, I can go ahead and add a 0 and drop that down to make 10. I then divide 10 by 5 to get 2 with no remainder. No remainder means I am done. So my answer is High five, Mr. Prof! Some things to remember about division are that we need to move the decimal point over until the decimal number we are dividing by is a whole number. However many spaces we move the decimal point in that number, we have to move the decimal point over the same number of spaces in the other number. Once we set up our problem, we place our decimal point on top in the answer field in line with the decimal point in the divisor  the number inside the division bracket. We then go ahead and divide like we normally do. Lesson Summary To recap what we've learned, decimal numbers are those numbers that have a decimal point in them. A decimal point is that dot placed before the part of the number that shows it has a portion that is less than a whole 1.
75 When adding and subtracting decimals, we set up our problem by lining up the numbers according to the decimal point and not according to the last digit. Right away, we place our decimal point in our answer in line with the other decimal points. Then we go ahead and add and subtract the way we normally do. We remember to always subtract the smaller number from the larger even if the problem is asking us to subtract a larger number from a smaller. If that is the case, then our answer will be negative. When multiplying, we set up the problem like normal, lining up the numbers by the last digit. We go ahead and multiply the way we normally do. And to find where the decimal point goes in the answer, we count the number of decimal places in the numbers we are multiplying together and we count that many spaces from right to left in our answer to find the location of our decimal point. When dividing, if we are dividing by a decimal number, we need to move the decimal point over to the right as many spaces as needed to turn that number into a whole. We then move the decimal point over the same number of spaces in the other number. We then set up our problem in long division form using the new numbers that we came up with. We place the decimal point on top in our answer field in line with the decimal point inside the division bracket. We then divide as we normally would. Solving Problems Using Decimal Numbers Chapter 3 / Lesson 4 Problems involving decimal numbers might sound like a way to torture you, but learning how important these types of problems are in the real world is important. In this lesson, you will see a couple areas in life where you routinely encounter decimal number problems. Decimal Number Problems Problems using decimal numbers are important in daily life; believe it or not. They're not just a way to torture you and your brain. Keep watching and you will see how you can apply your newfound skills to help you succeed as a student. Problems where you will see decimal numbers, numbers that contain a decimal point, include word problems, as well as straightforward math problems that just show you two numbers with the operation they want you to perform. Don't worry about the
76 word problems so much; I'm going to show you a way you can think about them to make them easier for you. Believe it or not, you might be doing these types of problems in your head already without even knowing it. How to Solve Them When you see a straightforward math problem such as =?, there are no questions about what needs to be done. What you see is what you do. But sometimes, it's hard to do the calculations because the numbers look confusing. When this is the case, think of areas in life where you see similar numbers and where you have succeeded in solving these types of problems. Then, take your success and apply it to your current problem. For example, if you are good with money and know right away whether something adds up or not, you can think of decimals in terms of money. So, for the problem , you can think of it as $ $3.15. If dealing with money is easier for your brain to handle, then think of ways to turn your problem into a money problem you can solve. If another method works, use that method. For word problems, the best way to solve these types of problems is to picture the problem in your mind and imagine the scene. What is going on and what are the important decimal numbers you need to work with? What operation is needed to solve the problem? What information is just extra fluff that you can ignore? Let's try solving a couple problems to see how this process plays out. The Change Problem Our first problem involves shopping. If you shop and pay for your own things, you might already be doing problems like this in your head. The problem we are given is this one: 'Sarah is shopping at the mall. She likes this one little shop, but today, the shop's computer system is down and everything has to be done on paper. Sarah has a $20 bill on her, and she wants to buy this gift set that costs $ Because the computers are down, the cashier has to figure out Sarah's change by hand. But Sarah, being careful of her money, has already figured out the change she is supposed to get. What is it?' After reading this problem, I first ask myself, what is going on and what numbers do I need to concern myself with? I can picture the scene in my head and I see that what I need to be concerned about is the $20 and the $18.97.
77 Now I ask myself, what operation needs to be performed to solve the problem? Sarah is looking for change, so my operation needs to be subtraction. What do I subtract from what? Sarah has $20, and she is going to spend $18.97, so I need to subtract $18.97 from $20. Everything else is just fluff that I can ignore now. The problem's looking pretty good now. I know how to subtract decimals, so that's what I will do. Since I am subtracting a decimal number with two decimal places, I will add a decimal to the end of my 20 and add two 0s to it so I can subtract properly. I place my decimal in line with all the other decimal points in my answer line. Then, I go ahead and start subtracting from right to left. I subtract 7 from 0, but since 0 is less than 7, I need to borrow a 1 from the number to the left of the 0, so I am subtracting 7 from 10. But since the number to the left of the 0 is also a 0, I need to keep going to the left until I see a number I can take 1 away from. I reached the 2, so now I can change the 2 to a 1 and the following 0s to 9s. Essentially, what I have done is to take 1 from 200, so I have 199 on the very top row now. So 107 is 3. Then, I subtract 9 from 9 to get 0. Then, I subtract 8 from 9 to get 1, and finally, 11 is 0, which I can ignore since it's a leading 0 to the left of the decimal point. So my answer is $1.03, which means Sarah needs to get that much in change back from the cashier. The Winning Score Problem Let's try another problem: 'James is a runner on the school's track team. In a recent event, he scored seconds on a sprint contest. There were four other runners competing in that event as well. If the other runners each scored , , , and , respectively, what place did James win?' So first, I see that I only need to concern myself with the scores of the runners, including James. What I need to do is to compare James' score to the scores of the other runners to see if James won first, second, or third place. In order to do that, I need to order my decimal numbers. Everything else in the problem I can ignore now. I know how to order decimals numbers by comparing the numbers starting at the decimal point and going from left to right. The process is similar to the way I would alphabetize two names. So, I go ahead and order all the decimal numbers and I get , (which is James), , , and Wow, what a close race!
78 I marked James' time so I can pick out his time easier in the lineup. I see that James is in second place. So that's what he won: second place. And that's my answer. Lesson Summary In review, we've seen how useful decimal number problems are in the real world. Recall that decimal numbers are the numbers that have a decimal point in them. Two places where you will see these types of problems are when shopping and in sports. To solve straightforward math problems, it might be helpful to relate the decimal numbers to something you can understand easier, such as money. For word problems, the key to solving them is to visualize them and figure out the important numbers you need to be concerned about. Then, think about what operation is needed to find your answer. The rest of the problem just becomes fluff at this point. What is a Percent?  Definition & Examples Chapter 3 / Lesson 5 Thinking in terms of pizza is an excellent way to think about percentages. Learn how it can help you understand percentages in this video lesson. You will also learn about the uses of percentages in the real world. What Is a Percent? Picture a big pizza with your favorite toppings on it. Now, I like pepperoni, so mine has lots of pepperoni slices. Now picture the pizza being sliced into fourths, then eights, then sixteenths. Yum! My stomach's rumbling  is yours? Picture holding a slice of pizza in your hand and you're holding a percentage of the pizza. So a percentage, or percent, is a part of a whole. The word has its roots in the Latin language, from the phrase 'per centum,' which literally means 'per hundred.' What Does It Mean? We can think of a hundred as a whole or all of something. A hundred percent of a pizza is the whole pizza. Half a pizza would be 50 percent, or half of a hundred percent. The percentage tells you how much of the whole you have.
79 If we divided our pizza into 100 little slices, then each slice would represent 1 percent of the pizza. This is another way to visualize percentages. You can take a whole of something and divide it into 100 little pieces and then figure out how many little pieces are in the portion you are interested in. If I wanted a quarter of the pizza, I would see that it would require 25 little slices out of 100, or 25 percent of the pizza. It is definitely possible to have a percentage that represents more than a whole or more than a hundred of something. What if you had a friend with a very good appetite and he happened to eat 2 whole pizzas? Let's visualize this. If we divided each pizza into our little 100 slices, how many slices did he eat? Yes, he would have eaten 200 little slices. That sounds like a lot, but your friend has a big appetite and he's won several eating contests because of it. Anyways, if he ate 200 little slices, then he would have eaten 200 percent of one pizza. We know that a hundred percent is a whole of something. Percents less than a hundred mean that they are less than a whole. Percents more than a hundred mean that they include a whole of something plus more of the something. Writing Percentages There are three ways to write our percentages. We can use the mathematical percentage symbol or we can write it in either decimal or fraction form. Let's take a look at how we can write 50 percent. Using the percentage symbol, it will look like 50%. The percentage symbol looks like two little zeroes with a slash separating them. In decimal form, 50% is In decimal form, 1 is the same as a 100%. In fraction form, 50% is 1/2. 1 in this form also means a 100%. If you divide 1 by 2, you will see that you get the decimal form of the fraction, Let's try writing 25 percent in the three different forms. Using the symbol, it's 25%. In decimal form, it is 0.25 and in fraction form, it is 1/4. Can you guess what 1 divided by 4 equals? Uses Now that we know how to write percentages, let's talk about where in the real world you will encounter them. The first that I'm sure you are familiar with is in school, when the teacher hands back a graded test. What do you normally see in red? Yes, you see your grade or your test score. Most times, you will either see the percentage
80 symbol or the fraction form. If you did really well on your test, you might see a score of 97% or 97/100 if there were 100 questions and you got 97 of them right. Another place that you will see percentages is when you go shopping, either at a physical store or online. Many stores will have sales or special discounts that'll save you money. You might see your favorite shirt on sale for a 20% discount. Or you might even see your favorite shoes on sale for 80% off. Which discount, the 20% or the 80%, will save you more money? That's right, the 80%, because that sale or discount means that you only pay 20% of the cost of the shoes, where you would have to pay 80% of the cost of the shirt. The higher the percentage discount, the more you save. Lesson Summary To review, percentage literally means 'per hundred;' you can also think of it as part of a whole. 100% is a whole of something. If you had more than a hundred, then you have more than a whole. You have a whole plus something more. If you had less than a hundred, then you only have a piece of the whole. The three ways to write your percentage are with the percentage symbol (%), in decimal form and in fraction form. Changing Between Decimals and Percents Chapter 3 / Lesson 6 Decimals As we wait in line for our roller coaster ride called Decimals and Percents, we see a couple of signs for us to look over. The first one talks about decimals. What makes a decimal a decimal? It's easy to spot a decimal. What you look for is a decimal point, because a decimal is any number that has a decimal point. A decimal point is that dot or point that you see in some numbers, especially those numbers you see on the price tag of your favorite pair of shoes. The number 3.14, for example, is a decimal because we see our decimal point, the dot between the 3 and the 1. Other examples of decimals include 0.99, and These are all decimals. Do you see all the decimal points?
81 Percents Our roller coaster line is moving along now, and we stop in front of our second sign that talks about percents. 'Percent' means 'per 100.' You write it as a number with either the word 'percent' after it or with the percent symbol (%) after it. And yes, percents can also be decimals if they have the percent symbol or the word percent after it. 50% is a percent, as is 50.5%. We're almost ready to board! Let's get going on our roller coaster ride now. This particular roller coaster has two big valleys with one big hill in the middle, and our roller coaster goes back and forth on this track. Our starting position is high up on the left side, and we're going to be headed down to the right side going down quickly into our first valley, over the hill, down into the second valley and back up to the end of the track on the right side. From Decimal to Percent We have boarded our roller coaster now and we see a number pop up overhead, a huge 4.12! What could this possibly mean? We hear the roller coaster announcer guy and he says we are about to embark on our journey from decimals to percents. Yes, this is the first leg of our roller coaster ride as we see how to go from a decimal to a percent. If our roller coaster is the decimal point, then our movement shows the movement of the decimal point as we make the conversion. Our starting point is We ride down our first valley and end up high on our hill. Our decimal has moved from 4.12 to We keep going and ride down our second valley and back up to the end of the track on the right side. Our decimal has moved again from 41.2 to 412. And look, we can now add either the percent symbol or the word to the end to get 412%. What we have just done is we have moved the decimal point two spaces to the right. Behind the scenes, the math we have just done is multiplication by 100. So to go from a decimal to a percent, we multiply by 100 so that our decimal point moves two spaces to the right.
82 From Percent to Decimal Whoa, we're moving backwards now! The roller coaster announcer guy says we are now leaving the world of percents and going back to the land of decimals. Now we're going to see how to go from a percent to a decimal. We see our percent, 412%, as we begin. But then as we start moving, we see the symbol disappear. We go back down the second valley and end up on the hill in the middle again. Whoa, our number has changed from 412 to But we haven't finished yet. We have one more valley to go before we reach the end of our ride. After going down our first valley again and stopping at the end of the ride, we see that our number has now changed from 41.2 to That was interesting. If our roller coaster is our decimal point, to change from a percent to a decimal, we move the decimal point two spaces to the left. Mathematically, what we are doing is dividing by 100. So to go from a percent to a decimal, we divide by 100 so that our decimal point moves two spaces to the left. Lesson Summary To recap our words, while the word 'percent' means 'per 100,' to identify them, we look for numbers with either the word 'percent' or the percent symbol (%) after them. For decimals, we look for numbers with a decimal point. For both conversions, you can either do the math or just remember the roller coaster as you move your decimal point two spaces to the right for changing from decimal to percent or two spaces to the left for changing from percent to decimal. And remember that only the percent will have the symbol. Decimals won't have the percent symbol. Solve Problems Using Percents Chapter 3 / Lesson 7 Percents are your friend. Think of all the cool things percents do for you. They save you money and they tell you how well you are doing in school. Learn how you can solve problems with percents in this video lesson.
83 The Awesome Percent Percents are awesome! Why are they awesome, you ask? Because they can tell me whether I will save loads of money or whether I have aced an exam. A percent, you may recall, literally means 'per one hundred.' You can also think of a percent as a part of a whole. When solving problems, working with percents can be a challenge. But let me show you a systematic approach that you can take to these problems so that they don't become confusing. To begin, let me show you some of the typical problems you may encounter. You might see some problems like these: What is 75 percent of 30? What percentage is 60 out of 130? I call problems like these straightforward ones because you can pretty much translate them directly into math symbols. Keep watching and I'll show you how. You might see slightly more complex problems, such as this one: If a certain shoe has been discounted by 55% and the original price is $218, what is the amount you will end up paying for the shoe after the discount? I'll also show you how to think about these problems so you can solve them easily, so keep watching. Before I explain how to do these problems, I want to tell you about the shortcuts that you should remember so that you can quickly translate your word problems into math problems. The Shortcuts The first shortcut is that when you see a phrase such as 'a percentage of something,' that 'of' translates into multiplication. So if you see a problem such as '75 percent of 20,' you should immediately translate that into 75 percent times 20. The second shortcut requires you to see the details in your problem. If you see a phrase such as 'a number out of another number' or 'a number of another number,' this 'of' translates into division. So if you see something like '65 out of 80,' you would write it mathematically as 65 / 80, or 65 divided by 80.
84 Just to recap, if you see 'a percent of something' you should write out the problem as the percent times the something. But if you saw 'something of something else,' then you would write it as something divided by something else. Remember these hints, because they can be a real lifesaver during tests. Now, let's see how we can use these shortcuts to solve the straightforward problems. Straightforward Problems Let's say that our problem is this one: What is 75 percent of 30? We can see the phrase 'a percent of something,' so we know that this 'of' means multiplication. We go ahead and rewrite the word problem into something mathematical that we can use. We write 75 percent times 30, but we can't just leave the percentage like that. We need to rewrite the percentage as a decimal. To change the number from a percentage to a decimal, all we need to do is move the decimal point two places to the left. So 75 percent becomes Our problem is now 0.75 * 30. This we can easily figure out. Our answer is So 75 percent of 30 is equal to Before trying another problem, I want to briefly talk about the usefulness and awesomeness of fractions. We can think of this problem as telling us the number of questions we would have to get right in order to pass a test. If our test had 30 questions, then to get a passing score of 75 percent, we would need to get at least 22.5 questions right. If our teacher didn't give half scores, that means we would need to get at least 23 questions right. Okay, let's try another problem: What percentage is 60 out of 130? Does this 'of' mean multiplication or division? Well, we have a number out of another number, so this 'of' means division. We rewrite our problem as 60 / 130. Ah, much better. We can easily solve this one and get our answer. After dividing, we find that our answer is This is a decimal, but our problem is looking for a percentage. We will need to translate our decimal into a percentage. We do that by moving the decimal point two places to the right. Doing so, we get 46 percent. So that means that 60 out of 130 is 46 percent. Here again, if our test had 130 questions and we only got 60 answers right, then that means our score is only 46 percent. We can use our knowledge of percentages and
85 how to solve percentage problems to figure out if we are likely to get a good score or not by the number of answers we think we are getting right as we're taking a test. Let's move onto to more complex problems. More Complex Word Problems We are looking at a test and we see this problem: If a certain shoe has been discounted by 55 percent and the original price is $218, what is the amount you will end up paying for the shoe after the discount? This looks long, but with some logical thinking on our part we can successfully tackle problems like this one. After reading this problem carefully, the first thing we want to do is to figure out exactly what they want. In this problem, what they want us to do is tell them how much we will end up spending. So they are not asking for the amount that was discounted. If that is the case, then they are not interested in the 55 percent but in the percentage that will make the 55 percent whole. To figure that out, we will take the 55 and subtract it from 100 since 100 percent is the whole. So, is 45 or 45 percent. So our actual problem is 45 percent of 218. Now this looks like a straightforward problem. We see the phrase '45 percent of 218,' so we know we can write it as 45 percent times 218. Again, we can't leave the percent like that; we need to rewrite it as a decimal. We remember that to switch from a percent to a decimal, we move the decimal point two spaces to the left. So now our problem is 0.45 * 218. Our answer, then, is 98.1, or $ What this tells us is that if a pair of $218 shoes is on sale at 55 percent off, then that means we are paying 45 percent of the $218, which is $ a pretty good deal. The most important part that I want you to take away from this is that you really need to think about what the problem is asking you to do. Sometimes the problem wants you to calculate the discount or the amount you would save; other times it can ask you what you would end up paying. If the problem had asked for the discount, we would have had 55 percent of $218 as our problem. Lesson Summary To summarize, a percent is a part of a whole and literally means 'per one hundred.' When working with them in solving problems, there are two shortcuts you should
86 remember. The first one is when you see a phrase such as 'a percent of a number,' the 'of' translates to multiplication. For example, 45 percent of 80 translates into 45 percent times 80. The second shortcut is that when you see a phrase such as 'a number out of another number,' the 'of' here means division. For example, the phrase '124 out of 516' translates into 124 / 516, or 124 divided by 516. And finally, to solve problems involving percentages, you first need to figure out what is being asked of you and then you can go about translating your problem into math. Changing Between Decimals and Fractions Chapter 3 / Lesson 8 Decimals and Fractions Decimals and fractions don't have to be a part of your worst math nightmare. Being able to switch between the two will actually help you solve problems faster. Let's begin our little trip into the world of decimals and fractions by briefly going over what decimals and fractions are. If we see something like 5.4 or even 1.12, I can immediately tell you that those are decimals because decimals are the numbers that have a decimal point in them. Do you see the dot or point in both those decimal numbers? If, on the other hand, I see something like 1/2 or 51/100, I can tell you that those are fractions because fractions are the division of two whole numbers. Fractions always have that slash between two regular numbers. Why Useful So, why is being able to change from one to the other useful? The biggest reason is when you are solving math problems. If a particular math problem wants the answer in decimal form, then it will be easier to work out the problem using decimals. If, on the other hand, the math problem wants the answer in fraction form, then working the problem out using fractions will be easier. Another reason is that sometimes an answer is easier to understand if it is in one form over the other. For example, if a teacher told you that you got 0.97 answers right on the test, would that mean much to you? Or would it mean more if the teacher told you that you got 97/100 answers right on the test?
87 Decimal to Fraction We've seen how useful being able to convert between the two is. Now let's talk about how to actually do it. The process of changing a decimal into a fraction is a twostep process. The first step is to convert the decimal part into a fraction. The second step is to simplify that fraction and add it to the part before the decimal point. A good way to remember these two steps is to view the decimal number in two parts. View the part before the decimal point as a whole number telling you how many pies you have. Then view the part after the decimal point as telling you how much of a whole pie you have. So, a decimal number essentially gives you the number of whole pies plus a portion of another pie. To change this into a fraction, you would turn the partial pie into a fraction and add it to the number of whole pies to get your total pies. Let's see how this works with a real decimal number. Let's turn the decimal 1.25 into a fraction. We are going to picture our decimal in two parts: the part before the decimal point and the part after the decimal point. We have 1 before and 0.25 after. So, if we change these into pies, we have 1 whole pie and 0.25 of another pie. How do we turn the 0.25 into a fraction now? What we need to do now is to use our knowledge of place values to help us out. We know that the place values after the decimal point start at tenths, then goes to hundredths, then thousandths, and so on. For our 0.25, the 2 is in the tenths place and the 5 is in the hundredths place. Because the decimal ends at the hundredths place value, we're going to use this information to turn our decimal into a fraction. Hundredths means divided by a 100, so we will divide our decimal by a 100 without the decimal point. So we will do 25 / 100. Look at that! Doesn't that look like a fraction? Yes, it does. Our next step is to simplify this fraction. Does 25 divide evenly into 100? Yes, it does. How many times? 25 goes into 100 four times, so we can simplify our fraction to 1/4. Because 25 divides evenly into 100, we can divide both the top and the bottom by divided by 25 is 1 and 100 divided by 25 is 4. We then add our simplified fraction to our whole number to get 1 + 1/4. At this point, we can stop if the problem wants the fraction in mixed form and write 1 1/4 as our answer. Otherwise, if the problem wants the answer in standard form, we will have to add the two together. We will need to find common denominators and then add the
88 fractions together. Our common denominator in this case is 4. Changing the 1 so that we have a denominator of 4 gives us 4/4. 4/4 + 1/4 gives us 5/4 as our standard form answer. Now we are done. There are a couple things I want to draw your attention to. The first is the part where we turn the decimal into a fraction by dividing by the place value. We always divide by the place value of the last digit in the decimal. So, if the last digit is in the tenths place value, then we divide by 10 and if the last digit is in the thousandths place value, then we divide by 1,000. The second thing is that if we have a decimal that never ends and whose numbers never repeat, then we can't turn it into a fraction. We can only turn decimals that end or decimals with a repeating pattern into fractions. A number like pi cannot be turned into a fraction because it has no ending. It keeps going and going and going. The approximation of pi, 3.14, can be turned into a fraction because it has an ending. It ends at the 4. Fraction to Decimal Now that we know how to turn a decimal into a fraction, let's now see how easy it is to turn a fraction into a decimal. There is only step involved in this process and that is to divide. The easy way I use to remember this step is to just look at the slash symbol. What does that tell me to do? It tells me to divide. So I will. For the fraction 1/4, to turn that into a decimal, I will simply divide 1 by 4. I can either use a calculator or do it by hand using long division. My answer is The fraction 1/4 is 0.25 as a decimal. Lesson Summary In review: Decimals are numbers with a decimal point and fractions are two whole numbers with a slash in between them. To convert from a decimal to a fraction, we view the decimal number in two parts, the part before the decimal point as a whole number and the part after the decimal point as a part of a whole. We then turn the part of the whole into a fraction by dividing by the last place value of the decimal and simplify. Then we add this part to the whole part to get our answer in standard form.
89 One important thing to remember is that we can't convert a neverending decimal whose digits never repeat into a fraction. To convert a fraction into a decimal, simply look at the slash and do as it tells you to do. Divide the two numbers to get your decimal. Use either a calculator or do it by hand using long division.
90 Evaluating Square Roots of Perfect Squares Chapter 4 / Lesson 1 The Language of Math Mathematics has its own special language. To understand and succeed in math, you need to be able to translate the language of math into English and back again. Just like you wouldn't move to France without knowing how to communicate at least the basics, you shouldn't venture into the study of math without knowing how to translate the terms you are using. Squares and Square Roots The section of math dealing with radicals, or squares and square roots, is one place where knowing the translations of the terms is critical to understanding. To square a number means to multiply that number by itself. It is notated by a superscript number 2 after the main number. It can also be written with a caret (^) between the number being squared and the 2. When you see this symbol, you say that the number is squared. For example, 3^2 is = to 3*3 or 9 and 10^2 is = to 10*10 or 100 The square root operation is the inverse of the squared function and is notated by the symbol you see on the screen:
91 This means that finding the square root of a number is the same as finding the opposite of a number squared. For example is 10 because 10^2 is 100 and 9 is 3 because 3^2 is 9 Perfect Square A number is a perfect square if it is an integer that is the square of another integer. An integer is a number that does not contain a fraction or a decimal. For example  5^2 is 25 So 25 is a perfect square. And 17^2 is 289 Which means 289 is also a perfect square. Finding the Square Root of a Perfect Square You can use a calculator to find the square root of a perfect square, but, if a calculator is not available, there is a way to calculate the square root. Here are the steps to calculating the square root of a perfect square: 1.) Factor the number completely. An easy way to factor a number is by using a factor tree. A factor tree can be created by writing down the number you want to factor and drawing two lines coming down from that number. Then, write two factors of that number under the lines. Continue on until only prime numbers remain. A prime number is one that cannot be reduced any smaller. The purpose of the factor tree is so you can easily find the square root of large numbers if you don't have a calculator handy.
92 By looking at this factor tree, you can see that the factors of 225 are: 3 * 3 * 5 * 5 Step number 2 is to match up pairs of factors. In this case, we have a pair of 3s and a pair of 5s: (3 * 3) * (5 * 5) Then step number 3 to calculating the square root of a perfect square is to multiply one number from each pair of factors together to get the answer. In this case: 3 * 5 = 15 Fifteen is the square root of 225. For example, let's find the square root of 144. Your first step is to factor 144. The prime factors of 144 are (2*2) * (2*2) * (3*3) Your second step is to match up the pairs. There are 2 pairs of 2s, and 1 pair of 3s. Then, for step 3, take out 1 number from each pair, and multiply them together. A 2*2*3, which equals 12 Therefore, the square root of 144 is = to 12. Therefore 144 is = to 12 Let's try another one. Solve 400 Again, your first step is to factor 400 by using a factor tree. The result you get is (2*2) * (2*2) * (5*5) Secondly, match up pairs of numbers. There are 2 pairs of 2s, and 1 pair of 5s. Finally, multiply together the 2*2*5 to get 20 So the 400 = 20 Conclusion Squares and square roots are inverse operations. A perfect square is the square of a number that does not contain a fraction or decimal. To determine the square root of a perfect square, all you need to do is factor the number, combine pairs of factors and
93 then multiply one number from each pair together. You can always check your work by performing the inverse operation, or squaring the number to see if your answer matches up with the original square root. Estimating Square Roots Chapter 4 / Lesson 2 Square Roots In mathematics, the square root operation is the opposite of squaring a number. To square a number, you multiply that number to itself. To find the square root of a number, you need to find the number that when multiplied to itself equals the original number. The symbol for square root looks like this: Perfect Square Roots A perfect square root is a number whose square root is an integer. An integer is a number that is not a fraction or a decimal. For example, the square root of 25 is equal to 5 because 5^2 is 25, and the square root of 121 is equal to 11 because 11^2 is 121. Square Root of an Imperfect Square A number whose square root is not an integer is an imperfect square. When finding the square root of these numbers, the answer will not be an integer, but will be a fraction or decimal. For example, the square root of 27 is equal to 5.196, and the square root of 215 is equal to 14 2/3. To determine the square root of an imperfect square, you can use a calculator. However, you can also estimate the square root of a number.
94 Estimating the Square Root Here are the steps to estimating the square root of a number to two decimal places: One  Determine which two perfect squares your number falls between. The answer you are looking for will fall between these two numbers. Step Two  Take a guess about the number just after the decimal place. Step Three  Divide your guess by the number whose square root you are trying to determine. Step Four  Find the average between your guess and the answer to the division problem you did in Step Three. And lastly, Step Five  Repeat Steps Three and Four until the two numbers you are averaging are the same. Let's do a few examples; that should help everything make sense: Estimate the square root of 10 to two decimal places. Step One  The square root of 9 is equal to 3 and the square root of 16 is equal to 4, so our answer to the square root of 10 will be between 3 and 4. Step Two  10 is much closer to 9 than it is to 16, so my first guess on the answer will be 3.2. For Step Three, I divide 10 by the guess of 3.2, which gives a result of For Step Four, I take the average of 3.2 and 3.13, and I get Now (for Step Five) I repeat Step Three with divided by 3.17 is equal to And then I average 3.17 and 3.15 to get Going back again to Step Three, I divide my original number 10 by my latest average of 3.16, and when I do that, I get 3.16 again. Because this division gives us the same result, our estimation is finished and the answer to our problem of 'what is the square root of 10' is Try this second example for yourself: Estimate the square root of 22 to two decimal places. Did you get an answer of 4.69? If you did, great! If not, don't despair. Here are the steps we took to get there:
95 When I solved this problem, my first guess was 4.5 because 22 is between 16, which is 4^2 and 25, which is 5^2. Next, I divided 22 by 4.5 to get Then I took the average of 4.5 and 4.89 to get Going back and dividing 22 by 4.70, I got Then the average of 4.70 and 4.68 is When you divide 22 by 4.69 you get Therefore, the answer to the question, 'what is the square root of 22' is Remember, you can always check your answer either on a scientific calculator or by multiplying your answer to itself and see if you end up with the original number. Lesson Summary Not all numbers have perfect square roots; some are imperfect, containing fractions or decimals. You can estimate the imperfect square root of a number using a guess and check method. It can be used on both small and large numbers. It is fairly quick and can be done using just some simple math. Simplifying Square Roots When not a Perfect Square Chapter 4 / Lesson 3 Rules of Mathematics The study of mathematics is universal. It is studied the same whether you are in America, Russia, India or any other corner of the globe. To make it easy for all these mathematicians who speak different languages and come from different backgrounds, there are universal rules that apply to writing mathematical equations and the way in which all math is formatted. This way, mathematicians in Korea can understand the work that is being done in Canada. One of these rules relates to how the square roots of imperfect squares are written. The proper mathematical way to write imperfect square roots is by simplifying them
96 as much as possible without using a fraction or a decimal. This will result in numbers like the following: 4 * sq. root of 2 and 2 * sq. root of 7 Before we can work out how to simplify imperfect square roots, we need to do a bit of review about perfect and imperfect square roots. Perfect Square A perfect square is a number whose square root is an integer. An integer is a number that does not contain a fraction or a decimal. The table on the screen shows the first 10 perfect squares. Imperfect Square As you can see, only a few numbers are perfect squares. The rest can be classified as imperfect squares. Imperfect squares are numbers whose square roots contain fractions or decimals. For example: Sq. root of 20 = 4 ½ and Sq. root of 77 = 8.77 Evaluating the Square Root of an Imperfect Square A calculator will not simplify the square root of an imperfect square properly. It will give a result, but it will be in decimal form. So, to evaluate and simplify an imperfect square, you need to follow these steps: 1. Factor the number completely. An easy way to factor a number is by using a factor tree. A factor tree can be created by writing down the number you want to factor and drawing two lines coming down from that number. Then, write two factors of that number under the lines. Continue on until only prime numbers remain. A prime number is one which cannot be reduced any smaller. The purpose of the factor tree is to determine which numbers can be removed from under the square root symbol.
97 2. Match up pairs of the same number. Any numbers with a partner are perfect squares, and you can take the square root of these numbers. 3. Numbers without a partner remain under the square root. These numbers cannot be simplified further. For example, let's simplify the square root of 48. First, we factor 48. Then, we match up the pairs. As you can see, there are two pairs of 2s. We can simplify both of these as 2 * 2 * 2 * 2, which is 16, and the square root of 16 = 4. This means that there will be a 4 outside the square root symbol when we're done. The 3 does not have a partner, so it will remain under the square root symbol. Therefore, the square root of 48 simplifies to 4 * sq. root of 3. Let's try another example. Simplify the square root of 450. By writing out a factor tree, you can see that 450 can be factored to (3 * 3) * (5 * 5) * (2). As you can see, there is one pair of 3s and one pair of 5s. The pair of 3s and pair of 5s, when multiplied together, equal 225, which has a perfect square root of 15. This means that a 15 will be removed from the square root symbol. Because the 2 does not have a partner, it will remain under the square root symbol. Therefore, the square root of 450 simplifies to 15 * sq. root of 2. Conclusion The proper mathematical way to write imperfect square roots is to simplify them as much as possible without using a fraction or decimal. This is done by factoring the number as much as possible and then moving those numbers with a partner. This can be done because the square root of a number is defined as a number that, when multiplied by itself, gives you the original number. Any number without a partner cannot be simplified and will remain under the square root symbol. Simplifying Expressions Containing Square Roots Chapter 4 / Lesson 4
98 Math Is Universal No matter where you live, the language of mathematics is the same. You might not be able to order dinner in a restaurant in South America, but by using the unique language that is math, you can work problems with anyone from any country. This is why it is very important that mathematical equations get treated the same no matter who is writing them. The Language of Radicals The term 'radical' is just another way to say 'square root.' When writing square roots in correct mathematical language, it is important that every radical is written in its simplest form. This applies to both the numbers and variables that are under the square root symbol. On the screen you see some examples of radicals that contain both numbers and variables. Some of them are simplified and others aren't. Can you tell which examples need to be simplified further? Of these four examples, numbers 1, 3 and 4 can all be simplified further. Simplifying Square Roots of Numbers Here are the steps to simplifying a square root with a number: 1. First, factor the number completely. An easy way to factor a number is by using a factor tree. A factor tree can be created by writing down the number you want to factor and drawing two lines coming down from that number. Then, write two factors of that number under the lines. Continue on in this manner until only prime numbers remain. A prime number is one that cannot be reduced any smaller. The purpose of the factor tree is to determine which numbers can be removed from under the square root symbol. 2. Match up pairs of the same number. Any numbers with a partner are perfect squares and you can take the square root of those numbers. 3. Numbers without a partner remain under the square root symbol. These numbers cannot be simplified further.
99 Let's return to the examples from earlier and look only at the number portion of some of them. The first example is 81x^4. Looking at just the number portion, factor 81. 9*9 is 81, and then 3*3 is 9; therefore the factorization of 81 is 3*3*3*3, which is 2 groups of 3. Each group means a 3 will be removed from underneath the radical, which means that the square root of 81 is 3*3, or 9. The 13xy is the second example. Since 13 is a prime number, it cannot be factored and therefore is as simple as it can be and no changes can be made. Simplifying Square Roots of Variables Before we can talk about finding the square root of a variable, we should probably review what exactly the square root is. The square root operation is the opposite of squared. This works for both numbers and variables. So x*x = x^2, and by performing the opposite operation, x^2 = x. Simplifying square roots of variables works about the same way as it does with numbers. Just like you can factor numbers, variables with exponents can also be factored. For example, x^4 is the same as x*x*x*x Then, you can proceed the same way as simplifying the square root of numbers. 1. First, match up pairs of the same variable. Using our example from before, grouping pairs of xs gives us (x*x)*(x*x)  two groups of xs. 2. Any letters with a partner are perfect squares, and you can take the square root of them. In this case there are two pairs of xs, so there will be two xs removed from under the square root symbol. 3. Variables without a partner remain under the square root. They cannot be simplified further. Going back again to our examples from earlier, let's look this time at the variable portion. The first example is x^4. If we factor x^4, we get x*x*x*x.
100 Break that into pairs to get (x*x)*(x*x). One x from each pair is taken out from under the square root symbol. This leaves us with an x^2 outside of the square root. The final simplification of 81x^4 is 9x^2. The second example 13xy, and as you can see, there is only one x and one y under the square root symbol. This means that it cannot be simplified any further; it is as simple as it can get. Put It All Together Let's try the last two examples to put it all together. Example number 3 is (13(x^6')y^2). As before, the first step is to look at the number  in this case, 13. Since 13 is a prime number, it cannot be factored. Next, we look at the variables. x^6 can be factored to (x*x)*(x*x)*(x*x). Because there are three groups of xs, three xs will come out from underneath the radical symbol. y^2 is y*y, which is one group of ys. Therefore, one y can be removed from under the square root symbol. So the simplification of this problem is (x^3)y 13 Let's try the last one: ((8x^2)(y^4)z). Starting with the number, factor 8 to its smallest parts, which is 2*2*2. One pair of twos means that a 2 will come out of the square root symbol. The lone two will stay under the square root symbol. Next, we move on to the variables. x^2 is x*x, which is a pair of xs, meaning that one x will come out of the square root. y^4 is y*y*y*y. The two pairs of ys mean that 2 ys will come out of the square root. Because the z is alone, it will stay underneath the square root symbol, which means that the simplification of ((8x^2)(y^4)z) is 2xy^2 (2z). Lesson Summary In the international world of mathematics, it is important that there is a universal language so that problems can be understood all over the world. Part of this language
101 means that radicals, or square roots, are simplified in a certain way. To simplify a square root, first factor the numbers or variables, then pair up the like terms. Finally, remove one of each pair from under the square root symbol. Any numbers or variables without a partner remain under the square root symbol.
102 Statistical Analysis with Categorical Data Chapter 5 / Lesson 1 Learn why percentages are so important when analyzing categorical data in this video lesson. Watch as the data is turned into a data table and a visual bar graph as ways to analyze the data. What Is Categorical Data? When you take a survey or fill out application forms at various places, you come across categorical data. So, what exactly is categorical data? It is the kind of information that can be categorized. For example, your race, gender, and occupation are all different types of categorical data. Your answer for race can be categorized into groups such as Asian, Caucasian, etc. For occupation, your answer can be categorized into groups such as teacher, student, artist, etc. Data as Percentage With this type of data, part of the analysis process involves changing your data into percentages. Let's work through an example scenario to see how the analysis process works. Our scenario is that we have just surveyed a group of 100 people about their natural hair color. After going through all the data, we found that 30 people had brown hair, 20 people had blonde hair, 40 people had black hair, and 10 people had red hair. Notice how we were able to group the people in our survey into just a few groups. For each person that answered a certain way, we added a 1 for that group. Now that we have this information, we need to analyze it and present it in such a way that makes it easy to understand and use. Having just our numbers doesn't do much for us. But, if we change the numbers to percentages, we can gain a better understanding of what's going on.
103 To change our numbers to percentages, we take the number from each group and divide it by the total number of data and then we convert this decimal into a percent by multiplying by 100. For our brown hair, we divide 30 by 100 to get multiplied by 100 is 30%. For blonde hair, we get 20%. For black hair, we have 40% and for red hair we have 10%. Data Table Now that we have our percentages, we need a way to present it to others for it to make more sense. One way we can present it is with a data table, which is a way to organize the information into rows and columns. We will present it with a title row and our information in two columns. The title row will state what each column is for. The first column is for 'Hair Color' and the second column is for the 'Result'. We write our groups in the column for Hair Color and we write our respective percentages in the next column. Hair Color Result Brown 30% Blonde 20% Black 40% Red 10% We can glance at our finished table and quickly gain the information we need. We can look at it and easily see that 40% of our population has black hair. We can use this information for business purposes if we wanted to market hair accessories for people. If we know that the majority of our population has black hair, then we will produce more accessories that match with black hair. Bar Graph Another way we can present our information, which will make it easy to analyze, is with the use of a bar graph, a graph that shows our data using bars. The way we create a bar graph is by writing our various groups on the xaxis and then we draw bars of various heights to correspond with the number of people in each group. Our first group we write down is Brown and our bar has a height of 30 because that's how many people are in this group. We do likewise with the rest of the groups. Blonde has a bar height of 20, Black has a bar height of 40, and Red has a bar height of 10. We
104 can easily look at this and see which group is more popular and which group is the minority. In my bar graph, I decided to keep the numbers instead of using percentages. Why? Because in a bar graph, it is easy to see which bar is taller than another. In table form though, it is easier to understand the information if it is presented using percentages. You could also use the percentages for your bar heights in your bar graph. You can use whichever one you feel will convey the information in the best manner. You can graph both to compare and then choose the one that is easiest to read. Lesson Summary In review, categorical data is data that can be categorized into groups. Examples include gender, occupation, and race. The two ways to analyze this information is with the use of either a data table, information presented in rows and columns, or a bar graph, a graph with bars of various heights. For a data table, while you can report your data using the numbers for each group, most times it makes more sense to report the groups using their percentages. For the bar graph, you can graph it using either the percentages or the numbers for each group. You choose the one that makes more sense for your situation. Graph both to see which one is easier to read and understand. Understanding Bar Graphs and Pie Charts Chapter 5 / Lesson 2 In this lesson, we will examine two of the most widely used types of graphs: bar graphs and pie charts. These two graphs can provide the reader with a comparison of the different data that is displayed. Understanding Bar Graphs & Pie Charts One of the more fascinating ways to record data is to put the data in a graph. Two of the more popular graphs are bar graphs and pie graphs. In this lesson, we will examine how to read and interpret the information that can be found on these graphs.
105 Bar Graphs A bar graph is a graph that can be used to compare the amounts or frequency of occurrence of different types of data. Bar graphs are helpful when comparing groups of data and comparing data quickly. Let's look at the important parts of a bar graph. All bar graphs have a graph title, which gives the reader a brief overview of the type of data that the graph contains. The graph title in this graph is 'Percentage of U.S. Population That Is ForeignBorn.' Bar graphs also contain two axes that are labeled and also have a scale. The axis that is labeled at the bottom of the bars is referred to as the group data axis. This axis details information about the type of data that is displayed. The other axis in a bar graph is called the frequency data axis. This axis has a range of values that measures the frequency that the data occurs. The most important part of a bar graph is the bars. These bars provide an instant comparison of data in a graph. The reader can easily compare the data to see which measure occurs the most often and if any data points have similar frequencies. Let's look at a different bar graph to sharpen our skills on labeling its parts. Looking at this new graph, there is one obvious difference. This bar graph is going horizontal instead of vertical like the first example. The title of this graph is 'Sports Shoes Sales' and is located at the top of our graph. The group data axis is located along the left side, which is directly under the bars. The title of the axis is 'Types of shoes.' The frequency data axis is located at the bottom of the graph and is labeled 'Average cost.' The bars of this graph are running horizontally. Reading a Bar Graph When reading a bar graph, it's important to pay attention to the intervals used on the frequency data axis. An interval is the amount of data that occurs between each section or tick mark. Looking at the graph 'Sports Shoes Sales,' we can see that the frequency data axis is located at the bottom of the graph. Along this axis, we can see each tick mark represents $10. That is important, because there is data between each of these tick marks. Data located between two marks is often estimated to the closest value. For example, look at the bar for 'Gym Shoes.' The bar for 'Gym Shoes' ends between the $20 and $30 marks. The bar appears to be more than halfway between these two
106 points, so a good estimate would be either $26 or $27. To read a bar graph, find the point where each bar would meet the frequency data axis. Let's look at the 'Favorite Syndicated Programs, ' This graph is also displayed horizontally, so the frequency data axis is located at the bottom of the graph. Which program had almost nine percent of TV households viewing? The answer is Jeopardy. Looking at the Jeopardy bar, it is slightly more than nine percent, so a good estimate would be about 9.5%. Which program had the least percentage of TV households viewing? The answer is The Jerry Springer Show. The Jerry Springer Show had the shortest bar and lies between three percent and six percent. A good estimate would be 4.5%. Approximately what percent of TVviewing households watched Friends? Friends had slightly more than six percent of viewers. A good estimate for this would be 6.2% of viewers. Pie Charts Shifting gears slightly, let's look at another popular type of graph used to display and compare data. Pie charts (also referred to as a circle graph) are circular graphs used to show the relationship of a part to a whole. A pie chart displays its data in sectors, which are parts of the circle and are proportional to the other parts displayed in the graph. Pie chart values are represented by percentages, with each chart representing 100%. Let's examine a pie chart and identify the important parts. Looking at this pie chart, the first important part is the graph title. The graph that we are looking at is titled 'What Tops Pizza.' It compares the most popular pizza toppings in relationship to one another. The title and the creative picture used for the pie chart give the reader information about the type of data displayed in the chart. Another important part of this pie chart is the sectors. These sectors are differentsized areas that divide up the circle into proportional parts. This is such a tasty graph, let's now examine how to read the information that it displays. Reading a Pie Chart Each sector in a pie chart is labeled with the sector title and its numerical data. In this chart, each sector is labeled with its percentage. All pie charts represent a whole
107 amount that equals 100%. If you were to add up the percentages of each sector, it would equal 100%. For example, let's look at the pizza pie chart. The percentages listed in each sector are 13%, 14%, 19%, 43%, four percent and seven percent. By adding all of these percentages together, 13% + 14% + 19% + 43% + 4% + 7% = 100%. The size of each sector is also proportional to its percentage. For example, the largest sector is 'Pepperoni' at 43%. Since the 'Pepperoni' sector is the largest percentage, its sector is also the largest size. Using the data in the pie chart, we can calculate other values. For example, there were 1000 people surveyed to create this chart. We can calculate the number of people who preferred each sector. In this pie chart, 14% of the people surveyed preferred mushrooms. To calculate the number of people who chose mushrooms in this survey, we would set up a proportion. The proportion would show x number of people out of 1000 chose mushrooms and 14% out of 100%. To solve this equation we will crossmultiply. Multiply 100 * x = 100x and 14 * 1,000 = 14,000. So, 100x = 14,000. To solve this equation, divide by 100: 14, = 140. There were 140 of 1,000 people surveyed who selected mushrooms. Examples and Practice Problems Now let's look at a few more examples of these two important types of graphs. Let's first look at a fascinating bar graph about the bestselling games in the U.S. titled 'Game Crazy.' Which game is the most popular game in U.S. history, and how many units have been sold? Looking at the graph, we can see that Uno has the tallest bar, meaning it is the most popular game. The bar for Uno is almost two million. A good estimate for the number of Uno games sold is 1.9 million. Which game has sold almost 1.3 million games? Looking at the bar on the left, 1.3 million would almost be between one million and 1.5 million. As I follow a line over between these two values, the Jenga bar appears to be closest to 1.3 million. The other type graph that we have talked about is a pie chart. Let's look at a pie chart that examines the earnings of college students beyond the classroom. The title of this pie chart is 'College Students' Earnings.' How many students of the 2,000 students that were surveyed have no job? To calculate this total, we will set up a proportion x students over 2,000 students surveyed equals 33% over 100%.
108 To solve this equation, we will crossmultiply. 100 * x = 100x and 33 * 2,000 = 66,000. So 100x = 66,000. To solve this equation, divide by 100: 66, = 660. There were 660 students out of 2,000 surveyed who did not have a job. Lesson Summary Let's review these two types of graphs. Bar graphs are graphs that can be used to compare the amounts or frequency of occurrence of different types of data. To read a bar graph, find the spot on the frequency data axis where the bar would meet. A pie chart can also be called a circle graph. A pie chart is a circular graph used to show the relationship of a part to a whole. All of the sectors in a pie chart add together to equal 100%. The data in a pie chart can also be used to calculate other data. Summarizing Categorical Data using Tables Chapter 5 / Lesson 3 Categorical Data Categorical data is data that can be categorized or grouped. You see examples of categorical data all the time whenever you fill out applications for anything. When it asks you for your gender and occupation, it is asking for categorical data. Once you answer those questions, whoever reads the application can place you into a group based on your gender or occupation. You can be grouped into the student group or the groups can incorporate both sets of data and you can be grouped into the female student group if you're a girl or the male student group if you're a boy. While the answers to a piece of categorical data may not be numerical in nature, once a survey is done, the various groups can be counted to see how many are in each group. It is this information that we will focus on. Prepping Your Data Let's see how we go about summarizing our categorical data with a little scenario. Imagine that we have just surveyed a random group of people about their favorite junk food. Each person we survey chooses his or her favorite junk food from a list of
109 five choices. Our choices are: potato chips, pizza, hamburgers, hot dogs, and fries. After our survey is done we have a stack of papers of everyone's choices. To prep our data so we can summarize it, we now need to count how many are in each group. We go through our stack of responses and we separate them into various groups based on their answers. We then count the number of papers in each group. We found that 15 people chose potato chips, 23 people chose pizza, 18 people chose hamburgers, 8 people chose hot dogs, and 10 people chose fries. Data Table We can take our group counts and input them directly into a table. Our table will have two columns: one for the type of junk food and the other for the result. Our top row will be our title row with 'Junk Food' and 'Result' as our titles for each respective column. Junk Food Result Potato chips 15 Pizza 23 Hamburgers 18 Hot dogs 8 Fries 10 We can reorder our data table so that the most popular is listed first and the least popular is listed last. Junk Food Result Pizza 23 Hamburgers 18 Potato chips 15 Fries 10 Hot dogs 8 This is one way we can use a data table to summarize our information. We can clearly see which junk food is the most popular out of our survey group and which is the least favorite. This does give us good information, but if we wanted to generalize our
110 information to the general public, we would need to report our results in percentage form. Using Percentages In order to change our data to percentages, we need to know the total number of people we surveyed. We can find out our total by adding up the number of responses: = 74. To find out the percentage of each group, we take the number of each group and divide by our total. We then convert this decimal to a percentage by multiplying by 100. So, for pizza, our percentage is 23 / 74 * 100 = 31%. For hamburgers, it is 24%. Potato chips has 20%, fries has 14%, and hot dogs has 11%. We'll leave our data table the same, but just switch out our numbers for the percentages. To check that we've converted properly, the total percentage must equal 100%. Let's check: 31% + 24% + 20% + 14% + 11% = 100%. Our percentages are correct. If our total is one percent off, we'll need to go back and check which one we can round up by 1%. Choose the one that is closest to being rounded up. For example, if two of our choices were 31.3% and 31.4%, we would choose to round the 31.4% to 32% to account for the 1% difference. Junk Food Result Pizza 31% Hamburgers 24% Potato chips 20% Fries 14% Hot dogs 11% These percentages allow us to generalize our information. We can say that 31% of people will choose pizza over the other four choices and we can say that only 11% of the people find hot dogs to be a favorite.
111 Lesson Summary We've learned that categorical data is data that can be grouped. We can summarize our information using a data table in two ways. We can use the counts from each group or we can convert our count into percentages. While we don't necessarily have to order from greatest to least, we can do so to make our data table more readable. How to Calculate Percent Increase with Relative & Cumulative Frequency Tables Chapter 5 / Lesson 4 In statistics, one way to describe and analyze data is by using frequency tables. This lesson will discuss relative and cumulative frequencies and how to calculate percent increase using these two methods. Statistics Are on the Rise When you decide to move to a new home, whether it is in the same city or across the country, one of the first things that you might do is look at the crime statistics for that area. Are they on the rise, or are they declining? You might also look at the schools. How do they stack up against other schools? Are their scores on the way down, or are they rising year after year? All of this information, and information on almost any topic you can think of, can be found using statistics. Statistics is the study of the collection and analysis of data. Relative and Cumulative Frequency In mathematics, frequency refers to the number of times a particular event occurs. There are two types of frequency: relative and cumulative. Cumulative frequency is the total number of times a specific event occurs within the time frame given. Relative frequency is the number of times a specific event occurs divided by the total number of events that occur. Let's use an example: Your soccer team ended the season with a record of 15 wins and 3 losses. The cumulative frequency of your wins is 15 because that event occurred 15 times. The
112 relative frequency of wins is 15 divided by 18, or 83%, because out of the 18 total games (or events) your team won 15. Relative frequency is a good way to predict how often an event might occur in the future. If you know that your soccer team has won 83% of the time in the past, you can reasonably assume that you will win 83% of the time in the future. Frequency Table One of the best ways to tally and organize frequency data is to use a frequency table. A frequency table is a table that lists items and shows the number of times they occur. Here you see an example of a frequency table that describes the different ways students travel to get to school. You can also include a column that gives the relative frequency of each event. Percent Increase Let's go back to our example of moving to a new town for a minute. If you were to read that the area you wanted to move to had a 2% increase in crime in the last year, you might think twice about moving there. At the very least, you would do some extra research. What if you discovered in your additional research that the test scores at the local elementary school increased by 23% in the last year? That statistic might be more important to you than the small increase in crime in the same area. The percent increase represents the relative change between the old value and the new value. In order to calculate percent increase, you must have collected data about the same event, just at a different time. Calculating Percent Increase To determine the percent increase between two sets of data, you can use a frequency table and the following formula: Percent Increase = Frequency 2  Frequency 1 / Frequency 1 * 100. Let's try an example: Data for crime in a certain area was recorded over two years. The following table shows the occurrences of three different types of crime over a
113 twoyear period. To calculate the percent increase, take each row individually and plug the numbers into the equation. Percent increase of robbery = ((3733) / 33) * 100 = (4 / 33) * 100 = 0.12 * 100 = 12% Percent increase of murder = ((82) / 2) * 100 = (6 / 2) * 100 = 3 * 100 = 300% Percent increase of assault = ((1615) / 15) * 100 = (1 / 15) * 100 = 0.07 * 100 = 7% Then, you can complete the table. Here we have another example: 100 people were asked how often they ate dinner at home in a typical week. Then, they all took a class on how to cook healthy meals at home and after six months were asked again how often they ate at home. What was the percent increase for eating at home 5, 6 and 7 nights a week? Take a minute to try this one on your own. And here you see the answers. The percent increase for eating at home 5 nights a week was 347%, 6 nights a week  350% and 7 nights a week  125%. Lesson Summary Determining the frequency and percent increase of events can be very helpful when trying to interpret certain sets of data. The cumulative frequency of a certain event is the number of times that event occurs in a certain time frame. Relative frequency refers to the percentage a certain event is of the total amount of events that occur. You can determine the percent increase of an event from time to time by using the formula: Percent Increase = Frequency 2  Frequency 1 / Frequency 1 * 100. What is a TwoWay Table? Chapter 5 / Lesson 5 Do you believe in Martians? Do you watch football on television? A TwoWay Table or Contingency Table is a great way to show the results of all kinds of survey questions. In this video we will learn how to read a twoway table.
114 TwoWay Table Do you believe in Martians? Here is a survey of 100 college students asking that very question, Do you believe in Martians?. Gender Yes No Male Female Total This type of table is called a twoway or contingency table. A twoway or contingency table is a statistical table that shows the observed number or frequency for two variables, the rows indicating one category and the columns indicating the other category. The row category in this example is gender  male or female. The column category is their choice, yes or no. There is a lot of information we can learn from this small table. Let's look at a couple questions you could see on a test. How many males were asked? Let's look across the male row. Ten said 'yes' and 32 said 'no.' That would be a total of 42 males. How many college students believe in Martians? Look down the Yes column. Ten males and 38 females said 'yes.' That would be a total of 48 college students. Example #2 A recent survey of 100 college students asked if they prefer to drink tea, coffee, or an energy drink during finals week. Here is the table created from that survey. Gender Tea Coffee Energy Drink Male Female Total Let's look at a couple questions you could see on a test.
115 How many college students drink tea given that they are a female? In this question, we're looking for tea drinkers that are female. Let's look at the female row and tea column. The answer is 18. How many students drink tea or are male? In this question, we're looking for all tea drinkers united with all males. To figure out the answer, let's highlight the Tea column and Male row. We do have an overlap at Tea and Male, so be sure not to add that twice. So our answer is Total Tea drinkers 20 plus Total Males 56 minus the overlap of = 74. There are 74 students that are tea drinkers or male. How many coffee or tea drinkers are not female? Let's look at each part. Let's highlight the coffee and tea columns. Now, let's highlight the male row. Remember, the question asks for drinkers that are not female, so they have to be male. So our answer will unite the coffee and tea drinkers with the contingency that they are males. Males that drink tea are 2, males that drink coffee, = 17. There are 17 males that drink coffee or tea. Lesson Summary A TwoWay or contingency table is a statistical table that shows the observed number or frequency for two variables. The rows indicate one category and the columns indicate the other category. Here are some tips for you when you are doing a twoway or a contingency table: Highlight the column or row that is in the question. This will keep you focused on the question and not all of the extra stuff in the table. Also, beware of highlights that are overlapped. Don't count anything twice! Read the question carefully. Tests try to trick students using words such as 'not' and 'or'. As in our example, if you are not female, you have to be male, and, drink tea or coffee. We need to count all of the tea and coffee drinkers. Make Estimates and Predictions from Categorical Data Chapter 5 / Lesson 6 Categorical data can be estimated but not predicted. Learn why in this video lesson along with how to read and gather information from a bar graph of categorical data.
116 Categorical Data Categorical data is data that can be grouped. Examples include age group, favorite color, dog breed, etc. These are all categorical in nature because if you ask different people about their favorite color, you can begin to separate the people into groups. You can separate them into groups that like blue, groups that like yellow, etc. The same goes for age group and dog breed. These categories can all be grouped. Bar Graph A bar graph, a graph with bars of varying heights, is a good visual way to represent the categorical data from a survey for analysis. What you will usually see is your survey option on the xaxis and the results on the yaxis. Let's look at one bar graph to see how we can make estimates and predictions using the data we see. This particular bar graph shows the results of a survey we did asking people who kept Bettas as pets what color Betta they had. These are the results we got from our survey. We got 20 red Bettas, 30 blue Bettas, 15 purple Bettas, 4 white Bettas and 2 yellow Bettas. Betta Color Result Red 20 Blue 30 Purple 15 White 4 Yellow 2 So, what kind of estimates and predictions can we make based on this information? Estimates We can certainly make some estimates based on this information. From our graph, we see that blue is the tallest bar, so blue is the most popular color. We can estimate that the majority of Betta owners keep blue Bettas. We see that yellow is the shortest, so we can estimate that yellow Bettas are not as popular among Betta owners.
117 The majority of Betta owners keep blue, red and purple Bettas. So, if I had a Betta store, I would do best by selling these three colors. Yellows and whites are very few, so I wouldn't do so well by selling those. These are the kinds of estimates I can make based on what I see from our bar graph. Predictions What about predictions? The thing about categorical data is that because they are groups, there is not much to say about other groups that aren't listed on the graph. If you don't have data for other groups, there is not much you can predict about them. With mathematical data, you can continue the pattern and make a prediction on what may happen with data that is outside the range of the data you have. But with categorical data, you can't do that because the groups you have data for are not connected to other groups. Looking at our graph, can we make a prediction about colors that aren't listed? No, because we can't say for certain what color group will come next. And there is no pattern to draw out. Lesson Summary What have we learned? We've learned that categorical data is data that can be grouped. A good visual way to represent categorical data is with a bar graph, a graph with bars of varying heights. We can make estimates about what is most common and what is least common, but we can't make predictions on categorical data. What is Quantitative Data?  Definition & Examples Chapter 5 / Lesson 7 What Is Quantitative Data? What's the difference between having seven apples and saying that they are delicious? Well, for one, we can count or measure the seven apples, but we can't put a number to how delicious they are. Those apples might be delicious to one person and be completely sour to another person.
118 What does this have to do with quantitative data? It has everything to do with quantitative data because it shows you what is considered quantitative data and what is not. Saying you have seven apples, because they can be represented numerically, is a piece of quantitative data. But saying that they are delicious is not because you can't write that using numbers. There are two types of data that quantitative data covers. They are data that can be counted and data that can be measured. Let's talk about what each data type looks like. Data That Can Be Counted Another way of saying that the data can be counted is to call it discrete data. Having the seven apples, for example, is discrete data because you can count seven apples. If you were to count the number of apples each tree produced in an apple orchard, that data is quantitative since the apples can be counted. Other examples of discrete data include the number of girls in a math class, the number of boys who come to eat ice cream at three pm, and the number of kittens that a particular mom cat has. All of these are discrete and quantitative data because they can be represented by a mathematical number and you can physically count them. Data That Can Be Measured Quantitative data is also data that you can measure. In math lingo, this is called continuous data. The weight of seven apples is continuous data because you can put the apples on a scale and weigh or measure them. Other examples of continuous data include the height of your mom, the length of a football field, and the weight of a wolf. All of these are continuous data because you can measure them and represent them in a numerical manner. Uses of Quantitative Data All this information is very useful. What can you do with all this quantitative data? You can turn a collection of quantitative data into a report. By counting the apples of each apple tree in an orchard, you will be able to pinpoint any problems the orchard has. If some trees are not producing, you can locate them and find out why. If all the trees are producing very little, you can start to figure out what the orchard is doing
119 wrong. And if all the trees are producing a lot, you can find out what the orchard is doing right. Without quantitative data, you wouldn't know any of this. Quantitative data helps you to gather information in a way that can be represented mathematically and can usually be graphed so you can see the data in a visual way. For example, by collecting data on the number of dogs each household has in various countries, you would be able to compare the countries to see which country favors dogs more than others. This data can be graphed and you can use this data to help you decide where best to begin a business focused on dogs. Lesson Summary Quantitative data is numerical data. It includes data that is discrete (can be counted) and data that is continuous (can be measured). Examples of discrete data include anything that can be counted, such as the number of girls in a class, the number of apples in a fruit basket, and the number of boys who eat ice cream every day. Examples of continuous data include anything that can be measured, such as the height of your mom, the length of a football field, and the weight of a wolf. Quantitative data is useful because a collection of it allows you to make educated guesses based on patterns that are observed. For instance, if you take quantitative data on the number of dogs each household has in various countries, you can use that data to find the best country in which to start a dogrelated business. What is a Histogram in Math?  Definition & Examples Chapter 5 / Lesson 8 What Is A Histogram? In math, a histogram is a visual way to display frequency data using bars. A feature of histograms is that they show the frequency of continuous data, such as the number of trees at various heights from 3 feet to 8 feet. They are not used to show categorical data, such as the population of dogs in Norway, Finland, and Sweden. Another feature of histograms is that the data can be grouped into ranges. For example, the height of the trees can be grouped into 3 to 4 feet, 5 to 6 feet, and 7 to 8 feet.
120 When to Use How do you know when to use a histogram? You can decide this by looking at your data. Ask yourself, 'Is the data continuous or can I group the data into ranges?' What you are looking for is a group of data that is continuous. What this means is that the data covers a range of values that does not jump. For example, the range of tree heights is 3 to 8 feet. The data does not jump from 4 to 6 feet. There is no gap. The countries of Norway, Finland, and Sweden, though, do jump. There is no continuity between the countries. They are separate entities. For histograms, you need continuous information and not just categories that jump. Histograms are great at showing results, such as the frequency of the number of trees at various heights from 3 to 8 feet. So, if you have a set of continuous data, then you can use a histogram to show the results visually. So, let's say that I surveyed a hill by my house and I counted the number of trees at various heights. I found that at 3 feet, my hill only had 2 trees. At 4 feet, my hill had 4 trees. At 5 feet, I only had 1 tree. At 6 feet, I had 7 trees. At 7 feet, it was 8 trees. Then at 8 feet, I had 2 trees. The histogram for this information would have the xaxis showing the range of heights from 3 to 8 feet. Then the yaxis would have the number of trees. Then I have several options for showing the data. I can show the data for each height separately, or I can group them into ranges, such as 3 to 4 feet, 5 to 6 feet, and then 7 to 8 feet. I can combine the data into any number of different ranges. I can also group them into ranges, such as 3 to 5 feet and 6 to 8 feet. It is your choice based on the kind of information you want to show. If you want to be specific, then use smaller ranges or don't use ranges at all. If you want to be more general with your data and just show an overview, then you can use larger ranges to group your data. To create the histogram, what I have done for every tree measured is increase that tree height's bar on the histogram by 1. So, if my tree measures 3 feet, then I increase the bar for 3 feet by 1. After measuring 2 trees that are both 3 feet high, my bar at 3 feet should be at the number 2 since there are 2 trees that are 3 feet high. I continue doing this until I have all of my data on the histogram. How to Read
121 By looking at the tree height histogram, I can tell how many trees are in each data range. For example, for the tree height histogram that shows each height, I see that I have 2 trees that are 3 feet high, and 7 trees that are 6 feet high. Reading a histogram is fairly straightforward. The yaxis tells you the frequency of the data. By looking at the whole thing, I can get a sense of which part of the data is more popular or is more important for answering my question. For the tree height histogram, I see that my hill contains the most trees that are 7 feet high. So, I can say that on my hill, 7foot trees are more common than trees of other heights. Lesson Summary A histogram is a visual way to show the frequency of data using bars. The histogram's data must be continuous. It cannot have a jump in it or be categorical. The data can be grouped into ranges for more general data. The bar shows the frequency of each data point or range. To create a histogram, you increase the bar for each data point or range by 1 whenever you see data at that point or range occur. For example, when I see a tree with a height of 4 feet, I increase the bar at 4 feet by 1 to account for this particular piece of data. To read a histogram is a matter of looking at the bar, then at the xaxis to see what the data represents, then looking at the yaxis to see how often that particular data occurs. For the tree height histogram, if the bar at 7 feet goes up to 8 on the yaxis, it means that I have 8 trees that are 7 feet high. What are Center, Shape, and Spread? Chapter 5 / Lesson 9 Center, shape, and spread are all words that describe what a particular graph looks like. Watch this video lesson to see how you can identify and explain each. Center When we talk about center, shape, or spread, we are talking about the distribution of the data, or how the data is spread across the graph.
122 The center of a distribution gives you exactly what it sounds like. It tells you the center or median of the data. When you look at a graph, it will be the value where approximately half of your data is on one side and the rest of your data is on the other side. The median point of your data set is the middle number if you were to put your data in ascending order. Let's say we are taking surveys of different groups of people and their donut eating habits. For the first group of people, we have this graph. We see that our center is 5 because half of the people are to the left and the other half are to the right. Another way to describe the center is to take the mean or average of all your data. When you describe your center in terms of mean and median, you might find that they are slightly different. Your mean might be more or less than your median. We will discuss what skewed means in just a little bit, but as far as the center is concerned, if your graph is skewed, then you will want to use the median as your center. Shape Depending on the group of people we survey about their donut eating habits, we will get different sets of data. When graphed, we can get different looking graphs. We use shape to describe the different types of graphs we will see. There are four different ways in which we can describe a graph's shape. 1. We can say a graph is symmetric if the left and right sides of the graph are mirror images of each other. This graph, for example, is symmetric because the left side is a mirror image of the right side. We see that at either end of the distribution, only 1 person chooses to eat 3 donuts and 7 donuts. Going closer to the center, we see that 2 persons choose to each 4 donuts and 6 donuts. They are mirror images of each other. 2. Sometimes, our graph will look like a rollercoaster and will have a number of peaks, or areas where the graph is higher than the surrounding areas. If there is only one peak, then we call it unimodal. If this one peak occurs at the center of the graph, it is also called bellshaped. Doesn't this look like a bell? If it has two peaks, then we will call it bimodal. 3. If our graph has more data on one side rather than the other, we call it skewed. If there are more to the right, we call it skewed left. For our donuts eaten survey, this would mean that more people choose to eat more donuts and fewer people choose to eat just a few. If our graph has more data to the left, then we would say that our graph is skewed right. For our donuts survey, it would mean that more people prefer to eat fewer donuts. A good way to remember this is to view the graph as a slide. If you
123 slide down to the right, then it is skewed right and if you slide down to the left, then it is skewed left. 4. If our survey of people's donut eating habits showed that for each amount of donuts eaten, the same number of people would choose that amount, then our graph will look flat all across the top, then we call it uniform. A uniform shape has no peaks nor is it skewed. Spread Let's move on to spread now. The spread of a distribution tells you the range of your data. If your spread is small, then your data covers a short range. If your spread is large, then the data covers a larger range. For our donuts, a small range would mean that people cluster together with their choices being very close to each other. A larger range means that people aren't afraid to choose differently from others. The small range of donuts is 2 to 4 donuts eaten and the larger range is 2 to 9 donuts eaten. Putting It All Together Now that we've covered center, shape, and spread, let's put it all together. Let's go take another survey of people's donut eating habits. This time, we are going to this part of town where people are known for their unique eating habits. Let's see what we find. Looking at the data and the resultant graph, I see that my center is 5 because half of my data is to the right and the other half is to the left. Okay, so far so good. What about my spread and shape? My spread ranges from 2 to 8 donuts, so it's rather large. What about my shape? What kind of shape does my graph have? It's not skewed one way or another, but it does look like a rollercoaster with two peaks. So it must be bimodal. But wait, the left side looks like a mirror image of the right, so it is also symmetrical. So, my data distribution is a symmetric bimodal with a range of 2 to 8 and a center of 5. Nice! That wasn't so bad. Lesson Summary What we've learned in this lesson is that center, shape, and spread are ways to describe the graph of a data distribution. The center is the median and/or mean of the data. The spread is the range of the data. And the shape describes the type of graph.
124 The four ways to describe shape are whether it is symmetric, how many peaks it has, if it is skewed to the left or right, and whether it is uniform. A graph with a single peak is called unimodal. A single peak over the center is called bellshaped. And a graph with two peaks is called bimodal. We can use all the information we have just learned to describe a graph. How to Calculate Mean, Median, Mode & Range Chapter 5 / Lesson 10 Measures of central tendency can provide valuable information about a set of data. In this lesson, explore how to calculate the mean, median, mode and range of any given data set. Measures of Central Tendency In this lesson, we're looking at the measures of central tendencies. The measures of central tendency provide us with statistical information about a set of data. The four primary measurements that we use are the mean, median, mode and range. Each one of these measurements can provide us with information about our data set. This information can then be used to define how the set of data points are connected. To really examine these data points, let's take a look at a football game between the Green River Ducks and the Southland Bears. Mean The first measure is the mean, which means average. To calculate the mean, add together all of the numbers in your data set. Then divide that sum by the number of addends. For example, let's look at the Bears' first offensive series. They had plays of 16, 14, 12 and 18 yards on their way to scoring a touchdown. To find the average number of yards, or the mean, we would first add all four of these values together: = 60. Since there were four numbers in our data set, you would divide that sum by 4: 60 4 = 15. So, the average number of yards that the Bears gained was 15 yards. The mean is used to show us the true average of a set of data.
125 Median Another measure of central tendency is the median, which is the middle number when listed in order from least to greatest. You may have heard the word median before, and it was likely on a highway. On a highway, you have opposite lanes of traffic. In the middle of the lanes, there is typically a grassy area or a turning area. This area in the middle of the highway is referred to as a median. Let's return to our game in progress and see how the Green River Ducks are doing. On the Ducks' first offensive series, they had plays of 10, 6, 19, 21 and 4 yards before scoring a touchdown. Let's find the median number of yards gained by the Ducks. The first thing you need to do with this list of yardages is put the numbers in order from least to greatest. So, in order from least to greatest, we would have 4, 6, 10, 19 and 21. Now that your yardages are in order from least to greatest, find the middle number. Since there are five numbers, the middle number would be the third value. The median value of this set of data is 10. On the first offensive series for the Ducks, their median yards gained were 10 yards. Occasionally there may be an even number of values, which would provide you with two numbers in the middle. If this occurs, you will need to average the two values. At halftime of our game, the Bears quarterback has passes of 3, 8, 9, 12, 12 and 15 yards. Let's find the median pass thrown by the Bears quarterback. The first step is to make sure your numbers are in order from least to greatest, which they are in this problem. The next step is to find the middle number. Since there are six numbers in this set, the middle numbers would be the third and fourth values. Since there are two numbers in the middle, you will average them together. Add the two numbers together, = 21. Then, divide by 2: 21 2 = The median of this set of data is In the first half, the Bears quarterback had a median passing yardage of 10.5 yards. When looking at a data set, the median is used when there is an outlier, which is a number that is significantly greater or smaller than the rest of the data. In the second quarter, the Ducks had plays that were 21, 24, 26, 20, 56 and 20 yards. You can see that the value 56 is significantly larger than the other values. 56 would be an example of an outlier. Compared to the other yards that the Ducks gained, 56 yards was much greater than their other gains.
126 Mode The mode is another measure of central tendency that tells us the number that occurred the most often in your data set. When looking for the mode, there can be more than one mode or no mode. The mode can tell us the most popular choice. The Bears threw the ball to the following jersey numbers in the third quarter: 5, 6, 6, 3 and 4. You can see that there was only one receiver that had the ball thrown to him more than once. The mode of this data set would be 6. The Bears receiver #6 was the most popular choice to throw the ball to in the third quarter. The Ducks threw the ball to the following receivers: 12, 13, 15, 17, 19 and 20. You can see that none of these receivers caught more than one pass. This data set has no mode. Entering the fourth quarter, the Bears had scored the following points: 6, 7, 3, 0, 7, 3, 7 and 3. You can see there are two values that repeat three times each. The mode of this data set is both 3 and 7, which can sometimes be referred to as bimodal. This means that the most popular scoring values for the Bears were 3 and 7. Range The last measure of central tendency is the range. The range is the difference between the highest and lowest values. Simply put, find the largest and smallest numbers and then subtract them. The range tells us the distance between the values in our data set. At the end of the game, the Ducks' kickers had kicked field goals of 10, 14, 17, 19, 21 and 30 yards. Find the range. The smallest value is 10 and the largest value is 30. To calculate the range, subtract the two values: = 20. The range of this data set is 20. Example Let's put our new skills into practice with an example. Let's find the mean, median, mode and range of how many medals the U.S. has won over the last six summer Olympics. To find the mean of this data set, we would add , and then divide by 6 because there are six values. So,
127 108 = 618. And, = 103. So, over the past six Summer Olympics, the United States has been awarded an average of 103 medals. To find the median, we must first put the data in order from least to greatest. So our numbers in order from least to greatest would be 94, 101, 101, 104, 108, 110. The middle of this data set is actually two numbers (101 and 104). To find the median, we will need to add these two numbers together and divide by = 205, then dividing by 2 makes the median Looking at this data set, we can see that there is only one number that repeats itself, which is 101. This means that the mode of the data set is 101. The range of this data set is found by taking the largest value (110) and the smallest value (94) and subtracting. So, = 16. The range of this data set is 16. Lesson Summary In this lesson, we've discussed four measures of central tendency. These measurements can provide you with important information about a set of data. These four measures are the mean, median, mode and range. The mean means average. To find it, add together all of your values and divide by the number of addends. The median is the middle number of your data set when in order from least to greatest. The mode is the number that occurred the most often. The range is the difference between the highest and lowest values. Describing the Relationship between Two Quantitative Variables Chapter 5 / Lesson 11 What Is a Quantitative Variable? A quantitative variable is something that can be measured and written out as a number. For example, age, height, and life expectancy are all quantitative variables because all of these can be written as a number. If you asked someone to give you their age, they would give you a number. It's the same with the others.
128 You will see quantitative variables used in surveys. When you have more than one quantitative variable in a survey, relationships between the various variables can be seen. But why is this important? Why is it helpful to describe the relationships between quantitative variables? Why Describe Relationships? Many times, one variable is related to another. Knowing how one variable relates to another gives you valuable information you can use. For example, the two quantitative variables age and height are related to each other. We know that as children age, they also grow taller. We can do surveys on different populations to find out each group's average agetoheight information. Knowing this can help us. For example, if we are a doctor and we meet a child who happens to be much taller than expected for his age, based on the information we have, we can then proceed to inform his parents and figure out why he is much taller. There could be a medical condition that is causing it. And if there is, he needs to be treated for it before it causes him damage. If we didn't have our information about the relationship between age and height, then we wouldn't be able to provide this level of help to that child. Knowing how one variable relates to another is important and it can even save lives. Let's see how we can come up with a relationship on our own given two quantitative variables and some data collected from a survey. Two Quantitative Variables Our two quantitative variables are level of happiness on a scale of 1 to 10, with 10 being the highest level of happiness, and the number of chocolate bars eaten. Both of these are quantitative variables because each is represented by numbers. We take our little survey to the really nice chocolate bar store at the mall and we start asking people to rate their level of happiness as they keep eating chocolate bars. We were able to survey 20 people and this is their data: Level of Happiness Chocolate Bars Eaten
129 How Are They Related? The best way to figure out how these two variables are related is to use a scatter plot, a graph where you place an individual dot for each data point. We will use the xaxis for our level of happiness and the yaxis for the number of chocolate bars eaten. Just remember that the level of happiness scale goes from 1 to 10, with 10 being the highest level of happiness. So if someone put down a 10, that means they enjoyed their chocolate bars the most. If someone put down a 1, then they really weren't happy with the chocolate bars. Let's go ahead and plot our data and see what we find. This looks rather interesting. I love chocolate, but I do know that if I eat too much, I'll get sick and won't want to see chocolate for a while. The data that we got seems to
130 show that too. I see that the unhappiest people are the ones that ate a lot chocolate bars, around 10. The happiest people ate only 1 or 2 chocolate bars. Taking a step back and looking at the whole graph, I see that my dots are loosely following a straight line. This tells me that happiness is related to the number of chocolate bars eaten. If my dots were scattered all over the place, it means that my two quantitative variables are not related to each other. But if they seem to form a line of some sort, then that means they are related in some way. For my data, the relationship between the level of happiness and chocolate bars eaten is that the more chocolate bars you eat, the unhappier you will get. If you want to be happy with your chocolate bars, just eat 1 or 2. Lesson Summary What have we learned? We have learned that a quantitative variable is something that can be represented by a number. Age, height, and life expectancy are all examples of quantitative variables. The best way to determine how two variables relate to each other is by plotting the data points on a scatter plot, a graph where each data point is plotted individually. If all the points seem to follow a line of some sort, then we can interpret that relationship. If all the points are randomly scattered, then there is no relationship between the two variables. Reading and Interpreting Line Graphs Chapter 5 / Lesson 12 Line Graphs Line graphs are cool! They are cool to look at and they provide so much useful information once you know how to read them. You know when you are looking at a line graph when you see a graph with a line connecting the data points. The line can be straight, or it can be curved, or it can look like a connectthedots line. Whatever the case, there is so much information to be gained from these graphs.
131 Reading a Line Graph Let's look at this line graph that shows us the number of correct math problems a particular friend of ours gets right in one minute over the course of a week. Our friend, Joe, is taking a timed math test every day for a week. He is given a minute to solve as many basic arithmetic problems as he can. At the end of the minute, his work is graded and the number of correct answers is noted down. Let's see what information we can gather from the line graph that was produced from his results. We see that on day 1, the first day, Joe didn't do very well, with only 1 correct math problem. I see this by looking at the bottom x axis first to locate day 1, drawing an imaginary line straight up, and then looking to find where our actual line crosses the imaginary line for day 1. I find the point where they cross and then I draw another imaginary line across to see where it hits on the y axis. I see that it hits the y axis at 1, so Joe only got 1 correct math problem. I keep reading the graph to gain more information. I look at day 2. How many math problems did Joe get right that day? I see where my line crosses day 2 and I see that it is where the y axis equals 5, so Joe got 5 correct math problems on day 2. On day 3, Joe got 8 correct answers. Day 4, he got 15. Day 5, it is 22. Day 6, Joe got 34 correct answers. And at the end of the week, on day 7, Joe was able to get 50 correct answers in one minute. Now that we've read our graph, what kind of interpretations can we make about Joe? Interpreting a Line Graph Reading a bunch of mathematical information from a graph is kind of meaningless unless we can interpret that data. What do all the numbers we read from the graph tell us about Joe? Looking at the line, we see that it keeps getting higher and higher. That is good for Joe. We can see that as the line gets higher, Joe's number of correct answers also gets higher. That means that Joe is doing really well. He is progressing in his math ability every day. He started out a bit slow, but then he started understanding more quickly. We can see this by the way the line curves slightly upwards. We can also make an interpretation of what may happen to Joe's mathematical ability if he keeps going. It would make sense for the line to keep following the same pattern of going up, so if Joe kept this up, he'd be able to do more math problems in a minute
132 on day 8 and day 9. Whether this will happen, we can't be sure, but based on the information we have, it seems likely. We just can't say that it will definitely happen. A RealWorld Line Graph Now, let's look at how we can read and interpret a line graph from a realworld scenario. We will see how being able to read and interpret the data will help us in our decisions. This particular line graph shows the cost of a cell phone plan that includes 250 minutes. This particular cell phone plan includes 250 minutes per month along with unlimited texting. The basic cost of the plan is $40 per month. Now, I read my line graph and see that after 250 minutes, my cost rises with the number of minutes. I went from $40 for 250 minutes to $65 for 300 minutes. That's a lot of money! I see that the cost gets higher and higher the more minutes I use, all the way to $115 for 400 minutes. Doing a little math, I figure out that for every 50 minutes I go over my 250 minutes, I have to pay $25 more. If I divide $25 by 50, I figure that for each minute over 250 minutes, I am paying $0.50. I don't want to pay that much for a minute of talking if I go over my allotted time for the month. I stop and think about this for a minute. I have to think about how much I really use this cell phone. Will I use more than 250 minutes per month with the cell phone? 250 minutes per month works out to a little over 4 hours of talk time. Do I talk that much on the phone every month? Or do I text more? I like the $40 a month cost, but how likely am I to go over the 250 minutes? Answering these questions will help me decide whether I should take this plan or upgrade to the next higher plan. After some thought, I decide I will take this plan because I don't talk on the phone that much. Because I text more than I talk, this plan will work for me. Lesson Summary What have we learned? We have learned that a line graph is a graph with a line connecting the data points. To read the graph, we look at the x axis to find one part of what we are looking for. We then draw an imaginary line straight up. We look for the actual line to see where it crosses our imaginary line. Then we draw another imaginary line across from the point where our line crosses our imaginary straightup line to see where it meets with the y axis.
133 We now know the information for that point. We can keep repeating the process until we have read all the points we are interested in. After gathering the data and looking at the line, we can make some interpretations about the data. If our line goes higher and higher, we can say that whatever it is keeps growing. If the line keeps going down, then it's shrinking. Making Estimates and Predictions using Quantitative Data Chapter 5 / Lesson 13 Quantitative Data To begin, quantitative data is something that you can measure and write down using numbers. Examples include age, height and weight. These are all considered quantitative data because you can measure all of them and write down what they are in numbers. Scatter Plot The best way to represent quantitative data is with the use of a scatter plot. A scatter plot is a graph that plots each data point individually on it. You end up with a bunch of dots on the graph. Usually, you have one measurement of quantitative data on the xaxis and another on the yaxis. If a pattern emerges, then we can see that there is a relationship between the two pieces of quantitative data. If no pattern emerges and the dots look like they have been randomly placed on the graph, then there is no relationship and nothing more can be said about it. Let's look at an example to see how we can read a scatter plot. Our scatter plot shows some data we collected from our hot chocolate cart business. We wanted to find out if there is any relationship between the outside temperature and hot chocolate sales. We've plotted all of our data and we see that the points do form a pattern. It looks like a line that is slanting downwards. Let me go ahead and draw a line that is roughly in the middle of all the data. This is the line I can use to make my estimates and predictions.
134 Making Estimates Of course, any estimates I make will not be exact. As you see, the actual data can fluctuate slightly from the line in the middle. But my estimate can give me an idea of what to expect. For example, looking at my graph and the line that I've drawn through the middle of the dots, I see that if the outside temperature is 30 degrees Fahrenheit, then I can estimate that my hot chocolate sales will be around $500. I can look on the xaxis to find an outside temperature I am curious about and make an estimate about how much I can expect to make in sales at that temperature. You try. What can you expect to earn if the outside temperature is 60 degrees Fahrenheit? By locating 60 degrees Fahrenheit on the xaxis, you can see that the point where the line reaches 60 degrees Fahrenheit is roughly at $350, so you can expect to earn around that much. Making Predictions In addition to making estimates along the line in the middle of the data, I can also extend that line to predict what may happen at other temperatures. Remember, you can only make estimates and predictions for quantitative data that have a pattern to them. If you can't draw a line of some sort through the data, then you can't make estimates or predictions about it. I was able to draw a line through my hot chocolate sales data, so I can make estimates and predictions on it. To make predictions, I need to extend my line. Extending my line in both directions, I can see that as the temperature gets warmer, my sales get lower. But when the temperature gets even colder, my sales increase even more. I can predict that at 100 degrees Fahrenheit, it is possible that I could earn about $100. But what about when the temperature is at 0 degrees Fahrenheit? What kind of sales can I predict? By looking at my extended line, it looks like I can expect to make about $700. Lesson Summary In review, quantitative data is data that can be written down as numbers. The best way to represent quantitative data is with the use of a scatter plot, a graph where each data point is plotted individually. In order to makes estimates or predictions, you have to be able to draw a line through the middle of the data.
135 If the data is scattered all over the place, then you can't draw a line through it and you say that the data has no relationship. If you can draw a line through the data, then you can make estimates about the data and predict what may happen outside of the data range.
136 What is a Variable in Algebra? Chapter 6 / Lesson 1 Did you ever encounter something like a + b = c and wonder how all these letters snuck into a math problem? In this lesson, we'll learn about variables, as well as other parts of equations, constants and coefficients. Variables Have you ever played poker? Sometimes when I play, I may look at my cards and be one card short of a good hand. Maybe I have three jacks. That's pretty good, but if I just had one more jack, I'd have four of a kind, which is great. Unfortunately, a deck of cards only has four of each card, so the odds of getting all four jacks in your hand are small. But what if we were playing with wild cards? A wild card is a card that can be whatever you want it to be. It could be a jack for me or an ace for someone else. It's a card that has an undetermined value. This is just like a variable in algebra. A variable is a symbol that represents an unknown number. It will be equal to a number, but you don't know what that number is yet. Until you figure it out, you use the symbol. The variable you most often see is x. But a variable can really be any letter or symbol that doesn't otherwise have a number associated with it. Think again about wild cards. A wild card has something printed on it, but its real meaning is determined by the other cards in your hand. In an algebraic equation, a variable's real meaning is determined by the things around it.
137 Constants and Coefficients Before we go any further, let's learn about those other things around our variables. Here's an equation: 2x + 1 = 7. What are 1 and 7? Numbers! Yes, they're numbers. But we also call them constants. A constant is a fixed quantity that cannot change. It's constant. Constants are our nonwild cards. Just like a 3 of hearts is always a 3 of hearts (supposing that card is never wild), and a 7 is always a 7. By the way, we can't use a number as a variable  since they're constants, that would be too confusing. Back to our equation, there's also that 2. It has a special name. It's called a coefficient. A coefficient is a number that multiplies a variable. If you have 5y, your coefficient is 5. If you found out that y = 6, then 5y is 5*6, or 30. Why Use Variables? You'll often be asked to solve equations with variables, like x  2 = 5. But you should know that variables don't just exist for algebra quizzes. They have very practical purposes. Let's say you're at your favorite local burger place. They sell double bacon cheeseburgers for $4 each. You want to get as many as you can, but you only have 13 dollars. Plus, if you're going to buy a bunch of double bacon cheeseburgers, you're also going to need a drink, which costs a dollar. How can you figure out how many burgers you can get? The number of burgers is your variable. Maybe it's 1, maybe it's 7. You don't know. So let's call it x. Each burger costs $4, so 4x is the cost of a burger times the number of burgers. 4x + 1 is the cost of the burgers and the soda. That has to equal 13, so our equation is 4x + 1 = 13. You just used a variable to help you solve a crucial problem and help you maximize your double bacon cheeseburger intake. Practice Problems But how do you solve this equation? When you're trying to solve an equation that has a variable, you're trying to find out the value of that variable.
138 Let's first try a simpler example: x  5 = 10. Our variable is x, so that's what we want to figure out. 5 and 10 are our constants. How do we figure out x? Well, our equation has two parts: x  5 and 10. They're separated by an equals sign. In order to learn what x is we need to get it alone on one side of the equation. Here, we want to add 5 to both sides. This will get us x = 15. So what is x? It's 15! We can check ourselves by putting 15 where x was in the original equation. So instead of x  5 = 10, we have 155 = 10. Is ? Yes! Let's try one with a coefficient: 3y = 27. This time, our variable is y. Does it matter what letter it is? No! It could be w or j or that thing Prince was when he wasn't Prince for a while. It's just a symbol for an unknown number. What's a constant in this equation? 27. And our coefficient? 3. To solve this one, we still want to get our variable by itself. That means we need to divide both sides by 3. That will get us y = 9. So our variable was standing in for 9. Let's check that. Does 3*9 = 27? Yes! Okay, we built up our appetite. Let's figure out that burger equation. It was 4x + 1 = 13. Let's start by subtracting 1 from both sides to get 4x = 12. Then we divide both sides by 4. That gives us x = 3. Do you remember what x was symbolizing? Double bacon cheeseburgers, of course. So you can afford 3 double bacon cheeseburgers and a soda with your $13. Oh man, thank you algebra! Lesson Summary In summary, a variable is a symbol that represents an unknown number. Though it can be anything, you often see letters like x or y as variables in algebraic equations. The other parts of equations include constants, which are fixed quantities that cannot change, and coefficients, which are numbers that multiply a variable. Expressing Relationships as Algebraic Expressions Chapter 6 / Lesson 2 What do you do when you don't know what a number is but you do know how it relates to something else? You use an algebraic expression. In this lesson, we'll learn how to express relationships as algebraic expressions.
139 Expressions and Variables Do you ever have trouble putting what you want to say into words? Maybe you're trying to explain something and it's like you're speaking in another language that the other person can't understand. I think this is how my high school chemistry teacher felt while teaching me. Or maybe you're trying to tell someone you, you know, like them, or like like them, but he or she just isn't hearing you. It's frustrating, I know. Or, I mean, you know, I've heard. Life would sometimes be simpler if we could just use math to speak for us all the time. Fortunately, there are endless reallife situations we can express using algebra. We just need a few tools. First, of course, are numbers. This is math, after all. We also need variables. A variable is a symbol that represents an unknown number. Then we'll need some operators, like addition and division. With these tools, we can say all kinds of things with ease. Algebraic Expressions We use the tools to build algebraic expressions. An algebraic expression is a mathematical phrase that may include numbers, variables and operators. It's basically like a sentence. But you're substituting these numbers, variables and operators for words. Algebraic expressions can look like x + 1, 17y, 4a  3 or q/6. The most common and useful application of this idea is in solving word problems. We need to take the real situations that are in regular language in the problem and translate them into algebraic language to better understand them. These can involve several different types of operations. Addition and Subtraction Let's start with addition and subtraction expressions. Here's one: There are two competing lemonade stands run by siblings April and Mike. We want to describe the relationship between the prices for a cup of lemonade between the two stands. April is selling her lemonade for 50 cents less than Mike's. What do we do? First, we need a variable for the cost of April's lemonade. Let's call that a. We use that in place of a number we don't know. Now, if Mike's lemonade is 50 cents more than
140 April's, we can describe his as a That a + 50 describes the cost of Mike's lemonade relative to the cost of April's. We can test our expression by substituting a number for our variable. Let's say a = 75 cents. a + 50 = $1.25. Is $ cents more than 75 cents? Yes! So we know that we have the correct expression. We could also use subtraction here. We could use m as the cost of Mike's lemonade. Since April's is 50 cents less than Mike's, hers would be m Again, we could test this. Let's say m = 80 cents. So m  50 = 30 cents. Is cents less than 80? It is. It's also a very cheap cup of lemonade. Multiplication and Division Next, let's look at multiplication. Let's say it's the holiday season and you're buying gifts for your siblings. Since you live in an algebra problem, let's say you have 15 siblings. That's a lot, so you have to keep these gifts small. Let's say you can spend d dollars on each gift. How do you describe the total cost of the gifts? We could add them together and say d + d + d +... well, 15 total ds. It would be simpler to just say 15 times d, or 15d. That's the cost of one gift times 15. So if you spent $10 per gift, you'd spend 15 times 10, which is $150. Here's one that involves division. Let's say you're working with a study group. You all get hungry and order some pizza. There are three of you. If everyone's going to get the same amount, how much pizza can you eat? Well, you don't know how many slices there will be. So let's use s to stand in for the total number of slices. Since there are 3 of you, each of you gets s/3 slices. That's the total number of slices divided by 3. If there were 9 slices, you'd get 9 divided by 3, or 3 slices each. Multiple Variables So far, all of our expressions used just one variable. But life isn't always that simple. Let's say you're going on a date. It's dinner and a movie. You want to express how much this date will cost. Entrees cost e and movie tickets cost t. How do you write this? Well, there are two of you, so two entrees will be 2e. And supposing your date doesn't bail on you before the movie, you'll need two tickets, or 2t. The total cost is 2e + 2t.
141 Sometimes we need to mix operators as well as variables. Don't worry. You can handle this. Let's say you don't have one date this week, but three. How do you show that? Just take our expression, 2e + 2t, and multiply the whole thing by 3  so 3(2e + 2t). Lesson Summary In summary, an algebraic expression is a mathematical phrase that may include numbers, variables and operators. We use variables in place of the numbers we don't know. Then, using our operators, we explain the relationships between different values. Evaluating Simple Algebraic Expressions Chapter 6 / Lesson 3 In this lesson, we'll learn how to evaluate algebraic expressions, which involves substituting numbers for variables and following the order of operations. By the end of the lesson, you'll be an algebraic expression expert. Algebra in Real Life We should always remember that algebra has a basis in real life. When we have all these variables and exponents and things, that's easy to forget. But the purpose of algebra is to help us find answers to real situations. For example, let's say that you have a parttime job as a dog walker. You earn $15 each time you walk a dog. If d is the number of dogs you walk, your total income is $15d, or $15 times the number of dogs. That's an algebraic expression. And what if you had 6 walks scheduled this week? How much will you earn? That's where we get to evaluating algebraic expressions, our topic for this lesson. Evaluating Algebraic Expressions Evaluating algebraic expressions is when you substitute a number for each variable and then solve the expression. These types of problems typically look like this: Evaluate 15d when d = 6. So this is our dog walking scenario. And you can see there are two parts to the question. First, there's an algebraic expression. This is a mathematical sentence of
142 sorts that contains one or more variables. Here, it's 15d. Second, there's a sample value for the variable. In this case, we have d = 6. To solve this problem, we take the value and plug it into our expression. So we'd replace the d in 15d with a 6. We complete the problem by multiplying 15 times 6, which gets us 90. That's our answer! You'll earn $90 this week  plus you'll get lots of exercise, which is kind of its own payment, right? Well, okay, the money's good, too. Order of Operations Evaluating algebraic expressions can be pretty straightforward. Whether you have one variable or ten, it's really just plugging them in and solving. Usually, the trickiest part is remembering the order of operations. The order of operations is a system that determines which procedures occur first in an expression. As you'll recall, the mnemonic PEMDAS gives us our order. PEMDAS stands for parentheses, exponents, multiplication, division, addition and subtraction. You might remember PEMDAS with the phrase Please Excuse My Dear Aunt Sally. This mnemonic makes inoffensive Aunt Sallys everywhere cringe. Offensive Aunt Sallys don't care what you say. But why do we care about Aunt Sally? Well, if we had 42 * 3, would we do 42, which gets us 2, then multiply that by 3, which gets us 6? Or would we multiply 2 * 3, which gets us 6, then subtract that from 4, which gets us 2? Those are two totally different answers. PEMDAS tells us multiplication precedes subtraction, so 2 is the correct answer. Also, remember that multiplication and division are linked, as are addition and subtraction. Or, 'my dear' is a package deal, as is 'Aunt Sally.' Remember Parentheses Ready to try some practice problems? Almost. First, a warning. It's critical that you always use parentheses. Did you ever go on a carnival ride where they tell you to keep your arms inside the ride? Parentheses are the ride, and minus signs are the numbers' arms. They even kind of look like arms, which is a bonus. Here's why this matters. Let's say you have x^23 and you want to evaluate it when x = 2. Okay, no problem. That's 2^23. Exponents are first, so I square the 2 and now I
143 have 43. That's easy: 1. You might be thinking, 'I don't need parentheses. I solved that problem without them just fine.' Okay, what if x = 2? So we have 2^23. Again, I square the 2 and I get That will be 7. And that's wrong. You should have squared not just 2 but 2. Then you'd get 43 and again get 1. That would've been easy to remember if you'd written the expression as (2)^23. So consider this a safety reminder: Always use parentheses. Don't let your numbers lose their arms. Practice Problems Okay, let's practice evaluating some algebraic expressions. Let's start simple: Evaluate y  2 when y = 10. Just plug 10 in for y: Then solve. Our answer is 8. That means that when y = 10, y  2 equals 8. Here's another: Evaluate a^2 + 5 when a = 3. If we plug 3 in, we get 3^ Now, remember PEMDAS. Exponents come before addition. So we first square the 3 to get And is 14. So when a = 3, a^2 + 5 = 14. Let's get more complicated: Evaluate 4xy^312 when x = 2 and y = 3. Okay, don't worry. We can handle this. The first step is to plug in our x and y values. We get 4(2)(3)^312. Never forget those parentheses. Keep your arms and minus signs inside the ride. Now, it's 'please excuse my dear Aunt Sally' time. We'll start with the exponent. 3^3 is what? 3 * 3 = 9 and 9 * 3 = 27. So now we have 4(2)(27) The next step is multiplication. 4 * (2) = 8. And 8 * 27 = That gives us That's That's our answer! I think you can handle one that's even harder: Evaluate 2(a^2 + b^3)/(ab) when a = 4 and b = 2. Okay, that's a big messy one. But just follow these steps. First plug in our a and b values. We get 2((4)^2 + (2)^3)/((4)(2)). And what about Aunt Sally? Okay, parentheses are first, so let's tackle that stuff inside the parentheses. 4^2 is 16 and 2^3 is... 8 or 8? It's 8. So we have 16 + (8), which is 168, or 8. Let's look at where we're at. 2(8)/((4)(2)). Let's simplify that to 16/(8). And that'll just be 2. That big messy problem? It's just 2. Not bad! Always remember, if you can please excuse my dear Aunt Sally, you can evaluate algebraic expressions.
144 Lesson Summary In summary, evaluating algebraic expressions is when you substitute a number for each variable and then solve the expression. Usually, the trickiest part is remembering the order of operations. For this, we use PEMDAS. This stands for parentheses, exponents, multiplication, division, addition and subtraction. You can remember the acronym with the phrase Please Excuse My Dear Aunt Sally. The Commutative and Associative Properties and Algebraic Expressions Chapter 6 / Lesson 4 The order of operations is important and useful, but a few mathematical properties highlight cases where order doesn't matter. In this lesson, we'll learn about the commutative and associative properties, which may save you time and effort. Math Rules Math is full of rules. You have to divide before you subtract has to equal 4. You can only eat pie after you finish your vegetables. Working with pi will always make you think of pie. Fortunately, there are a few rules that actually make math simpler. These are like the Casual Fridays of mathematics. They're rules, yes, but they define how you can loosen up a bit and lose that tie. Commutative Property Let's say you want to know what is. Do you have to add 8 to 3? Or could you add 3 to 8? It doesn't matter, right? And the same thing is true if you want to know 2 * 5. That's the same as 5 * 2. These examples illustrate the commutative property, which states that the order of the numbers when you add or multiply doesn't affect the sum or product. In other words, a + b = b + a and ab = ba. The name comes from the word 'commute.' When you commute, what are you doing? You're moving from one place to another. When you get to the end of your commute,
145 you're still the same person. Well, unless you're commuting using a malfunctioning teleporter. Note that the commutative property doesn't work for subtraction or division. 38 does not equal 83. And 10/5 does not equal 5/10. But it does apply with addition and multiplication. Commutative Practice Let's practice. Let's say you have 2 * 5 * 3. That's 30. What if we rearrange it to 3 * 5 * 2? Still * 2 * 3? Still * 3 * 5? Yep, still 30. Here's one with addition: That's 26. If we rearrange it to ? Still 26. You could use the commutative property to justify eating your dessert first at dinner. After all, no matter which order you eat the food, it all ends up in your stomach. So why not polish off that ice cream before getting to the broccoli? Well, that's sort of the same thing, but not exactly. When you add 10 and 7 and 9, you're always dealing with constant numbers. If you had all the parts of your meal laid out, and you were sure to have room for all of them in your stomach, then order really doesn't matter. Granted, parents everywhere may still not approve. Associative Property There's another law that's similar to the commutative property. To understand this one, let's imagine the world's saddest yard sale. You're selling three things: a broken hair dryer for $1, a threelegged chair for $4 and a box of old VHS tapes for $2. Let's say your neighbor Mrs. Lake buys the hair dryer. Then your other neighbor, Mr. Rivers, buys the chair and tapes. You just made $1 from Mrs. Lake and $4 + $2 from Mr. Rivers  that's $7. While that won't buy you nicer stuff, it will buy you a burrito with guacamole. But what if Mrs. Lake bought the hair dryer and the chair? And then Mr. Rivers bought just the tapes? You'd then make $1 + $4 from Mrs. Lake and $2 from Mr. Rivers. You'd still get $7. And you'd still get that burrito.
146 This sad yard sale illustrates the associative property, which states that the way you group numbers when you add or multiply doesn't affect the sum or product.. In other words (a + b) + c = a + (b + c) and a(bc) = (ab)c. Whether Mrs. Lake buys two items and Mr. Rivers buys one or Mrs. Lake buys one and Mr. Rivers buys two, you still get $7. That was an addition example, but it works the same with multiplication. Let's say you have this: (5 * 2) * 3. If you remember the order of operations, you need to handle the stuff inside the parentheses first. That gets you 10 * 3, which is 30. But the associative property says that (5 * 2) * 3 like this is the same as 5 * (2 * 3) like this. That latter format gets you 5 * 6, which is, yep, also 30. Note that I said the property works for addition and multiplication. The associative property doesn't work for subtraction and division. (74)  2 does not equal 7  (42). With (74)  2, you first subtract 74 to get 3. Then you do 32, to get 1. In 7  (42), you start with 42, which is 2. You then do 72, which is 5. Associative Practice Let's try a few of these. Here's one: (5 + 10) + 7. Again, the order of operations says we need to do that first. But since everything here is addition, the grouping doesn't matter. So you could add the 10 and 7 first. In other words (5 + 10) + 7 = 5 + (10 + 7). No matter how you group it, you get 22. How about this one: 6 * (2 * 5)? If you do 2 * 5 first, you get 6 * 10, which is 60. But the associative property tells us that we could go (6 * 2), which is 12, then multiply that by 5, which still gets us 60. Lesson Summary In summary, the commutative property states that order doesn't matter when you are performing only addition or only multiplication. a + b = b + a and ab = ba. Your terms can commute around without changing the result. The associative property states that the way you group numbers when you add or multiply doesn't matter. (a + b) + c = a + (b + c) and a(bc) = (ab)c. Your terms can associate with whomever they'd like.
147 The Distributive Property and Algebraic Expressions Chapter 6 / Lesson 5 The order of operations is great. But sometimes we need to bend the rules to simplify an equation. Fortunately, the distributive property gives us a scenario in which this is okay. Learn all about it in this lesson. Distribution What do you think of when you hear the term 'distribution center'? I think of the mail. When you mail a letter or a package, you might bring it to the post office or put in a mailbox. People all over your town are doing the same thing. All the items from your town get collected and go to a distribution center. Once they're there, they get sorted and distributed into different trucks depending on where they're going. Then they leave the distribution center and head off to other parts of your town, your state, or even other parts of the world. The distribution center is the place where everything is organized into logical groups. All the mail for Wyoming goes in one place, and all the mail for Japan goes in another. And this is essentially what the distributive property is all about. Distributive Property The distributive property is a handy math rule that says when you are multiplying a term by terms that are being parenthetically added, you can distribute the multiplication across both terms, then sum their products. That was totally confusing, I know. The distributive property is much easier to show, and it's much simpler than it sounds. Think of it this way: a(b + c) = (ab) + (ac). Let's prove it with real numbers. If we have 5(3 + 4), the order of operations tells us we start with the parenthesis. So we do = 7, get 5(7) and then end up with 35. That's all well and good. But the distributive property tells us that in this situation, we can instead do (5 * 3) + (5 * 4), where we distribute the 5 across the parenthesis. Does it work? We get Is that still 35? Yep. It is. For most purposes, the distributive property is limited to multiplication. And while I said that the parenthetical terms must involve addition, remember than something like (72) is really just (7 + (2)), so this rule still works.
148 If you're wondering why (5 * 3) + (5 * 4) is in any way easier than 5(7), well, it isn't really. It would help to look at some examples of when this is particularly helpful. Practice Problems What if you can't add what's inside the parentheses? Look at this one: 7(3x + 5y). You can't simplify 3x + 5y. But you can distribute the 7 and get 21x + 35y. In fact, that's when you'll most often use this rule  when you have variables. Here's another one: 5(6 + 2x) Don't forget that negative sign. If we distribute the 5, we get 5 * 6, which is 30, and 5 * 2x, which is 10x. Put that together and our simplified expression is x. Here's one with a minus sign inside the parenthesis: 4a(62a). Remember, 62a is really just 6 + (2a), so our two terms are 6 and 2a. 4a * 6 is 24a. And 4a * 2a is 8a^2. So our simplified expression is 24a  8a^2. Let's try one that's a little more complicated: 2x(x  8y). Again, pay attention to those negative signs. 2x * x is just 2x^2. Okay, that's not so bad. And 2x * 8y? Wait  remember, it's 8y. Okay, 2x * 8y. You can't add x + y, but you can multiply them. We get positive 16xy. So our simplified expression is 2x^2 + 16xy. How about one more? (5a  3b). What's that negative sign hanging out in front of the parenthesis? It's really a 1. So we need to distribute the 1 across the terms. 1* 5a is 5a. And 1 * 3b is positive 3b. So our simplified expression is 5a + 3b. Lesson Summary In summary, the distributive property can be expressed as a(b + c) = (ab) + (ac). All we're doing is distributing the a across the terms inside the parenthesis. This is especially useful when we're dealing with variables that can't be added. The distributive property gives us the power to simplify our expression. Combining Like Terms in Algebraic Expressions Chapter 6 / Lesson 6 When you have an algebraic expression that's much too long, it would be great if you could simplify it. That's when knowing how to combine like terms comes in. In this lesson, we'll learn the process of combining like terms and practice simplifying
149 expressions. Puzzles I love jigsaw puzzles. There's something very gratifying about first finding all the edge pieces, then building that border. Then I like to sort the other pieces by color. Maybe it's a landscape painting, and a good third of the top of the puzzle is all blue sky. If I can't differentiate the pieces by color, I'm matching by shape. One way or another, the picture reveals itself. Building a jigsaw puzzle is very much like combining like terms with algebraic expressions. Combining Like Terms The phrase 'combining like terms' is kind of a puzzle itself. Let's look at the pieces of that phrase and then put them together. First, let's talk about terms. You're probably familiar with things like constants, which are just ordinary numbers. And then there are variables, like x or y. These are just symbols used in place of numbers we don't yet know. When you start putting these together, you get terms. In algebra, terms are constants, variables and products of constants and variables. So terms can be 1, 38 and They can also be x, a^2 or 98y. Let's connect 'like' and 'terms.' Like terms are individual terms that have the same variable. For example, 3x, 95x and 17x are all like terms. They all have a single variable, x. 4y^2 and 12y^2 are also like terms since they share the y^2. Constants are also considered like terms. These are numbers without variables, like 2, 5.4 and Our puzzle is almost complete. Let's add 'combining' to 'like terms.' Combining like terms is the process of simplifying expressions by joining terms that have the same variable. Adding Variables You know you can add That's a form of combining like terms. When we have variables, we do much the same thing. The key is to pay attention to the exponent. You can only add variables if they have the same exponent.
150 For example, if we have x + 5x, we can add those to get 6x. But if we had x^2 + 5x, we couldn't add those. Why not? Remember, the variable is just a symbol. In our example, x is standing in for a number we don't know. What if x = 2? Then x + 5x would be 2 + 5*2, which is , or 12. When we added x + 5x, we got 6x. If x = 2, what is 6x? It's still 12. But what about x^2 + 5x? If x = 2, x^2 + 5x is 2^2 + 5*2. That's , or 14. How could you combine x^2 and 5x? Would it be 6x^2? Well, then if x = 2, 6x^2 would be 6(2)^2, which is 6*4, or doesn't equal 14. Also, note that we can only add similar variables, like if we have two xs. But we can't add different variables, like x + y. x + y does not equal 2x and it doesn't equal 2y. When we have different variables, that means they're potentially representing different numbers. Practice Problems Okay, let's try some practice problems and get comfortable with combining like terms. Here's one: x^ x^ Think of this like a jigsaw puzzle. We need to put the like terms together. Which terms are like each other? First, we have two numbers with no variables: 2 and 3. Those can be combined to give us 5. So now we have x^2 + 6x^ And those x terms  do they have the same exponent? They do. They're both x^2. So we can add them to get 7x^2. That makes our simplified expression 7x^ Let's try another: 2x^25x + 3x + 8x^2 + y. Okay, again, put the puzzle pieces together. That 5x and 3x share the same exponent, so let's combine those. But wait  don't forget that that 5x is really 5x, so 5x + 3x = 2x. And that 2x^2 and 8x^2 both have an x^2 in them, so we can combine them to get 10x^2. What about that y? There are no other terms with a y in them, so we can't do anything with that. That means our simplified expression is 10x^22x + y. Okay, it's not a landscape painting or a picture of cats playing with yarn, but it is a simpler expression than what we started with. Let's try a different kind of algebraic expression: [(5y + 8y)  (6y + 2)]  [(3y y) + 9y]. The trick with this one is to not lose track of those negative signs. Let's start with what's inside parentheses. We can combine this first 5y and 8y to get 3y. Now, what about that 6y and 2? They're not like terms, so we can't combine them. If we distribute the minus sign across the parenthesis, this first section becomes 3y  6y  2.
151 Okay, there's more we can do there. We can combine the 3y and 6y  remember that it's a 6y  and get 3y  2. Next, let's look at the second part. We can combine 3y and y to get 2y. And we can add that 2y to this 9y to get 11y. It's time to bring these two sections together. So we have 3y  2 and 11y. But remember that there's this minus sign before the 11y. So it's 3y y. Anything else we can combine? Yep, the 3y and 11y. That becomes 14y. So 14y  2 is our final expression. We took an expression with 7 terms and, by combining like terms, got it down to just 2. That's pretty good for matching puzzle pieces! Lesson Summary In summary, combining like terms is just the process of simplifying expressions by joining terms with the same variable. Terms include constants, or numbers, and variables, like x or y. We can only add variables if they have the same exponent, and we can't add different variables together. Practice Simplifying Algebraic Expressions Chapter 6 / Lesson 7 In this lesson, we'll practice simplifying a variety of algebraic expressions. We'll use two key concepts, combining like terms and the distributive property, to help us simplify. Practice Makes Perfect If you've ever played a sport, you know the importance of practice. No track star just shows up at a meet and expects to run a recordbreaking 100 meters without training. The more challenging the concept, the more practice you need. If you ever watch professional football, you know that even with an incredible amount of practice, athletes can still make mistakes sometimes; a receiver may forget a route or a punt returner may drop the ball. But we know we can get better at whatever we're trying to accomplish with practice. What's true in sports is also true in algebra. Simplifying algebraic expressions can be as tricky as mastering a play in football. Fortunately, when we practice algebra, we're
152 unlikely to get knocked to the ground  unless you're playing fullcontact algebra. But let's not do that here. Simplifying Algebraic Expressions Here we're going to practice simplifying algebraic expressions. Simplifying algebraic expressions is more or less defined by its title. It involves distributing terms across parentheses and combining like terms in order to make an expression simpler. By simpler, we usually mean shorter, or more condensed. Why is this useful? Algebraic expressions can get cumbersome with all their various bits and pieces. Think of it like food. What if every time you wanted a cookie, you had to ask for it by its parts  flour, sugar, butter, eggs, etc.  as opposed to just saying 'I want a cookie'? That would be tedious, and it would interfere with your cookie eating. When we simplify an expression, we're combining what we can so we're just dealing with cookies, not the things that make up cookies. There are two main skills involved in simplifying algebraic expressions. First, there's combining like terms. This is the process of simplifying expressions by joining terms that have the same variable. So if you have x + 2x, you can combine them to get 3x. If you have 5x^2 + 3x^2 + 9x, you can only combine the 5x^2 and 3x^2 since they are the only terms that share the same exponent. But we can still make that expression simpler by saying 8x^2 + 9x. Second, there's the distributive property. This helpful law tells us that a(b + c) = (ab) + (ac). So, let's say we have 7(x + 2y). Since they're not like terms, we can't add that x and 2y. But the distributive property tells us we can distribute the 7 across the parentheses, giving us 7x + 14y. Practice Problems Okay, time for some practice. Let's start simple: 2y + 4y + 9. How can we simplify this? Well, we have two like terms: 2y and 4y. Both of these terms have the same exponent, y. Let's combine them to get 6y + 9. Can we go any further? No. The 6y and 9 don't share an exponent, so that's as far as we can simplify this one. Here's a good one: 9 + 3t  5. In this one, all we can combine are the 9 and the 5. So our final expression is 4 + 3t. That's it.
153 Those first two were a good warmup. Let's try a longer one. What if we have 3x^2 + 4x + x^ x? A good first step is to get like terms next to each other. What are our like terms? 4x and 11x both have an x. What about 3x^2 and x^2? They are like terms as well. If we move things around, we get 3x^2 + x^2 + 4x + 11x + 2. Now we just need to combine the like terms. We add 3x^2 and x^2 to get 4x^2. Then we combine 4x and 11x to get 15x. So our simplified expression is 4x^2 + 15x + 2. That's much better! Up to this point, we've only dealt with one variable. That's kind of like flag football. Let's jump to the NFL by using two: 9m + 8n + 3mn + 4m 2mn + n. It's a little trickier with multiple variables, isn't it? But let's do the same thing we did before  moving like terms next to each other. There are two terms with just one m: 9m and 4m. Then there are two with one n: 8n and n. What else? Those two with an mn? They're like terms, too. So with a little shuffling, we have 9m + 4m + 8n + n + 3mn  2mn. 9m + 4m is 13m. 8n + n is 9n. And 3mn  2mn is just mn. That gives us 13m + 9n + mn. Let's do one with some serious exponent work: (5x^2y)^3. First, let's handle that 5 cubed. That's 125. And what do you do with an exponent raised to an exponent? You multiply them together. So that x^2 to the third will be x^6. And the y will just become y^3. So our simplified expression is 125(x^6)(y^3). Okay, here's one that involves the distributive property: 4ab + a(3b + b^2). Remember, we can distribute that a across the parentheses. That gets us 4ab + 3ab + ab^2. And do we have any like terms? 4ab and 3ab. Combine those to get 7ab + ab^2. That's as far as we can take this one. It's like in a chocolate chip cookie  the flour, eggs and whatnot just become cookie dough, but the chocolate chips, represented here by the ab^2, are still chocolate chips. They're just tastier when baked into cookies. I think we're ready for a bigger challenge: 6x(x + 2y) + 3y(2x  y) + 4(x^2 + y^2). Okay, lots to do here. First, note that nothing inside the parentheses can be simplified. They all involve addition or subtraction with different variables. So let's use the distributive property and distribute the terms outside the parentheses. First, 6x * x is 6x^2 and 6x * 2y is 12xy. Next, 3y * 2x is 6xy and 3y * y is 3y^2. Then, 4 * x^2 is 4x^2 and 4 * y^2 is 4y^2. That gives us 6x^2 + 12xy + 6xy  3y^2 + 4x^2 + 4y^2. Let's do some shuffling and get 6x^2 + 4x^23y^2 + 4y^2 + 12xy + 6xy. 6x^2 + 4x^2 is 10x^2. 3y^2 + 4y^2 is just positive y^2. Then, 12xy + 6xy is 18xy. Okay, that means our final, simplified expression is 10x^2 + y^2 + 18xy. That's much simpler than where we started! Cookie metaphor or not, I think we've earned a treat.
154 Lesson Summary In this lesson, we practiced simplifying algebraic expressions. We became experts at combining like terms and utilizing the distributive property. With these methods, we made long, complicated expressions much simpler and easier to understand. We also talked football and cookies, because all things should involve exercise and baked goods. Negative Signs and Simplifying Algebraic Expressions Chapter 6 / Lesson 8 Don't let negative signs get you down. In this lesson, we'll stay positive as we practice simplifying algebraic expressions that have those tricky negative signs. Negative Signs Negative signs  they have incredible power. They can reduce a number to not just nothing, but less than nothing. It's one thing if you had 3 oranges and now you have 0. What if you had 3 oranges? That's like having 3 black holes where your oranges were. It's weird. It's like negative signs create Bizarro World numbers of regular numbers. That negative sign is thrown on them like the goatee on evil Spock. You know you can't trust someone with a goatee. Well, except me. Or am I an evil version of myself? Oh man. Anyway, negative signs. They can also wreak havoc in algebraic expressions. They can pop up anywhere and take an otherwise straightforward problem and make it confusing. But fear not. We can handle a few negative signs in some algebraic expressions like the algebra superheroes we are. Let's learn how. Simplifying Algebraic Expressions First, we should familiarize ourselves with simplifying algebraic expressions, or making expressions simpler by using the distributive property and combining like terms. The distributive property, of course, is one of our algebraic expression superpowers. It's when we take a(b + c) and make it equal to (ab) + (ac). This law keeps multiple variables inside parentheses from slowing us down. If you see something like 3(2m +
155 5n), use the distributive property and distribute that 3 across the parentheses, leaving you with 6m + 15n. Then there's combining like terms. This is when we join terms with the same variable. You know how at the end of superhero movies the villain is defeated, but the city is also kind of destroyed? It takes a special kind of superhero to put those buildings back together. You need to know where everything goes and match up the right pieces. When you have 2x + x^2 + 6x + 3x^2, we match up 2x and 6x to get 8x. Then we match x^2 and 3x^2 to get 4x^2. That makes our rebuilt city, and our simplified expression, 8x + 4x^2. Okay, we know what we need to do. Let's go handle some negative signs. Let's start with some distributive property ones. Distributive Property Practice Here's one: (3s + 2t). Can we simplify this? We can't add 3s to 2t. But we can distribute the number outside the parentheses. 'Wait,' you might say. 'There is no number there. There's just that negative sign.' Remember, that negative sign is really a 1 with some cloaking powers, and we can totally distribute it. 1 * 3s is 3s. And 1 * 2t is 2t. We put that together and we get 3s + 2t, which is 3s  2t. Here's another: (5p  2r). Okay, this looks just like the last one, but with one important difference. Yep, that minus sign. I think the easiest way to handle this is to treat the 5p, 2r and minus sign as unique parts. We multiply 1 * 5p and get 5p. Then we multiply 1 * 2r and get 2r. We put those back in our expression and we have 5p  (2r). What's  (2r)? + 2r. So our final expression is 5p + 2r. You could also think of the original expression as 1(5p + (2r)). Either way, don't lose track of that minus sign! Like Terms Practice Okay, how about one practice with combining like terms? Here's one: 3y  4y. Here we have two terms, 3y and 4y. And they just happen to be like terms  awesome. What do we get when we take 4 from 3? 1. So what's 3y  4y? y. That's it.
156 That wasn't really superherolevel, was it? How about this: 3p  9p^26p^ p + p^2. We want to get the like terms next to each other by shuffling things around. But when we have a mix of plus and minus signs, we need to be very careful that we don't lose any. What like terms do we have? 3p and 2p. So let's move the 2p over here. But wait, it's a 2p. That's better. We also have this 9p^2, 6p^2 and +p^2. Let's move the +p^2 over here. We still have that 4, but we can't do anything with that. Now let's combine the like terms. 3p  2p is just p. 9p^26p^2 is 15p^2. A common mistake is to just see the 9p^2 and do 96 to get +3p^2. If you do that, you're letting a negative sign get away. Then you're just going to have to deal with it in the sequel. And nobody wants to see the same villain twice. So we have 15p^2 + p^2. What's ? 14. So our simplified expression is p  (don't forget that minus) 14p^ Now we've practiced both the distributive property and combining like terms. Let's put them together for a final, epic battle. Additional Practice [2(3x^25xy)  3x(x + 2y)]  [x(4x + y)  y(32x)] Whoa. That's a monster. Let's first see if there are any like terms inside parentheses that we can combine. Not here, or here, or here, or here. Okay, it's time to bring out the distributive property. Now, there are a lot of minus signs. Let's take it slow and not miss any. First, we distribute this * 3x^2 is 6x^2. 2 * 5xy is +10xy. Now the 3x. 3x * x is 3x^2. 3x * 2y is 6xy. Don't forget the negative sign here, which makes it 3x^26xy. So the first half of our expression is 6x^2 + 10xy  3x^26xy. Let's look at the second half. x * 4x is 4x^2. x * y is xy. y * 3 is 3y. y * 2x is 2xy. So we have 4x^2  xy and 3y  2xy. But don't forget this sneaky minus sign here. So it's 4x^2  xy  3y + 2xy. Before we put these two halves together, remember that they're joined by a minus sign. So we also need to distribute that across the second half. That will give us +4x^2 + xy + 3y  2xy. So now we have, wait for it, 6x^2 + 10xy  3x^26xy + 4x^2 + xy + 3y  2xy. Our monster is apparently a shapeshifter. Well, we've been conquering it to get here. Let's put our like terms together and finish it. Let's move this 3x^2 and this +4x^2 over here with this 6x^2. When we do that, we have this +10xy next to this 6xy and +xy. Let's drag this 2xy over with these. And now it's time to combine.
157 6x^23x^2 is 9x^2. Add 4x^2 and we have 5x^2. Now 10xy  6xy is 4xy. If we add xy and subtract 2xy we have +3xy. That makes our simplified expression 5x^2 + 3xy + 3y. Remember what we started with? Yeah, I think we won that battle. Lesson Summary In summary, negative signs can wreak havoc on complicated algebraic expressions. The principles of what we're doing when we simplify, though, don't change. Remember, simplifying algebraic expressions is making expressions simpler by utilizing both the distributive property and combining like terms. The distributive property can be boiled down to a(b + c) = (ab) + (ac). And combining like terms is joining terms with the same variable. When our expressions have negative signs, watch them closely. As long as you never lose sight of them, there's no expression you can't simplify.
158 What Are the Five Main Exponent Properties? Chapter 7 / Lesson 1 We'll look at the five important exponent properties and an example of each. You can think of them as the order of operations for exponents. Learn how to handle math problems with exponents here! Understanding the Five Exponent Properties We are going to talk about five exponent properties. You can think about them as the order of operations for exponents. Just like the order of operations, you need to memorize these operations to be successful. The five exponent properties are: The Quotient of Powers property Product of Powers Power to a Power Quotient of Powers Power of a Product Power of a Quotient Let's look at the first one.
159 Product of Powers Here's the formula: (x^a)(x^b) = x^(a + b). When you multiply exponentials with the same base (notice that x and x are the same base), add their exponents (or powers). Let me show you how that works. Let's say I have (x^2)(x^3). Well, x^2 is x times x, and x^3 is x times x times x. When we add all those xs up, we get x^5, which is the same thing as adding Power to a Power The Power of a Product property We can see from the formula we have (x^a)^b. When you have a power to a power, you multiply the exponents (or powers). Let me show you how this one works. If I have (x^2)^4, which would be x^2 multiplied four times, or x^2 times x^2 times x^2 times x^2. Once again, we add all the exponents and get x^8, and x^8 is the same as x^(2 * 4), which is 8. Not too bad, right? Quotient of Powers Remember, 'quotient' means 'division'.' The formula says (x^a) / (x^b) = x^(a  b). Basically, when you divide exponentials with the same base, you subtract the exponent (or powers). Let me show you how this one works. Let's say I had (x^4) / (x^3). In the top (or numerator), we have x times x times x times x. In the bottom (or denominator), we have x times x times x. Hopefully, you remember that x divided by x is 1, so the xs cancel. So, x divided by x is 1, x divided by x is 1, and x divided by x is 1. So, when
160 we cancel them, what are we left with? That's right: x^1, or just x. So (x^4) / (x^3) is just x^(43), which is x^1. The Power of a Quotient Property Power of a Product The formula says (xy)^a = (x^a) and (y^a). When you have a product of a power, you give each base its own exponent. Think about it as distribution putting the exponent with each base. Let me show you how this one works. Let's say we had (xy)^2. That means we take xy and multiply it twice that means xy times xy. Well, that would give us two xs, or x^2, and two ys, or y^2. That is the same as if I distributed 2 to the x, getting x^2, and 2 to the y, getting y^2. Power of a Quotient When we look at this formula, we have x/y, or a fraction raised to the a power, this gives us (x^a) / (y^a). When you have a quotient to a power, you give each base its own exponent. We think of it as the exponent being distributed to each part of the fraction, just like the last one, power of a product. Let me show you how this one works. Let's say I have (x / y)^3. Remember, that means I'm going to take x / y and multiply it three times. That would be x / y times x / y times x / y. If we look at the top (or the numerator), we have x times x times x, or x^3. If we look at the bottom (or denominator), we have y times y times y, or y^3. That would give us (x^3) / (y^3), which is basically distributing 3 to the x and y.
161 Lesson Summary Let's take a final minute to review the five properties. Product of a Power: When you multiply exponentials with the same base, you add their exponents (or powers). Power to a Power: When you have a power to a power, you multiply the exponents (or powers). Quotient of Powers: When you divide exponentials with the same base, you subtract the exponents (or powers). Power of a Product: When you have a product of a power, you give each base its own exponent (or distribute the exponent to each base). Power of a Quotient: When you have a quotient to a power, you give each base its own exponent. How to Define a Zero and Negative Exponent Chapter 7 / Lesson 2 The zero and negative exponent properties are two you will use quite a lot in mathematics. The negative exponent property can be confusing, but when you remember a couple fun ideas, you will get it right every time! Negative Exponent Formulas for negative exponents In math, we like to write exponents with a positive number. So what happens if I get a negative exponent? What about a zero exponent?
162 Before we get started, I need to tell you something important here: x^a does not mean x^a. The negative exponent has nothing to do with positive or negative numbers. Let's look at the two formulas. x^a = 1/(x^a) 1/(x^a) = x^a I want to make sure that you understand that x^a does not mean x^a. Once again, the negative exponent has nothing to do with positive or negative numbers. If the exponential is negative in the numerator, or the top, it tells us the exponential is actually positive in the denominator. If the exponential is negative in the denominator, or the bottom, it tells us the exponential is actually positive in the numerator. If you see a negative exponent, flip it to a positive. That is, if the exponent is negative in the numerator, flip it positive to the denominator. If the exponent is negative in the denominator, flip it positive to the numerator. Let's say I have x^4. Now remember, x^4 can be written as a fraction (x^4)/1. Remember, if it's negative in the numerator, you flip it positive to the denominator. So that's the same thing as saying 1/(x^4). In this next example, x^7 is in the denominator. We're going to flip it positive to the numerator. x^7 is the same thing as 1/(x^7). Zero Exponents The formula for zero exponents I think the zero exponent is really fun! Basically, the formula says that anything  any number, any letter  raised to the zero exponent is always one. So when I have x^0, it's 1. Let's say I had something funny like 999,999^0. It's still 1!
163 The Quotient of Powers Rule says that when we divide exponentials with the same base, we subtract their exponents. So, I have (x^3)/(x^3). Remember, we subtract their exponents. So that would be x^(33), which is x^0. But where does the 1 come from? We have (x^3)/(x^3). That is, (x*x*x) on the top, or numerator, and (x*x*x) on the bottom, or denominator. x/x is 1, x/x is 1, and x/x is 1. When we multiply those together (1*1*1), we get 1! So that pretty much proves it: x^0 = 1! How to Simplify Expressions with Exponents Chapter 7 / Lesson 3 Review of the Properties of Exponents Let's review the exponent properties: Product of Powers: (x^a)(x^b) = x^a+b Power to a Power: (x^a)^b = x^a*b Quotient of Powers : (x^a) / (x^b) = x^ab Power of a Product : (xy)^a = x^ay^a Zero Property: x^0 = 1 If you're having problems memorizing these properties, I suggest using flash cards. Flash cards are a fantastic and easy way to memorize topics, especially math properties.
164 Split up these expressions using multiplication after putting like terms on top of one another Practice Problem #1 So, let's get started! Simplify is the same as reducing to lowest terms when we talk about fractions. Simplifying these terms using positive exponents makes it even easier for us to read. Our first expression has x^3y^8 / y^3x^7. The first step I like to do is put the like terms on top of each other. On the top, I have x^3y^8. In the denominator, I want the xs over each other and the ys over each other, so I write x^7y^3. My next step is to split these up using multiplication. This step is important when you first begin because you can see exactly what we are doing. Splitting the multiplication gives us x^3 / x^7 times y^8 / y^3. Next step  look at each part individually. Since we have x^3 divided by x^7, we subtract their exponents. This gives us x^37. Since we have y^8 divided by y^3, we subtract their exponents. This gives us y^83. This will give us x^37, which is 4 and y^83, which is 5. Solution to the first practice problem Remember, we're simplifying using positive exponents, so we need to change x^4. We know from our exponent properties that x^4 is 1 / x^4 times y^5. Well, 5 is positive, so we don't need to change it. My last step is to multiply. Our final, simplified answer is y^5 / x^4. This is our simplified answer with positive exponents.
165 Practice Problem #2 Let me show you another one. This time we have 5x^2y^9 / 15y^9x^4. Let's rewrite this with like terms over each other: 5/15 times x^2 / x^4 times y^9/y^9 We start at the beginning. 5/15 reduces to 1/3. Next, x^2 divided by x^4 is x^(24). y^9 divided by y^9 is y^(99). Let's keep simplifying. We have 1/3 times x^(24), which is 2, times y^(99), which is y^0. This gives us 1/3 times 1/x^2 times 1. Multiplying straight across, our final answer is 1/3x^2. Steps in solving practice problem #2 This is our answer simplified using positive exponents. Practice Problem #3 There are a lot of letters and numbers here, but don't let them trick you. If we keep separating the terms and following the properties, we'll be fine. Our first step is to simplify (2p)^3. We distribute the exponent to everything in the parenthesis. This will give us (8p)^3q^4 in the bottom or denominator, but our top or numerator will stay the same. Next, we separate them into multiplication: 16/8 times p/p^3 times q^2 / q^4 times r^9. Here's the fun part, simplify. 16/8 is 2/1 times p^(13) times q^(24) times r^9. We're almost done: 2 times p^(13) is 2, times q^(24), which is q^(2) times r^9. We are asked to simplify using positive exponents: p^(2) is the same as 1/p^2; q^(2) is the same 1/q^2.
166 Solution for the third practice problem Finally, our last step  multiplying the fractions straight across. Our final answer is r^9 / p^2q^2. This is in simplified form using positive exponents. Remember, it will take time and practice to be good at simplifying fractions. Rational Exponents Chapter 7 / Lesson 4 Quick Review Now, remember, a rational number is any number that can be written as a fraction. If that's true, what is a rational exponent? A rational exponent is an exponent that is written as a fraction. Here are a couple examples: x^(2/3), y^(1/2). A rational exponent is an exponent that is written as a fraction You'll notice that the exponent is a fraction. So what is the fraction telling us? The fraction refers to a radical. Let's look at a radical expression.
167 Rational to Radical First, we have the radical symbol. The b is called the index or root number. Inside the radical is the radicand. For our example, x^a is the radicand. How does that convert to a fractional exponent? Here's what the radical and rational exponent look like together. As you can, see the b in the denominator is the same as the index for the radical. The a in the numerator is the same as the exponent in the radicand. There are two ways to rewrite a rational exponent: x^(a/b) = bth root of (x^a) = (bth root of x)^a. These all mean the same! A Few Examples Let me show you how this works. So, we have y^2/3 is the same thing as saying the cube root (it's the cube root because we have a 3 here) of y^2. Well, that's the same thing as saying (cube root of y)^2. Let's look at another one. Here we have the fifth root of x^3. The fifth root is rewritten as the denominator in the rational exponent fraction. The exponent 3 is written as the numerator in the rational exponent fraction. So, the fifth root of x^3 is rewritten as x^(3/5). Here's the next example: the seventh root of the quantity (2x)^5. How would that look as a rational exponent? The quantity (2x)^(5/7). Remember, the denominator of the rational fraction is the radicand exponent and the numerator is the index number. Without parentheses around the 4x, only the x is raised to the 1/2 power Mathematicians love to throw a wrench into the system. Let's say I had 4x^(1/2). Do you see the 4x is not in parenthesis like the last one we looked at? Since the 4 and the
168 x are not in parenthesis, only the x is raised to the 1/2 power. The 4 is all by its self. So how would that look as a radical? Since 4 is not raised to a power, it will sit in front of the radical. As far as the radical is concerned, the numerator is 1, so the exponent inside the radicand is 1. The denominator is 2, so this would be our index. When we see a radical symbol without a number in the index, we always assume it's 2. Lesson Summary As you've seen, the fraction tells us we will be rewriting the exponential as a radical expression. The numerator of the fraction is the exponent inside the radicand and the denominator is simply the index. Simplifying Expressions with Rational Exponents Chapter 7 / Lesson 5 Simplifying expressions with rational exponents is so easy. In fact, you already know how to do it! We simply use the exponent properties but with fractions as the exponent! Expressions with Rational Exponents Rational exponents follow exponent properties except using fractions. Review of exponent properties  you need to memorize these. Just can't seem to memorize them? Have you tried flashcards? They work fantastic, and you can even use them anywhere! 1. Product of Powers: x^a*x^b = x^(a + b) 2. Power to a Power: (x^a)^b = x^(a * b) 3. Quotient of Powers: (x^a)/(x^b) = x^(a  b) 4. Power of a Product: (xy)^a = x^ay^a 5. Power of a Quotient: (x/y)^a = x^a / y^a 6. Negative Exponent: x^(a) = 1 / x^a 7. Zero Exponent: x^0 = 1 Putting the exponent rules to work with exponent properties...
169 Example #1 y^(1/2) * y^(1/3) For this one, we're going to follow the product of powers. Remember, when we multiply, we add their exponents. 1/2 + 1/3 = 5/6 So the answer is going to be y^(5/6). Example 2 uses the quotient of powers property to find the solution Example #2 Simplify: x^(3/5) / x^(2/3) For this one, we're going to use the quotient of powers. Remember, when we divide, we subtract their exponents. So, we're going to have: x^(3/52/3) 3/52/3 = 1/15 So our answer is x^(1/15). Example #3 Simplify: x^(2/7) For this one, we're going to use the negative exponents property. Remember, when we have a negative exponent, we flip it. If it's in the numerator, we flip it to the denominator, which is in this case.
170 So our answer is going to be 1 / (x^(2/7)). Example #4 Simplify: (x^(4/5))^(3/4) In this one, we have power to a power. We're going to have (x^(4/5))^(3/4), so we're going to multiply 4/5 * 3/4 which is 12/20. We need to reduce our fractions when we're going to get our final answer. 12/20 reduces to 3/5. So our answer is going to be x^(3/5). After reducing 12/20 in example 4, the final answer is x^(3/5) Example #5 Putting multiple exponent rules to work with exponent properties... Simplify using positive exponents. Always reduce the fractions to lowest terms. ((p^(1/2)q^3)^(2/3)) / (( pq)^(1/2)) First we're going to simplify the power to a power. So now we'll have: (p^(2/6)q^(6/3)) / (p^(1/2)q^(1/2)) Write like terms over each other, if necessary. Well, we already have the p's over the p's and the q's over the q's. There's no need to simplify fractions now. We're going to go right to simplifying quotient of powers. Remember, when we divide, we subtract their exponents. So we have: p^(2/61/2) * q^(6/31/2) That gives us: p^(1/6) * q^(9/6)
171 Next we need to reduce the fractions because we're almost to our answer. So we'll have: p^(1/6) * q^(3/2) We want to rewrite these using positive exponents. Remember, if it's negative in the numerator, it flips to the denominator. So our final answer's going to be: q^(3/2) / p^(1/6) Example #6 The solution for example 6 after applying the quotient of powers property Simplify using positive exponents. Always reduce the fractions to lowest terms. We're going to have: ((2m^(2/3))^3 / (8m^(1/6))^2 We're going to simplify power to a power. So we'll have: 2^(3)m^(6/3) / 8^2m^(2/6) Remember, power to a power means to multiply the exponents. Next, let's write like terms over each other. We already have 2^3 over 8^2 and m^(6/3) over m^(2/6). So let's move to the next step. There's no need to simplify fractions just yet, so we're going to simplify quotient of powers. Remember, when we divide, we subtract. So now we're going to have: 8/64 * m^(6/32/6) Well, 8/64 is 1/8. m to the 6/32/6 is m to the 10/6. So it turns out that our final answer is: m^(5/3) / 8
172 We won't touch the improper fraction in this video. We're just simplifying rational exponents. Example #7 Simplify using positive exponents. Always reduce fractions to lowest terms. ((p^(3/5)q^(1/2) / ( p^(2/5)q^(1/2))^(1/2) First we're going to simplify power to a power. Remember, power to a power means to multiply the exponents. That'll give us: (p^(3/10)q^(1/4)) / (p^(2/10)q^(1/4)) Next, if we need to, write like terms over each other. There's no need to simplify fractions now. We're going to move to quotient of powers. Remember, when we divide, we subtract their exponents. So that's going to give us: p^(3/10  (2/10))q^(1/4  (1/4)) So let's keep simplifying. p^(5/10)q^(0) The zero exponent says q^0 equals 1. Now we need to reduce our fraction 5/10. That's going to give us our answer: p^(1/2) The radical to rational fraction formula used in example 8 Radical to Rational Fraction Formula Review radical to rational fraction formula... The bth root of x^a = x^(a/b)
173 The index is the denominator. The exponent is the numerator. What happens when the expression has radicals? 1. Change the radicals to rational exponents 2. Follow exponent rules Example #8 (third root of x)(fifth root of x^4) First we need to change to rational exponents, so we're going to have: x^(1/3) * x^(4/5) Did you remember the denominator is the index number and the numerator is the radicand exponent? Following our exponent rules, we're going to do a product of powers. Remember, when we multiply, we add their exponents. So we're going to have: x^(1/3 + 4/5) Well, 1/3 plus 4/5 is 17/15. So our answer's going to be: x^(17/15) Remember, we need to change the rational exponent back into a radical expression. x^(17/15) = 15th root of x^17 Lesson Summary Rational exponents follow the exponent rules. Remember to reduce fractions as your final answer, but you don't need to reduce until the final answer. For operations on radical expressions, change the radical to a rational expression, follow the exponent rules, then change the rational expression back to a radical expression.
174 What is a Linear Equation? Chapter 8 / Lesson 1 Most cars won't be able to run for more than 250,000 miles, so how much longer will your car live? Linear Equations are the most basic kind of algebraic function and can help you answer questions exactly like this. Learn about what they look like, how they come up in your life and why they are powerful tools. Defining a Linear Equation This lesson is on what a linear equation is. And the answer to that question is essentially a linear equation is any pattern of numbers that is increasing or decreasing by the same amount every step of the way. This means that the only two things that we need to define a linear equation are where the pattern begins and what that pattern moves by. What that leaves us with is the slopeintercept form of the linear equation, y = mx + b, where the m value is the slope, and the b value is the yintercept. Examples of linear equations in slopeintercept form Some examples of linear equations in slopeintercept form: You could have y = 2x + 1; you could have y = 3x; and you could have y = (2/3)x  6. In each equation, the number in front of the x represents the slope, or the number that it's moving by. The
175 number on the end represents where it begins, or the yintercept. If there is no number after the x, that implies that the yintercept is zero. Building a Linear Equation What's the point? Why do I care about a linear equation? Well, linear equations are the most common patterns of numbers we see, and they can be used to describe all sorts of situations you see around you. One example has to do with the first car I bought. As soon as I turned 16, my parents helped my buy a used car. When I got my car it had 27,000 miles on it, and I've owned it for a while now. Every year, I drive it about 12,000 more miles. What we end up with is a linear equation to represent the situation that looks like y = 12,000x + 27,000, because I drive it 12,000 more miles every year, and that's how much the pattern moves by. And the pattern began at 27,000 miles when I first bought the car. So, after year one, it had 39,000; after year two, it had 51,000; and after year three, it had 63,000. And so on. Plotting a Graph Drawing a graph of this linear equation can help us make a lot of predictions about what this pattern will do in the future, and it can also give us a better sense of the pattern that may not be easy to see just from looking at the numbers. When I want to graph a linear equation, I use the yintecept, the b value as the beginning value on my graph. So, before I had driven the car a single day, it already had 27,000 miles on it because I got it used. After year one, I'd been driving it along, and after that year I was up to 39,000 miles. I kept driving it and driving it, so after year two I was up to 51,000 miles. So, every time I go over one year, I go up by 12,000 miles. What we end up with are a bunch of points that are exactly in a row, and these points form a line. All linear equations come across as lines when you graph them  which makes sense, since linear and line are almost the same word. While I was drawing the graph, you may have noticed the triangle that I drew underneath where I went over one year and up 12,000 miles. This triangle is something called a slope triangle, and it helps us determine and draw the slope of the graph. Again, the slope is another name for how much the pattern is moving by. The slope is also what we call the rise over the run; that basically means how it goes up and down divided by how much it goes left and right.
176 Graph of a linear equation This graph is really nice because it tells me how many miles will be on my car at any one point in just one picture, but these are really only estimates. If I wanted to know how many miles were on my car after 1 year, 7 months and 6 days, I'd have to go up to this point and read it over. We could guess that it's around 50, but we don't really know. And that's where the algebra comes in to it. Solving a Linear Equation So, it's my understanding that most cars last around 250,000 miles; I think I'd be happy if my car made it that far. My question is how much longer do I have? Will my car last as long as I think it will, or am I going to have to get another car pretty soon? We can use the equation to answer that question, where y is the number of miles and x is the number of years that have gone by: 250,000 = 12,000x + 27,000. At this point I can undo the things that have been done to x to isolate x and get the number of years, which will give my answer. I know that x first gets multiplied by 12,000 and then it gets 27,000 added to it, so I need to undo those things backwards. The first thing I have to do is subtract 27,000 from both sides. Next, I'll need to divide both sides by 12,000. So, I undo multiplication with division; I divide both sides by 12,000. And I find out that x is about 18 years. Summary Linear equations are patterns of numbers that either increase or decrease by the same amount each step of the way. They can be represented in slopeintercept form, y = mx + b, where m is the slope and b is the yintercept. On a graph, the yintercept tells you where to begin, and the slope tells you how much to go up and down over how much to go left and right.
177 While the graphs provide us with a good overview and a good way to get a lot of estimates, the specific linear equation can provide us with specific numbers and the exact amount we're looking for to help us make predictions of things far into the future. How to Write a Linear Equation Chapter 8 / Lesson 2 Simply knowing how to take a linear equation and graph it is only half of the battle. You should also be able to come up with the equation if you're given the right information. A Mathematician's Needs Equation in slopeintercept form where m is the slope and b is the yintercept Mathematicians like to be thorough and find all the connections between things. So just being able to go from an algebraic rule to drawing the graph isn't really good enough for them. They want to be able to go the other way, too. Going from the graph to the rule isn't too bad because you can just pick out the information you need. But sometimes the picture isn't always going to be given and only small pieces of information will be given to you, and it's up to you to fill in the gaps. Finding a Linear Equation There's a lot of different kinds of information that could be given. The most common is that you'll be given two points. Maybe they'll ask you to find the equation of the line between the points (3,4) and (6,2).
178 For any linear equation, the two things we need to know are the slope and the yintercept. The slope is every time going to be the first thing we should try to find. Once we know the slope, it's a lot easier to find the yintercept. I personally like to find the slope when I'm given two points just using the logic that I know because I know the slope is the rise over the run, which means how much it goes up and down over how much it goes left and right. I know the rise is the change in y. So in this one, I see the y starting at 4 and going down to 2, which means the rise is 6. The run, the left and right, is x, and I see it start at 3 and go up to 6, which is a change of 3. And so I can get the slope right from this, which is 2. Parallel lines have the same slope A lot of people like using the formula, which is the x1, y1, x2, y2 formula. If you do it this way, you'll get the exact same answer. You plug in numbers y2 minus y1 over x2 minus x1 (y2  y1 / x2  x1). You do 2 and get 6. And 63 = 3. And again, we end up with a slope of 2. Like I said earlier, the two things we need are the slope and the yintercept. We now know what the slope is, which means the yintercept is the only thing left to find. Because I know m (2) and because it gave us a sample x and y (3,4), I can substitute in everything I know and I'm left with an equation with only one variable in it (4 = 2 * 3 + b). First, I do the operation it asks us to: 2 * 3 = 6. Then, I have to get the b by itself, which means I undo the 6. I undo subtraction with addition because of inverse operations. I end up with b = 10. Now that I know m and I know b, I'm done, and my answer is y = 2x That means that this linear equation begins at 10 and is decreasing by 2 every step of the way. So that' s just one example of some information you may be given. But there's a lot of different things that could be given to you and we've only touched on one of those things.
179 Parallel and Perpendicular Another very common thing to see is that instead of you getting two points, you only get one point, but they tell you that your line is either parallel or perpendicular to other sample line that they tell you. In order to solve this question, you have to know what the deal with parallel and perpendicular lines is. Parallel lines are two lines that kind of look like train tracks. They go in the same direction that ends up meaning that they have the exact same slope. They go over and up the exact same amount. So parallel lines have the same slopes. Perpendicular lines, on the other hand, are two lines that intersect each other at right angles. Perpendicular lines' slopes are what we call opposite reciprocals. One line is going over a lot, and up a little and one line is going over a little and down a lot. That has to do with the reciprocal part, which essentially means that if you have a fraction (1/4), you flip it (4/1). Perpendicular lines have slopes that are opposite reciprocals So opposite reciprocals are numbers where one is positive (1/4) and one is negative (4/1), and one is one fraction and the other is the flipped fraction. Don't forget to do both. It's really common to just make it negative or just make it positive, or forget to flip it or flip it but forget to change the sign. It's both. Finding a Parallel Equation Now that we know about parallel and perpendicular lines, we can answer questions like this that ask us, 'What is the equation of a line that is parallel to y = 3x + 2 and through the point (3,6).' Again, the two things we need to find are m and b, the slope and the yintercept. We always want to find the slope first, so now I have to go about finding the slope in a
180 slightly different manner than we did earlier. I can no longer use the slope formula, because it only gives us one point, so I have to use what I know about parallel lines. Luckily, we just learned that parallel lines have the same slope, which means that my slope is the exact same as this one, so I can just take the slope from this line (3) and just steal it and I already know the slope; all my work is done and I have half the answer to my question (y = 3x + b) The other half that I still need is to find the b and I'm going to do this in the exact same way that we just did. I now know a sample x (3), a sample y (6), I know what m is (3), all I have to do is find b. I can substitute in what I know (6 = 3 * 3 + b). I can solve the equation for b by doing inverse operations to get the b by itself, and we find that b = 15. Now that we know b and m, we have our answer, which is y = 3x Using slopeintercept form to solve a perpendicular equation Solving a Perpendicular Equation Similarly, you could be asked to find the equation of the line that is perpendicular to this one through this point. We've got the same equation (y = 3x + 2) and the same point (3,6), but now instead of it being parallel, it's perpendicular. I still need to find the m and the b, just like before, and again, I cannot use the slope formula because I only have one point. That means we use what we know about perpendicular lines and how their slopes relate. I know that the slope of y = 3x + 2 is 3, but because it's perpendicular, I can't simply steal that. I have to first take the opposite reciprocal of it. The opposite part means that it changes from positive (3) to negative (3). The reciprocal part means that it changes from 3/1 to 1/3. And now, we have the slope (1/3) of our line, which means that we have half our answer (y = (1/3)x + b) and all that's left to find is b.
181 We find b in the exact same way we found b in every other problem in this video. We substitute in our sample x (3), our sample y (6), what we know m is (1/3) and we solve the equation for b (6 = (1/3)(3) + b). I multiply what I can and I undo to get b by itself, and this time we find that b = 5. This means my solution is y = (1/3)x + 5. Lesson Summary To review, we learned that for any equation, we need to know what is the slope and what is the yintercept. Once we know those two things, we're done and we can substitute them into y = mx + b. The way we actually find m and b differs depending on the information that's given. We like using the slope formula (y2  y1 / x2  x1) if it gives us two points. Or, we might have to use what we know about parallel and perpendicular lines. We always find m first and then substitute in to find b. You need to remember that parallel lines have slopes that are the same, whereas perpendicular lines have slopes that are opposite reciprocals. Problem solving using Linear Equations Chapter 8 / Lesson 3 From sale prices to trip distances, many real life problems can be solved using linear equations. In this lesson, we'll practice translating word problems into linear equations, then solving the problems. Real World Math A train leaves Chicago at 7 a.m., traveling at 70 mph to New York, which is 800 miles away. Another train leaves New York at the same time, traveling on a parallel track to Chicago at 85 mph. When will they meet? The question is: why do we care so much about trains? Well, I like trains, but I still feel a little nervous when I read a math problem that starts with a train. If I'm going to have to translate a real world scenario to an algebraic equation, can't it be something I might actually encounter in my life? I mean, I've ridden trains between Chicago and New York, but I've never plotted when my train will pass another.
182 In this lesson, we'll not only practice solving problems that can be translated into linear equations, we'll also focus on problems you may encounter in your life  problems not involving trains passing each other. Linear Equations As a reminder, a linear equation is just an algebraic expression that represents a line. These equations typically have one variable and look like 3x = 9 or y + 4 = 10. In these equations, we're trying to figure out the variable, which involves getting it alone on one side of the equals sign. Simple Problems There are simple problems that involve linear equations. For example, the sum of 35 and a number is 72. What is the number? The thing we don't know is our variable. Let's use x here. We know x + 35 = 72, so that's our equation. If we solve for x by subtracting 35 from both sides, we get x = 37. Now we know our number. They can be a bit more complex, like this: 15 less than four times a number is 57. What is the number? Again, let's use x for the number. Four times that number is 4x. 15 less than that is 4x 15. So our equation is 4x  15 = 57. To solve for x, we add 15 to both sides. Then we divide by 4, and x = 18. Practice Problems Let's take that knowledge and look at some real life situations. Let's start with money. We all like money, right? Let's say you're a little short on cash and need a loan. Your cousin agrees to loan you money and you agree that you'll repay him in full plus 4% interest. We're going to ignore the questionable judgment he displays in loaning money to family. If he loans you $500, how much interest will you need to pay? Our variable here is the amount of interest, so let's call that x. The interest will be the amount of the loan, $500, multiplied by the interest rate, 4%. To multiply with a percent, we convert it to a decimal. So x = 500*.04. What is 500 *.04? 20. So you'll owe him $20, plus the original $500.
183 You decide that you want to be better about saving money. Your cell phone company is promoting a text message plan that costs $10 each month plus five cents per text. You currently pay $20 each month for an unlimited plan, but you want to save a few dollars. If you want to try the new plan and spend only $15 each month, how many texts can you send? Ok, let's use t for texts. It costs five cents per text, so that's.05t. Plus, there's that $10 fee. So you want.05t + 10 to equal 15. First, subtract 10 from both sides. Then divide by.05. So you could send 100 texts each month. You've been averaging way more than that, so maybe this isn't a great plan. But then you get a new job and suddenly you have some extra cash. You decide that you want to save up for new bike. You find one you like that costs $400. If you can save $35 each week, how many weeks will it take you to get the bike? So we want to know weeks, or w. You're saving $35 each week, so that's 35w. If we want to save $400, then our equation is 35w = 400. This one's simple. Just divide 400 by 35 and you get So it'll take you just over 11 weeks to get that bike. After you get the bike, you decide to have some fun. Let's say you and two friends go bowling. At the end of the night, your bill is $42. If you played 3 games and paid $3 each for shoes, how much did you pay per game? Let's call the cost per game g. You played three games, so that's 3g. You also paid $3 each for shoes, and there were three of you, so that's 3*3, or 9. That means 3g + 9 = 42. That's our equation. First, subtract 9 from both sides. 3g = 33. Now divide by 3. g = 11. So each game costs $11. Ok, before we go, why don't we try that train problem, you know, just to show that we can. How did it go? A train leaves Chicago at 7 a.m. traveling to New York, which is 800 miles away, at 75 mph. Another train leaves New York at the same time, traveling on a parallel track to Chicago at 85 mph. We want to know when they'll meet. We're trying to find how much time it will take, or t. The first train is traveling at a rate of 75 mph, so the distance it covers in t time is 75t. The second train is going 85 mph for t time, or 85t. We want to know when 75t + 85t = 800. In other words, when do the two distances add up to the total distance, 800 miles. We add 75t and 85t to get 160t. Now we divide both sides by 160 and we get t = 5. So they'll meet 5 hours into their respective trips.
184 Hopefully, the passengers will have finished their linear equation word problems and look up in time to wave. I mean, two trains passing each other at 75 and 80 miles per hour won't see each other very long. Lesson Summary In summary, we learned how to translate word problems in linear equations, or algebraic expressions that represent lines. We looked at simple examples, where the problem describes a number in terms of details about it, like the sum of twice a number and 52 is 174. Then we looked at problems that involve reallife scenarios, from loaning money to bowling. We focused on defining the variable, or the unknown quantity, in terms of what is known, then solving for the variable. Oh, and we solved the dreaded algebra train problem. Nice work! Solving Linear Equations with Literal Coefficients Chapter 8 / Lesson 4 In this lesson, we'll literally learn about literal coefficients. We'll look at how to solve linear equations that contain literal coefficients and practice solving several problems. Literally This lesson is literally the greatest thing you could be watching right now. Literally. Does it bug you when people use 'literally' when they shouldn't? I mean, it's literally the worst thing ever. Well, except for maybe a few things  I don't know, war, climate change, the lack of InNOut Burgers where I live. As much as it's literally misused, there are literally good uses of the word 'literal.' And we're about to literally learn about one in algebra. Literally. Literal Coefficients Let's say you're walking down Algebra Street and you bump into this: ax + 2 = b  5. Whoa. Hold on. What are you supposed to do with all those letters? And what kind of street is this where linear equations come to life and wander the streets? That's literally weird.
185 Your first thought is, 'I gotta move to a new neighborhood. Geometry Street has some coollooking houses.' But wait, before you pack up, let's look again at this algebraic pedestrian. This equation has literal coefficients. A literal coefficient is a symbol that represents a constant, or a fixed number. Wait  aren't variables just symbols used to represent numbers? Yes! And literal coefficients are in many ways similar to variables. But in a linear equation, we treat literal coefficients more like numbers, and we're still trying to solve for the variable. Solving Literal Equations Let's look at how this literally works. Remember that stranger from Algebra Street? ax + 2 = b  5. We just want to solve for x. And how do we do that? We get x alone on one side of the equation. First, subtract 2 from both sides. ax = b  7. If that a were a number, like 7, we'd just divide by that number. We do the same thing with the literal coefficient. If we divide by a, we get x = (b  7)/a. And that's our answer. We can't go any further. We're basically defining x in terms of b and a. When you think about that, since we can't do anything with those literal coefficients, there's actually less math to do. If a and b were numbers in that equation, we'd have to keep solving until we got a final number. This makes literal coefficients literally pretty cool. Note that our literal coefficients here were a and b; we usually use the letters from the beginning of the alphabet for our literal coefficients, like a, b, c, and d. You may have seen those same letters used as ordinary variables, as in 3a = 15. So how do we know when we have a variable and when we have a literal coefficient? Literally the easiest way is just to look at what the problem says. Problems with literal coefficients will usually say something like, 'if ax = 15, then x = what?' In this case, by the way, we'd divide both sides by a and get x = 15/a. Practice Problems Let's get a little practice. Here's one: 6y  c = b. We want to solve for y. First, we move that c over by adding c to both sides. Then we just divide both sides by 6 to get y = (b + c)/6. That was simple. Maybe Algebra Street isn't going to be so bad after all.
186 What about this one? 12x + 5 = a + 4x. We need to isolate the x. But there are xs on both sides. That's okay. We just subtract 4x from both sides to get 8x + 5 = a. Now let's subtract 5 from both sides. Finally, divide by 8. So, x = (a  5)/8. In this lesson, we'll literally learn about literal coefficients. We'll look at how to solve linear equations that contain literal coefficients and practice solving several problems. Literally This lesson is literally the greatest thing you could be watching right now. Literally. Does it bug you when people use 'literally' when they shouldn't? I mean, it's literally the worst thing ever. Well, except for maybe a few things  I don't know, war, climate change, the lack of InNOut Burgers where I live. As much as it's literally misused, there are literally good uses of the word 'literal.' And we're about to literally learn about one in algebra. Literally. Literal Coefficients Let's say you're walking down Algebra Street and you bump into this: ax + 2 = b  5. Whoa. Hold on. What are you supposed to do with all those letters? And what kind of street is this where linear equations come to life and wander the streets? That's literally weird. Your first thought is, 'I gotta move to a new neighborhood. Geometry Street has some coollooking houses.' But wait, before you pack up, let's look again at this algebraic pedestrian. This equation has literal coefficients. A literal coefficient is a symbol that represents a constant, or a fixed number. Wait  aren't variables just symbols used to represent numbers? Yes! And literal coefficients are in many ways similar to variables. But in a linear equation, we treat literal coefficients more like numbers, and we're still trying to solve for the variable. Solving Literal Equations Let's look at how this literally works. Remember that stranger from Algebra Street? ax + 2 = b  5. We just want to solve for x. And how do we do that? We get x alone on one side of the equation. First, subtract 2 from both sides. ax = b  7. If that a were a number, like 7, we'd just divide by that number. We do the same thing with the literal coefficient. If we divide
187 by a, we get x = (b  7)/a. And that's our answer. We can't go any further. We're basically defining x in terms of b and a. When you think about that, since we can't do anything with those literal coefficients, there's actually less math to do. If a and b were numbers in that equation, we'd have to keep solving until we got a final number. This makes literal coefficients literally pretty cool. Note that our literal coefficients here were a and b; we usually use the letters from the beginning of the alphabet for our literal coefficients, like a, b, c, and d. You may have seen those same letters used as ordinary variables, as in 3a = 15. So how do we know when we have a variable and when we have a literal coefficient? Literally the easiest way is just to look at what the problem says. Problems with literal coefficients will usually say something like, 'if ax = 15, then x = what?' In this case, by the way, we'd divide both sides by a and get x = 15/a. Practice Problems Let's get a little practice. Here's one: 6y  c = b. We want to solve for y. First, we move that c over by adding c to both sides. Then we just divide both sides by 6 to get y = (b + c)/6. That was simple. Maybe Algebra Street isn't going to be so bad after all. What about this one? 12x + 5 = a + 4x. We need to isolate the x. But there are xs on both sides. That's okay. We just subtract 4x from both sides to get 8x + 5 = a. Now let's subtract 5 from both sides. Finally, divide by 8. So, x = (a  5)/8. Did you ever go on a quick run to the grocery store and know you need three things, but when you get there, you can only remember two? That's like what we're doing here. It's like x is the list you should've written down but didn't. You know it's something like milk, eggs and something else. That something else is our literal coefficient. If we knew what that was, we'd know everything. Well, not literally everything. But we wouldn't get home from the store and say, 'I got the milk, eggs and this letter a!' Okay, let's try another: 3az + 7z = 4. This looks trickier, but we're still following the same protocol. How do we get z alone? Well, 3az + 7z can be factored. If we pull out the z, we have z(3a + 7). And guess what? Now we just need to divide both sides by 3a + 7. So our final answer is z = 4/(3a + 7).
188 How about one more? bx + 5 = ax  2. Let's first subtract ax from both sides. Then let's subtract 5 from both sides. Okay, bx  ax = 7. Let's factor out that x. Now, we just divide by b  a. So, x = 7/(b  a). And we're literally done! Lesson Summary In summary, literal coefficients are literally not scary at all. They're just symbols representing constants, or fixed numbers. In equations with literal coefficients, we treat these symbols like numbers and solve for the variable. Literal coefficients are usually represented by letters from the beginning of the alphabet, like a, b, c, and d. That's all you need to know about solving linear equations with literal coefficients. Literally. Solving Linear Equations: Practice Problems Chapter 8 / Lesson 5 With practice, linear equations can be straightforward to solve. In this lesson, we'll define linear equations and learn how to solve them. We'll look at multiple practice problems and walk through solving each one. X Marks the Spot X. X is everywhere. It's the checkbox on the user agreement we never read. It marks railroad crossings and superhero uniforms. And it's at the end of every pirate map. In all these examples, x stands for something. For example, on that pirate map, x is where you'll find the buried treasure. In algebra, x is sort of like a marker for treasure, if by treasure we mean a number. That's because x is a common variable. And a variable is a symbol used to represent a number. Here we're going to solve for x in linear equations. It's just like following a pirate map, where we follow the clues until we know where x is. Fortunately, with linear equation solving, you're much less likely to lose a leg or an eye. And scurvy is very rare. First, let's do a quick review of what these equations are.
189 Linear Equations A linear equation is simply an algebraic expression that represents a line. These equations commonly contain one or two variables, usually x or y. These are called firstdegree equations because the variable's exponent is always one. We won't see anything like x^2 or x^3. Those may get you lines like what you'd actually see on a pirate map. But our pirate map has straight lines. It's much easier that way. You also won't see things like x times y, x over y or the square root of x. That's for pirates who travel through time and space. Oh, and the linear equations we'll be solving here only have one variable, not two or more. I mean, if we had both x and y, how do we know which one has the treasure and which one is a trap? Pirates are big into traps. Solving Linear Equations As I mentioned before, to solve one of these equations, we're trying to solve for x. If you have x  4 = 10, then 10 isn't where the treasure is. It's where x  4 is. And that's just not what we need. To solve this equation, we need to get x alone on one side of the equation. Here we do that by adding 4 to each side. That gets us x = 14. So that's our treasure. I know that's a modest haul, but it was a basic equation. For all of these equations, we'll always just do whatever we can to isolate x. If we have 2x = 6, we divide by 2 to get x = 3. Just keep your focus on x. Practice Problems Now that we know what to do, let's go after some serious treasure. Let's start with x + 2 = 9. This looks like the first one we saw. This is like Pirate Treasure Hunting 101. Let's subtract 2 from both sides to get x = 7. It's another small treasure, but it is ours. What about 3x  4 = 11? Let's add 4 to each side. Then divide by 3. We get x = 5. That's a little more gold. We could buy ourselves a parrot. Here's another: 4x  9 = 2x  3. We have xs on both sides, so we need to move them around. Let's subtract 2x from both sides. Now add 9 to both sides. Then divide by 2. That's x = 3. I think we can get a pretty sweet sword with that.
190 Okay, now let's try x/4 + 2 = 5. First, subtract 2 from both sides. And how do we get rid of that 4? We multiply both sides by 4. So x = 12. Maybe we should put some of this gold in our Pirate College Fund. You know, pirates don't do that well in school. They're always in the high Cs. Okay, that was bad. Let's go after more treasure. What about 2(x + 4) = 22? We first need to distribute that 2 across the x + 4. We get 2x + 8. Next, subtract 8 from both sides. Then divide by 2. So x = 7. Maybe we can buy a better pirate joke book with that. Here's one: 6x x = 264x. Let's combine all those xs. 6x  x is 5x. Then we add 4x to both sides to get 9x over here. Add 1 to both sides to get 9x = 27. Divide by 9 and x = 3. Now we're getting some gold. Let's try a tougher one: 3x/4 + 2/3 = 5/6 + x. Oh, man. What do we do here? We need the lowest common multiple of 4, 3 and 6. That's 12. So we multiply both sides by 12. That gets us 12(3x/4) + 12(2/3) = 12(5/6) + 12(x). That last one is just 12x. What about the others? 12(3x/4) is 36x/4. That reduces to 9x. 12(2/3) is 24/3, which is 8. And 12(5/6) is 60/6, or 10. Okay, 9x + 8 = x. Let's subtract 12x from both sides to get 3x over here. Then subtract 8 from both sides to get 2 over here. Finally, divide by 3. So x = 2/3. Now that's a treasure. We could buy our very own pirate ship with that. Lesson Summary In summary, being a pirate is all about finding the right parrot. Wait, no. That's not what we learned here. We learned about solving linear equations. A linear equation is an algebraic expression that represents a line. We focused on single variable equations, where we try to get x alone on one side of the equation. Once we find x, we have our treasure! What is a System of Equations? Chapter 8 / Lesson 6 So what happens if we want to compare more than one equation? Welcome to a 'system' of equations! Learn what one is, how to solve them and when they come up in real life.
191 Comparing and Contrasting Comparing and contrasting is something that comes up in all school subjects and also in real live too. Maybe you're taking an English class that asks you to write and essay that compares and contrasts 'The Wizard of Oz' and 'Huckleberry Finn.' Or maybe you're taking a world history class that asks you to talk about the similarities and differences between World War I and World War II. Math is the same way; sometimes one equation can be good enough for all the information we're trying to find, but it's often true that we want to compare multiple equations at the same time. Any time we do have more than one equation in a single problem, it's called a system of equations, and that's what this lesson is all about. Comparing the Speed of Two Runners So I'd like to give you an example of a system of equations, but I'm going to start with a little background information first. I like to run. I go out every once in a while and I've even been in a few races. I actually did a triathlon a few years ago too. But my girlfriend is actually a huge runner and she runs almost every day. We decided it would be fun to do a race together so we started running with each other to get ready for it, but we quickly realized that she was way faster than I was. So to keep it interesting for both of us, we decided to give me a little bit of a head start and see if she could catch me. I'm able to run about 1 mile every 9 minutes, but she can do 1 mile every 7 minutes. So if we were going to practice for a half marathon, which is 13 miles, and I got a 2mile head start, would she be able to catch me? This represents a system of equations because we have two equations  one that represents me and one that represents her. When we solve it, we're trying to figure out when these equations are the same. Now we can do this two ways, either with a graph or with algebra. Just like always, the graph is going to provide us with a good visual estimate, but the algebra is going to do a much better job of telling us an exact answer.
192 Graphing a System of Equations So let's go ahead and start with the graph first so we can get an idea of what's going on and maybe make a guess about what we think, and then we'll use the algebra later to check our guess. If we start by graphing me first, I'm going to start already 2 miles ahead even after zero minutes have gone by, so my first point is up here at 2 miles. Then every 1 mile I go up, I have to go 9 minutes over, so my next point would be right here. Then I would go up another mile over 9 minutes and my next point would be right here. We can keep going up 1 mile over 9 minutes and we get a bunch of points in a row. But I'm not just magically teleporting between points; I'm kind of slowly making it there. So between these points there are a bunch of little dots. I'm kind of slowly making it to that point and if you put enough little dots in a row, they end up turning into a solid line. What we end up with is a straight line that says exactly where I am after a certain number of minutes. The reason it's a straight line is because we're assuming that I can go the same speed the whole time. I never slow down and I never speed up. It's a linear equation. I'm always increasing by the same amount every time. Graph of a system of equations Her equation begins down at zero because she doesn't get a head start. She starts down here at zero, but then every 1 mile she goes up, she only has to go over 7 minutes. So if we continue that pattern, we get a bunch of little points in a row; we connect all the points and we get a line for her as well. And what we're looking for is where she catches me, which is where the lines intersect; the spot where they're in the same place. It appears to be right here. So it looks like she's going to beat me, but let's go ahead and check back with the algebra.
193 Setting Up a System of Equations So because both of these equations are lines, that means they're linear equations, which means I can write them in slopeintercept form (mx+b). So the only two things I need for find for each equation is the slope (how much we're moving by) and the yintercept (where we begin). If we do my equation first, my beginning point is at 2 miles because I get a 2mile head start. So my b value is 2. The slope, which is the rise over the run, is up 1 mile and over 9 minutes. So my slope is 1/9, which means my equation is y = (1/9)x+2. She, on the other hand, doesn't get a head start, so her yintercept is zero. We could put a 'plus zero' on the end or we could just not write it at all. Her slope, how much she's moving by (the rise over the run), is she's going up 1 mile and over 7 minutes each time. Her slope is 1/7. Her equation is y = (1/7)x. Solving for x in the system of equations She, on the other hand, doesn't get a head start, so her yintercept is zero. We could put a 'plus zero' on the end or we could just not write it at all. Her slope, how much she's moving by (the rise over the run), is she's going up 1 mile and over 7 minutes each time. Her slope is 1/7. Her equation is y = (1/7)x. Solving for x in the system of equations Solving a System of Equations Since we're trying to find where these two equations are the same, we just want to know at what point are the ys and the xs the same. So if I want the ys to be the same  if I want my y to be the same as her y  what I can do is simply substitute what I know my y is in for her y. So where I see her y, I put what my y is. This means I take the expression (1/9)x+2 and put it where her y was. I get a new equation, which is
194 (1/9)x+2 = (1/7)x. Now I have a linear equation with one variable that I can solve using inverse operations. I can undo the (1/9)x with subtraction over to the other side. I have to do some fraction subtraction, so I have to do (1/7)x minus (1/9)x, which means getting a common denominator. (9/63)x minus (7/63)x gives me (2/63)x. On the other side of the equation, we still have our 2. We now need to undo a fraction. I can undo a fraction by multiplying by its reciprocal. So I multiply both sides by 63/2. The x is now by itself and 2*63/2, using some fraction multiplication, tells us that x is equal to 63. If we refer back to our graph, we see that the xaxis is minutes, and so we have just found that she is going to catch me in 63 minutes. But we're not exactly sure if that's before or after we've finished the 13 miles, so to find out where on the track that was, we have to substitute 63 back in to either equation. It actually doesn't matter which one because we're in the same spot, so we'll get the same answer either way. Solving the system reveals the exact point where the runners meet up I'll choose to plug it into hers because it's a little bit easier (we don't have to add the 2 on the end). And I get y = (1/7)*63. Doing another quick fraction multiplication problem, we end up with y = 9 miles. This means it took her 63 minutes to catch me and we were 9 miles into the race, so she totally beat me. Lesson Summary To review, a system of equations is any problem that has more than one equation in it. When we solve a system of equations, we're finding the point where those equations are the same. That means that if we graph the system, and we try to solve it using the graph, we find where the two lines intersect. The intersection point is our solution.
195 Whereas if we want to try solving it with algebra, we can use what's called the substitution method. If we know what one variable is, we can substitute the expression for one variable in for that same variable in the other equation and then solve the resulting equation. How Do I Use a System of Equations? Chapter 8 / Lesson 7 There are a few classic algebra word problems, such as the one about two trains traveling at different speeds. In this lesson, you'll learn how to take a word problem and convert it into the system of equations that will allow you to find the answer using either substitution or elimination. Building a System of Equations Probably the best known algebra word problem (which often seems to have been a traumatic experience for people) has to do with two trains leaving the station going opposite directions. It usually asks something along the lines of how fast they were going or how far apart they were. But I like to make math as untraumatic as possible, so we're going to stay away from any train problems and instead, we're going to look at a difference classic word problem. Let's say you're hosting a Super Bowl party and have been given $68 by all your friends that are coming to spend on pizza. You're going to get either cheese pizza or pepperoni pizza. Cheese pizzas cost $8 each, and the pepperoni pizzas cost $11 each. Since you might as well spend all the money (it's your friends', not yours) and you've also decided that 7 pizzas is the perfect amount of food, the question is how many of the seven pizzas should be cheese, and how many of them should be pepperoni? We could try to guess and check our way through this problem, which might work for this one, but that's probably not going to work for most problems like this. So we'd rather use this example as a chance to learn how to do it with math. But in order to do it with math, we have to first change the words into mathematical equations. This can oftentimes be the hardest part. In order to set up those equations, we need to figure out what the variables we're looking for are. In order to that, we're going to have to go back to the problem and figure out what it's really asking us for. This problem is asking us for how many cheese pizzas and how many pepperoni pizzas should we buy. Which means our
196 variables should be the number of cheese pizzas (C) and the number of pepperoni pizzas (P). When we solve for those two things, we'll know how many we should buy. So now that we have our variables, we have to come up with equations that represent the situation. This problem is going to need two equations to solve it because there are two variables, which means that this is a system of equations, because there's going to be more than one. The easier equation to come up with uses the fact that we know we have to buy a total of 7 pizzas. We've decided that 7 pizzas is the perfect amount. Since C is the number of cheese and P is the number of pepperoni, and we know that the total has to be 7, my first equation is the number of cheese (C) plus the number of pepperoni (P) must equal 7: C + P = 7. The harder equation to come up with has to do with the price and how much money we have and how much money different pizzas cost. What I want to do is set up an equation that equals 68. It should say that we are going to spend $68. This means I have to figure out how much money I'm spending on my cheese pizzas and how much money I'm spending on my pepperoni pizzas. Using the substitution method to solve for C Well, if we just think about this logically, if I bought one cheese pizza, it would cost me $8. And if I bought two, it would cost me $16. If I bought three, it would cost me $24, and so on. This means that the total money I spend on my cheese pizzas is simply 8 times the number of cheese pizzas I buy (8C). A very similar expression could be derived for the amount of money you spend on pepperoni pizzas by doing 11 times the number of pepperoni pizzas you buy (11P). So 8C represents the money you spend on cheese pizzas and 11P represents the money you spend on pepperoni pizzas. If you add those two amounts together, that's the total amount we spend. And that has to be $68. So our second equation, the little more complicated one, is 8C + 11P = $68.
197 Solving a System of Equations with Substitution Now that we have the system set up, we can actually begin to solve it. We're going to go over 2 ways of solving a system of equations, the first of which is called substitution. It is called this because we're going to take one expression from one equation and substitute it in for one of the variables in our second equation. For example, we can take the equation that says C + P = 7, telling us the total number of pizzas we're getting, and we can solve this equation for P by undoing C from both sides with subtraction. We find an equivalent equation that says that P = 7  C. Now that I know what P equals, and since I want these two equations to be the same, I can substitute into the second equation what I now know what P is. So instead of writing 8C + 11P = $68, I can write 8C + 11(7  C) = $68. Now I only have one variable in one equation, which is solvable for that variable. Setting up equations to use the elimination method I can use inverse operations to get the C by itself. First, you distribute the 11 to both terms: 8C C = $68. Then, you can combine like terms, grouping the Cs together because they're on the same side of the equals sign: 3C + 77 = $68. Next, you undo an addition of 77 with a subtraction of 77: 3C = 9. Last but not least, you undo a multiplication of 3 with a division of 3 to both sides. You find that C, the number of cheese pizzas we're going to get, is 3: C = 3. Now that I know the number of cheese pizzas, it's a pretty easy process to just take that and substitute back into the first equation: P = 7  C. I now know that C is 3, which means that P = 73. This means that P is 4. I now know how many pepperoni pizzas we should buy.
198 Solving a System of Equations with Elimination So that's the substitution method to solving a system of equations. But there's another method called the elimination method, which is really nice to use when both of my equations are in standard form. This means that both of my equations have the C's and the P's on the same side. The elimination method requires us to add the two equations together in order to eliminate one of the variables. Right now, if I add these two equations together, none of my variables are going to eliminate. I'm going to end up with 9C and 12P and that doesn't really get me anywhere. But if I first multiply the top equation by 8, and I have to distribute it into everything, it changes that equation into: 8C  8P = 56. If I leave the bottom equation the same and now add the two equations, my C's cancel. What I've done is that I've made the coefficients (which is a fancy word for the numbers in front of the variables) on my C variable opposites. The 8 and the 8 cancel. I just get 0C. On the P's, I get 8P plus 11 P which is 3P. On the other side of the equals sign, I get 56 plus 68, which is 12: 0C + 3P = 12. Solving for P using the elimination method Now I have one equation with one variable. I can solve pretty quickly by undoing times 3 by dividing by 3, and we find that P = 4. The number of pepperoni pizzas is 4. Again, the same answer. Once we have one answer, we do the same step as we did in the substitution method. We take our one answer and we substitute it back into one of the original equations. We now know that P = 4, so I know that C + 4 = 7. I undo adding 4 by subtracting 4 and I find that C = 3. We need 3 cheese pizzas.
199 Lesson Review To review, step one to solving a word problem is to identify what the variables are by checking: what is the problem asking for? Once you know what you want your variables to be, you need to set up the equations. Is there one variable? Then there's going to be one equation. Are there two variables? Then there are going to be two equations, and so on. Step three, the last step, is to actually solve the problem using the substitution method or the elimination method. Substitution is nice when one equation is already solved for a variable. For example, when one of the equations already says x = or y = or c =. It requires you to take that one expression and substitute it in to the other equation for that same expression, then solve for the variables using inverse operations. Elimination is nice when neither equation is solved for a variable, which means that both equations are in standard form (the x's and the y's are on the same side). It is then your job to make either pair of coefficients opposite numbers by multiplying one or both equations by something that will turn the coefficients into opposites. Once they're opposites, you can add the equations together, in order to eliminate one of the variables. Then, again, solve for the remaining variable using inverse operations and substitute it back in to find the second one. Solving a System of Equations with Two Unknowns Chapter 8 / Lesson 8 When you have two variables and two equations, is solving the problem four times as hard? No! Not when you use the substitution or elimination method to solve systems of equations. We'll practice both in this lesson. Life Is Complex If there was always just one unknown thing or one problem at a time, life wouldn't be so tough. Let's say you get invited to a wedding and all you need to figure out is if you want the chicken, fish or vegetarian entrée. That's not so bad, right? But life isn't that simple. At one point or another, it's your own wedding. And then there isn't just chicken or fish, there's this reception hall or that one, this best man or that one, this florist or,
200 geez, when did this town get so many florists? And since when are there so many types of paper and different fonts for invitations? Systems of Equations Systems of equations can seem just as overwhelming. A system of equations is a group of two or more equations with the same variables. Multiple equations? Multiple variables? It's enough to make you want to elope. Fortunately, though, solving systems of equations is much more straightforward than it seems. In this lesson, we're going to practice the two most common methods of solving systems of equations. First, there's the substitution method. The substitution method is when you solve one equation for either variable, then substitute the solution into the other equation. Then there's the elimination method. The elimination method is when we add or subtract equations together to solve for a variable. Let's try out each method as we solve some equations. Substitution Practice Let's start with the substitution method. Here are two equations: y  2x = 1 and 5x  2y = 3 Let's take the first one and solve for y. We add 2x to get y = 1 + 2x. Next, we substitute 1 + 2x for y in the second equation. So we get 5x  2(1 + 2x) = 3. Now we solve for x. First, we distribute this 2 and get 24x. Then 5x  4x is just x. We add 2 to both sides to get x = 5. Now we have our x value. Let's plug that in to either equation and get a y value. Just pick the one you think will be easier. Let's use the first one. y  2(5) = 1. y  10 = 1. Add 10 and we get y = 11. A good check is to put both variables back in both equations. If they don't work, you know you made a mistake somewhere. Let's try that here. y  2x = 1 becomes 112(5) = 1. That's = 1. And yep, 1 = 1. And 5x  2y = 3 becomes 5(5)  2(11) = 3. That's = 3. Yep again! 3 = 3. We're good! And no future inlaws were insulted in the solving of this problem. Let's try one more. Here are two equations:
201 y = 3x  4 and 2y  5x = 2 In this one, we already have one solved for y, so let's just plug 3x  4 in for y in the second equation. We get 2(3x  4)  5x = 2. 2 * 3x is 6x and 2 * 4 is 8. 6x  5x is just x. Then we add 8 to both sides and we get x = 10. Now plug 10 in for x in that first equation. y = 3(10)  4. y = 304, or 26. Okay, let's check our work. We have y = 3x  4, we get 26 = 3(10)  4. That's 304, which is 26! And 2y  5x = 2 becomes 2(26)  5(10) = 2. That's = 2. Success! Elimination Practice Let's try some elimination practice. Here are two equations: x + y = 5 and 5y  1 = 2x Think of this like making a sandwich. You need your ingredients to line up. If your tomato is over too far, it's going to slide right out. So let's line everything up. x + y = 5 5y  1 = 2x Great! Now we need to get coefficients balanced for one of the variables so they'll cancel out. This is like making sure your amounts of mayonnaise and mustard are in balance. Here, we can multiply the top equation by 2 so we get 2x on top to match the 2x on the bottom. Now let's subtract. The 2x's cancel out. We're left with 3y = 9. Divide by 3 and y = 3. Now plug 3 in for y in the first equation. x  3 = 5. So x = 8. That's our sandwich! But wait. Let's check our work before we eat it. In x + y = 5, we get 83 = 5. That works out. In 2x + 5y = 1 we get 2(8) + 5(3) = 1. That's = 1. Success! Let's try another: 2x  y = 34x + 3y = 1 Hey, look  our terms are neatly stacked. Thanks, sandwich gods! And we can multiply this top equation by 2 to get rid of the x terms. So it becomes 4x  2y = 6. Here, we want to add. 2y + 3y is just y. And is 7. So, y = 7. Now plug 7 in for y in one of the equations. 2x  7 = 3. 2x = 10. x = 5.
202 Let's check our answers. 2x  y = 3 becomes 2(5)  7 = 3. That's 107 = 3, so that works. And 4x + 3y = 1 becomes 4(5) + 3(7) = 1. That's = 1, so that works too. We did it! Lesson Summary In summary, weddings are hard. Also, we practiced solving systems of equations. There are two common methods of solving systems of equations. The first is the substitution method. This involves solving one of the equations for one of the variables, then substituting the result for that variable in the other equation. The other method is the elimination method. This method involves stacking the equations and multiplying the terms in one so that when you add or subtract the equations, one of the variables disappears. Solving Problems involving Systems of Equations Chapter 8 / Lesson 9 Did you ever encounter a problem where you seem to have lots of information, but still have a couple of critical numbers missing? In this lesson, we'll practice solving word problems that require us to set up systems of equations. Systems of Equations Lots of things have systems. For example, there's a knitter's system of organizing yarn by color or a chef's system of chopping and organizing ingredients before cooking. Even when society breaks down, systems still matter. Let's say it's the zombie apocalypse. You'll stand a better chance of surviving if you use systems. This includes not only systems of avoiding zombies. It also means systems of equations. A system of equations is a group of two or more equations with the same variables. There are many different kinds of word problems involving systems of equations. In this lesson, we'll focus on a few of the most common types of these problems. Remember, though, that the principles at work are the same in all of them.
203 Counting Practice Let's start with one that could save your life. It's the zombie apocalypse. You try to take shelter at Farmer Zed's place, but his barn is full of zombie animals. You know he had 20 animals, a mix of chickens and pigs. Your recon scout tells you he counts 50 legs, but can't see what they are. You hope there are more chickens, because zombie chickens are easier to kill. Zombie pigs? Not so much. Okay, math can help. First, we need our variables. What don't we know? The number of chickens and pigs. Let's use c for chickens and p for pigs. Next, we need our equations. We know c + p = 20. That's all the chickens and pigs combined. We know the chickens should have 2 legs and the pigs should have 4, supposing none were gnawed off by zombie ducks. So 2c + 4p = 50. To solve, let's rearrange this first equation to c = 20  p, then substitute 20  p for c in the second problem. This is the substitution method. So that's 2(20  p) + 4p = 50. We get 402p + 4p = 50. That becomes 2p = 10. So p = 5. Let's plug in 5 for p in c + p = 20. So c = 15. Let's check our math by plugging p = 5 and c = 15 into the second equation. That's 2(15) + 4(5) = = 50. Okay, so 5 zombie pigs and 15 zombie chickens. Money Practice Let's try a problem involving money. In this postapocalyptic hellscape you call home, you need two things most of all: bullets and cookies. Look, it may be the end times, but you still have a sweet tooth. Fortunately, you encounter some delusional retail workers who are convinced the monetary system should still function. I guess the zombies aren't much different than Black Friday shoppers to them. Unfortunately, their cash registers no longer work, so you have no receipt. But you know you spent $ on a total of 135 $3.00 bullets and $8.00 cookies. How many of each did you get? Let's make our variables b for bullets and c for cookies. That's good enough for me. What are our equations? You got 135 total items. So b + c = 135. And with a money problem, we can multiply the cost of each item times the number of the item to get its total cost. So 3b is the cost of $3.00 bullets and 8c is the cost of $8.00 cookies. You spent $615.00, so 3b + 8c = 615.
204 Let's use substitution again. We can make b + c = 135 into b = c. Then substitute into the section equation to get 3(135  c) + 8c = 615. That simplifies to 4053c + 8c = c  3c is 5c. And is 210. So 5c = 210. Divide by 5 and get c = 42. Plug 42 in for c in b + c = 135 and b = 93. Let's check for zombies. No zombies? Okay, let's check our work in the second equation. 3b + 8c = 615. That's 3(93) + 8(42) = * 93 is * 42 is is 615. So we have 93 bullets and 42 cookies. Hmm, that might last a few days. Percent Practice Let's try a percent problem. These can seem like tricky systems of equations questions, but they're really just using the same principles. Let's say you meet up with a chemist who has the key to stopping the zombification process. She needs a waterbased solution with 14% vinegar. Yep, vinegar. Vinegar does all kinds of useful things. Unfortunately, you only have vinegar in 8% and 24% solutions, and it only works at 14%. If you want 300 gallons of 14% solution, how much 8% and 24% will you need? Let's use x and y. x can be the number of gallons we'll need of the 8% solution and y is the same for the 24%. We know we want 300 gallons. So x + y = 300. That's our first equation. We also know we want the amount of vinegar in the 8% solution plus the amount in the 24% solution to equal 14% in the 300 gallons. We can rewrite what I just said as.08x +.24y =.14 * 300. That's the number of gallons of 8% times the 8%, which is.08, plus the number of y gallons times its concentration at 24%, equaling the final 300 gallons at 14 zombiestopping percent. Let's make that first equation x = y and substitute it into the second equation to get.08(300  y) +.24y = 42. Simplify that to y +.24y = 42. Let's get the ys together and move the 24 over to get.16y = 18. Divide by.16 and get y = Plug that into x + y = 300 and x = Again, check for zombies. Good? And cookies? Running a little low, but if we get this finished, more cookies await. Ok, let's plug and into our second equation. That's.08(187.5) +.24(112.5) = * is 15. And.24 * is ? 42. Zombies? Vanquished!
205 Lesson Summary In summary, a system of equations is simply a group of equations with the same variables. Problems involving systems of equations are useful whether you're organizing your yarn collection or battling the zombie threat. Just identify your variables, set up your equations, then solve for your variables. And watch out for zombie ducks  they like to gnaw! What is an Inequality? Chapter 8 / Lesson 10 Deciding how to spend your money can be a tricky thing. Should you save it, invest it, or enjoy it? Learn how inequalities can help you make your decision! What is an Inequality? The symbols that can be used in inequalities There are a lot of complicated topics in math, so it's really nice when you get to one where there's an easy explanation for what it is. That fact makes me happy to be writing this lesson on the question, 'what is an inequality?' Well, it's simple! An inequality is just an equation with one of these guys instead of an equal sign! The second you see a greaterthan or lessthan symbol, you know you've got an inequality. They're actually really useful in our lives because it's pretty rare to find a situation where there's only one specific solution. You don't need exactly an 80% on that history test to get a B; you need at least an 80%. You don't need to be driving exactly 25 mph to be following the law; you need to be driving 25 mph or less. These are just
206 a few examples of where inequalities come into our lives. Let's take a more indepth look at another one that everyone can relate to: 'How should I spend my money?' TwoVariable Inequalities I know that when I got my first teaching job right out of college I was really exited for my first payday. I had had parttime tutoring jobs before, but nothing like this. This was a real job! Too bad I didn't realize that along with a real job I now had real bills. It ended up being the case that I had to spend basically all my money on rent, food, and other bills. I was bummed to say the least, but it meant that when I heard about a summerschool teaching opportunity later that year, I was all for it. That June, I got an extra $3,000 to spend on whatever I wanted! I thought a lot about it and decided to spend the money on two things, video games and plane tickets. During my busy school year, I hadn't gotten to play as many video games as I usually do, and the thought of spending my summer days gaming all day and night was too enticing to ignore. But I was also living in California while my girlfriend was in Minnesota. We didn't get to see each other too often and this was a great way to make that happen. So now I had a problem. I had $3,000 to split between $60 video games or $300 plane tickets, but I also thought that it'd probably be a good idea to put some of it into my savings account. What should I do? Should I just get a few video games and one plane ticket and save the rest? Should I get as many plane tickets as I can? I thought about making a list of all the different possibilities, but I quickly decided that my list would have taken up pages and pages of paper and wouldn't really help me make my decision. What I needed was one nice visual that showed me all the different choices I could make. 'You know what that sounds like?', I thought to myself, 'a graph!' Example of a twovariable inequality
207 But before I could make the graph, I decided to write an inequality to represent the situation. While it would have been possible to create the graph without the inequality, I wouldn't have an easy way of checking which combinations of games and tickets are allowed without the inequality. This would have made me less organized and much more likely to make a silly mistake. So when I sat down to write the inequality, I knew that no matter what, I couldn't spend more than $3,000. But I also didn't have to spend it all; I could save some of it. Therefore, the money I spend had to be less than or equal to $3,000. What I still needed was a mathematical expression for how much money I was going to spend. Because each video game was $60, 60v would be how much I spent on v number of video games. If I bought two, I would just do 60(2) = $120, and that would be how much I spent on video games. Using the same logic, 300p is how much I spent on p plane tickets, so then 60v + 300p would represent the total amount of money I spent. So we have our inequality that allows us to quickly plug in combinations of games and tickets to see if I can afford it. But we still want the graph in order to see what my different options are in one nice picture. Graphing TwoVariable Inequalities To get my graph, I knew I needed two points. So I thought to myself, 'if I decide to be a terrible boyfriend and not buy a single plane ticket, but instead spend every penny I have on video games, how many could I get?' Substituting these values into our equation and quickly solving for v using inverse operations tells us that I would be able to purchase 50 video games! At this point we can go ahead and put a point on our graph where v = 50 and p = 0. But we need two points to find a line, so I also needed to ask myself the question, 'If I put love first and simply fly out to Minnesota as many times as possible, how many times could I go?' Substituting in these slightly different values and instead solving for p tells us that I would be able to fly out to Minnesota ten times. This gives us another point on our graph, one at v = 0 and p = 10. Now, simply connecting these two points gives us the line 60v + 300p = 3,000. That means that all the points on this line are possible combinations of video games and plane tickets for me to choose. But we need to be careful, because this is where 60v + 300p equals 3,000, which means that all the points on this line  our combinations of video games and plane tickets  have me spending all my money.
208 The shaded area on the graph represents solutions to the inequality It's probably a good idea to save some of it, right? So maybe instead of getting ten plane tickets I only get eight. Or maybe I'd get one video game and six plane tickets. Or maybe I'd only get five planet tickets but get ten video games, or maybe I'd get only one plane ticket but 30 video games. As it turns out, any point underneath our line is a possible solution, so we shade the entire area underneath the line to indicate that any point within this space is an acceptable choice for me to make. What we have now is the graph of the inequality 60v + 300p is less than or equal to 3,000 that helped me decide how to spend my summer school paycheck! If you're curious, I decided to go with three plane tickets and two video games. This was an example of a twovariable inequality. All twovariable inequality graphs will look something like this: a line, and shading on one side of that line. But there are also onevariable inequalities as well. To see what is different about a onevariable inequality, and to learn more about the nuts and bolts behind solving and graphing inequalities, check out the other inequality lessons. Lesson Summary Inequalities are equations with greaterthan or lessthan symbols instead of equal signs. Instead of just having one answer, like equations do, there can be multiple answers to an inequality. This is because instead of having to be equal, they can simply be anything that is less or greater than. Finally, graphs of twovariable inequalities have a line and half of the graph shaded, where the shaded part depends on whether it was lessthan or greaterthan.
209 Solving Linear Inequalities: Practice Problems Chapter 8 / Lesson 11 Linear inequalities may look intimidating, but they're really not much different than linear equations. In this lesson, we'll practice solving a variety of linear inequalities. Greater or Lesser Greater than and less than. These two symbols can be quite controversial. It all depends on what you put on either side. What if I said hot sauce is greater than ketchup? Or cats are less than dogs? Or the Denver Broncos are greater than the New England Patriots? These are debatable points. I mean, I think they're all true, and I'd argue them passionately, but they're really just opinions. But what if I said 6 > 7? Well, that's just wrong. 6 < 7. And so are other numbers, including 5, 4, 3... well, the list just goes on from there. There's no debate. And in this lesson, we're going to practice handling these types of situations. Linear Inequalities You've already seen linear equations like this: x  2 = 0. You solve for x and get x = 2. Our variable, x, has a single value that we can determine. That's straightforward but also a little dry. It's like finding out that hot sauce is hot sauce and not comparing it to any potentially inferior condiments. But then there are linear inequalities. A linear inequality is a linear expression that contains relational symbols. That means that instead of this (=), you'll see >, <, >, or <. So instead of our variable standing in for a single value, it's standing in for a relational value, as in x > 2, where x is all values greater than 2. Basic Practice Let's solve some basic linear inequalities, then try a few more complicated ones. Just as with linear equations, our goal is to isolate the variable on one side of the inequality sign. First, what about this: x + 5 > 9. We treat this just like we would if we had x + 5 = 9. We subtract 5 from both sides. Now we have x > 4. On a number line, that would look like this, where x is all numbers larger than, but not equal to, 4.
210 Note what x + 5 > 9 looks like. Here, the phrase 'x + 5' can be shown as all numbers greater than 9. So all we did was take that 5 away, which shifted our line like this. Here's another: 18 < 12 + x. Okay, again, get the variable alone. Subtract 12 from both sides and we have 6 < x. We could flip that around to say x > 6. Just remember that if you do that, don't forget to flip the inequality sign! That one looks like this. Here's one: x  7 < 1. Let's add 7 to both sides to get x < 8. This graph looks like this. Note that we fill in the circle around the 8 because x isn't just less than 8, it's less than or equal to 8. x could be all these values as well as 8  no sense making 8 feel left out. Let's do one more basic one: x + 11 > 14. This time, we subtract 11 from both sides, which gives us x > 3. If we graph that, we get this line. Again, we have a solid circle because x is greater than or equal to 3. Advanced Practice That's enough of the basic ones. Those are all like saying chocolate ice cream is greater than vanilla. Pretty straightforward, right? Okay, maybe in some circles. Here's a trickier one: 3x  6 < 9. Let's start by adding 6 to both sides. Now we have 3x < 15. What do we do? Remember, it's just like a linear equation. If we had 3x = 15, we'd divide both sides by 3. We do the same thing here. That gives us x < 5. That's it! Here's another kind we haven't seen before: 7x + 2 < 8 + 5x. It looks more complicated, but the principle of getting that variable all alone hasn't changed. First, let's subtract 2 from both sides. Now, let's move this 5x over by subtracting 5x from both sides. Now we have 2x < 6. Just like the last one, we divide. So we have x < 3. And we did it. There is one key rule that makes solving linear inequalities different than solving linear equations: if you multiply or divide by a negative number, you must reverse the inequality. Let's see this in action: 2x < 8. This time, we divide by negative 2. So we flip the sign. Now we have x > 4. Does this make any sense? Well, if x > 4, what could x be? How about 1? If x = 1, is 2x < 8? Let's see. It'd be 2(1), which is 2. Is 2 < 8? Yes. We could keep trying this with all of the infinite numbers greater than 4, or you could trust me that it will always work. I say trust me. Otherwise this lesson will be really long.
211 Let's solve one more: 12x  3 < 19  x. Let's start by adding 3 to both sides. 19 becomes 22. Now let's add this x to both sides. 12x becomes 11x. Finally, we'll divide by 11. That means flipping the sign. And 22 becomes 2. It's important to not forget that negative sign. We're flipping the sign, but otherwise the normal rules of solving equations still apply. So we have x > 2. Lesson Summary In summary, solving a linear inequality is just like solving a linear equation. The big difference is that instead of your variable being equal to a single number, like x = 10, it's defined in relation to a number. We use four symbols: >, <, > and <. When solving, we treat it as though the inequality sign were an equal sign. There's just one exception: If you're multiplying or dividing by a negative number, the sign gets flipped.
212 What is an Absolute Value? Chapter 9 / Lesson 1 When we're talking and comparing numbers, we often don't care whether its positive or negative, just how big it is. This is often called the magnitude of a number and we find it by taking the absolute value. Learn all about it here! The Distance from Zero One February, during Super Bowl XLV, which was the Packers vs. the Steelers, I made a bet with a friend over how much money it costs to get a 30second commercial on during one of the commercial breaks. Now I'm a huge Packers fan and he was a Steelers fan, so we decided that the loser would have to wear the opposing team's jersey for the rest of the game. I thought it would cost around $1,250,000 and my friend was convinced that it cost $5,500,000. I wasn't sure if I was exactly right or not, but I was pretty confident that he was way off, so I decided to make the bet. Well, a few internet searches later, we discovered that a 30second commercial that year cost $3,000,000. So who won? Well, a simple calculation of $3,000,000 minus $1,250,000 gave us the answer that I was $1,750,000 off the actual answer. Doing the same for him, $3,000,000 minus $5,500,000 told us that he was $2,500,000 off. This means that I was definitely closer, so I won and my Packer pride stayed intact. But a closer inspection into this reveals a situation that actually comes up a lot in mathematics. What my friend and I cared about was purely how far away our guess was from the actual answer. We didn't care about whether we were over or under, or whether the number we did after the little subtraction problem was positive or negative. All that we cared about was the magnitude of the answer, or how large the number was regardless of the sign in front of it. In mathematics, this operation of taking the number and only looking at the size of it and ignoring the sign, is called the absolute value.
213 We often use absolute values when talking about distance because there is no such thing as negative distance. For example, just because the store you're going to is a block behind you doesn't mean it's negative one blocks away. It's just one block, right? Notation of Absolute Values Just like how the winner of our bet was the one whose guess was less far away from the answer, the absolute value of any number can be thought of as how far away that number is from zero. For example, the absolute value 7 (written as 7 )? Still 7. It's 7 units away from zero. The absolute value of 7 (written as 7 )? 7, because it's still 7 units away from zero. Notice that we use these straight up and down bars around the number to indicate that we're taking the absolute values. And, also, the important thing to note here is that the absolute value turns negative numbers positive but leave positive numbers exactly the same. Solving Equations with Absolute Values There are a few things to be wary of when evaluating absolute values. Let's take the expression we've got right here: First, we need to know that the absolute value bars act as parentheses. That means that order of operations say that we have to first do whatever is on the inside and then worry about the stuff hanging out on the outside. Also, because they're kind of like parentheses, when you have a number in front of them, it implies multiplication. The second thing you need to be careful of is that just because absolute values make things positive does not mean that we are going to change the subtraction sign that's going on on the inside into an addition sign. The absolute value operation only happens once everything else on the inside has been done. This means that in this problem, we would first simply do 38 and get 5. Now that we're finished with everything on the inside, now we can take the absolute value of 5, which turns it into 5. Because the number 4 was simply sitting in front of the absolute value, it implies multiplication, and 4 times 5 is 20.
214 Lesson Summary To review, absolute values are used when we don't care about the sign of the number, simply its magnitude or how large of a number it is, regardless of whether it's positive or negative. You can think of the absolute value of any number as its distance away from zero. Absolute values are denoted with two straight up and down parallel bars (e.g., 8 ) and make negative numbers positive, while leaving positive numbers alone. How to Evaluate Absolute Value Expressions Chapter 9 / Lesson 2 Solving by Substituting Variables So we know that absolute values make things positive, but how does that effect how we evaluate absolute values at specific values? Let's take a look at an example: Evaluate m 52n +7, when m equals 2 and n equals 10. This problem is actually no more different than any other problem that asks us to substitute in values as long as we don't fall into the biggest absolute value pitfall; the absolute value bars DO NOT change subtraction symbols into addition ones. Yes, they do make things positive, but only after you've completed whatever operations are going on on the inside. Treat the absolute value symbols like parenthesis and start solving inside them So if we first substitute in 2 for m and 10 for n, we treat the absolute value bars like parentheses and we begin on the inside of them. Multiplication comes before subtraction, so I do 2*10 and I get 20. Then I do 520 and end up with 15, and only at that point, after I've finished all the different things on the inside of the absolute value, do we actually take the absolute value, making 15 positive 15. Again, because
215 absolute value bars are kind of like parentheses, when you have a number in front it means multiplication, and 2*15 is 30. Last but not least, is 23. Solving by Replacing Groups of Variables Now, where these problems can get kind of tricky is when the values of m and n aren't given to us quite so nicely. Take this example: Evaluate 2m2n + nm when mn is equal to 4. So instead of simply telling us what n and m are individually, it only gives us what their difference is. It's going to be up to us to manipulate the given equation to make it look like what we want. So what do we want? Well, the two expressions that would be nice to be able to substitute in are 2m2n (that first absolute value) and nm (that second one). If we knew both of those, we could simply plug in the numbers that we knew and be two steps away from our answer. This means that it's up to us to learn what 2m2n and nm are from only the info we've got, which is mn is 4. So we need to turn mn into 2m2n. Luckily, we know that equations can be manipulated by doing the same thing to both sides to give us equivalent statements. So all I really need to do here is multiply everything in this equation by 2 because I want to turn m into 2m and I want to turn 1n into 2n. So multiplying the whole thing by 2 and distributing in gives us that 2m2n is equal to 8. And we've already got one of our two expressions. Multiplying by 1 provides an equivalent statement to use as the second expression The other thing we need to do is make mn into nm; we need to switch the order around. This basically means I want to make the positive m into a negative one and make the negative n into a positive n. Again, I can manipulate equations by doing the same thing to everything, so if I just multiply everything by 1 and distribute it out,
216 my given information turns into m+n equals 4. Because addition can be written in either order as long as the sign stays with the value, that's the same thing as nm is equal to 4. We've found our second expression. Now that we've changed our given information into what we really want, all I have to do is substitute in the two new equations we came up with. So I now know that 2m2n is 8 so I can substitute that in for the first part of my absolute value in the original question; I know that nm is now 4 so I can now substitute that in. What I end up with is just Absolute values leave positive numbers alone so the 8 stays the same; they turn negative numbers positive so the 4 becomes positive, and 8+4 is 12. Lesson Summary To review, absolute value signs do not change addition to subtraction, they just make the end value positive. You have to do whatever operations are on the inside of the absolute value and then make it positive in the end if you have to. Substituting values into absolute value expressions is no different than any other algebraic expression. When the given information is not exactly what we need, we can manipulate it to become what we do need. How to Solve an Absolute Value Equation Chapter 9 / Lesson 3 Once you get familiar with any new operation, the next step in any algebra class is to learn how to solve equations with that operation in them. Absolute values are no different. Solve absolute value equations here! Review of Absolute Values Absolute value involves the magnitude of the specified numbers
217 Alright, so you're a pro at taking absolute values. 10 =? 10! 10 =? Still 10! Negative numbers turn positive again because all an absolute value cares about is how 'large' the number is or what the magnitude is, not what sign it is. You even know that absolute values can be thought about as how far away that number is away from zero. But all that has just been to get you ready for this: solving absolute value equations. Now this is a really common theme in algebra classes; simply learning a new skill isn't usually good enough and you're almost always going to end up having to learn how to apply that skill in order to solve an equation. Solving Basic Equations with Absolute Values So what are we talking about here? How about solve the x = 15. Well, any time we are asked to solve, it's our job to find a value for the variable that makes the equation true. So what number can I substitute in for x that will make the absolute value of it 15? Or you could think about this as what number is 15 units away from zero. 15 seems like the obvious choice, but we can't forget that 15 is also 15 units away from zero, which means that we actually get two answers here. x could be 15 or 15. This makes sense, right? If I just plug 15 into the absolute value, I just get 15 back out. Or if I plug 15 into the absolute value, I, again, just get 15 back out. Absolute value equations result in two answers This brings up a really important point: absolute value equations give us two answers. x = 20? x is 20 or 20. x = 1,000? x is 1,000 or 1,000. x = 5,000? x equals 5,000 or negative  wait! Absolute values can't be negative numbers, so no matter what I plug in for x, there's no way I'm going to get 5,000. So anytime an absolute value is set equal to a negative number, there is no solution.
218 Splitting an Equation into Two Let's take the difficulty up a bit now. What if it asks you to solve x5 = 20? Now it can be really tempting to simply undo the subtraction with addition because they're inverse operations. So we add 5 to both sides and we get x = 25 and then we say, 'hey, two answers! 25 and 25 and we're done.' But you can't forget that absolute value bars kind of act like parentheses, which means you have to do everything on the inside before you take the absolute value. So when you first add 5 to both sides like that, you're kind of breaking that rule, which means doing that is a big nono. You can't do that. So if we take a step back and look at the big picture and say, 'okay, we're going to do some stuff to our x here, and after all that's done, we take the absolute value and get 20.' It's kind of like we can just cover up what's on the inside of the absolute value with our finger and say, 'I don't really care what's on the inside there; I take an absolute value when I get 20, which means that whatever is on the inside there must be equal to 20 or whatever is on the inside there must be equal to 20.' After you split it up into two equations, then you can substitute back in what's on the inside of the bars by taking your finger away and saying, 'oh yeah, there's an x5 there.' So x5 = 20 or x5 = 20, and those are the correct two equations we should end up with. Solve for x to get the two answers for the problem Now we can undo the subtraction with addition. Adding 5 to both sides gives us our two answers that x = 25 or x = 15. When you're solving equations, you can always check your answers by substituting them back into the original equation. I can do that here just to make sure I did everything okay. If I plug in 25, 255 is 20 and 20 stays the same and I get 20, so it checks out. I put in 15, is 20. Now I take the
219 absolute value, it turns positive again and it checks out as well so we know we got the right answer. Just to remind us of another common misconception; when splitting up the equation into two separate ones, do not change it into x5 = 20 or x+5 = 20. Again, absolute value bars do not change the operations going on inside of them, they only change the end result to be a positive number. So we still have to do the 5 and then after we do that, we change the number back. So changing it into two equations where one is a minus and one is a plus is another thing you've got to be careful not to do. Lesson Summary To review, absolute value equations give us two answers. Absolute values that equal negative numbers have no solution because absolute values are always positive. Absolute value equations with operations on the inside must be split up first before solving for the variable. Solving Absolute Value Practice Problems Chapter 9 / Lesson 4 There are many easy mistakes to make when solving absolute value equations. Learn how to avoid those mistakes here by working on examples of absolute value equations with operations on the inside and the outside of the absolute value. Review of Absolute Values Substituting in 3 for x in the beginning would lead to the answer 7 When it comes to solving absolute value equations, the one most important thing to remember is that they always give you two answers. We get those two answers by
220 splitting the equation up, but how and when to split it up is where it starts to get a little more complicated and also where the mistakes start happening. In order to avoid those mistakes, it really helps to have a conceptual understanding of what 'solving an equation' means. So let's first ask ourselves a question that you probably don't hear every day; what does a mathematical equation have in common with an onion? Now, I suppose your answer could be 'they both make you cry,' but if that was your answer, I hope this video can change your mind. Well, just like an onion has a very center and then a bunch of layers on the outside, an equation has a very center (the number) and then a bunch of operations being done to that number. Using Onions to Solve Equations So for example, an equation might tell us that we start with some number (lets call it x), we multiply it by 5, then we subtract 1, then we divide all of that by 2, and after we've done that, we apparently end up with 7. This can be expressed as a normal mathematical equation, or using this onion analogy where the very middle of the onion is the number we start with and each thing we do to that number is a new layer on the outside. So when we're asked to 'solve' this equation, it's simply our job to figure out what was that number in the very beginning that made it that after we do all these things we end up with 7. It's our job to peel off the layers of the onion one at a time to work our way all the way back to the very center  the x. Solving the two resulting equations in example #2 Well the first layer I peel off is this divided by 2. After I divide it by 2, apparently I ended up with 7, which means that before I divided by 2, I must have had 14 because 14 divided by 2 is 7. So the way I can undo division is by multiplication because those are inverse operations. So by putting the little times 2 by this divide by 2, they undo each other  they cancel each other out  and over on the other side of the equation, I
221 get 7 times 2 and I get 14. So that last step has been undone and now I have an equation with one less step. The next layer I need to undo is the minus 1. We can undo subtraction with addition and by adding 1 to both sides of my equation, I can end up with 5 times x equals 15. Now the last layer I need to undo is multiplication by 5; I can undo multiplication with division. Those things undo and I'm left with 15 divided by 5, which is 3. This means that if we would have substituted 3 back into the very beginning and done all those different things to it, we would have ended up with 7. Onions and Absolute Values Absolute value equations are done the exact same way. Le'ts take a look at the example 2x+1 = 9. If we draw our onion analogy, we start with a number (x), we multiply that by 2, then we add 1, and after we've done those things, we take the absolute value and we end up with 9. Because we're being asked to solve the equation, we have to figure out what was that number x that we plugged in way back in the beginning, and since the absolute value was the last thing to be done, it's the first thing we have to undo. I undo an absolute value by splitting it up into two separate equations; one that equals 9 and one that equals 9. Now it's just a matter of solving these two separate equations one step at a time; the first thing I have to undo is addition by 1. I undo addition with subtraction because those are inverse operations. I take one away from 9, one away from 9 and I end up with 8 or 10. Then I have to undo multiplication by 2; I undo multiplication with division and I end up with either x is equal to 4 or x is equal to 5. The onion for the complex absolute value problem in example #3
222 Onions and Complex Absolute Values The last example will show us the final way we can complicate absolute value equations; having operations on the inside and the outside of the absolute value. Take 2 x4 +3=1. Looking at the onion picture for this problem shows that we start with x, we subtract 4, we take the absolute value, then we multiply by 2 and then we add 3. So the difference between this problem and the previous one is that the absolute value is not the very last thing being done to my variable, which means it's not the first thing we have to undo. A lot of people, when they see this problem, get so concentrated on the fact that there's an absolute value in it that they just think, 'split it in two equations right away!' That's not our job until we get to that layer. First, we have to undo addition with subtraction so I minus 3 from both sides and I get 2 times the absolute value of x minus 4 is equal to 2. Now I have to undo multiplication; I undo multiplication with division and I divide both sides by 2. Now I'm simply left with the absolute value of x minus 4 equals 1. Now the absolute value is the outermost layer of our onion, so it's time to undo it by splitting the equation into two. One where x minus 4 equals 1 and another where x minus 4 equals 1. After I've undone the absolute value, the last step to undo is the subtraction by 4; again, I undo subtraction with addition and I get my two answers that x is equal to 5 or 3. Lesson Summary Finding the solution for example #3 To review, an equation is just a list of steps being done to an unknown value and the answer you get after those things are done. You can undo an absolute value by breaking the equation into two separate ones  one equal to the positive answer and one equal to the negative.
223 If there are no steps on the outside of the absolute value, it's the first thing you can undo by splitting it into two equations. But if there are steps on the outside of the absolute value, you must to undo those first before you split the equation up. How to Solve and Graph an Absolute Value Inequality Chapter 9 / Lesson 5 Taking two skills and combining them to make a more complicated problem is a classic tactic in mathematics. If you feel confident with absolute values and inequalities, see if you can tackle them with their powers combined here! Reviewing Inequalities and Absolute Value Mathematicians love to experiment, so it makes sense that once they came up with the separate ideas of absolute values and inequalities they thought 'let's see what happens when we put them together!' But before we try to figure that out for ourselves, let's quickly go over what those two things are individually. Inequalities mean 'greater than' or 'less than' and are good for expressing situations where something can be at least or at most something else. We can have 1variable inequalities that have graphs on number lines and 2variable inequalities that have graphs on the coordinate plane. We can solve inequalities, just like equations, with inverse operations as long as we remember the one extra rule: you must flip the inequality symbol around when multiplying or dividing both sides by a negative number. Example of an or compound equality Absolute values are important when we only care about the magnitude of a number, or how 'large' it is regardless of whether it's positive or negative. Because of this, the
224 absolute value turns all numbers positive whether or not they were positive to begin with. Graphing absolute values gives us what look like Vs, because when the line would normally keep getting negative, the absolute value instead bounces it back up to the positives. When we solve absolute value equations, we can still use all the same inverse operations to get variables by themselves, but we must remember that the way to 'undo' an absolute value is by splitting the equation up into two new ones: one that equals the positive and one that equals the negative. Therefore, it's usually the case that absolute value equations give us two answers. 1Variable Absolute Value Inequalities So now that the review is out of the way, let's combine those two skills into one as simply as we can. Solve and graph x > 5. Without learning anything new, let's try to solve this problem just like we would before. If I'm being asked to solve, and I've got an absolute value, I need to undo it by splitting the inequality up. This leaves me with two inequalities, x > 5 or x > 5. Well, that wasn't too bad. I am now left with a compound inequality because I have two inequalities in one, but that's okay. Let's check our answers to make sure it's really that easy. x > 5 says that numbers like 6 or 100 should work in the original problem, so we can substitute those in, take absolute values, and, sure enough, they check out. Example of an and compound inequality Okay, now x > 5. That says that numbers like 4 or 1 should work. So we can substitute those into the original problem, take absolute values and... huh, they don't. Since the absolute value makes them positive, 4 and 1 don't work because their magnitude isn't as big as 5. So what we really need is actually 'large' negative
225 numbers, like 10 or 100, because then the absolute value would change them positive again and we'd be in business. So instead of x > 5, what we actually need to have is x < 5 because numbers like 10 and 100 are technically less than 5, although they are 'larger' or highermagnitude numbers. What we've just discovered is the Golden Rule of solving 1variable absolute value inequalities. When splitting the inequality into two new ones to undo the absolute value, we need to flip the sign on the one we set to the negative. I personally remember this by thinking about it like the multiplying and dividing by a negative rule, where you have to flip the sign. It's kind of similar. Anyways, now that we've got our correct answer as x > 5 or x < 5, we can graph the compound inequality using our previous knowledge. Doing one inequality at a time and putting an open circle at 5 because it's just 'greater than' and not 'or equal to' and drawing an arrow to the right to indicate all the numbers bigger than 5 takes care of half of the graph. Then putting another open circle at 5 and drawing our arrow to the left to indicate the smaller numbers takes care of the rest. Use translation skills to graph the boundary for the equation Looking at the graph shows us that what we have is an 'or' compound inequality because the two arrows are going in opposite directions. That means that numbers bigger than 5 or ones smaller than 5 will both work in our original inequalities. Absolute value inequalities can also give us 'and' compound inequalities as well. For example: solve and graph x < 5 Now, instead of looking for very 'large' numbers that will need to be bigger than five after they've been turned positive, we are looking for a select group of 'small' numbers that will still be smaller than five even after they have been turned back positive. Using our new skill of flipping the inequality symbol when undoing an absolute value gives us our solved inequality as x < 5 and x > 5. Remember I had to flip it.
226 The reason this is an 'and' compound inequality is that in order for our solution to work, it has to satisfy both conditions. Not only does it have to be less than 5, it also has to be greater than 5. It won't work if only one of those things is true and not the other one. Take, for example, 10: 10 is greater than 5, but it is not less than 5. Therefore, when we substitute 10 back into the original inequality, you'll find that it doesn't work. Any compound inequality where both conditions have to be true in order for your solution to work is an 'and' compound inequality. Once we graph the inequality, we get this, and we see that only the 'small' numbers between 5 and 5 will still be less than 5 after being changed positive. 2Variable Absolute Value Inequalities Lastly, we can also graph 2variable absolute value inequalities, say: y < x Just like other 2variable inequality graphs, we need to begin by graphing the boundary line right where y = x Using a dotted line shows it is not part of the solution This requires us to use our translation skills: use the 2 on the inside of the absolute value to shift the vertex to the right and the positive 1 on the outside to move the vertex up 1. Because there is no variable in front of the absolute value, we can assume there's a little imaginary number 1 in there, and that tells us the slope of the V is 1, which means from my vertex, I go up 1 and then over 1 in both directions to find the V. We now need to decide which side of the V is the 'greater than' side and which is the 'less than' side. Substituting in the origin (0,0) to the original inequality is still a good strategy, and doing so gives us 0 < Simplifying this down gives us 0 < 3, which is a true statement. Therefore, the area of the graph that the origin is in is the part of the graph that we should shade. This means that we shade everything on the outside of the V  on the left, below it and to the right.
227 Lastly, because the inequality was strictly 'less than' and not 'equal to', we make the V a dotted line to indicate that it is not part of the solution. After all the steps are done, we are finished and we have this graph as our answer. Lesson Summary When solving 1variable absolute value inequalities, there is one new rule to remember: when undoing an absolute value by splitting the equation into two, you must flip the inequality symbol on the inequality that is set to the negative (kinda like the whole 'flipping the sign when multiplying or dividing by a negative' thing). 1variable absolute value inequalities give us compound inequalities. An absolute value that is greater than something gives us an 'or' compound inequality, whereas an absolute value that is less than something gives us an 'and' compound inequality. Graphing 2variable absolute value inequalities has no new rules to remember, and your answers should look like Vs, with either the inside or the outside of the V shaded. Solving and Graphing Absolute Value Inequalities: Practice Problems Chapter 9 / Lesson 6 Inequality Review There is one big difference between absolute value equations and absolute value inequalities that we should quickly review before jumping into some practice problems, and that is: when we split an inequality into two different ones in order to undo an absolute value, we need to remember to flip the sign on the one we match with the negative. Other than that, all the rules are the same and we should be good to go. I also want to mention that because this is a practice problem video and we'll be doing about four practice problems, I encourage you to pause the video when you see the problem. Try it on your own and see how far you get. If you get stuck, or if you finish the whole thing and want to check your answer, watch the video to see if I did it the same way you did it and if you got the right answer. That way you'll be able to focus
228 in on what you did wrong, or you can skip through it quickly if you already know you got it right. OR Compound Inequality Graph of an OR compound inequality We'll practice that one difference in this first question here: Solve and graph x + 4 > 5. Because there is nothing going on the outside of the absolute value, we can begin by splitting the inequality up to undo the absolute value. That leaves us with two inequalities: one, x + 4 > 5, and another, x + 4 < 5. So not only did we set it to 5, we also flipped it around to a 'less than' sign. We still need to solve each inequality for x, which means undoing the + 4 on both of them. You can undo addition with subtraction, and doing that for both inequalities gives us our solved inequality, x > 1 or x < 9. This is now a compound inequality, because there are two inequalities in one problem. Graphing this compound inequality is as easy as putting both graphs on the same number line one at a time. We can begin by putting an open circle at 1 and drawing an arrow to the right. It's an open circle because it's only 'greater than,' not 'equal to', and the arrow goes to the right because we want all numbers that are bigger than 1. Putting the other one on there means an open circle at 9 and an arrow going to the left. We see that our graph is complete, and it looks like it's an OR compound inequality because the graphs are going in opposite directions. That means that either being bigger than 1 or being less than 9 is enough to satisfy this inequality. It does not have to be both; it can be one or the other. We actually could have known that it was an OR inequality from the beginning just by realizing first that all onevariable absolute value inequalities give us compound inequalities and that problems where the absolute value is greater than something lead us to OR examples.
229 AND Compound Inequality Our second problem might at first seem like it will also be an OR compound inequality, but there is a difference that will make it end up not being exactly as it might initially seem. The problem, solve and graph 2 x 1 > 9, also has an absolute value and a > symbol, but this time there are mathematical operations on the outside of the absolute value. This means before we split the inequality into two, we need to undo the 1 and the times 2 on the outside. The outermost step is the 1, so we undo that with addition. We next undo multiplication of 2 with division of 2, and now we must remember the rule of inequalities that tells us to flip the symbol whenever multiplying or dividing by a negative number. This leaves us with the resulting inequality as x < 4, and now we realize that this will end up being an AND compound inequality because the absolute value is now less than 4 instead of greater than like in the beginning. Splitting the inequality and flipping one of the symbols leaves us with our solved inequality: x < 4 and x> 4. To graph this, I can again do one at a time and put a closed circle (because it's 'or equal to') at 4 and draw an arrow to the left, then put another closed circle at 4 and draw an arrow to the right. Because the arrows are pointing toward each other, we can just connect the two dots and we end up with our AND compound inequality graph looking like so. Graph of an AND compound inequality Negative Numbers and Inequalities If at any point you are solving a problem like one of these two and you end up with a statement where an absolute value of anything is either greater than or less than a
230 negative number, you know something is up. For example, take 7x  1 > 5. It doesn't matter what is happening on the inside of the absolute value, because we know that it will eventually be turned into a positive number. Therefore, this absolute value will always be bigger than 5; the inequality will always be true no matter what we substitute in for x, and therefore there are an infinite number of solutions to this problem. But on the other hand, if instead we had 7x  1 < 5, the same logic tells us that it is impossible for us to make an absolute value smaller than 5, and therefore there are no solutions to this one. System of Inequalities We'll end this practice problem video with a system of twovariable absolute value inequalities. Graph y > (1/2) x  5 and y < 2. This is a system of inequalities because there is more than one inequality, and at least one of them has two variables. We can solve this problem simply by putting one inequality on the graph at a time and then figuring out where they overlap. First, let's start with graphing y > (1/2) x  5. To do this, we'll have to remember how to use translations and also how to graph an absolute value. All absolute value graphs look more or less like 'V's, but this one is going to have a few differences. First off, the 5 on the end will pull the vertex of the graph down five places, so instead of it being at the origin (0,0), it will be pulled down to the point (0,5). Secondly, the 1/2 in front of the absolute value will make the slope of the 'V' 1/2 instead of 1. That means that, starting from our vertex, we'll go up one and then over two in each direction to determine how steep the 'V' is, and we can sketch it in like this. We can leave the line of the 'V' solid because it is 'or equal to' from the inequality, but we still need to determine which part of the graph to shade. Substituting in (0,0) give us the inequality 0 > (1/2)(0)  5, which can simplify down to 0 > 5, which is true. That means we can shade the area of the graph with (0,0) in it, and we fill in everything on the inside of the 'V' to get that twovariable inequality.
231 Graphing a system of inequalities But this was a system of inequalities, so we've still got another piece to add to this graph, y < 2. Just like we began the first half of this problem by graphing the line where y was equal to the absolute value to get our 'V' and then determining which side to shade, we'll start graphing y < 2 by graphing where y=2 and then deciding which side of that line to shade. Graphing y= lines on a coordinate plane leaves us with horizontal lines, so y=2 looks like this. We need to make it a dotted line to indicate that it is strictly 'less than,' not 'equal to,' and then shade below it to indicate that y=0 works when I substitute it into the inequality (0 < 2). Now we've got a graph with a ton of shading everywhere. Because the solution to a system of inequalities is only where the shaded regions overlap, when we lay the y < 2 graph on top of our absolute value 'V', we find that our solution is only the triangular region in the middle of the graph, inside the 'V' but below the horizontal line. Lesson Summary To review, when splitting an absolute value inequality into two new ones to undo an absolute value, you must flip the inequality symbol on the one with the negative. Operations outside an absolute value must be undone before undoing the absolute value itself. Finally, systems of inequalities can be done with absolute values just like other lines, one graph at a time, where the solution is only the area where the shading overlaps.
232 What are Polynomials, Binomials, and Quadratics? Chapter 10 / Lesson 1 Polynomials, Binomials, and Quadratics This is one area of math where you can relax a little bit. Polynomials, binomials, and quadratics are not tricky things you have to learn. They are names of mathematical expressions that are easy to work with. Knowing what they have in common and how you can identify each of these will help you work through your math problems easier. In this video, we're going to go through what each of them is and how you can identify each. The first thing I can tell you about all three of these is that they are all polynomials. So, picture a big balloon with all your polynomials floating around inside. Inside the balloon, you will find two smaller balloons with binomials and quadratics inside. Let's talk about polynomials and what makes them unique. Polynomials Polynomials literally means many terms. They are easy to work with because they have three restrictions. What are the restrictions? First, the variable in each term cannot be a negative exponent. Second, the variable in each term cannot be in the denominator. Third, the variable in each term cannot be inside a radical. In other words, the variable's exponent must be a whole number. If you see any of these, then you can zap them out of your balloon as they are not polynomials. Even though polynomials literally means many terms, you can have just one term or just two terms. A term means the multiplication of a number with a variable. The variable's exponent can be any positive whole number. All terms are
233 separated by either a plus or a minus symbol. Of course, you can have as many terms as you want in a polynomial. There is no limit. All of these are polynomials that belong inside the balloon. Notice how none of the variables have negative exponents, are in the denominator, or are inside a radical. All of the exponents for the variables are positive whole numbers. Also notice how all the terms are separated by either a plus or a minus. All of these polynomials are written in what is called standard form, starting with the largest exponent and working my way down. Now let's talk about binomials. Binomials So, inside the balloon are two smaller balloons. One of the smaller balloons is for binomials. Binomial means two terms. So, what kinds of polynomials do you think binomials cover? That's right; binomials include all polynomials that only have two terms. So, from what we see inside the big polynomial balloon, the two polynomials that are binomials are the x + 7 and the x^4 + 3x. Both of these only have two terms and that is all that is needed for them to go inside the smaller balloon. Now what about quadratics? Quadratics We have another small balloon inside the large balloon and it is for quadratics. Quadratics include polynomials whose highest exponent is 2. We can have quadratics with only one term, two terms, or three terms. But three terms is the maximum limit. We can't have more than that. The one polynomial that we see that fits this criteria is the x^2 + 3x  1. Other quadratics that we can add to our little balloon includes these. Do you see how all of these are polynomials whose highest exponent is 2 and that all of these are either one term, two, or three terms in length? They all fit the criteria for quadratics. It is definitely possible that you can have a polynomial that can belong to two or all three balloons. Lesson Summary To summarize everything we have learned, polynomials means many terms, binomials means two terms, and quadratics means polynomials whose highest
234 exponent is 2. All of these are polynomials with binomials and quadratics being special cases. The three restrictions that make polynomials easy to work with are that: 1. The exponents cannot be negative. 2. The variables cannot be in the denominator. 3. The variables cannot be inside a radical. How to Add, Subtract and Multiply Polynomials Chapter 10 / Lesson 2 Adding, subtracting and multiplying polynomials are, basically, the same as adding, subtracting and multiplying numbers. They only difference is that we have a pesky variable to worry about, but this video will show you that's no problem, so no worries! This method has worked for many of my students, and I think it will work for you, too! Adding Polynomials Use shapes to organize like terms in the polynomials. Adding polynomials is very easy! There are many ways to add polynomials that have been taught, but here's my favorite. I look at the problem and put a circle, square or triangle around the like terms so I don't get them confused. Let's look at this example. (2x^2 + 3x + 4) + (x^25x + 7) I like to start from the left. I circle 2x^2. Now, I look for another like term to 2x^2. Of course, that's x^2, so I circle it. Then I add them: 2x^2 + x^2 is 3x^2, and then I write that down as part of my answer.
235 The next term is 3x. I put a square around that one. Then look for another one. Of course, we have 5x, so I put a square around that one too. Then add them: 3x + 5x is 2x, and I write that down as the next part of my answer. You always want to double check that there aren't any more. Finally, we have 4. I continue the same idea, but I put a triangle around it. Now I look for another like term to 4. Of course, that's 7, so I put a triangle around that one. Then I add them: is 11, and I write that down as the last part of my answer. It's always good to double check that you have everything in a circle, square or triangle. When you get longer polynomials, it's easy to miss terms! So here's my final answer: 3x^22x 'Okay, Kathryn, do I have to use circles, squares and triangles?' No! If you're good at adding polynomials, you can cross off as you go, but for those that haven't had enough practice, this is definitely my suggestion. 'Kathryn, what about problems that have more terms? What other shapes or ideas do you have?' Great question! You can use different colors to circle like terms. I don't use it here for the benefit of students who are color blind, but you could have used red to circle the x^2s, blue to circle the xs and green for the numbers. I've also used the method of underlining like terms too! Whatever you choose is how you will distinguish the different terms! Distributing 1 into the second expression turns the subtraction problem into an addition one. Subtracting Polynomials In subtraction, let me show you the underlining method.
236 (3x^22x + 5)  (2x^26x + 7) First, I am going to distribute the 1 into the second expression. That will make this an addition problem! The first expression stays the same: 3x^22x + 5. We will distribute the negative, like this: 1 * 2x^2, which is 2x^2; 1 * 6x, which is a positive 6x; and 1 * 7, which is 7. This gives us our new subtraction problem: (3x^22x + 5) + (2x^2 + 6x  7). Remember, I am going to show you the method of underlining instead of circling to add the expressions. We look at the first term 3x^2 and underline it. Now, I continue to look for a like term. Here it is, 2x^2, and I underline it. Now I add them: 3x^2 + (2x^2), and we get x^2. That's going to be the first term of our answer. The second term is 2x, and this time, I put a squiggly line under it. Now I continue to look for a like term. Here it is: 6x, and I put a squiggly line under that one. I add 2x + 6x, and I get 4x. That's the second term in our answer. The last term is 5, and this time I put a jagged line under it. Now, I continue to look for a like term. Here it is: 7, and I underline it. Now, I add them: , and we get 2. This will be my last term in the answer: x^2 + 4x 2. Multiply the first term in the product, x, by everything in the second expression. Multiplying Polynomials Now, we get to multiplication. This problem won't quite work like addition or subtraction, and we can't use FOIL because these are larger than a binomial times a binomial! (x+5)( x^2+3x2)
237 First, multiply the first term in the product: x times everything in the second expression. I like to draw arrows to remind me which multiplication I've done; otherwise I tend to get lost. This is how it will look: x(x^2) + x(3x) + x(2) Let's multiply. x(x^2) = x^3 + x(3x) = 3x^2 + x(2)= 2x. This isn't our final answer; we need to multiply everything in the second expression by 5! So we'll have 5(x^2) + 5(3x) + 5(2) Are you ready for the final answer? We simply add the like terms together! x^3 + 3x^22x + 5x^2 + 15x  10 Start from the left, and circle x^3. It looks like there aren't any like terms for x^3, so we write that down as our final answer. Put a square around 3x^2. I look and find 5x^2, so I put a square around that term, too. I don't see any more, so 3x^2 + 5x^2 = 8x^2. 8x^2 is written next to x^3 as part of our final answer. Before getting the final answer, multiply all terms in the second expression by 5. Put a triangle around 2x. I look and find 15x, so I put a triangle around that term too. Why? Well, they're like terms. I don't see any more like terms for x, so 2x + 15x = 13x. 13x is part of our final answer, and I'm going to write it next to 8x^2. Finally, I see 10. I underline this term and look for another one like it. I don't see one, so 10 is written in my final answer.
238 So, what is the final answer then? x^3 + 8x^2 + 13x Lesson Summary Adding: No trick here. Just add like terms together. If you get lost at first, use circles, squares, triangles, lines...anything to separate the different terms. Subtraction: Multiply the second expression by 1, and add the two expressions together. Multiplication: Multiply each term in the first expression by each term in the second expression. Add like terms together and you're done! Multiplying Binomials Using FOIL and the Area Method Chapter 10 / Lesson 3 From the distributive property, to FOIL, to the area model, to happy faces and claws, there are many different ways to learn how to multiply binomials. In this lesson, you'll learn how to use all of them and get to pick which one you like the most. What is a Binomial? Do you know this phrase: 'We all put our pants on the same way: one leg at a time.' It was created to help remind you that you have something in common with all different types of people  movie stars, superstar athletes, homeless people, CEOs; we all put our pants on the same way. But I kind of like it because I like the idea that something is so obviously the right way to do it that everyone does it that way. Sure, you could put your pants on by diving in with both legs right away (and to be honest, I've probably done this in a tent camping or something) but it's clearly the wrong way to go about it. The fact that this phrase has become so well known makes it clear: One leg at a time is without a doubt the best way to put your pants on. So right around now you might be wondering, isn't this an Algebra lesson on multiplying binomials? Why is he talking about pants? Well the reason is that the process for multiplying binomials is the exact opposite of putting your pants on  humans cannot agree on what the best way to do it really is. Instead of there being one obvious way to do it, there are actually tons of different ways. If there was a best way to do it, we wouldn't have
239 come up with so many different strategies. But before we jump right into the strategies, let's start with the basics. What does it mean to multiply binomials? Multiplying binomials involves 4 mini multiplication problems Well, a monomial is an algebraic expression with only one term, which makes a binomial an algebraic expression with two, like this: x + 6. We can even have trinomials, expressions with three terms, and bigger ones too. Although the fancy names stop at trinomial and we just start calling them all polynomials no matter how many terms they have, once they have more than three. Multiplying Binomials So when we're asked to multiply two binomials, we just need to do something like x + 2 * x  1. Because parentheses next to each other imply multiplication, we can change this into (x + 2)(x  1).Okay so how do we do it? Well all we really need to do is basically a fancy version of the distributive property. So if I was to look at the problem we have in front of us, but I was to cover up that first x, what we would have is just 2(x  1), which we could use the distributive property for. It tells us to not only multiply the 2 and the x, but also distribute that 2 to the 1 as well. Whatever is on the outside of the parentheses needs to be multiplied to everything on the inside. So, when I uncover the x and bring it back into my problem, I have to do the exact same thing with it that I just did with the 2: distribute it to both things in the second parentheses. That means first multiplying the x by the x and second, multiplying the x by the 1. This means that anytime you multiply two binomials, it's really just four mini multiplication problems. First I can do x * x to get x^2, then I do x * (1) to get x, now move on to the 2 and do 2 * x to get a positive 2x and finally, 2 * (1) to get 2.
240 The area method is ideal for visual learners We have now successfully multiplied two binomials, but we are usually going to want to simplify our answer by combining like terms. Grouping the x's together gives us our final answer as x^2 + x  2. And that's it! All the methods we'll go over in the rest of this lesson are simply different ways of remembering how to do all four of those mini multiplication problems. It seems a little silly, but it actually can be easy to forget which mini multiplication problems you've done and which ones you still need to do. So all these little tricks have been developed to help you organize your work and develop a pattern that will prevent you from making those silly mistakes. The FOIL Method Let's start with the most well known method: FOIL. This is the method that uses an acronym to help you remember, and it stands for 'firsts, outsides, insides, lasts'. If you say this in your head as you multiply the first terms, the outside terms, inside and last terms, hopefully you'll remember to do all four mini multiplication problems without doing any of them twice. Once we've done the multiplying, we again combine like terms in the end to get our answer. The claw method for multiplying binomials
241 The Area Method If you're more of a visual learner like me, you might like the area method more than FOIL. Out of the methods we'll go over, this is probably the safest and surest way to make sure you don't make any mistakes. But it's also probably the most time consuming. Instead of having you remember an acronym like FOIL, the area method asks you to draw a chart. Because each binomial has two terms, we'll draw a box for the problem (2x + 3)(3x  1) with two sections on each side. This gives us a box with four regions, one for each mini multiplication problem that we're about to do. Labeling each side with one of the binomials and giving each term its own section lays out all the mini multiplication we need to do. Now we just need to come up with what goes on the inside of the four regions by looking across and above to see which terms are labeled on the outside and multiplying those two things together to get our four mini multiplication problems. Once all four regions have been filled, we can rewrite the expression outside the chart, 6x^2 + 9x 3, and combine like terms to get our final answer: 6x^2 + 7x  3. The Claw and Face Methods The last two methods we'll go over are simply cute ways we can draw lines right on top of the problem to help us remember which four mini multiplication problems to do. My favorite of these is what I call the claw. By drawing in lines from the first binomial to the second, where each line represents one of the mini multiplications we need to do, we get what looks like a lobster claw. If you draw those same four lines a little bit differently, we can end up with what kinda looks like a smiley face. Each line still represents a mini multiplication problem to do, and once all four have been drawn in, you still must combine like terms to get your final answer.
242 The face method is yet another visual option for solving problems So even though we all put our pants on the same way, we all multiply our binomials a little bit differently. That's fine as long as whichever way you pick, you remember to do the correct four mini multiplication problems and then combine your like terms. Lesson Summary Binomials are algebraic expressions with two terms, and there are many ways to help you remember how to multiply two of them together. FOIL helps you remember which four mini multiplications to do by being an acronym that stands for 'firsts, outsides, insides, lasts'. The area model helps you remember by giving you a chart with four blank spots to fill in. While the claw and the happy face are pictures you can draw on top of your problem, where each line is one of the mini multiplications you'll need to do. They all work and get you the same answer, so pick your favorite and go for it! Multiplying Binomials Using FOIL & the Area Method: Practice Problems Chapter 10 / Lesson 4 There are a few mistakes that are easy to make when multiplying binomials with FOIL and also a few ways to complicate problems like this, so why not make sure you're brushed up on your skills? You'll also learn a shortcut and how to use the area method to multiply even bigger polynomials. Multiplying Binomials Kind of like learning your times tables, multiplying binomials is one of those skills that can start out slow but with some practice become really fast and easy. This lesson is here to help you move from slow and unsure to fast and confident! Because there are so many different ways to multiply two binomials, we'll do each problem in this lesson two ways. I won't explain both ways wordforword in each problem, but we'll have both ways going on the screen at once so you can pick which of them you like more and also see how they're basically the same thing. Feel free to pause the lesson when I introduce a problem and give it a shot on your own before you watch me do it. That way you can either focus on the specific mistake you make or quickly skip through the problem if you got it right!
243 Example #1 Let's start with a pretty basic one. Multiply (x + 6)(x  6) We'll use the FOIL method on the left side of the screen and the area method on the right. Remember that both are simply procedures that help you remember the four minimultiplication problems that these examples will break down into. FOIL uses an acronym to help you remember which things to multiply, while the area method uses a chart to do the same thing. FOIL is a little bit quicker but also easier to make mistakes with, while the area method takes a little bit more time because you need to draw out the chart, but it's much better at preventing silly mistakes. Because the numbers aren't too bad on this one, I'll go ahead and use FOIL. That means I begin by multiplying the First terms, x and x, to get x^2. Then we do the Outer terms, x and 6, to get 6x. The Inner terms, 6 and x, give us 6x. And finally we have the Last terms, 6 and 6, leaving us with a 36 on the end. x^26x + 6x  36 Using the two methods of multiplication to solve the problem We're already basically done, but we'd like to simplify our answer by combining like terms. I've got two groups of xs here, and when we do 6x + 6x, they will actually cancel each other out, leaving our answer as simply: x^236 Now, there is actually a shortcut to answering problems like this where you are multiplying two conjugates together. Two binomials are conjugates if they have the
244 same two terms but opposite signs on the second one like (a + b) and (a  b). Any time we multiply two conjugates together, the two middle terms will drop out, just like we saw, giving us the shortcut: (a + b)(a  b) = a^2  b^2 Example #2 Let's move on to another example: Multiply (2x + 5)^2 I bring this example up for one important reason: it's good to learn that this problem should be done like the other ones in this video, even though it doesn't look like it. If you're not aware of this, it can be very easy to make this simple mistake: 2x^2 + 5^2 Distributing the exponent like this is not allowed! Here's the reason. Squaring something means multiplying it by itself. That means taking 2x + 5 and multiplying it by 2x + 5. Aha! Two binomials! If you make the mistake of distributing the exponent, you are only doing the first and last steps of FOIL and skipping the middle two  not good! But now that we know how it should be done, let's do it the right way really quick. I'll use the area method for this one just to mix it up a little bit. Because we are multiplying two binomials, we draw our chart with two segments on each side, making four boxes on the inside. I now label each side with one of the binomials, each term getting its own segment of the side, like so. 2x 5 2x 5 Now we just look above and across to decide what to multiply for each box. Doing so gives us this: 2x 5 2x 4x^2 10x 5 10x 25 4x^2 + 10x + 10x + 25
245 and now we can combine like terms to end up with our answer: 4x^2 + 20x + 25 Example #3 Our last example will stray away from the title a little bit. You might call it an extra credit question  a preview into a future concept. The reason I do this is to show you that the skills we're mastering now can help you do more than just multiply binomials. Take (x  1)(x^2 + 5x + 3) for example. These aren't two binomials  this is a binomial times a trinomial. While the FOIL acronym no longer quite applies, it's still the same idea and the area model still works great. This time, though, instead of making our chart have two segments on each side, we'll make a rectangular chart with two segments on one side and three on the other. The (x  1) goes on the side with two segments, while the (x^2 + 5x + 3) goes on the side with three. And we're good to go! x^2 5x 3 x 1 We simply look above and across to decide what to multiply for each box, fill the entire chart in, and combine like terms. Just like before! x^2 5x 3 x x^3 5x^2 3x 1 x^25x 3 x^3 + 5x^2  x^2 + 3x  5x  3 = x^3 + 4x^22x  3 Area Method and Other Polynomials Need to multiply two trinomials? Make your chart three by three! A polynomial with five terms multiplied by one with six? A five by six chart! The area method will help you multiply any two polynomials together no matter what size they are. Hopefully you're starting now to feel like a polynomial pro.
246 Lesson Summary Let's review. Conjugates are binomials that have the same terms, only the sign has been changed on the second one. When you multiply them together, you end up with the square of the first term minus the square of the second one: (a + b)(a  b) = a^2  b^2. When squaring a binomial, remember not to distribute the square to the two terms and instead rewrite another binomial next to the first one and FOIL it out: (a + b)^2 = (a + b)(a + b). The area model can be used to multiply any two polynomials together, not just binomials. Simply make your chart have dimensions equal to the number of terms in each polynomial. How to Factor Quadratic Equations: FOIL in Reverse Chapter 10 / Lesson 5 So, you know how to multiply binomials with the FOIL method, but can you do it backwards? That's exactly what factoring is, and it can be pretty tricky. Check out this lesson to learn a method that will allow you to factor quadratic trinomials with a leading coefficient of 1. Third Grade Review Probably the most fundamental concept of algebra is the use of a variable (like x) to represent any number we want. That means that a lot of the stuff we do in an algebra class is actually stuff you learn in elementary school; we just use variables instead of specific numbers. I've found that one of the best ways to teach algebra is to first review the skill with specific numbers and then apply that same skill to our new algebraic problem. What is 3 * 5? I think I need my calculator for this one. Ah! 15. Okay. Another third grade question, but this one might actually stump you because you'll need to remember some vocabulary: factor 15. So, the question really becomes, what does it mean to factor something? Turns out that it simply means to split that number up into what you can multiply to get it. So when I factor 15, I just turn it into 3 * 5. That
247 means factoring is just doing multiplication in reverse. A little more vocab before we move on here. In the statement 3 * 5 = 15, 3 and 5 are called the factors of 15, while the answer to a multiplication problem is called the product. Multiplying Binomials Find two numbers that have a product of 24 and a sum of 11 So, back to algebra class. Let's do the same set of problems that we just did  but with variables this time around. So, instead of 3 * 5, how about (x + 3)(x + 5)? There are a lot of different ways to multiply binomials just like this, but the most wellknown way is the FOIL method. FOIL is an acronym that helps us remember to multiply the first terms, the outer terms, inner terms and finally the last terms. Once we've done that, we can combine our like terms to get our answer, x^2 + 8x Following the same order of questions from the third grade review, the next thing I'll ask is this: factor x^2 + 8x We now know that this means to break up our number into the two things that I can multiply to get it (its two factors). Since we just did that, I know that the answer is going to be (x + 3)(x + 5), but what if we hadn't just done it? How would we know exactly how to break this trinomial up into its two separate binomials? This question is what factoring quadratic expressions is all about, and it can be pretty tricky. But if we look closely at the example we just did, try to find some patterns and then generalize those patterns, we'll be able to come up with a method that will help you factor most quadratic trinomials you'll encounter.
248 Factoring Quadratic Expressions Let's look closely at that last problem we just did. We said we could factor x^2 + 8x + 15 as (x + 3)(x + 5). What do the numbers from the trinomial (the product) have to do with the numbers from the two binomials (the factors)? Two things, actually! gives us 8, which was the coefficient from the middle term from the trinomial. And 3 * 5 gives us 15, the constant on the end of the trinomial. This means that if you can find two numbers that add to the middle term of your trinomial and multiply to the constant on the end, those are going to be the two numbers in your factored expression. Use FOIL to check your answer Let's see if we can apply this idea to a different problem; maybe factor x^2 + 11x So, for our method to work, we need to find two numbers that add up to the middle coefficient (11) and that will also multiply to the constant on the end (24). Sometimes the answer is going to be obvious, but sometimes it will be hard to see. When it's hard to see right away, I like to write out all the different ways we can multiply to get the constant on the end and then see which one of those will add up to the middle. Let's do that here to practice. We'll always be able to do 1 times the number. In this case, (1 * 24). 24 is even, so 2 goes into it as well, so (2 x 12), I think 3, (3 x 8) and 4, (4 x 6), and that's it. I could do the other ones, (6 * 4), (0 * 3), but those are just the same just backwards, so those don't matter. So we have these four options  which ones add up to 11? Hey, 3 and 8! That means that x^2 + 11x + 24 factors into (x + 3)(x + 8). If you'd like to check your answer, you can quickly multiply it out and make sure it ends up back where we started! If we do that here, sure enough, we're all good! One last example: factor 2x^36x^280x. This one looks quite a bit different. x^3? We don't know how to do that. Sometimes, there are ways (maybe you'll learn those
249 in a different class), but if you see that in this class, there's probably something tricky going on. Sure enough, looking at each term in our trinomial here, there's an x in it. Whether it's an x^3, an x^2 or just a plain old x, there's an x in each term, which means I can divide it out (or factor it out) of each term by doing the distributive property backwards. When I pull that x out of each term, I can now write it in front and leave what I'm left with after I divide it out on the inside of the parentheses. So, I no longer have 2x^3, but I have 2x^2; instead of 6x^2, I have 6x; and instead of 80x, I just have plain old 80. But now there's another different thing going on. There's a 2 in front of my x^2. In a later lesson, we'll get to examples where you're going to have to deal with that 2, but it's also worth checking to see if I can do the same thing. Can I divide that 2 out of each term, just like I divided the xs out? Sure enough, again, we're going to be able to do that because 2, 6 and 80 are all even numbers and therefore divisible by 2. Doing the distributive property backward and factoring a 2 out of each, bringing the 2 in front and then putting what I'm left with in parentheses, gives me 2x(x^23x  40). Now, even though I have a 2x out front, what's inside the parentheses is something we know how to do; it's a trinomial, it's got 1x^2 and I can try to use my pattern. That pattern says I need two numbers that add up to 3 and multiply to 40. So we actually have some negatives this time, but that won't actually change too much. Again, if the two numbers aren't obvious to me right away, I can write out all the factors of 40, like so. But since we need to multiply to 40, one of the numbers is going to have to be positive and one's going to have to be negative. So now we're looking for a pair of numbers that have a difference of 3, not a sum. Which set of two can I subtract to get 3? It looks like 58 would do the trick. Because I'm doing a +5 minus an 8, the two numbers that are going to work are +5 and 8. The 2x in front of the parenthesis was factored out of the original trinomial That means I factor the x^23x  40 part as (x + 5)(x  8), and the 2x is still hanging out in the front, so we get 2x(x + 5) (x  8) as the answer.
250 The patterns we've talked about here are going to help you factor quadratic expressions that have a leading coefficient of 1. Like I mentioned before, sometimes we're going to have a 2 or 3 in front of the x ^2s. For those, you're going to have to use a different method. To find out about that method, you'll have to check out a future lesson. Lesson Summary When you are asked to factor something, it's your job to break that number up into its factors, the numbers that multiply to your original number. This means that factoring is basically multiplication in reverse. If the quadratic expression you're factoring has only 1x^2 in front, you only need to find the pair of numbers that fit the following pattern. They have to add up to the middle coefficient, the number in front of the xs. That pair of numbers needs to multiply to the constant, the number on the end. Factoring Quadratic Equations: Polynomial Problems with a Non1 Leading Coefficient Chapter 10 / Lesson 6 Once you get good at factoring quadratics with 1x squared in the front of the expression, it's time to try ones with numbers other than 1. It will be the same general idea, but there are a few extra steps to learn. Do that here! Factoring Quadratic Equations Factoring a quadratic expression involves turning a trinomial into multiplication of
251 two binomials. This lesson will be on advanced factoring techniques, so I'm going to assume that you know a few things: first, that factoring is the process of breaking up a number into the things that we can multiply together to get that number. This means that factoring a quadratic expression is the process of taking a trinomial and turning it into multiplication of two binomials  basically FOIL backwards. We do this by looking for a pair of numbers that have a product equal to the constant on the end of the trinomial and a sum equal to the coefficient on the x's. For example, turning x^23x  10 into (x + 2)(x  5) by realizing that 2 * 5 = 10, and = 3. So, where do we go from here? What else is there to know? Well, the method I just described only works for quadratic trinomials where the leading coefficient (the number in front of the x squared) is equal to 1. As soon as we're asked to factor something else, like 2x^25x  3, we're going to be in trouble. The goal is still the same  split the trinomial into a product of binomials  and we'll still find a lot of the same patterns, but now we'll have to make two slight changes in the process in order to end up with the correct answer. Let's go ahead and take a look at the example I just mentioned. Adjusting the Pattern Factor 2x^25x  3. An example of factoring quadratics We'll start this problem very similarly to the simple factoring problems by looking for two numbers that fit the pattern. The thing is, it's not going to be quite the same pattern. One part stays the same, and that is the fact that we need to find a pair of numbers that will add up to the middle coefficient  in this case, 5. But the second part is going to be different. Instead of our two numbers needing to have a product equal to the constant on the end, we now need the product of our pair of numbers to
252 be equal to the constant on the end times the leading coefficient. This was actually true for the easier factoring problems as well, but the leading coefficient was just 1, so multiplying by 1 didn't change what the number was. The next step of the problem should be familiar. Find a pair of numbers that has a sum of 5 and a product of 6. Making a quick list of the factor pairs of 6 and keeping in mind that we'll need one to be positive and one to be negative, makes it clear that 6 and +1 are our two winners. And now we come to the only other difference in our process. Instead of simply being able to say 6 and +1 go in our binomials and we're done, there is an extra step before we can be sure of our answer. While there are multiple ways of doing this step, I recommend using the area method to work your way backward to the answer. Putting the 2x^2 and the 3 in one diagonal of the chart and the two values we came up with using our pattern attached with x's in the other diagonal, gets us one step away from our answer. 2x^26x x 3 If we can find what terms must have been on the outside of this chart to get multiplied in and give us what we have here, we'll be done. We do this by dividing out the greatest common factor from each row and column of our chart. So, if I look at the top row of this chart, I have a 2x^2 and 6x. I need to ask myself what do those things have in common? Well, 2 and 6 are both divisible by 2, so I can take out a 2. But they also both have an x, which means I also take out an x, so I can pull a 2x to the outside of that row. Going down a row to the bottom, 1x and 3, they don't have any factors with the numbers in common, and they also don't have any variables in common, which means the only thing I can divide out is a 1. Now, we go to the columns. Let's start in the left column, 2x^2 and 1x. The numbers don't have anything in common but the variables do, which means I can take out 1x. Finally, the column on the right: 6x and 3. Both share a 3, which means I divide that out and write it on top. What we now have on the left and above our little area model is our factored answer. The terms that are on the same side are the terms that go in parentheses together to make up our two binomials, and I end up with (2x + 1)(x  3). You can always quickly multiply out your answer to make sure you got the right thing, and if you do that here, looks like we're good. Factoring quadratics like this take a while, and they're not always simple, so practice is key. Let's do one more example during this lesson.
253 Multiply the answer to make certain it is correct. Another Example Factor 9x^24. Not only is this example another one with a non1 leading coefficient, it's also an example of a special quadratic that often messes students up. Why? Because there are no x's in the middle! It's not a trinomial! But that's okay. Just think of this problem as being this one: Factor 9x^2 + 0x  4. Now we're ready to go. We begin by finding the two numbers that fit the pattern. For this one, they'll have to add up to zero (the middle term) and multiply to 36, which was the 9 in front times the 4 on the end. Quickly looking through our options here and knowing that we're going to have to add up to zero makes it pretty obvious that 6 and 6 are going to be our winners. Now that we have these two values, we can fill in our area model. We put the 9x^2 and the 4 in one diagonal and the 6x and 6x in the other, but it actually doesn't matter which diagonal or even which order you put them in; they're all going to work! Lastly, we need to divide out the greatest common factors from each row and column to find out which binomials exist on the outside of this chart. 9x^26x 6x 4 Looking at the top row, it looks like they both have a 3 and an x. The bottom row, both numbers are divisible by 2. The left column, I see a 3 and an x again, and the right column, they both have a 2 so I can pull that out. Rewriting the terms from each side together in parentheses as binomials says that the factored form of this is (3x + 2)(3x  2).
254 Lesson Summary Factoring problems with a leading coefficient that isn't 1 have two differences from their simpler counterparts. First, the pattern we use to determine the pair of numbers that will help us find our answer now requires you to find two numbers that have a product equal to the constant times the leading coefficient, instead of simply being the leading coefficient itself like before. Secondly, once you come up with the pair of numbers that fit the pattern, you must substitute those numbers into an area model and factor out the greatest common factors to determine the answer. If you're asked to factor a quadratic that does not have an x term, just pretend that there is a 0x term in the middle there, and continue to do the problem just like you normally would. How to Divide Polynomials with Long Division Chapter 10 / Lesson 7 Arithmetic long division and polynomial long division are very similar. Yes, it's a long process, but once you have the rhythm you will get every problem correct! Polynomial Long Division Polynomial long division is very similar to the long division we did as kids, except now we have numbers and xs. It takes only two steps that are repeated until you're done. 1. Divide the first terms. 2. Multiply the result from the divisor, then subtract it from the dividend. 3. Repeat! Let's review the parts of a division problem. We have the divisor divided into the dividend, and of course our answer is the quotient. Example #1 Here's our first example: (x^2 + 7x + 12) / (x + 3).
255 To fill in our long division, x^2 + 7x + 12 is the dividend, so it goes under the long division symbol. x+3 is the divisor, so it goes to the outside, in front of the long division symbol. Now we're ready to start the division. The steps are the same every time. It's easier to show and talk about the steps than listing them, so let's take a look. To find the first term of the quotient, we take the first terms from the divisor and the dividend and divide them. I like to write them as a fraction; it's easier to divide or reduce: x^2 / x = x. x is the first term of the quotient; we write it above the long division symbol. To figure out what we will subtract from the dividend, we multiply x times the divisor, x+3: x(x+3) = x^2 + 3x. x^2 + 3x is written under the dividend, matching like terms. Here's where I like to make life easy. To subtract the polynomials, I simply change their signs and add! So x^2 x^2 = 0, and 7x  3x = 4x. Just like we did in long division, we bring down the next term, which is 12. Take the first terms from the divisor and dividend and divide them. Guess what? We do the exact same step! To find the next term of the quotient, we take the new first terms and divide them, so we'll have 4x / x. I like to write them as a fraction; it's easier: 4x / x = 4. 4, or positive 4, is the next term of the quotient. Remember, that gets written next to the x above the long division symbol. To figure out what we will subtract from the dividend, we multiply 4 times the divisor, (x + 3). (4)(x + 3) = 4x x + 12 is written under the dividend, matching like terms. Next, I like to change the signs and add down: 4x x  12 = 0. It turns out our remainder is zero. What does our answer look like? Just look above the division symbol: x + 4. Are you thinking to yourself 'That wasn't too bad'? Let's try one a little longer. Don't panic. Follow the exact same steps, repeat and we'll have an answer. Example #2 Divide (15x^2 + 26x + 8) / (5x + 2).
256 To fill in our long division, 15x^2 + 26x + 8 is in the dividend, so it goes under the long division symbol. 5x + 2 is the divisor, so it goes to the outside in the front of the long division symbol. Now we're ready to start the division. To find the first term of the quotient, we take the first terms from the divisor and dividend and divide them: 15x^2 / 5x = 3x. 3x is the first term of the quotient; we write it above the long division symbol. To figure out what we will subtract from the dividend, we multiply 3x times the divisor, 5x + 2: 3x(5x + 2) = 15x^2 + 6x. 15x^2 + 6x is written under the dividend, matching like terms. I like to change the signs and add straight down: 15x^215x^2 = 0, and 26x + (6x) = 20x. Just like we did in long division, we're going to bring down the next term, which is 8. Bring down the 8 and divide again. And we do exactly the same step! To find the next term of the quotient, we take the new first terms and divide them: 20x / 5x = 4. 4, or +4, is the next term of the quotient. Remember, that gets written next to the 3x above the long division symbol. To figure out what we will subtract from the dividend, we multiply 4 times the divisor, 5x + 2: (4)(5x + 2) = 20x x + 8 is written under the dividend, matching like terms. I'm going to change their signs and add straight down: 20x + (20x) = 0; 8 + (8) = 0. It turns out our remainder is zero. To find your answer, or the quotient, just look above the division symbol: 3x + 4. Okay, I know what you're thinking: 'What happens when I have a remainder?' Let's try a long division problem that has a remainder. No magic here! We're going to follow the same steps, except our final answer will look a little different. Example #3 (4x^2 + 8x  5) / (2x +1). To fill in our long division, 4x^2 + 8x  5 is the dividend, so it goes under the long division symbol. 2x + 1 is the divisor, so it goes to the outside in front of the long
257 division symbol. Now we're ready to start division. To find the first term of the quotient, we take the first terms from the divisor and dividend and divide them: 4x^2 / 2x = 2x. 2x is the first term of the quotient; we write it above the long division symbol. To figure out what we will subtract from the dividend, we multiply 2x times the divisor, 2x+1: 2x(2x + 1) = 4x^2 + 2x. 4x^2 + 2x is written under the dividend, matching like terms. I like to change their signs and add straight down: 4x^2 + (4x^2) = 0, and 8x + (2x) = 6x. Just like we've done before, we're going to bring down the next term, which is 5. Change the signs, add straight down, and bring down the 5. We're going to do the exact same step again! To find the next term of the quotient, we take the new first terms and divide them: 6x / 2x = 3. 3, or +3, is the next term of the quotient. Remember, that gets written next to the 2x above the long division symbol. To figure out what we will subtract from the dividend, we multiply 3 times the divisor, 2x+1. (3)(2x + 1) = 6x x + 3 is written under the dividend, matching like terms. Once again, I like to change the signs and add straight down, so 6x + (6x) = 0; 5 + (3) = 8. It turns out our remainder is 8. We're going to write the remainder as a fraction: 8 / (2x + 1). So our final answer is going to look like (2x + 3) + 8/(2x + 1). Lesson Summary Dividing polynomials using long division takes only two steps that are repeated until you're done! 1. Divide the first terms. 2. Multiply that quotient by the divisor and subtract it from the dividend. 3. And repeat! Remember how much work you did with long division? How much paper you used? The same is true for polynomial long division. Don't skip steps, use your paper and
258 don't try this in your head. Multiplechoice answers look very similar for a reason! One small mistake and the whole problem is wrong. There isn't partial credit on a multiplechoice test. How to Use Synthetic Division to Divide Polynomials Chapter 10 / Lesson 8 Synthetic division is a 'shortcut' way of dividing a polynomial by a monomial. You still need to know long division, sorry, but this method is way fun when you're dividing by a monomial! Synthetic Division Many students prefer synthetic division over long division and even argue why they have to learn long division. You need to know long division because synthetic only works when you are dividing by a first degree binomial, for example, (x + 3). If you are dividing by a longer polynomial, say (x^22x + 5), you need to use long division! Example #1: How to Synthetically Divide In this first example, x^2 + 5x + 7 is in descending order. (x^2 + 5x + 7) / (x + 2) At this point, I have seen synthetic division written two ways. The first has the divisor in a halfbox on the upper left; the other looks like a division symbol. Both ways are exactly the same, but I prefer the division symbol.
259 First, if the dividend is not in descending order, we need to do that first. I'll talk about missing parts in the next example. But for now, (x^2 + 5x + 7) is in descending order. The divisor also must be in descending order. And it is: (x + 2). Write your usual long division symbol. Place the coefficients of the dividend under the symbol, just like in long division, but do not write the variables. Make sure to leave space between each number. You don't want to get them confused. Next, look at x + 2. We're going to write 2 to the left of the long division symbol. You always take the opposite of what you see in the divisor. So inside the long division symbol, we're going to have 1, 5 and 7. On the outside, we're going to have 2. Our first step is to bring down the leading coefficient, 1. Multiply 1 * 2, and place the result underneath the next coefficient, 5. So you have 2 * 1, which is (2) is 3. (Write that next to the leading coefficient, 1.) Multiply 1 times 2 and place the result under the next coefficient, 5. Multiply 3 times 2 and place the result underneath the next coefficient, 7. So 2 * 3 is 6. We add straight down and get 1. (So we have 1 written next to the 3.) In synthetic division, the degree of the final polynomial answer is one less than the dividend polynomial. Since x^2 + 5x + 7 is degree 2, our answer will be degree 1. What will this look like? Starting from the left, we'll have 1x + 3 with a remainder of 1. The remainder will be written the same as if we had done this problem as long division  a fraction. So our answer is going to look like: x /(x + 2)
260 Example #2: Missing Parts In this example, let's look at a dividend that is missing terms when we write it in descending order. (x^4 + 81) / (x  3) These are both written in descending order, but we don't have an x^3, x^2, or x. In this case, we need to put in a 'place holder' for them. Since we don't have a number, we're going to write 0x^3, 0x^2, and 0x. Write the long division symbol. 3 will be our divisor  that's the number that goes in front of the long division symbol. 1, 0, 0, 0, 81 is our dividend. Now, we do synthetic division just like the last one. We bring down our leading 1. We multiply 3 * 1, which is 3, and we add down = 3, and we do it all over again. 3 * 3 is 9. Add down: is 9, and do it again. 3 * 9 is is 27. Finally, 3 * 27 is 81, and when we add down we get 162. When we add down, we get 162. Now, let's write our final answer. Remember, the answer degree is one less than the dividend polynomial. So our answer is: x^3 + 3x^2 + 9x /(x  3)
261 Example #3: Larger Problems (x^ x^ x^224 x  320) / (x + 8) Place the coefficients of the dividend under the symbol. Next, look at x + 8. We're going to write 8 to the left of the long division symbol. You always take the opposite of what you see in the divisor. Now we're ready for synthetic division. Our first step is to bring down the leading coefficient, * 1 is 8. So you have 8 written under 15. Add 15 + (8), which is 7, and I'm going to write that next to my leading coefficient of 1. Multiply 8 * 7, which is 56. So we have 56 written under 58. We're going to add down, 58 + (56), which is 2, and I'm going to write the 2 next to the 7. Multiply 8 * 2, which is 16. So we have 16 written under the (16) is 40, and I write that next to the 2. Finally, we have 8 * 40, which is 320. I'm going to take , which is 0. It turns out our remainder is going to be 0. The degree of the final polynomial answer is one less than the dividend polynomial. Since (x^ x^ x^224x  320) is degree 4, our answer is going to be degree 3. What is this going to look like? So we have (1x^3 + 7x^2 + 2x  40) with a remainder of 0. Lesson Summary Things to remember: Make sure your problem is in descending order. Create the division by writing only the coefficients. Remember, the number that goes outside of the division symbol is the opposite of the original. Bring the leading coefficient down. Multiply the coefficient by the number outside the division symbol. Add that number to the next coefficient until you have no more coefficients to multiply. The degree of the final polynomial answer is always one less than the dividend polynomial.
262 And don't forget to write the remainder as a fraction! Dividing Polynomials with Long and Synthetic Division: Practice Problems Chapter 10 / Lesson 9 Let's look at some more polynomial division problems. We will use long division and synthetic division, but this time we will have a couple of more involved problems. So, get out some paper and a pencil and let's begin! Polynomial Division We have looked at some basic polynomial long division and synthetic division. Now let's have a look at a couple more complicated problems. They work exactly the same, except we have larger divisors or parts of the dividend are missing. Example #1 Here's our first: (x^39) (x^2 + 3). When missing a term, use 0 as the coefficient. Did you notice that we don't have an x^2 or x term in the dividend? We need to put something in as a place marker. Whenever we have a missing term, we will put in 0 as their coefficients. For example, in this problem we will have (x^3 + 0x^2 + 0x  9). Why zeros? Because anything multiplied by zero is zero, so we aren't adding anything to the problem!
263 To fill in our long division, we now have (x^3 + 0x^2 + 0x  9) as the dividend, so it goes under the long division symbol. The divisor is (x^2 + 3). Do you notice it's also missing a term? Just like the dividend, we need to put in a place marker. So, (x^2 + 0x + 3) goes to the outside, in the front of the long division symbol. Now we're ready to start the division. The steps are the same every time. It's easier to show and talk about the steps than just listing them. To find the first term of the quotient, we take the first terms from the divisor and dividend and then divide them: x^3 x^2. I like to write them as a fraction. It's easier to divide or reduce that way: x^3/x^2 = x. x is the first term of the quotient, so we'll write it above the long division symbol. To figure out what we will subtract from the dividend, we multiply x times the divisor, (x^2 + 0x + 3): x(x^2 + 0x + 3) = x^3 + 0x^2 + 3x. x^3 + 0x^2 + 3x is written under the dividend, matching like terms. Here's where I like to make life easy. To subtract the polynomials, I just change the signs of its terms and add: x^3 + x^3 = 0 0x^2 + 0x^2 = 0 0x + 3x = 3x Just like we did in long division, we bring down the next term, which is 9. Now we have a predicament; x^2 doesn't go into 3x  9! This turns out to be our remainder. What's our answer then? x + ((3x  9)/(x^2 + 3)). There isn't a need to put 0x as part of the denominator in the remainder. Divide the first terms of the divisor and dividend to find the first term of the quotient
264 Example #2 (2x^ x) (x + 3) This one is almost ready for synthetic division. The divisor is a firstdegree binomial with a leading coefficient of 1. Place the coefficients of the dividend under the symbol, just like in long division, but do not write the variables! Make sure to leave space in between each number. You don't want to get them confused. Wait! Did you write it as you see it? Oh no! The dividend and divisor must be in descending order. What does that mean? We need to list the terms from the largest exponent to the smallest. Even if we're missing one, we need to put it in as a place marker. So now we have (2x^3 + 0x^214x + 10) (x + 3). Next, look at (x + 3). We write 3 to the left of the long division symbol. Remember, you always take the opposite of what you see in the divisor. Now we're ready to start synthetic division. Our first step is to bring down the leading coefficient, 2. Multiply 3 times 2 and place the result underneath the next coefficient, 0. So we have 6 written underneath the 0. Add and write that next to the leading coefficient, 2. Multiply 3 times 6 and place the result under the next coefficient, 14. So we have 18 written under the 14. Add and write the sum of 4 next to the 6. Multiply 3 times 4, and place the result underneath the next coefficient, 10. So we have 12 written underneath the 10. Add and write the sum of 2 next to the 4. Here is our final answer with the remainder. How do I know there's a remainder? Because our last term is not a 0. But what is the answer going to look like? In synthetic division, the degree of the final polynomial answer is one less than the dividend polynomial. Since 2x^ x is degree 3, our answer will be degree 2.
265 When the last term is not a 0, you have a remainder Starting from the left, we will have 2x^26x + 4 with a remainder of 2. The remainder will be written the same as if we had done this problem as long division  a fraction. So our answer will look like 2x^26x (2/(x + 3)). Example #3 Here's our next example: (6y^3 + 7y^25y + 5) (3y  1). To fill in our long division, (6y^3 + 7y^25y + 5) is the dividend, so it goes under the long division symbol. (3y  1) is the divisor, so it goes to the outside, in the front of the long division symbol. Now we're ready to start the division. To find the first term of the quotient, we take the first terms from the divisor and dividend and divide them: 6y^3 / 3y. I like to write them as a fraction. It's easier to divide: 6y^3 / 3y = 2y^2. 2y^2 is the first term of the quotient. We write it above the long division symbol. To figure out what we will subtract from the dividend, we multiply 2y^2 times the divisor, 3y y^32y^2 is written under the dividend, matching like terms. To subtract the polynomials, I just change the sign of the terms and add. 6y^36y^3 = 0 7y^2 + 2y^2 = 9y^2 Just like we did in long division, we bring down the next term, which is 5y. Guess what? We do the same exact step. To find the next term of the quotient, we take the new first terms and divide them: 9y^2 / 3y. I like to write them as a fraction: 9y^2 / 3y = 3y. 3y is the next term of the quotient. Remember, that gets written next to the 2y^2 above the long division symbol. To figure out what we will subtract from the dividend, we multiply 3y times the divisor, 3y y(3y  1) = 9y^23y. 9y^23y is written under the dividend, matching like terms. To subtract the polynomials, I just change the signs of the terms and add: 9y^29y^2 = 05y + 3y = 2y Just like we did in long division, we bring down the next term, which is 5. You'll notice that the dividend is now 2y + 5. When we divide to find the next term of the
266 quotient, we take the new first terms and divide them. 2y/3y. Do you see we end up with a fraction, 2/3? Well, we don't like fractions in the quotient except as a remainder, so 2y + 5 turns out to be our remainder. This is what the final answer will look like: 2y^2 + 3y + ((2y + 5)/(3y  1)) Lesson Summary Before we end this video, let's review when to use long division and when to use synthetic division. My students really like synthetic division, but it isn't right for every division problem. How do I know the difference? Any division problem that has a divisor as a onedegree binomial is perfect for synthetic division. That's why we used it on (2x^ x) (x + 3). x + 3 is a onedegree binomial with a leading coefficient of 1. Okay. It is true we can use synthetic division with a divisor that does not have a leading coefficient of 1, but that'll give us a fraction. Fractions are messy, and students tend to make more mistakes when using them, so keep to a onedegree binomial with a leading coefficient of 1. Any other division problem needs to be done using long division. That's why we used it on (6y^3 + 7y^25y + 5) (3y  1) and (x^39) (x^2 + 3). One final rule to keep in mind as you do polynomial division: always make sure your polynomials are in descending order, largest exponent to smallest. If you're missing a term, put a zero to hold that place, like we did in example #1. How to Solve a Quadratic Equation by Factoring Chapter 10 / Lesson 10 If your favorite video game, 'Furious Fowls,' gave you the quadratic equation for each shot you made, would you be able to solve the equation to make sure every one hit its target? If not, you will after watching this video! Solve a Quadratic Equation with Factoring Welcome to level one of 'Furious Fowls,' the game that puts you in control of the birds that are trying to get their eggs back from those pesky pigs. Can you launch your
267 bird at the exact right angle to make sure your bird gets a direct hit? We're about to find out. The only difference between 'Furious Fowls' and other games out there that are similar is that you only get one shot to make your hit. So it better be worth it! But, because you only get one shot, we will give you a hint to help you out and make sure your first shot is always a direct hit. What kind of a hint am I talking about? Let's check out level one. So here we go; our bird's all set in our sling shot ready to go, and our pig is waiting very smugly over there, thinking it's safe. But, we want to make sure we get him. I can try to eyeball it by swinging my bird up and down here. Notice that as I swing it in different directions, the mathematical equation in the corner of the screen is changing. It's going to be your job to use that mathematical equation to make sure that your one shot hits its target. Factoring changes the equation from standard to intercept form. Example #1 Let's go ahead and try to line up a shot. Maybe like this; that looks about right. The game has told us that if we fire at this angle, our bird will fly in a path that follows the quadratic equation y = x^2 + 6x + 16 where y is the height of the bird and x is the distance it has traveled. Since we're trying to hit a pig that's sitting on the ground, we're curious when our bird will hit the ground, which is where y = 0. Substituting this value in gives us this equation. This equation is asking us the question, how far will our bird go before it hits the ground? We normally solve equations by using inverse operations to get the variable by itself, but in this case, we'd run into some problems with that strategy. Because there are xs but also x^2s in this equation, there isn't just one variable we can get by itself. This means we need a new strategy for solving quadratic equations. While there are a few
268 different ways to do this, the most basic is through factoring. While this way won't always work, it is probably the easiest way to do it, so knowing how to is really helpful. Factoring the quadratic will change it from standard form into intercept form, which will tell us exactly where the parabola crosses the xaxis (where its x intercepts are), or in this problem, where the bird hits the ground. To factor a quadratic, it's really nice not having a negative x^2 term, so let's divide that part out first. Pulling a negative from each term leaves us with this equation instead. Now we can begin to look for the two numbers that have a product of 16 and a sum of 6. The reason we're doing that is all about factoring. So if you're struggling with factoring, you should probably go back and check out a previous lesson to get the basics. Finding a pair of numbers that satisfy these two equations will be easiest if we write out the different factors of 16 and look for the pair of them that can add or subtract to 6. Doing that makes it look like 2 and 8 are our winners. We can therefore rewrite our equation as this: (0 = (x  8)(x + 2)). But why did we do that? How has that gotten us any closer to solving this equation for x? I still have two of them; I can't get one of them by themselves. Well, what we now have is a product that equals zero. This allows us to use what is called the zero product property, which says that anytime a product equals zero, at least one of the things we are multiplying must be zero. Another way of saying that is this: the only way to multiply two things and end up with zero is by having one of those two things be zero itself. What this allows us to do is split the equation up and say that either x  8 or x + 2 is equal to zero. Now that the xs are separated, we can again use inverse operations to solve for them. Undoing 8 and +2 tells us that the two places where our bird will hit the ground are at x = 2 or x = 8. Well, in our game here that 2 answer might happen if the slingshot broke, but we're more concerned with the answer of 8. And it looks now, if our bird flies out to 8, it will go right over the pig we're trying to hit, so making this shot would be a waste of a bird.
269 Example #2 Use the area method to find the factored expression in example #2. But if we slightly adjust it, maybe aiming up a little bit so that our bird flies higher in the air but doesn't go as far, we are told it will now fly the path of this parabola y = 2x^2 + 11x Let's see if this does what we want it to. We'll again substitute in y = 0 in order to find out when our bird will hit the ground, and then factor the quadratic to separate the xs and allow us to use the zero product property. We again begin by taking out the negative in order to help us factor, leaving us with this. We then need to find a pair of numbers whose product is 42 (the first term times the last term) and again, whose sum adds up to 11. Writing out the factors of 42 and comparing the pairs of numbers makes it appear that 3 and 14 are our winners. But since we have a 2 in front of our x squared term, we'll need to use the area method to figure out exactly how this all factors. Putting the two values we just found in one diagonal of our box as coefficients on xs, with the 2x^2 and the 21 in the other diagonal, allows us to work our way to the outside and find our factored expression. We do this by finding the greatest common factor in each row and column. In the far left column, both terms have a 2 and an x; in the right column both terms have a 3. In the top row, both terms only share an x; and in the bottom row, both terms only share a 7. This makes the factored form of (2x^211x  21) the same as 2x + 3 times x  7. Again, for more help on factoring, check out the previous video. Now that we've factored the quadratic, we can use the zero product property to split up the xs and say that either 2x + 3 = 0 or x  7 = 0. A few quick inverse operations later tells us that our bird will hit the ground at either 7 or 3/2. Our 3/2 answer doesn't concern us too much, but it looks like having our bird land at 7 is exactly what
270 we want. We can fire this bird feeling confident that we'll hit it on our first try and pass level one. Congratulations! You've just passed level one of 'Furious Fowls!' You can now move onto level two, where things are going to get a little bit harder. Lesson Summary Before you do that, though, let's quickly review what you've learned from level one. If you are trying to find when an object launched into the air will hit the ground, you must set the quadratic equation equal to zero and solve it. Solving quadratic equations with only inverse operations won't work because there are xs and x^2's. What we need a way to break them up first. One way to do this is with factoring and then using the zero product property. This allows you to set up two separate equations that can be solved with inverse operations to get our two separate answers. How to Solve Quadratics That Are Not in Standard Form Chapter 10 / Lesson 11 It isn't always the case that your equation will be set up nicely for you to solve. In this lesson, learn how to factor or use the quadratic formula to solve quadratic equations, even when they are not in standard form. Solving NonStandard Quadratic Equations You've moved on to level three of Furious Fowls, the game that gives you a mathematical hint but only one chance for you to get your bird to land on the target pig that has your precious eggs. You've gotten here by solving the quadratic hint to find the roots of the equation and predict the place where your bird will land. But here in the third and final level things are going to get pretty tricky. They're going to stop going easy on you by giving you the quadratic all nice and organized, and now the hint will be messy and confusing, but your goal will still be the same: solve the quadratic in order to find the roots and predict where the bird will hit the ground. You ready? Here we go!
271 Let's line up our shot and get it as close as we think we can... eh, that looks good. What's the hint that it gives us? 2x (x  1) x = 0 Yikes. Well, it's kind of nice that it equals zero, I guess, but other than that it's pretty much a mess. Standard Quadratic Form Both ways we know how to solve quadratics require that the equation is in standard form (y = ax^2 + bx + c), and this one definitely is not. But don't let that phase you! Don't panic and think you need to do something crazy. All we need to do is use our algebra skills to move things around and put this equation back into standard form. The area method can be used to factor in the first example Let's start by distributing the 2x to the (x  1) on the inside of the parentheses in order to change this expression only to addition. Doing that gives us this: 2x^2 + 2x x = 0, and now it's only a matter of combining like terms to end up with our standard form quadratic, 2x^2 + 3x + 5 = 0. We can now start thinking about solving this, and these numbers aren't too bad, so let's try factoring. This means first factoring out the negative from everything in the trinomial to give us this: (2x^23x  5) = 0, and then finding a pair of numbers that have a product of 10 and a sum of 3. Writing out the factors of 10 and looking for the pair that fit our criteria makes it look like 5 and +2 are our winners. That means we can use the area method to factor this out. Substituting those values into the four quadrants and then taking the greatest common factor out of each column and row gives us the factored form of the equation as (2x  5)(x + 1) = 0.
272 The Zero Product Property That changes the equation in our problem to this: 0 = 2x  5 or 0 = x + 1, which now allows us to use the zero product property. The zero product property says that any time we multiply two things together and get zero, one of the things we started with must have been zero. That means that in this case either 2x  5 = 0 or x + 1 = 0. Now it's only a few quick inverse operations to find that our two answers are 5/2 or 1. x = 1 is a valid solution but doesn't make much sense for our game here. What we want to look at is our 5/2 answer, and it looks like we're a little bit off. We need to get a little more distance. Let's try leveling off our shot and see if that will give us the result you want. Just a little bit lower... okay. But remember, this is the final level, so our new equation probably isn't going to look any nicer than the last one, and sure enough, it's kind of crazy: 2x (3x  5) = x + 30 The quadratic formula This one isn't even equal to zero, but that's okay! We still need to do the same thing as earlier  use algebra to change this quadratic back to standard form. First multiplying with the distributive property, then using inverse operations to move everything to the left side of the equation, and finally combining like terms gives us our standard form quadratic equation 6x^211x  30 = 0. The Quadratic Formula Let's go ahead and use the quadratic formula for this one because the numbers are so much bigger than the last one. That means we have to first correctly identify a, b, and c as the coefficients on our trinomial. A, the first coefficient, is 6, the next one, b, is 11, and the constant on the end, c, is 30. We now need to substitute these values in
273 to the formula, and now we need to follow the order of operations very carefully to come up with our two answers. Working through this one step at a time, the will turn into a +11, the inside of the square root (what's called the discriminant) has an exponent that we can do first. 11 * 11 is 121. Then 4 * 6 * 30 is Continuing the order of operations, doing the multiplication on the bottom (2 * 6) would give me 12. Doing a minus negative turns into plus a positive, which means we end up with 11 +/ the square root of 841/12, and it turns out the square root of 841 is 29, which means I can split the two answers up into 11 + (29/12) or 11  (29/12). Doing that and then simplifying our fractions gives us our two answers, 10/3 or 3/2. The final problem after substituting values into the quadratic formula We can again more or less ignore our negative root here simply as what would happen if the slingshot broke and focus on the positive root at 10/3, which looks like it's going to work. I feel pretty confident going for this shot. Let's do it and yes, we've got it! You have now beaten Furious Fowls. You are a solving quadratics expert and ready to solve any type of quadratic equation out there! Lesson Summary Let's review what Furious Fowls has taught us. We can solve quadratic equations two ways: factoring and using the zero product property or using the quadratic formula. Both of these ways require the quadratic expression to be in standard form, so if the expression isn't initially in standard form, use algebra to manipulate it into being in standard form first before you begin solving the equation.
274 How to Complete the Square Chapter 10 / Lesson 12 Completing the square can help you learn where the maximum or minimum of a parabola is. If you're running a business and trying to make some money, it might be a good idea to know how to do this! Find out what I'm talking about here. Mathematics, Economics and Yogurt I decided pretty early on in my freshman year of college to become a math major, but a little ways in I also discovered that I was really into economics as well. I thought about getting an econ minor but was told that the minor was only two classes less than the major. Well, if I was going to take the time to go through 9 econ classes, why not 11? So I went for it  double major, here I come! And I'm really glad I did, too. It ended up being a breeze because there was enough math in my economics classes to make it hard for everybody else but pretty easy for me. Anyways, I bring this up because I wanted to share with you a math problem that many of my econ classmates might have gotten stuck on. Here's the scenario: I've decided to hop on the bandwagon and open up a new frozen yogurt chain  you know, the kind where they just let you put whatever you want on it, and then you weigh it at the end? Well, I know that people are going to want to be able to choose from a lot of different toppings, so having more is probably better. But how many should I have? If I went overboard and bought like 200 different types of toppings, some of them would probably never get eaten and I'd end up losing money. So this is what economists would do. Using data and theory, they would set up a mathematical model that predicts how much money they'd make with different numbers of toppings. Let's say that, based on how previous yogurt shops had done and how the economy is doing right now, they came up with this: y = x^2 + 36x  224, where y is the amount of profit I'll make (in thousands of dollars) and x is the number of toppings I'm going to offer. Because the biggest exponent on the x is a 2, what we've got here is a quadratic equation, which means our graph will be a parabola. This quadratic is in standard form (y = ax^2 + bx + c), which means we can make a few observations. The c value will tell us our yintercept, which means if I were to offer no toppings, or substitute in x = 0, I would end up losing quite a lot of money. Next, the negative a value tells me that this will be a concave down parabola, so I'm guessing that our
275 graph will look something like this. So, this makes sense to me  as I offer more and more toppings, my profit will slowly go up. But eventually I'll reach a point where I'm offering too many toppings and they're not getting eaten, so I start to lose money. Graph of the concave down parabola The question is, where is the vertex of this parabola? Well, too bad this function wasn't given to us in vertex form instead of standard form. If it was, I'd know exactly what the profitmaximizing number of toppings was. So how can we change it into the form that we want? Well, that's exactly what completing the square is: taking a standard form equation like this one (y = ax^2 + bx + c) and changing it into vertex form, like this: y = a(x  h)^2 + k. The main part of the problem comes down to this little guy: (x  h)^2. We call this a perfect square binomial and need to remember that it's not equal to x^2  h^2, but instead equal to (x  h)(x  h). So if we're trying to change our standard form equation into vertex form, what we need to do is factor the standard form equation into something like this, a perfect square binomial. As is often the case in math, it's all going to come down to pattern. Finding and Completing the Square Let's step away from our yogurt problem real quick to a slightly different one with a little bit nicer numbers to make finding the pattern easier  say, (x + 5)^2. Well, that's just (x + 5)(x + 5), which we can FOIL out to be x^2 + 10x So what relationship do the numbers in the standard form expression have with each other? Well, it might be a little bit tricky to see, but it turns out that our b value, 10, divided by 2 (which is 5) and then squared turns into our c value, 25. Since we know that this trinomial can be factored into a perfect square binomial, (x + 5)^2, it will turn out that any trinomial where the middle value divided by 2 and then squared is our final value, the c value, will be able to be factored as a perfect square binomial. If we
276 want to say that same thing in math, we're looking for a trinomial with a c value that is equal to (b/2)^2. If that's the case, it can be factored as a perfect square binomial. So, in order to do this, we need our c value to be a specific number. But when we've got an equation that we can, say, add 20 to both sides of, we can make that c value whatever we want. This allows us to force the c value into being the exact number we need it to be in order to become a perfect square binomial. This process of changing the c value into the exact right number that makes it fit the pattern and become a perfect square binomial is what mathematicians call completing the square. Completing the square involves turning a quadratic equation in standard form into one in vertex form Yogurt Profit Let's see how we would do this with our yogurt problem in order to determine the right number of toppings to offer. We were here  y = x^2 + 36x but now we know that our c value needs to be something specific in order for it to work as a perfect square binomial. What it needs to be is the middle number, 36, divided by 2 (18) and squared (324). Right now our c value is 224, so it's not quite the right number. Our trinomial is not ready to be factored as a perfect square binomial... yet. What we can do is use inverse operations on both sides of the equal sign to turn it into that number that we need, 324. Now, this is actually kind of a strange example and because we have this negative sign, a negative a value, we actually need our c value to be So if I want the c value to be 324, and right now it's 224, I simply need to make it 100 smaller. That means we can subtract 100 from both sides of the equation, giving us this: y = x^2 + 36x Now that our c value is the exact right value that it needs to be, we can factor it into a perfect square binomial. First let's take this negative sign out of everything, dividing that out, putting it in front and then changing the sign on everything else. And now we need to factor the
277 trinomial on the inside of the parentheses, (x^236x + 324). Finding two numbers that multiply to 324 and add up to 36 might be a little hard to find, but eventually we'd come to 18 and 18. The fact that they are both the same number and also equal to one half of 36 means that we're on the right track! Once we've found those two numbers, we can factor the expression and rewrite it as this: y = (x  18)(x  18), and then because they're both the same, we can combine them into one and change it into this: y = (x  18)^2. We have now completed the square! We took a trinomial and turned it into a perfect square binomial. One last step to change it into vertex form means adding the 100 back over to the other side of the equation, and we now have our vertex form, y = (x  18)^ Remembering that the h and k values of vertex form tell us where the vertex is, the vertex of our parabola was at (18, 100). That means that 18 is the optimal number of toppings to sell, and I'll make $100,000 if I do that. Hmm, maybe I should rethink being a teacher... The h and k values, 18 and 100, represent the vertex Completing the square can be really hard to see at first, and it's probably one of the most complicated things you'll need to learn in an algebra class. Now that you know the general idea, it's probably a good bet to check out the later lesson where we'll focus on the process and do some practice problems. Lesson Summary Completing the square turns a quadratic equation in standard form (y = ax^2 + bx + c) into one in vertex form (y = a(x  h)^2 + k). A trinomial can be factored into a perfect square binomial as long as the c value is equal to (b/2)^2, which means that if your trinomial's c value isn't naturally equal to what you want it to be, you can complete the square by performing operations on both sides of the equation to force the c value into being exactly what you want it to.
278 Completing the Square Practice Problems Chapter 10 / Lesson 13 Completing the square is one of the most confusing things you'll be asked to do in an Algebra class. Once you get the general idea, it's best to get in there and actually do a few practice problems to make sure you understand the process. Do that here! Completing the Square Completing the square is one of the most difficult things you'll be asked to do in an Algebra class. Not only does it require you to learn a bunch of new skills but you'll also need to remember a lot of older ones as well. Before we dive right into some practice problems, let's quickly review the basics. Completing the square turns a quadratic equation in standard form into one in vertex form. A trinomial in standard form (y = ax^2 + bx + c) can be factored into a perfect square binomial as long as the cvalue on the end is equal to (b/2)^2. If your trinomial's cvalue isn't naturally equal to this exact right amount, you can 'complete the square' by performing operations on both sides of the equation to force the cvalue into being exactly what you want it to be. Completing the square turns a quadratic equation in standard form into one in vertex form. Example #1 So what does this look like in practice? Let's try changing this standard form equation (y = x^220x  25) into vertex form by completing the square. That requires us to turn this trinomial into a perfect square binomial. For this to work, the cvalue of the trinomial needs to be the exact right number: (b/2)^2. Let's check to see if this is already the right value. In this case, our bvalue is 20, so 20/2 = 10 and (10)^2 =
279 100, so we're not in good shape yet. Right now our cvalue is 25 but we need it to be 100. Luckily this is an equation, which means we can add or subtract something from both sides of the equals sign in order to change the cvalue into exactly what we want it to be. In this case, changing 25 into 100 means that we'll have to add 125 to both sides of the equation, leaving it like this: y = x^220x At this point, as long as we haven't made a mistake, we can now factor the trinomial. Factoring requires us to look for two numbers that add up to the middle term, 20, and multiply to the constant on the end, 100. Quickly looking through the factors of 100, and then finding a pair of them that adds up to 20. So, 10 and 10 are our two winners, which means I can factor the trinomial on the right as (x  10)(x 10). Which means the new equation will look like this: y = (x  10)(x 10). Because the x  10s are the same, I can combine them into one and write it as a square binomial, (x  10)^2, leaving our equation like this: y = (x  10)^2. We've now completed the square! All that's left to do is move the 125 over to the other side in order for the quadratic to be in vertex form. That makes y = (x  10)^2125 the same equation as before, just expressed differently, and it's our final answer. The final answer is the same equation as before, expressed differently. As a note: Completing the square is actually often used to solve quadratic equations as well. If instead, this problem had not said complete the square, but solve, and instead of y we had a zero, we would have gotten an equation that looks like this: 0 = (x  10)^ Now we can simply use inverse operations to solve for x by undoing the 125, undoing the square with a square root and undoing 1 with a +1 to get our answer, like this: 1 +/ the square root of 125 = x. This is often what you'll use completing the square for  a way to solve a quadratic equation that's not factoring or the quadratic formula. Although that may have seemed like a pretty long process, it was actually a straightforward example. Although the other problem we'll do during this lesson will
280 basically be the same thing, it will have a few complexities that make it a little bit trickier. Example #2 First off, the directions are going to be a little bit different: If y = 6x^2 + 30x + 18 is expressed in the form y = a(x  h)^2 + k, what is the value of k? Different words, but same idea. Now though, once we've changed it to vertex form, the answer is simply the kvalue and not the entire equation. But that's fine, same thing. What is the value of K? Alright, so let's begin. Right away there is something else that is different though. Our avalue isn't 1. Hmm, okay. Well, looking at the vertex form here, I see that the avalue is on the outside of the parentheses. That means we'll have to factor it out of the trinomial before we can turn it into a perfect square binomial. Thankfully, each term in this trinomial is divisible by three, so it won't be too bad, but sometimes this actually won't be the case. So, I need to divide out a 6 from each term on the righthand side. Doing that leaves us with this: y = 6(x^2 + 5x + 3). Now we're ready to check to see if our cvalue follows the pattern. It's going to need to be (b/2)^2, which in this case will be (5/2)^2. Uh oh, another hurdle. We're going to have a fraction or a decimal this time. Last time, our bvalue was an even number, so dividing it by 2 wasn't a big deal. But this time it's odd, making things a little bit trickier but still doable. (5/2)^2, or 2.5^2, gives us 25/4 (if we want to use fractions) or 6.25 (if we want to use decimals). So, now we know what our cvalue needs to be. We look at our problem as it is, and we realize it is not quite there yet (3 is not the same as 6.25).
281 Turn the current c value into Turning our current cvalue, 3, into 6.25 will require us to add 3.25 to it, but we're going to have to be careful. When we add 3.25 to the trinomial on the inside the parentheses, we're actually adding 6 * 3.25 to that whole side of the equation. Since whatever we do to one side of the equation we have to do to the other, we need to actually add 19.5 to the left, not just Doing that leaves us with this: y = 6(x^2 + 5x ). Alright, we're making progress! Our trinomial is finally ready to be factored, and this time let's use the shortcut. As long as we've done everything right, x^2 + 5x should be factored as (x + 2.5)(x + 2.5) because that 2.5 is half of our bvalue, 5. Then compressing the factored form as a square binomial, (x + 2.5)^2, and then undoing the 19.5 to write the equation in vertex form, like so, y = 6(x+2.5)^219.5, allows us to answer the question 'What is k?' Well, k is the constant on the end, so k is This last example is close to as tough as it gets for completing the square. Hopefully, you feel confident enough to try the examples in the quiz following this video all by yourself. Lesson Summary To review, to complete the square, we need to change the cvalue in our trinomial to be (b/2)^2. If you have an avalue that is not equal to 1, you'll need to first factor that value out of each term in the trinomial before you can proceed. This also means that you'll have to be careful to multiply by this factor when deciding how much to add to both sides when making your cvalue the correct amount. If you end up with an odd bvalue in your trinomial, you'll end up with fractions or decimals in your answer, but that's okay! How to Use the Quadratic Formula to Solve a Quadratic Equation
282 Chapter 10 / Lesson 14 When solving a quadratic equation by factoring doesn't work, the quadratic formula is here to save the day. Learn what it is and how to use it in this lesson. Solving with the Quadratic Formula The quadratic formula You just passed level one of Furious Fowls, the game that asks you to launch birds across the screen to get back at those pesky pigs that stole your eggs. You learned that we could figure out where our bird would hit the ground by solving the given quadratic equation where it equals 0. That required us to factor the equation and then use the zero product property to determine which two values would give us 0. But now we're on to level two! It's going to get harder, but your goal will still be the same: use the equation given to you on the screen to make sure that your bird will hit its target on the first try. Alright, so here we go; let's go ahead and make a guess at where we'll need to launch our bird this time. I think this looks about right. Now, let's check our equation to see if this will actually work out or not. The game tells us that our bird will fly in a parabola given by the equation y = x^2 + 4x + 7. Since we're again shooting for a pig on the ground, we're curious where this equation equals 0, so let's go ahead and substitute that in. Okay, now we need to try to factor in order to find x. Let's divide out the negative, leaving us with this: 0 = (x^24x  7). Now look for a pair of numbers that have a product of 7 and a sum of 4. Hmmm, this isn't looking good. I don't think this trinomial can be factored. So, what now? In level one this worked every time, so we never had to do anything else! Looks like we're going to have to learn some new skills.
283 The Quadratic Formula Situations just like this, where we're trying to solve a quadratic that is unfactorable, is what the quadratic formula was made for! It's a little long and messy, but if you can remember it, you'll be able to solve any quadratic equation out there! The resulting problem for example #1 after dealing with exponents and multiplication The Standard Form You'll have to first remember the standard form of a quadratic equation, y = ax^2 + bx + c. This will tell you what your a, b and c values are, but then it simply becomes a matter of putting the numbers into the right spots. Let's see if we can successfully do this to pass the second level of Furious Fowls. Identifying A, B and C Going back to the original equation and identifying a, b and c should be our first step. A is in front of the x^2, where there's only a negative symbol. That means that a is just 1. B is the number in front of the xs, which makes it 4, and c is the constant on the end, 7. Using the Quadratic Formula Now, we can substitute these values into the quadratic formula and evaluate the resulting expression to find our answers! While this is theoretically pretty straightforward, there are quite a few spots that can easily trip you up. Because this is our first time, let's evaluate this one step by step together and go over some of the
284 common mistakes. We're going to have to remember our order of operations and make sure we also keep track of all our negative signs. For this problem, then, the first thing we'll have to take care of is the exponent we see. The exponent I see is a 2, which means a square, which means multiply that number by itself, so we do 4 * 4 to get 16. Next in the order of operations will come multiplication. Doing the multiplication on the inside of the square root (4 * 1 * 7) gives us 28, then doing the multiplication on the bottom of the fraction (2 * 1) is simply 2. We're now left with x = (4 +/ sq. root ) / 2, and we're going to focus on the inside of the square root in the top of the fraction. The inside of the square root in the formula is b^24ac. This part is one of the hardest parts to evaluate and one of the easiest parts to make a mistake on. It also has a special name; it's called the discriminant. It has some important features that we'll talk about in other lessons, but for this lesson, you just need to know to be careful when evaluating it, specifically for the reason I'm about to mention. Notice that we ended up with a 28 after we did the multiplication, but I still have Minus a negative is plus a positive, so gives us 44. The most mathematically correct answer for example #1 Now that we have the square root of 44, we'd like to check to see if we can take the square root of 44 nicely. Is there a number that, multiplied by itself, gives us 44? Well, 7 * 7 is 49, which is pretty close, but not quite, and 6*6 is 36, so again, not quite. So, the square root of 44 is an irrational number; it's going to be a really long decimal. That means that what we have is actually a pretty good answer, but there are two mistakes that students tend to make without realizing that this is an okay answer. One of those mistakes is to divide the 4 by the 2 to get 2 +/ the square root of 44. This is not allowed because the + root 44 kind of gets in the way of you doing that division. Another mistake is to divide the 4 and the 44 by the 2 to give us (4/2) +/ (square root of 44/2). The problem with this is that dividing inside a square root like that isn't really allowed. That means that what we have here (x = 4 +/  the square root of 44/2)
285 is actually the most mathematically correct answer. But since we're estimating, let's change it to a decimal by rounding the answers off. The square root of 44 is in between 6 and 7; it's actually about That means we'll find our two answers by doing /2 and also /2. Doing that gives us our two estimated answers as or Our answer is what would happen if the slingshot broke, so we're not too worried about that one. But over at it looks like we're pretty close to our target but maybe a little short. Maybe if we leveled out our shot a little bit our bird might fly a little further. Maybe like this? Okay, apparently now our equation is y = 3x^2 + 15x + 29, but again we want to know when the bird will hit the ground, or what the roots of the equation are. That means we substitute in y = 0 and begin thinking about which way we want to solve. We could try the factoring method here, but those numbers seem pretty messy, and there's also a good chance it won't work anyways, so let's go straight into the quadratic formula. This means identifying a, b and c from our standard form quadratic. That makes a equal to 3, b equal to 15 and c equal to 29. Now it's just a matter of carefully substituting these values into the quadratic formula and then following the order of operations to come up with our two answers. Substituting the numbers in will look like this: x= 15 +/ (the square root of 15^24 (3) (29)) /2 (3). Using a Negative B Value Resulting example #2 problem after substituting values into formula One other common mistake I see students make is to mess up the b value in the front left of the fraction. In the formula, it's a b, so in this case 15 turns into 15. But often times your b value itself will be negative. When you have a b and you plug it into the formula,  b will turn it positive again, so you've got to be careful with that b in front.
286 Anyway, moving on to evaluating, we can start with the exponent on the inside  the b^2 part of the discriminant  and 15^2 gives us 225. Next, we do 4 * 3 * 29 to get 348, and we do 2 * 3 to get 6 on the bottom. Again, changing a minus negative into plus a positive will make our discriminant equal to 573. The square root of 573 is about Doing 15 plus this and dividing by 6 will give us , and doing 15 minus this divided by 6 will give us We can again focus on the positive answer even though the negative one is a valid mathematical solution, and it looks like this one's going to work! I'm feeling pretty confident; let's let our furious fowl fly... and heyo! We got it! You've now passed level two of Furious Fowls and are pretty good at solving quadratics, but are you ready for level three? Lesson Summary Let's quickly review what we learned from level two. Not every quadratic equation is factorable, so we need another way to solve them. Enter the quadratic formula. Using a, b and c from the standard form quadratic, we substitute these values into the formula and then use the order of operations to end up with our two answers. Using the Quadratic Formula to Solve Equations with Literal Coefficients Chapter 10 / Lesson 15 Equations with Literal Coefficients I, personally, like to see and work with equations with literal coefficients. Let me tell you why. Equations with literal coefficients are your typical equations written with letters instead of numbers. So, instead of seeing an equation like x ^ 2 + 4x + 6 = 0, we would see ax ^ 2 + bx + c = 0. A literal coefficient is a letter instead of a number. In our example equation, it is the letters a, b, and c. Notice how these letters take the place of the numbers that we are used to seeing. I actually like to work with these letters because I don't need to go through and do arithmetic calculations. Once I've put in my letters and have simplified my equation
287 as far as I can, I am done. I think this is a lot easier. Keep watching and you can decide for yourself. The Quadratic Formula We are going to work with the quadratic formula,which is defined as x = (  b +/ sqrt ( b ^ 24ac ) ) / 2a. I call formulas like this 'plugandplay' formulas because you can replace your letters with their appropriate number and then evaluate the formula to find your answer. There are no complicated rules you have to remember other than your order of operations. One other thing: the quadratic formula is only for solving quadratic equations. Quadratic equations are those equations made up of a polynomial whose degree must be 2. Quadratic equations generally look like our beginning example of x ^ 2 + 4x + 6 = 0. Notice how the degree, or the highest exponent, of the polynomial is a 2. Now, let's see about working with an equation with literal coefficients instead of numbers. You will see how easy it is. Standard Form Equation The first equation with literal coefficients that I want to solve with you is the standard form equation for quadratic equations. It is ax ^ 2 + bx + c = 0. Instead of numbers we have the letters a, b, and c. What do we know about the same letters in the quadratic formula? They are indeed based off of this standard form equation. Our a is the same a in the quadratic formula, our b is the same b, and our c is the same c. So, for this part, there is not much to do to solve this equation using the quadratic formula. We simply plug in our a for a, our b for b, and our c for c and we are done. Since the quadratic formula is already at its simplest form, we can't simplify it anymore and we are left with the formula as it is. That is our answer for solving the standard form equation using the quadratic formula. x = (  b +/ sqrt ( b ^ 24ac ) ) / 2a Even though the answer looks complicated, it is at its simplest form. I can't reduce it any further so I have nothing to worry about. I am totally done. No arithmetic to do, no adding, subtracting, multiplying, or dividing. I am totally done. See how easy it is?
288 Random Quadratic Equations It gets even easier with random quadratic equations. We only have three choices for our random quadratic equations. They are: ax ^ 2 + bx = 0, ax ^ 2 + c = 0, and ax ^ 2 = 0. I will show you how to solve one of them and you can follow the same pattern to solve the others on your own. The one we are going to solve together is ax ^ 2 + c = 0. What I see here is that I have a and c, but no b. Since I have no b, that means my b equals zero. So, going to the quadratic formula, I will put in a for a, 0 for b, and c for c. x = (  0 +/ sqrt ( 0 ^ 24ac ) ) / 2a I simplify this and I get: x = ( sqrt (  4ac ) ) / 2a At this point, I can't simplify any further. That means I am done and this is my answer. I think this is a lot easier than doing it with numbers because again, I didn't have to do any arithmetic. I just plugged in my letters and reduced as far as I can. To solve the other equations, you would follow the same steps and for each letter that is missing, you would plug in a zero into the quadratic. You would then simplify to get your answer. Lesson Summary In summary, a literal coefficient is a letter instead of a number and an equation with literal coefficients is an equation written with letters instead of numbers. To solve quadratic equations with literal coefficients, you plug in your letters into their respective slots in the quadratic formula, simplify, and then you are done. If your equation is missing a letter, you plug in a zero for it before simplifying.
289 Solving Problems using the Quadratic Formula Chapter 10 / Lesson 16 Quadratic Formula I used to hate working with the quadratic formula until I realized that it actually helped me with my problems. And sometimes, it even made my problem shorter as it gave me what I needed to know early on in the solving process. Right now, the quadratic formula might look like a beast, but think of it as a prince in disguise. It's here to help you. The quadratic formula allows you to solve quadratic equations in the form of ax^2 + bx + c = 0 and is defined as x = (b +/ sqrt(b^24ac)) / 2a. The a, b, and c letters come from the quadratic equation that you are solving. Let's see how this works. Two Solutions The first problem we're going to solve is x^2 + 5x + 6 = 0. How do we figure out our letters? Well, we compare this to the general form of a quadratic: ax^2 + bx + c. A good way to do this is to write the general form above your quadratic so you can see how everything lines up. What we are looking for is the location of our x^2, our x, and our number by itself. We can then figure out what our letters stand for. In our general form, the number attached to the x^2 is the letter a. So looking at my quadratic, I see that there is nothing in front of the x^2, which means there is a 1 there, so that means my a = 1. Good. Next, I look for what is front of the x. I see a 5, so that means my b = 5. The number by itself is a 6, so my c = 6. Now I have all my letters. If, though, you have missing terms, then that means the letter that matches that missing term equals to zero. At this point, the quadratic formula turns into a prince to help you. To solve, all you need to do is plug in all your letters where they belong and evaluate the formula. Let's do this. We plug in our numbers to get x = (5 +/ sqrt(5^24*1*6)) / 2*1. We begin evaluating by first calculating what is inside the square root. x = (5 +/ sqrt(2524)) / 2*1 x = (5 +/ sqrt (1)) / 2*1 Look at that. We have a 1 inside the square root. It is a positive number inside the square so we know we can calculate that down. If we know the square root off the top
290 of our head, we can go ahead and write that number down. Otherwise, you can use a calculator to find the exact square root. In our case, we know that the square root of 1 is 1, so we'll write that down. x = (5 +/ 1) / 2*1 Our formula includes a +/ so we have to split our problem into two problems, one for the plus and one for the minus. So, x = (5 + 1) / 2*1 and x = (51) / 2*1 Now, we can finish solving these two little problems. x = 4 / 2 and x = 6 / 2 So, x = 2 and x = 3 And yes, we have two solutions. Most times, you will end up with two solutions like this. If they aren't whole numbers, then they'll be decimal numbers. The way you can tell that you have two solutions like this is if you get a positive number inside your square root. There are two more cases that will give you different numbers of solutions. Let's see what happens when the square root ends up as a 0. One Repeated Solution This time, our problem is x^2 + 4x + 4 = 0. Comparing this to our general form of ax^2 + bx + c, we see that our a = 1, our b = 4, and our c = 4. Plugging these values into our quadratic formula gives us x = (4 +/ sqrt(4^24*1*4)) / 2*1. Evaluating the square root, we find that it equals zero. x = (4 +/ sqrt(1616)) / 2*1 x = (4 +/ sqrt(0)) / 2*1 So, x = (4 +/ 0) / 2*1 Now, we split up our problem into two and solve. x = (4 + 0) / 2*1 and x = (40) / 2*1
291 x = 4 / 2 and x = 4 / 2 So, x = 2 and x = 2 Hey, we have two solutions, but they're the same answer! Pretty cool, huh? So, if your square root ends up as zero, then you know that you will get the same solution twice. There is one more special case and that is when your square root is negative inside. No Real Solutions When your square root is negative inside, it actually makes your problem a lot shorter. Let's see how. Our problem is x^2 + 2x + 3 = 0. We see that our a = 1, our b = 2, and our c = 3. Plugging these into our quadratic formula, we get x = (2 +/ sqrt(2^24*1*3)) / 2*1. x = (2 +/ sqrt(412)) / 2*1 x = (2 +/ sqrt(8)) / 2*1 Evaluating inside our square root, we see that we have a negative inside. What do we know about square roots? We know that we can't take the square root of a negative number. What does this mean for us? It means that we have to stop here, because there's no such thing as the square root of a negative number in the real world. It means that our problem has no real solutions, and we are done. We put 'no real solutions' as our answer. Lesson Summary In this lesson, we've learned quite a few interesting things about the quadratic formula. We know that the quadratic formula lets us solve problems in the general form of ax^2 + bx + c = 0 and is defined as x = (b +/ sqrt(b^24ac)) / 2a. We've learned that the square root part of the formula tells us how many solutions we will be getting. If it is positive, then we will have two different solutions. If it is zero, then we have two solutions, but they are the same, just repeated. And if it is a negative inside the square root, then we have no real solutions. The quadratic formula is not such a horrible thing to work with especially when it helps us to know if we have the right number of solutions or not.
292 How to Add and Subtract Rational Expressions Chapter 11 / Lesson 1 Adding and subtracting rational expressions brings everything you learned about fractions into the world of algebra. We will mix common denominators with factoring and FOILing. Rational Polynomial Defined The word 'rational' means 'fraction.' So a rational polynomial is a fraction with polynomials in the numerator (top) and/or denominator (bottom). Here's an example of a rational polynomial: (x + 4) / (x^2 + 3x + 2) As we get started, let's remember that to add or subtract fractions, we need a common denominator. Try this mnemonic to help you remember when you need a common denominator and when you don't: Add Subtract Common Denominators; Multiply Divide None Auntie sits counting diamonds; Mother does not. Let's get started! Adding and Subtracting Rational Expressions 1. We need to factor. 2. Find a common denominator. 3. Rewrite each fraction using the common denominator. 4. Put the entire numerator over the common denominator. 5. Simplify the numerator. 6. Factor and cancel if possible.
293 7. Write the final answer in simplified form. There are quite a few steps, but let me show you how they work. Find the common denominator and use it to rewrite the fractions Example #1 Our first expression is (1 / (x  2)) + (3 / (x + 4)). The first step is to factor. Since we don't have anything to factor, let's move to the next step, writing down our denominators, (x2) and (x+4). This will be our common denominator: (x  2)(x + 4). Now we need to create our common denominator. Let's look at our first term, (1 / (x  2)). (x  2) is in the denominator. We need to multiply by (x + 4) to make our common denominator. But if we multiply by (x + 4) on the bottom, we need to multiply by (x + 4) on the top. For right now we are going to write it and not multiply yet. Let's look at our second term: (3 / (x + 4)). The denominator is (x + 4). We need to multiply (x  2) times (x + 4) to get our common denominator. But once again, if we multiply by (x  2) on the bottom, we need to multiply by it on the top too. So far, this is what we have: ((1(x + 4)) / ((x  2)(x + 4))) + ((3(x  2)) / ((x + 4)(x  2))) Don't FOIL the denominator  we may have to cancel as our final answer! Now let's write the entire numerator over our common denominator. (1(x + 4)) + 3(x  2)) / ((x  2)(x + 4)) Let's simplify the numerator.
294 1(x + 4) = x + 4 3(x  2) = 3x  6 (x x  6) / ((x + 4)(x  2)) Collect like terms in the numerator. Putting the numerator over the common denominator prepares the problem for simplifying (4x  2) / ((x + 4)(x  2)) Factor the numerator if possible. 4x  2 = 2 (2x  1) (2(2x  1)) / ((x + 4)(x  2)) There isn't anything to slash or cancel, so we distribute and FOIL for our final answer. (4x  2) / (x^2 + 2x  8) Example #2 ((2x) / (x^216))  (1 / (x + 4)) x^216 factors into (x  4)(x + 4). So let's put that into the expression. ((2x) / ((x  4)(x + 4)))  (1 /(x + 4)) Our next step is to write down all of our denominators. In the first term, we have (x + 4)(x  4), so we write those down. We continue to the next term and look at the denominator. We never duplicate denominators from term to term. Since we already have (x + 4) written as part of our
295 denominator, we don't need to duplicate it. So it turns out our common denominator will be (x + 4)(x  4). Now we need to create our common denominator. Let's look at our first term ((2x) / (x + 4)(x  4)). We already have our common denominator here, so we're going to move to the next term: (1 / (x + 4)). Here, we need to multiply (x  4) to make our common denominator. But if we multiply (x  4) on the bottom, we need to multiply by (x  4) on the top. For right now, we are going to write it and not multiply yet. So we have ((2x) / (x + 4)(x  4))  (1(x  4) / (x + 4)(x  4)). After collecting like terms, the resulting expression can be factored Let's write the numerator all over the denominator. ((2x)1(x4))/((x+ 4)(x  4)) Simplify the numerator (or top) and rewrite it over the denominator. Distribute the 1 into (x  4) = 1x + 4. Collecting like terms, 2x  1x= x. So now our expression looks like: (x + 4) / (x+ 4)(x 4) We can slash, or cancel, (x+ 4) over (x+ 4). This gives us 1/(x  4) as our final answer. Example #3 ((5x^23) / (x^2 + 6x + 8))  4 The first step is to factor.
296 x^2 + 6x + 8 = (x + 4)(x + 2) Our next step is to write down all of our denominators. In our first term, we have (x + 4)(x + 2), so we write it down. The denominator for the next term is 1. Therefore, our common denominator will be (x + 4)(x + 2). Now we need to create our common denominator. Let's look at our first term (5x^23)/((x + 4)(x + 2)). We already have our common denominator here, so we're going to move to the next term, 4. Here, we only have a 1 in the denominator, so we need to multiply by (x + 4)(x + 2) over (x+4)(x+2). This is what our new expression is going to look like: ((5x^23) / (x + 4)(x + 2))  ((4 (x + 4)(x + 2)) / ((x + 4)(x + 2))). The final step after using FOIL to simplify the expression Let's write the whole numerator (top) over the denominator (bottom). ((5x^ (x + 4)(x + 2))) / ((x + 4)(x + 2)) We can now simplify the top, or numerator. (x+4)(x+2) = x^2 +6x +8 Multiply 4( x^2 +6x +8) and we have 4x^224x Let's continue with the numerator and collect like terms, so our expression looks like: (x^224x  35) / ((x + 4)(x + 2))
297 The numerator does not factor without using the quadratic formula, so this is almost our answer, except we need to FOIL the bottom, or denominator. Here is our final answer: (x^224x  35) / (x^2 + 6x + 8) Lesson Summary As we have seen, the process to add or subtract rational expressions is: 1. We need to factor. 2. Find a common denominator. 3. Rewrite each fraction using the common denominator. 4. Put the entire numerator over the common denominator. 5. Simplify the numerator. 6. Factor and cancel if possible. 7. Write the final answer in simplified form. Practice Adding and Subtracting Rational Expressions Chapter 11 / Lesson 2 Adding and subtracting rational expressions can feel daunting, especially when trying to find a common denominator. Let me show you the process I like to use. I think it will make adding and subtracting rational expressions more enjoyable! Introduction Remember back when we added and subtracted fractions? Well, a rational expression is simply a fraction with 'x's and numbers. We follow the same process for adding and subtracting rational expressions with a little twist. Now we may need to factor and FOIL to simplify the expression. The process we will follow is: 1. Factor 2. Find the common denominator 3. Rewrite fractions using the common denominator 4. Put the entire numerator over the common denominator 5. Simplify the numerator
298 6. Factor and cancel, if possible 7. Write the final answer in simplified form As we get started, let's also remember that to add or subtract fractions, we need a common denominator. Try this mnemonic to help you remember when you need a common denominator and when you don't: Add Subtract Common Denominators, Multiply Divide None. Auntie Sits Counting Diamonds, Mother Does Not. Example #1 Let's look at our first example. (x + 4)/(3x  9) + (x 5)/(6x 18) First, we need to factor. (3x  9) = 3 (x 3) and (6x  18) = 6 (x  3) After we replace the factored terms, our new expression looks like: (x + 4)/3 (x  3) + (x  5)/6 (x  3) To find our common denominator, we simply write down our denominators. From the first term we have 3 (x  3) as our denominator. We write that down for our common denominator. When we look at the second expression's denominator, 6 (x  3), we notice that 6 = 3 * 2. So the second expression has 2 * 3 (x 3). We already have 3 (x  3) written, so the only piece not used is 2. We write that down multiplied by 3 (x  3). Our common denominator will be 2 * 3 (x  3) or 6 (x  3). Our next step is to multiply each piece of the expression so we have 6 (x  3) as our new denominator. In our first fraction, we need to multiply by 2 over 2. This will give me 2 (x + 4)/2 * 3(x  3). Looking at the second fraction, I notice I already have 6 (x  3) in the denominator, so I can leave this one alone. Now let's write the entire numerator over our common denominator: 2(x + 4) + (x  5)/6(x  3) Let's simplify the numerator. 2(x + 4) = 2x + 8 2x (x  5)/6(x  3)
299 Collect like terms in the numerator. 3x + 3/6(x  3) Factor the numerator if possible. 3x + 3 = 3 (x + 1) The 3 over 6 reduces to 1 over 2. There isn't anything to slash or cancel, so we distribute in the numerator and denominator for our final answer: x + 1/2x  6 Example #2 (x  2)/(x + 5) + (x^2 + 5x + 6)/(x^2 + 8x + 15) First, we need to factor. x^2 + 5x + 6 = (x + 3)(x + 2) x^2 + 8x + 15 = (x + 5)(x + 3) After we replace the factored terms, our new expressions looks like: (x  2)/(x + 5) + (x + 3)(x + 2)/(x + 5)(x + 3) To find our common denominator, we simply write down our denominators. From the first term, we have (x + 5) as our denominator. In the second term, we have (x + 5) and (x + 3). Since we already have (x + 5) written as part of our common denominator, we will just write (x + 3). So, our common denominator is (x + 5)(x + 3). Our next step is to multiply each piece of the expression, so we have (x + 5)(x + 3) as our new denominator. In the first fraction, we need to multiply by (x + 3) over (x + 3). This will give us (x  2)(x + 3)/(x + 5)(x + 3) as our first fraction. Looking at the second fraction, I notice I already have (x + 5)(x + 3) in the denominator, so I can leave this one alone. Now, let's write the entire numerator over our common denominator. ((x  2)(x + 3) + (x + 3)(x + 2))/(x + 5)(x + 3) Let's simplify the numerator by writing the numerator over our common denominator and FOIL. (x  2)(x + 3) = (x^2 + x  6) and (x + 3)(x + 2) = (x^2 + 5x + 6)
300 Collect like terms in the numerator. 2x^2 + 6x Factor the numerator if possible. 2x(x + 3) Our expression now looks like: 2x(x + 3)/(x + 5)(x + 3) We can slash, or cancel, (x + 3) over (x + 3). This gives us our final answer, 2x/(x + 5). Example #3 (x^2 + 12x + 36)/(x^2  x  6) + (x + 1)/(3  x) First, we need to factor. (x^2 + 12x + 36) = (x + 6)(x + 6) (x^2  x + 6) = (x  3)(x + 2) After we replace the factored terms, our new expressions looks like: (x + 6)(x + 6)/(x  3)(x + 2)) + (x + 1)/(3  x) To find our common denominator, we simply write down our denominators. From the first term, we have (x  3)(x + 2) as our denominator. In the second term, we have (3  x). I could write (3  x) as part of the common denominator, but I know that 1 * (x  3) = (3  x). So, now it will match with the denominator (x  3). Now, our expression looks like: (x + 6)(x + 6)/(x  3)(x + 2)) + (x + 1)/1(x  3 And that 1? It can be put into the numerator. Remember, 1/1 = 1/1 = 1. It doesn't matter where I put the 1 in the fraction as long as I have a +1 to match it. So, our common denominator is (x  3)(x + 2). In the first fraction, I already have the common denominator (x  3)(x + 2), so I leave that one alone. In the second fraction, I need to multiply by (x + 2) over (x + 2). This gives us the common denominator of (x  3)(x + 2). Our expression now looks like:
301 (x + 6)(x + 6)/(x  3)(x + 2) + (1)(x + 1)(x + 2)/(x  3)(x + 2) Let's simplify the numerator by writing the numerator over our common denominator and using FOIL, which is First Outside Inside Last. (x + 6)(x + 6) = x^2 + 12x + 36 and (1)(x + 1)(x + 2) = (1)(x^2 + 3x + 2) = x^23x 2 Collect like terms in the numerator. Our expression now looks like: (9x + 34)/(x  3)(x + 2) The numerator doesn't factor, so our last step is to FOIL the denominator. Our final answer is (9x + 34)/(x^2  x  6). Lesson Summary The process we follow is: 1. Factor 2. Find the common denominator 3. Rewrite fractions using the common denominator 4. Put the entire numerator over the common denominator 5. Simplify the numerator 6. Factor and cancel, if possible 7. Write the final answer in simplified form Simplifying rational expressions may feel like a daunting process right now, but with practice, you will get better. One tip from me to you: If you don't have the right answer the first time, don't erase the entire expression. Start from the beginning of your work, and look for little mistakes. Many of my students have the right idea, just a misplaced sign or a factoring error. How to Multiply and Divide Rational Expressions Chapter 11 / Lesson 3 Multiplying and dividing rational polynomial expressions is exactly like multiplying and dividing fractions. Like fractions, we will reduce. With polynomial expressions
302 we use factoring and canceling. I also give you a little mnemonic to help you remember when you need a common denominator and when you don't. Multiplying and Dividing Rational Expressions Factor in order to simply the polynomials A polynomial rational expression is a fraction containing polynomials. Example of a rational expression: (r  4) (r²  5r + 6) Today we're going to look at multiplication and division of rational polynomial expressions. Remember how we multiplied fractions? We multiplied straight across the fractions. The same rule applies to rational polynomial fractions. It's just like 1/2 * 3/5, where we multiply straight across to get 3/10. Division of fractions requires use to 'flip' the second fraction. You got it! That rule also applies to rational fractions. It's just like with 1/9 2/7, where we flip the second fraction and change it to multiplication, so we have 1/9 * 7/2. Multiply straight across, and we get 7/18. So is it really that easy? Almost. We need to add a couple new steps, but they aren't too bad. Example #1 ((y  1) (y²  7y + 10)) * ((y²  8y + 15) (y²  4y + 3) First, we need to get the polynomials in simplified form. That means factoring. Let's review factoring. Let's look at y²  7y We're looking for multiples of 10 that add to * 5 = 10, and = 7. So y²  7y + 10 factors into (y  2)(y  5).
303 Multiplying straight across provides the answer for example 1 How about y²  8y + 15? Well, 3 * 5 = 15, and = 8, so y²  8y + 15 factors into (y  3)(y  5). And then y²  4y + 3 factors into (y  1)(y  3). So it turns out we have: ((y  1) ((y  5)(y  2))) * (((y  3)(y  5)) ((y  1)(y  3))) The second new step is reducing, or what I like to call slashing! You can only reduce from top to bottom  never reduce from side to side! Once you have 'slashed' all of the like terms from the top and bottom, we multiply straight across. Don't multiply anything we slashed! We can cancel (y  5) over (y  5) and (y  1) over (y  1). Are you asking yourself, why we can cancel? Well, we know that 4 4 = 1; then (y  3) (y  3) also equals 1, so we slash those, too! Once we've 'slashed' all of the like terms from the top and bottom, we multiply straight across. Don't multiply anything we slashed! Why? Because those are now 1s! It turns out that ((y  1) (y²  7y + 10)) * ((y²  8y + 15) (y²  4y + 3)) = 1 (y  2).
304 Example #2 Remember to flip the second fraction in division problems Let's look at ((r  4) (r²  5r + 6)) ((r  3) (r²  6r + 9)). First, we need to get the polynomials in simplified form, and that means factoring. Let's review again factoring. Let's look at r²  5r + 6. We're looking for multiples of 6 that add to * 2 = 6, and = 5. So r²  5r + 6 factors into (r  3)(r  2). How about r²  6r + 9? Well, 3 * 3 = 9 and = 6, so r²  6r + 9 factors into (r  3)(r  3). Our expression with simplified polynomials now looks like: ((r  4) ((r  3)(r  2))) ((r  3) ((r  3)(r  3))) We aren't done yet. To divide fractions, we need to 'flip' the second fraction. So, we have: ((r  4) ((r  3)(r  2))) * (((r  3)(r  3)) (r  3)) The second step is canceling, or what I like to call slashing! Once we have 'slashed' all the like terms from the top and bottom, we multiply straight across, but don't multiply anything we slashed! So we're going to slash (r  3) over (r  3) and (r  3) over (r  3). It turns out that ((r  4) (r²  5r + 6)) ((r  3) (r²  6r + 9)) = (r  4) (r  2). Example #3 Let's look at ((m + 5) (m  9)) ((m² + 10m + 25) (m²  81)).
305 First, we need to get the polynomials in simplified form. That means factoring. Factoring m² + 10m + 25, we get (m + 5)(m + 5). Factoring m²  81, we get (m + 9)(m  9). So our expression now looks like: ((m + 5) (m  9)) (((m + 5)(m + 5)) ((m + 9)(m  9))) Next, flip! We flip the second fraction and change it to a multiplication. So our expression now looks like: ((m + 5) (m  9)) * (((m + 9)(m  9)) ((m + 5)(m + 5))). The solution to example problem 3 Before we have our final answer, we cancel, or 'slash,' like terms. I love this part! We slash (m + 5) over (m + 5) and (m  9) over (m  9). Once we have 'slashed' all of the like terms from the top and the bottom, we multiply straight across. Don't multiply anything we slashed! It turns out that ((m + 5) (m  9)) ((m² + 10m + 25) (m²  81)) = (m + 9) (m + 5). Lesson Summary Multiplication and division of rational expressions is easy once you remember these steps! Multiplication 1. Factor 2. Slash 3. Multiply Division
306 1. Factor 2. Flip 3. Slash 4. Multiply One final note  many of my students think they need to get a common denominator when multiplying or dividing fractions. No, no, no! Never! Only when you add and subtract, so please don't make that same mistake. Try this mnemonic to help you remember: Add Subtract Common Denominators, Multiply Divide None. Auntie Sits Counting Diamonds, Mother Does Not. Multiplying and Dividing Rational Expressions: Practice Problems Chapter 11 / Lesson 4 Let's continue looking at multiplying and dividing rational polynomials. In this lesson, we will look at a couple longer problems, while giving you some practice multiplying and dividing. Review Multiplication and division of rational polynomial expressions is easy once you remember the steps. For multiplication: factor, cancel or slash, and multiply. For division: factor, flip, cancel or slash, and multiply. Let's do some larger problems.
307 Example #1 In example #1, cancel out the like terms to find the solution ((q^211q + 24) / (q^218q + 80)) * ((q^215q + 50) / (q^29q + 20)) First, we need to factor. (q^211q + 24) factors into (q  8)(q  3). (q^218q + 80) factors into (q  10)(q  8). (q^215q + 50) factors into (q  10)(q  5). (q^29q + 20) factors into (q  5)(q  4). So, this is what our new expression is going to look like: ((q  8)(q  3) / (q  10)(q  8)) * ((q  10)(q  5) / (q  5)(q  4)) Next, we are going to cancel (what I like to call slash) like terms. We're going to cancel or slash (q  10) over (q  10), (q  8) over (q  8), and finally (q  5) over (q  5). Now that we have canceled or slashed all of the like terms from the top and bottom, we multiply straight across. Don't multiply anything we slashed because those are now 1's. It turns out, our answer is (q  3) / (q  4). Example #2
308 In example #2, flip the second fraction before changing the problem to a multiplication one ((y^29) / (2y + 1)) / ((3  y) / (2y^2 + 7y + 3)) Let's factor. (y  9) = (y  3)(y + 3) and (2y^2 + 7y + 3) = (2y + 1)(y + 3). Our next step is to flip the second fraction and change it to multiplication. Our new expression is going to look like this: ((y  3)(y + 3) / (2y + 1)) * ((2y + 1)(y + 3) / ((3  y)). The next step is canceling, or what we've been calling slashing. We can slash (2y + 1) over (2y + 1). In the numerator, we have (y  3)(y + 3) and (y + 3). In the denominator we have (3  y). If we multiply (3  y) by 1, we'll get 1(y  3). Guess what? We can cancel (y  3) over (y  3), but remember to leave the 1! So, our final answer's going to look like: (y + 3)(y + 3) / 1. But hold on a second! Let's multiply the top and the bottom by 1. This is going to give us 1(y + 3) (y + 3) / 1. When we FOIL, we're going to end up with an answer of 1(y^2 + 6y + 9) / 1. Well if we distribute the 1, we end up with (y^26y  9)! Example #3 Multiplying by 1 in example #2 removes the negative from the bottom ((x^2 + x  2) / (x^24x  12)) * ((x^29x + 8) / (x^22x + 1) We begin by factoring. (x^2 + x  2) factors into (x + 2)(x  1), (x^24x  12) factors into (x  6)(x + 2), (x^29x + 8) factors into (x  8)(x  1), and x^22x + 1 factors into (x  1)(x  1). Let's start canceling (or slashing)! We can cancel (x  1) over (x  1) and (x + 2) over (x + 2). Once we have canceled, or slashed, all of the terms from the top and bottom, we multiply straight across. That gives us a final answer of (x  8) / (x  6).
309 Lesson Summary Multiplication and Division of rational polynomial expressions is easy once you remember the steps! For multiplication: we factor, cancel or slash, and multiply. For division: we factor, flip, cancel or slash, and multiply. How to Solve a Rational Equation Chapter 11 / Lesson 5 Rational equation is one that contains fractions. Yes, we will be finding a common denominator that has 'x's. But no worries! Together we will use a process that will help us solve rational equations every time! Rational Equations A rational equation is an equation that contains fractions with xs in the numerator, denominator or both. Here is an example of a rational equation: (4 / (x + 1))  (3 / (x  1)) = 2 / (x^21). Let's think back for a moment about solving an equation with a fraction. 1/3 x = 8. We think of the 3 in the denominator as being a prisoner, and we want to release it. To set the 3 free, we multiply both sides of the equation by 3. Think of it as 3 letting both sides of the equation know he's leaving. 3 (1/3 x) = 8 (3). This process freed our denominator and got rid of the fraction  x = 24. It is also the process we use to solve rational equations with one extra step. In rational equations, sometimes our solution may look good, but they carry a virus; that is, they won't work in our equation. These are called extraneous solutions. The steps to solve a rational equation are: 1. Find the common denominator. 2. Multiply everything by the common denominator. 3. Simplify. 4. Check the answer(s) to make sure there isn't an extraneous solution.
310 Let's solve a couple together. Example #1 Example number one: solve. Remember to check for extraneous solutions. (3 / (x + 3)) + (4 / (x  2)) = 2 / (x + 3). Our first step is to figure out the terms that need to be released from the denominators. I look at 3 / (x + 3). I write down (x + 3) as one of my common denominators. I look at 4 / (x  2). I write down (x  2) as another part of my common denominator. I look at 2 / (x + 3). Since I already have (x + 3) written in my denominator, I don't need to duplicate it. Next, we multiply everything by our common denominator  (x+3)(x2). This is how that will look: ((3(x + 3)(x  2)) / (x + 3)) + ((4(x + 3)(x  2)) / (x  2)) = (2(x + 3)(x  2)) / (x + 3)) It isn't easy for the denominators to be released; there is a battle, and like terms in the numerator and denominator get canceled (or slashed). Slash (or cancel) all of the (x + 3)s and (x  2)s in the denominator and numerator. Our new equation looks like: 3(x  2) + 4(x + 3) = 2(x  2). In example #1, the first step is finding the common denominator. Distribute to simplify: (3x  6) + (4x + 12) = 2x  4. Collect like terms and solve. 3x + 4x = 7x, = 6. We end up with 7x + 6 = 2x  4. Subtract 2x from both sides: 7x  2x = 5x. Subtracting from the other side just cancels out the 2x, and we get 5x + 6 = 4. Subtract 6 from both sides: 46 = 10. Again, subtracting 6 will cancel out the +6, so we end up with 5x = Divide by 5 on both sides, and we cancel out the 5 and give us x =  2. It turns out x =  2. The reason we check our answers is that sometimes we get a virus, or, in math terms, extraneous solutions. To check, I replace all the xs with 2: (3 / (2 + 3)) + (4 / (22))
311 = (2 / (2 + 3)). Let's simplify: (3 / 1) + (4 / 4) = (2 / 1). Since = 2 is true, x =  2 is the solution! Example #2 Example number two: solve. Remember to check for extraneous solutions. (4 / (x + 1))  (3 / (x  1)) = 2 / (x^21). First we need to release our denominators. To release our denominators, we write down every denominator we see. I have found the easiest way to do this is to first factor, if needed, then list the factors. x^21 = (x + 1)(x  1). Our new equation looks like this: (4 / (x + 1))  (3 / (x  1)) = 2 / (x + 1)(x  1). I look at 4 / (x + 1). I write down (x + 1) as one of my common denominators. I look at 3 / (x  1). I write down (x  1) as another part of my common denominator. I look at 2 / (x + 1)(x  1). Since I already have those written in my denominator, I don't need to duplicate them. So my common denominator turns out to be (x + 1)(x  1). Kathryn, why aren't we using the factors of x^21? Great question! We already have (x + 1) and (x  1) being released. We don't need to do it twice. Now we multiply each part of the equation by the common denominator  (x + 1)(x  1). Think of this as the key to the prison: (4 (x + 1)(x 1) / (x + 1))  (3 (x + 1) (x  1) / (x  1)) = 2 (x + 1)(x  1) / (x + 1)(x  1). It isn't easy for the denominators to be released; there is a battle, and like terms get canceled (or slashed)! Slash (or cancel) all of the (x + 1)s and (x  1)s in the denominator and numerator. This leaves us with 4(x  1)  3 (x + 1) = 2. Like terms are cancelled out or slashed in the second example. Now we need to solve for x. Distribute 4 into (x  1) and 3 into (x + 1). (4x  4)  (3x  3) = 2. Collect like terms: x  7 =  2. Add 7 to both sides of the equal sign: x = 5.
312 It looks like our answer is 5, but we need to doublecheck. I replace all the xs with 5 and simplify. It turns out 5 works, and it is the solution to our equation. And so our solution checks! Lesson Summary The steps to solving a rational equation are: 1. Find the common denominator. 2. Multiply everything by the common denominator. 3. Simplify. 4. Check the answer(s) to make sure there isn't an extraneous solution. Rational Equations: Practice Problems Chapter 11 / Lesson 6 Mario and Bill own a local carwash and have several complex tasks that they must use rational equations to solve for an answer. Enjoy learning how they solve these equations to help them with some of their daytoday tasks. Example 1 Mario and Bill own a local carwash. Mario can wash, vacuum, and wax a car in 6 hours. Bill can wash a car, vacuum the car, and wax a car in 5 hours. They've decided to have a special event on the upcoming Saturday. The two are curious how long it will take them to wash, vacuum, and wax each car if they work together, so they decide to figure out how long it should take them. They decide to write a rational equation to represent their situation. A rational equation is an equation that contains fractions and has polynomials in its numerator and denominator. Mario knows that he can clean a car in 6 hours. He needs to use the rate of 1 car for every 6 hours. Bill knows that he can clean a car in 5 hours. He will need to use the rate of 1 car for every 5 hours. Since they do not know how long it will take them to clean a car together, they are going to use the variable w to represent that time. So the rate that they will clean a car together will be 1 car for every w hours.
313 Mario and Bill know that each of their rates added together will equal the amount of time it would take them combined, so they will use the equation Mario's rate + Bill's rate = their combined rate. Next, they will plug in each rate into their equation: 1/6 + 1/5 = 1/w. Mario and Bill are now ready to solve their problem to see how long it would take them together to clean each car. Clearing the Fraction Multiply by the LCM of each denominator to clear fractions from an equation Mario and Bill are both perplexed when looking at their equation. They both know that in order to solve an equation with fractions, they can clear the fractions from the equation. To do this, they will need to multiply each term by the least common multiple (LCM) of each denominator. The denominators in the problem are 6, 5, and w. The LCM of the two numbers (6 and 5) is 30. They must also multiply by the variable w since it is also a term. So the LCM of these three denominators is 30w. By multiplying each term in this equation by 30w, Mario and Bill can clear out all of the fractions in their equation. The reason that this is possible is because when they multiply each term, they can cancel out the common terms. When doing 1/6 times 30w, we can divide out a common factor of 6. So 1/6 times 30w would equal 5w. When multiplying 1/5 times 30w, we can divide out a common factor of 5. So 1/5 times 30w would equal 6w. When doing 1/w times 30w, we can divide out a common factor of w. So 1/w times 30w would equal 30. Now that Mario and Bill have multiplied each term by 30w, they now have the equation 5w + 6w = 30. Solving Example 1 The next step to solving this problem is to combine like terms: 5w + 6w = 11w. The equation now is a onestep equation: 11w = 30. To solve for w, divide each side by 11.
314 11w divided by 11 would equal w, and 30 divided by 11 would equal 2 and 8/11 hours. To find out exactly how many minutes 8/11 hours is, Mario and Bill would multiply 8/11 times 60 minutes. This would equal 43.6 minutes. Adding that to the whole number of 2, Mario and Bill realize that together they can wash, vacuum, and wax each car in 2 hours and 43.6 minutes. Example 2 Mario and Bill are very happy with the turnout during their Saturday sale. They had so many customers that they decided to work together to sanitize the inside of their carwash. Together, it took them 12 hours to completely sanitize the inside of their carwash. The last time the carwash was sanitized, Mario had cleaned it himself. The year before, Bill had cleaned the carwash by himself but took 3 times as long as Mario had taken by himself. How long had it taken each one of them to clean the carwash individually? We know that Mario cleaned the carwash by himself, but we do not know how long that took. Let's use the ratio 1 carwash in M hours to represent how long it took Mario to clean the carwash by himself. Also, Bill cleaned the carwash by himself. For Bill we will use the ratio 1 carwash in B hours to represent how long it took Bill to clean the carwash by himself. Together they cleaned the carwash in 12 hours. For this ratio we will use 1 carwash in 12 hours to represent how fast they cleaned the carwash while working together. We can now see that Mario's time plus Bill's time equals their combined time. So our equation is now 1/M + 1/B = 1/12. We now know that it took Bill three times as long as it took Mario to clean the carwash by himself. Therefore, instead of using B to represent the amount of time it took Bill, we can use 3M because 3M would represent three times the number of hours it took Mario. To solve this equation, we now need to clear the fractions from the equation by multiplying by the LCM. The least common multiple that will divide by our denominators M, 3M, and 12 is 12M. So we are going to multiply each term by 12M. This will allow us to get rid of our fractions. 1/M times 12M equals 12. 1/3M times 12M equals 4. And finally, 1/12 times 12M equals M. So our equation now looks like = M. To solve, add 12 and 4 together, and M would equal 16. Since M equals 16, we know that it took Mario 16 hours to clean the carwash by himself. It took Bill 3 times as long, so 16 * 3 = 48 hours. Bill cleaned the carwash himself in 48 hours. He must have had a long week.
315 Example 3 Mario and Bill are so pleased with their sparkling clean carwash. The two guys only have one more task to make their carwash completely stocked for the next day's opening. They need to fill the large soap tower. Mario and Bill can use two separate hoses to fill the soap dispenser. When both hoses are being used together, it takes 12 minutes to fill the soap tower. If used separately, one hose can fill the tower 10 minutes faster than the other hose. How long does it take each hose to fill the tower separately? For this problem, let's call hose A the faster hose and hose B the slower hose. The ratio for how fast hose A can fill the soap tower alone is 1 tower/a minutes. The ratio for how fast hose B can fill the tower alone is 1 tower/b minutes. Also, the ratio for how fast both hoses combined can fill the water tower is 1 tower/12 minutes. We now know that when we use hose A plus hose B combined it takes 12 minutes. So our equation will be 1/A + 1/B = 1/12. Since it takes hose B 10 minutes longer than hose A, we can substitute A + 10 in for term B. Our equation is now 1/A + 1/A+ 10 = 1/12. The easiest way to clear our equation of fractions is to multiply each term by the LCM, which would be all three denominators multiplied together. The LCM for this equation would be A times A+10 times 12. As we multiply each term, we can cancel out the like terms. As we multiply 1/A times A times A+ 10 times 12, the A terms will cancel out, leaving (A + 10) times 12. As we multiply 1/(A + 10) times A times A + 10 times 12, the (A + 10) terms will cancel out, leaving A times 12. And as we multiply 1/12 times A times A + 10 times 12, the 12 terms will cancel out, leaving A times (A + 10). Our equation is now ((A + 10) * 12) + (A * 12) = (A * (A + 10)). Next we need to simplify each term. 12 times (A + 10) would equal 12A times A would equal 12A, and A times (A + 10) would equal A^2 + 10A. Next, let's combine 12A and 12A on the same side of our equation to equal 24A. So our equation now is 24A = A^2 + 10A.
316 Equation written in standard form The equation now needs to be put into standard form so that we can factor out the possible solutions. Standard form is written as Ax2 + Bx + C = 0. To convert our equation to standard form, we will need to perform inverse operations. Our equation in standard form would be written as A^214A = 0. To solve this equation, we need to factor our trinomial. Look at our constant, We need to think of two numbers that multiply together to give us 120 but also add together to give us our middle term of 14. The two factors of 120 that that gives us a product of 120 and a sum of 14 are 20 and 6. When we factor the equation A^214A  120, we get (A  20) and (A + 6) = 0. We must solve each term by setting both equations equal to zero. We will solve the equations (A  20) = 0 and (A + 6) = 0. After solving both of these equations, we are left with two solutions  A = 20 and A = 6. Mario and Bill both know that it is not possible to have a negative hour, so they eliminate the answer A = 6. So Mario and Bill now know that hose A, which is the fastest hose, took 20 minutes to fill up the soap tower by itself. To find out how long it took hose B, we will need to add the 20 minutes and the additional 10 minutes that the slower hose took, so hose B took 30 minutes to fill up the soap tower. Mario and Bill are now pleased with their clean carwash and ready to start another busy week of working at the carwash. Lesson Summary Let's review some of the main points that we used to solve these rational equations. To solve a rational equation, you will need to pay close attention to the wording of the problem and write an equation to represent the ratios that are provided. The easiest way to solve a rational equation is by clearing the equation of fractions. This can be done by multiplying each term by the least common multiple (LCM). After you clear the equation of the fractions, you will solve the equation the same way as you would
317 any other multistep equation. Don't forget at the end of your problem to evaluate your answer to check if it makes sense. For example, we could not have 6 hours. Therefore, 6 was not a possible solution. Solving Rational Equations with Literal Coefficients Chapter 11 / Lesson 7 Literal Coefficients When I hear the phrase 'solving rational equations with literal coefficients,' it actually does scare me a bit. It sounds like a BIG problem, but you know what? Once I get into the whole process of it and remember my two basic steps, the big scary problem that used to be an elephant is now a mouse that I can easily handle. What I need to know first before solving this type of problem and what I want you to understand first is what a literal coefficient is. Simply stated, a literal coefficient is a variable used to represent a number. The number the variable represents can be either known or unknown. It can be our usual x or y or it can be other letters, such as a, b, or c. Although we can have more than one variable in an equation, we will consider the common case where there is only one literal coefficient. Rational Equations with Literal Coefficients We now know what a literal coefficient is but what about rational equations? An equation with a fraction made up of polynomials is a rational equation. You can identify them easily by looking for fractions with polynomials in the denominator and numerator. If you see one, then you know you have a rational equation. All of these are examples of rational equations: t/(t +2) + 1/2 = 1 s/22 = (s+3)/4 1/x + 4/(x+2) = 5/(x1) Don't get scared, but we are going to work with the largest problem here, the one with the x variable. I'll show you how to go about working with and solving problems like this so they don't become a headache for you.
318 Step 1: The Common Denominator 1/x + 4/(x+2) = 5/(x1) The first step is to find the common denominator. Our current problem has three denominators we need to look at. We have an x, an x + 2, and an x  1. Finding common denominators when you have variables involved is slightly different than when you only have numbers involved. I think it's easier when you have variables involved. Why do I say that? Let me show you. When we have variables in our denominators, to find our common denominator, all we have to do is to write down all the factors we see. For our problem, the first fraction has an x for the denominator, so our common denominator will have an x as well. The second fraction has an x + 2, so we will add that to our common denominator since we don't have that yet. If our common denominator already had it, we won't need to add it. Our common denominator is now x ( x + 2 ). The last fraction has an x  1. I look at my common denominator and see that I don't have that yet, so I will add that on, too. When adding factors to the common denominator, I multiply them. So our common denominator is x ( x + 2 ) ( x  1 ). If our denominators had numbers only, we would go ahead and find the least common multiple of the numbers involved. x(x+2)(x1) We are going to use this common denominator to help us solve our rational equation. What we are going to do is to multiply each term by our common denominator. Watch what happens when we do this. You'll like it. It simplifies the problem into something I know you can manage. Because we are multiplying each term by the common denominator, we can cancel like terms. The first term has a common x factor in both the top and bottom. The second one has a common x + 2 in both the top and bottom that can be cancelled and the third has a common x  1 that can be cancelled. Our problem now looks like this: (x+2)(x1) + 4x(x1) = 5x(x + 2) I consider this a lot more manageable than what we began with. Don't you agree? No more fractions! Yay! Now we get to use our algebra skills to solve for x.
319 Step 2: Solving the Rational Equation We will multiply the factors out making sure to follow the order of operations. (x+2)(x1) + 4x(x1) = 5x(x+ 2) Then we combine our like terms. 5x^23x  2 = 5x^2 + 10x We then move all our variables to the same side and simplify. I'm going to choose the left side. You can choose whichever side you want. 13x  2 = 0 I see that I'm left with a linear equation. This means I can move my number over to the other side and solve for x. 13x = 2 o x = (2/13) We have solved for x, but we are not done. Because a rational equation has fractions, we need to go back and plug in our answer to see if it causes division by zero in any of our fractions. We can do this by plugging our answer into each denominator separately to see if it gives us a zero in the denominator. If all the denominators are okay, meaning none end up being zero, then our answer is valid. Checking our answer of  2 / 13 and plugging it into each denominator, we get  2 / 13 for the first denominator,  2 / = 24 / 13 for the second denominator, and  2 / 131 =  15 / 13 for the third denominator. None of these are zero, so my answer is valid. Also, if we ended up with a quadratic equation after multiplying by the common denominator and simplifying, we would need to use our quadratic equation solving skills such as factoring or the quadratic formula to help us solve the problem. Lesson Summary In summary, solving rational equations with literal coefficients is not terribly bad. We recall that a literal coefficient is a variable used to represent a number and that a
320 rational equation is an equation with fractions made up of polynomials. Solving these problems is a twostep process. The first step requires us to find the common denominator and then to multiply the whole equation by it to simplify our equation so we can solve it easier. The second part requires our algebra skills in solving for the literal coefficient or variable. Solving Problems Using Rational Equations Chapter 11 / Lesson 8 Follow the steps you will learn in this video lesson to help you solve rational equation problems. See how the process turns what looks like a huge problem into a much simpler and easier to manage problem. Rational Equations What is a rational equation? The name sounds worse than what it actually is. A rational equation is just an equation that has fractions. These fractions might just be numbers, but they also may be polynomials with numbers and letters in them. Think of the equations you work with in algebra when you are solving for a particular variable. A rational equation will look similar to those except that it will have a fraction on one or both sides of the equation. What you see is an example of a rational equation. Let me show you some more. All of these are also rational equations. A good way to recognize rational equations is to look for fractions. If you see a fraction made up of only numbers and/or variables, an equal sign and other numbers and/or variables, then you are looking at a rational equation. The Problem What I want to show you in this video lesson is a way to solve any kind of rational equation. I know there are shortcuts out there, and you will find them if you search. But these shortcuts have limitations and can sometimes be only used on a particular type of equation. The process I want you to take home from this lesson is one that can be applied to all rational equations. It's not hard, and I will give you helpful tips on how to remember each step.
321 The problem we will use to go over this information is the very first one you saw in this lesson. 4/3 = x / 5 Real quick: how do you know that this is a rational equation? What is it that you are looking for to help you recognize it? That's right, fractions! And this one happens to have one fraction on each side of the equal sign. Dealing with the Denominator Remember, fractions are not your enemy. Yes, I said that. It's true. But don't worry  our very first step is to work at rewriting this equation without fractions. It can be done, and I will show you an easy way. Fractions are really not bad at all when you focus on the numerator and denominator separately. This is exactly what we are going to do. We are going to focus on just the denominator. A good way to remember this first part of solving the problem is to think about what identifies a rational equation. Yes, fractions, and fractions have denominators. Step 1 is to find the common denominator. I remember this step easily by thinking of the first step I usually take when adding or working with regular fractions. We always start by finding a common denominator. Because rational equations can involve variables, finding our common denominator is slightly different than when working with regular fractions. It's actually easier, I think. What you do is you look at the various denominators that you have, and all you need to do is make sure that your common denominator has all the denominators in it and multiplied together. Our problem has denominators of 3 and 5, so our common denominator will be 3 * 5. I'm going to leave it in the multiplication form because this will make my next step easier. Step 2 is to multiply the whole problem by the common denominator. Remember this step by remembering how with regular fractions you want all the fractions to have the common denominator, but in our case, we will multiply everything by the common denominator. When we do this, we can cancel like terms, and we end up with an equation that doesn't have fractions. (3 * 5) 4/3 = (3 * 5) x / 5 (5 * 4) = (3 * x) 20 = 3x
322 Freedom! Doesn't this equation look much easier to solve than the one we started with? You see, fractions really aren't all that bad. Finding Our Solution Step 3, now, is to solve the equation for our variable. You can remember this step by also thinking of regular fractions and what you do with them after you find the common denominator. You do the same and solve the problem, either getting the answer or solving a variable. We will continue to use our algebra skills to solve this problem. We will divide both sides by 3 to isolate our x variable. Once we have x by itself, we have found our answer. 20 = 3x x = 20/3 It looks like our answer is 20 / 3. It might seem like we are done here, but we have one more step to do. Step 4 is to check to make sure our answer does not produce a division by zero in our original problem. Remember this step by thinking of your teacher. You want to doublecheck your work so you can prove to your teacher that your answer is correct. To do this, we will look at our original problem, plug in our answer and see if it produces a division by zero anywhere. If it doesn't, then our answer is good to go. If it does, then our answer is not a valid one. When your answer produces division by zero, it doesn't mean that you did something wrong, it just means that this particular problem is not defined at that point. It is possible to have an answer of 'no real solutions.' I see that it doesn't produce any division by zero, so that means my answer checks out. My final answer is 20 / 3. Lesson Summary In summary, rational equations are equations with fractions in them. The most important things I want you to take away from this video are the steps to solve them. There are only four of them, and they are similar to the ones you follow to solve regular fraction problems. Step 1 is to find the common denominator. Step 2 is to multiply everything by the common denominator. Step 3 is to solve for the variable. And step 4 is to check the answer to make it sure it doesn't product division by zero,
323 which is undefined. Use the steps to working with regular fractions as a memory aid to help you remember these four steps. Fractions really are not your enemy. Finding Constant and Average Rates Chapter 11 / Lesson 9 What Is a Constant Rate? So, what exactly is a constant rate? Well, a constant rate is something that changes steadily over time. Picture the burning of a candle. That is a constant rate, because a candle burns down steadily over time. We can be confident that if a particular candle takes so many hours to burn down, and we make the candle twice as big, we will have a candle that will burn twice as long. What Is an Average Rate? Now, what about the average rate? An average rate is different from a constant rate in that an average rate can change over time. An average rate is actually the average or overall rate of an object that goes at different speeds or rates over a period of time. For this one, picture the flight of a bumblebee. The bumblebee sees a flower and rushes over to the flower at a quick speed. Once it's there, the bumblebee slows down and slowly buzzes around the flower to inspect it. Oh, but look! There's another even bigger flower in the distance. The bumblebee sees it and buzzes off at an even quicker speed. The average rate of the bumblebee would take all these different speeds and find the average speed the bumblebee had over his whole trip. Now that we've covered these definitions, let's see how we can find each of these rates. We will stick to our original visuals to help us out with our problems. Finding a Constant Rate Let's go back to our burning candle to help us with finding a constant rate.
324 We are taking a test, and we see this problem in front of us: Object A takes 1 hour to burn 1 inch, and Object B takes 2 hours to burn down 1 inch. What is the constant rate of each object? Okay, another dry math problem. But, that's okay! We have our useful visual we can refer to, so we won't be bored. We can picture two differently sized candles. We can get creative here and make the problem more fun. We can have one rainbow colored candle and another candle with a lightning bolt drawing on the side. Picture anything that keeps your interest. Now that we have our candles in front of us, we can get to the meat of the problem and find the constant rates of each candle. We will call the rainbow colored candle Object A, and we will call the lightning bolt candle Object B. I'm picturing my rainbow colored candle burning. I see that for each hour that passes, my rainbow colored candle goes down by an inch. To calculate my constant rate for this candle, this object, I recall the formula for rate, which is rate = distance / time. Okay. So, I don't have a distance, per se. But I do have the amount of candle burned, which is 1 inch. I also have a time, which is 1 hour. I plug these numbers into my formula to get my constant rate. I get 1 / 1, which gives me 1. So my constant rate for my rainbow colored candle, or Object A, is 1 inch per hour. That is part of my answer. The other part of my answer is the rate for the lightning bolt candle, or Object B. I will do the same and plug in 1 for the amount of candle burned, or the distance, and 2 hours for my time. My constant rate here, then, is 1 / 2. or 0.5 inches per hour. Now I'm done with this problem. Object A takes 1 hour to burn 1 inch and Object B takes 2 hours to burn down 1 inch. What is the constant rate of each object? I have found that my answer for this problem is 1 inch per hour for Object A and 0.5 inches per hour for Object B. Finding an Average Rate To help us out with an average rate problem, let's go back to our bumblebee visual. Our math problem might say something like this: An object has three parts to its travel from point A to point B. In part 1, the object travels at a speed of 10 km per hour for half an hour. In part 2, it travels at 2 km per hour for 15 minutes. And in part
325 3, the object travels at 15 km per hour for another half hour. What is the average rate for this object? This one's a little more complicated. We can picture our bumblebee for the object, and we can relate each part of the travel to each part of the bumblebee's journey. The first part can be our bumblebee going fast to its first flower; the second part is then our bumblebee checking out the flower; and the third part is our bumblebee flying fast to the second, larger flower. To find our average, I will use the same rate formula I did for finding the constant rate. The only additional information I need is the total distance traveled and the total time. But I can find this information from what was provided in the problem. Rereading the problem, I can see that the bumblebee traveled 5 km in half an hour for the first part of the trip, because if the bumblebee is flying at 10 km per hour, then at the halftime point, the bumblebee will have gone 5 km. For the second part of the trip, because the bumblebee is traveling at 2 km per hour, after 15 minutes, or a quarter of an hour, the bumblebee will have gone 0.5 km, which is a quarter of 2 km. Then, in the last part of the trip, because the bumblebee is traveling at 15 km per hour in half an hour, the bumblebee will have gone 7.5 km, or half of 15 km. Now that I have my distances and times for each part of the trip, I will add them up to get the total distance and total time. My total distance traveled is , which equals 13 km. My total time is , which equals an hour and a quarter, or 1.25 hours. Plugging these two numbers into my formula, I get 13 km / 1.25 hours, which equals 10.4 km per hour. That is my answer, and I am done. Lesson Summary In review, recall that a constant rate is something that changes steadily over time, and that an average rate is the overall speed or rate of something that travels at different speeds or rates over time. To find both rates, we use the rate formula, which is rate = distance/time. The distance in the formula is the amount of change in the problem over a period of time. To find the constant rate, you would plug in the change and divide by time. Because it is constant, this number won't change.
326 For the average rate, though, you would first have to find the total distance or change and the total time involved. You would then divide the total change or distance by the total time to find your average rate. Solving Problems Using Rates Chapter 11 / Lesson 10 What Is a Rate Problem? What is a rate problem? A rate problem is a problem involving a rate of some sort, such as speed, earnings, etc. A rate is anything that can be gained or lost over time. You can gain speed over time and you can gain money over time too if you work hard. Rate problems usually come at you in the form of word problems. I know that word problems are the worst because you have to set up everything yourself. But let me show you how you can easily solve a rate word problem in this video. Keep watching! So we are looking at our test paper and we see this problem: 'Two brothers want to purchase the nextgeneration entertainment system that costs $400. How many hours does the brother who makes $12 / hour have to work if the brother who makes $10 / hour can only work for 18 hours?' See how useful solving rate problems can be? If you have a job and are getting paid on an hourly basis, you can use what you learn in this video lesson to figure out how much you have to work to purchase something you want. The Rate Formula Before we can set up our problem, I need to show you what our formula for rate is. Do you remember what rate means? A rate is anything that can be gained or lost over time. There are two ways you can write this mathematically. You can write it using division like so  rate = gain or loss/time  or, you can write it using multiplication like this  gain or loss = rate * time. This also happens to be the most often used form of this formula. Both are the same formula, they're just written differently.
327 Your rate can be things like your speed or earnings. For our problem, we will use the formula with multiplication because it is the most commonly used form and it's easier for setting up our word problems. Setting up the Problem Setting up the problem is the most important step in our solving process. I know, this is the hardest part of word problems, but like I said earlier, don't worry. I'm going to show you an easy way to do it. What we are going to do is make a table showing all of our important information for the problem. For the table, we are going to follow the format of our rate formula, the one that uses multiplication. Since we are making money, our formula is gain = rate * time. We will call our gain 'earnings' to make it clear that the amount we earn is our gain. We will have a separate column for each item. We will have a column for gain, a column for our equal sign, a column for rate, a column for our multiplication sign, and a column for the time. We will add an extra column in the beginning to label our brothers. We will call the brother who earns $10 / hour Brother 1 and the brother who earns $12 / hour Brother 2. Our table now looks like this. Now that we have our table set up, we can fill it in with our information. We will use variables for the boxes we don't know or need to find out. The boxes under earnings are for the total amount earned over a period of time for each brother. We don't know this information, so we will put a variable e into the box for Brother 1. For Brother 2, we will put e because brother 2 only needs to make the difference between the cost of the entertainment system and the earnings of Brother 1. For the boxes in the rate column, we will put $10 for Brother 1 and $12 for Brother 2. Under the time column, we will put 18 for Brother 1 since the problem states he can only work for 18 hours. For Brother 2, we will put the variable t because we don't know how much time he needs to work yet. Look at that, we've filled out our table! That wasn't so bad, was it? Now that we have our table filled out, we can go ahead and solve the problem. Solving the Problem But what exactly is the problem asking for? Let's go back and see. 'Two brothers want to purchase the nextgeneration entertainment system that costs $400. How many
328 hours does the brother who makes $12 / hour have to work if the brother who makes $10 / hour can only work for 18 hours?' Reading it carefully, I see that what it is asking for is the number of hours the brother who earns $12 / hour needs to work. Hmmm. That would be our Brother 2. The number of hours would be our variable t. So I need to solve for that variable. I go back to my table and look at it again. I see my t in the row for Brother 2. I also see an e in that row. Hmmm. What do I do about that e? Looking at Brother 1, I see that e equals 10 times 18. Aha! I can put that information in place for the e for Brother 2. This step I just did is called substitution. Writing it our mathematically, I get this (10 * 18) = 12 * t. Hey! This looks like something I can solve using algebra skills. So let's see. I do my order of operations and multiply the 10 and the 18 first to get 180. Then I subtract that from the 400 to get 220. Then to solve for t, all I need to do is divide the 220 by 12. So then, my t would equal 220 / 12 = Brother 2 would need to work hours so the two brothers can purchase the nextgeneration entertainment system. Here's a word of caution. If you are given a multiplechoice test for this type of rate problem and the options didn't have hours, but you had an 18 and a 19, you should pick 19 hours since most likely Brother 2 has to work full hours and not just part of an hour. For Brother 2 to be able to purchase the entertainment system, he would have to finish the partial hour to make 19 hours. If he only worked 18 hours, it wouldn't be enough. Lesson Summary To review, a rate problem is any problem involving rates. Rates are gains or losses occurring over time, such as speed or earnings. The formula for rate is your gain or loss equals your rate times time. In speed applications, your gain or loss is your distance. In earnings, your gain or loss is the total amount of earnings over a period of time. And remember, gains are positive and losses are negative. To solve rate word problems, you first set up a table following your rate formula. You fill in the table with the numbers you know and variables for the numbers you don't yet know. Make the variables fit the situation, like we did in the earnings column for the cost of our entertainment system. Then you solve your problem using algebra and substitutions.
329 Perimeter of Triangles and Rectangles Chapter 12 / Lesson 1 Without realizing it, we calculate and use the perimeter of triangles and rectangles in regular everyday situations. Learn more about the perimeter of triangles and rectangles in this lesson, and test your knowledge with a quiz. The Perimeter As part of a new fitness program, Melody has to walk around her block three times every day. If each side of her block is 80 ft long, how many feet will she walk every day? To answer this question, we must find out how many feet are around her entire block, not just one side. This concept is known as the perimeter, which is defined as the distance around an object, shape or figure. To find the perimeter of any polygon, all we have to do is add up the length of all the sides. For this lesson, we will focus on the perimeter of triangles and rectangles, and we'll look at a few examples, beginning with Melody. Triangle and Rectangle Perimeters Recall that Melody has to walk around her block three times every day, and each side of her block is 80 ft long. To determine the total length of her walk, we must figure out how many feet are around the entire block. To do so, we'll add , since there are four sides to her block. Alternatively, since all of the sides of her block are the same length, we could also multiply 4 * 80. Using either method, we will see that the perimeter of Melody's block is 320. Since she has to circle the block three times, we will add to get a total of 960. So in conclusion, Melody will walk 960 ft every day for her fitness program. Let's look at another.
330 Daniel has 25 ft of decorative border for his rectangular bulletin board. If the width of his board is 3 ft and the length of his board is 10 ft, does he have enough border to line the entire board? To determine if Daniel has enough border, we must calculate the perimeter of his board. Since it's a rectangle, we know that both widths and both lengths will be the same. Therefore, to find the perimeter, we can either add or we can multiply 2 * 3 and 2 * 10 and then add No matter which method we use, the perimeter of the board is 26 ft. Since Daniel only has 25 ft of border, he does not have enough to line the entire board. Here's one last example. Christy is creating a triangular collage of pictures. One side of the collage is 12 in and the other two sides are 18 in. If she wants to frame the collage with ribbon, how much ribbon will she need? And if the ribbon costs $0.75 per foot, how much will the ribbon cost her? To determine the amount of ribbon that Christy needs, we must first figure out the perimeter of her collage. To do so, let's add up the length of all the sides of the triangle: Simplifying this calculation leaves us with a perimeter of 48. Therefore, Christy needs 48 in. of ribbon to frame her collage. Now, let's determine the cost. Recalling that 12 in = 1 ft, we can conclude that 48 in = 4 ft. Since the ribbon costs $0.75 per foot, let's multiply 4 * $0.75. Doing so gives a total of $3. In conclusion, Christy will spend $3 on ribbon for her collage. Solving with Perimeter In all of these examples, we were given the measure of each side to calculate the perimeter. Now, let's look at how we can use perimeter to find the measure of the sides. Here's rectangle DOTS, which has a perimeter of 56. If the length of the rectangle is 2x  3, and the width of the rectangle is x + 1, how long is the rectangle? For this problem, knowing the value of x will allow us to determine the length. So, we must first solve for x. Since this is a rectangle, let's multiply the length by two and multiply the width by two. Then, we will add them together to equal the perimeter. This process gives us 2(2x  3) + 2(x + 1) = 56. From distributing, we get 4x x + 2 = 56, and after combining like terms, we have 6x  4 = 56. We must now add 4 to
331 both sides to cancel it out of the equation, leaving us with 6x = 60. Our next step will be to divide each side by 6, giving us x = 10. Since we were only asked to determine the length of the rectangle, let's plug 10 into the equation for the length, which is 2x  3. By substitution, we have 2(10)  3. To calculate, we refer to the order of operations and multiply 2 * 10 before subtracting 3. In the end, the length of rectangle DOTS is 17. Here's one last example. Triangle PUD has a perimeter of 162 cm. Side PD is 4 less than twice a number, side PU is 8 less than the number, and side UD is 15 less than three times the number. What is the length of each side of triangle PUD? Since this problem does not directly give us the equation for each side, we will have to create them based on the descriptions. The length of side PD is 4 less than twice a number. Since we don't know what this number is, we will call it x. So, PD is 4 less than two times x, which means that PD = 2x  4. Side PU is 8 less than the number, so PU = x  8, and side UD is 15 less than three times the number, which means that UD = 3x We will add all three equations to equal the perimeter. So, our beginning setup is 2x x x  15 = 162. Combining like terms takes us to 6x  27 = 162. From here, we cancel out the 27 by adding it to both sides, leaving us with 6x = 189. Then, we will divide each side by 6 to conclude that x = We're ready to determine all of the side lengths for triangle PUD. Substituting 31.5 into each equation, we see that PD = 59 cm, PU = 23.5 cm, and UD = 79.5 cm. To check our work, let's add up all of these lengths. Doing so gives us , which is equal to 162. Since this sum is the same as the perimeter we were given, we know that we have correctly solved the problem. Lesson Summary In review, perimeter is defined as the distance around an object, shape or figure and is calculated by adding the lengths of all the sides of a polygon. Rectangles have two pairs of congruent sides. The lengths have the same value, and the widths have the same value. This allows us to use addition or a combination of addition and multiplication to calculate the perimeter. On the other hand, some triangles have congruent sides and some do not; therefore, we should only use addition to calculate the perimeter of triangles. With both shapes,
332 we can also use the perimeter to work backwards and determine the length of each side. Perimeter of Quadrilaterals and Irregular or Combined Shapes Chapter 12 / Lesson 2 Sometimes you need to figure out a shape's perimeter whether you have all the information you want or not. In this lesson, you'll learn how to find the perimeter of any quadrilateral as well as more complicated irregular and combined shapes. Finding the Perimeter Let's say you like running. Normally, you run on a track. You know the track is 400 meters long. If you think about it, that 400 meters is the perimeter of the oval. Perimeter, as you'll recall, is the path around the outside of a shape. But what if you get tired of running in endless ovals? Maybe you'd rather run around the block where you live. Or maybe you want to wind a bit through your neighborhood. How do you know how far you went? There's where knowing how to find a perimeter is essential. Quadrilaterals Let's start with you running around your block and wanting to know the distance, or perimeter. If you look at your block on a map, it has four sides. That makes it a quadrilateral. A quadrilateral is just a foursided shape. If you're lucky, you know how long each side is. North Street and South Street are each 250 meters long. East and West Streets are each 100 meters long. To find the perimeter of this block, just add up the sides. That's = 700 meters. That's almost two times the length of the track and you're already home when you're done! But what if you don't know how long each street is? Let's say you're staying at your cousin's house and want to run in that neighborhood. She tells you that she knows George Street is 100 meters and John Street is 50 meters. What about Paul and Ringo? Fortunately, you recognize that the corners on this block are right angles. Therefore, this is no ordinary quadrilateral; it's a rectangle. And you know that the opposite sides of rectangles are congruent, or equal in length. So if George Street is 100 meters, then
333 so is Ringo Street. If John Street is 50 meters, so is Paul Street. Therefore, the perimeter of this quadrilateral, and the distance you ran, is , or 300 meters. That's a small block, less than one lap around the track. If you know the length of all four sides of a quadrilateral, you can always find the perimeter by just adding them. If a side or two is missing, see if you can figure out the missing information using your knowledge of the shape's properties. Remember, with squares and rhombuses, all four sides are equal, so you only need to have one side given. With squares and parallelograms, the opposite sides are equal. Irregular Shapes But what happens when you get so good at running that you're ready to train for a marathon? That's 26.2 miles. You'll need to run a much bigger loop while you train, and it may not be such a tidy rectangle. It might look more like this. If you need to know the perimeter of an irregular shape like this, then you really do need to know the lengths of each side. Fortunately, you spotted helpful signs along your way and you noted how many miles each street was. To find the perimeter, just add up each length. In miles, this example is , or 26 miles. That's almost a full marathon! If you add in that.2 milelong detour you took to flag down an ice cream truck at the end, you just ran a marathon distance! Combined Shapes The next time you go for a long training run, the route doesn't have helpful signs to tell you how long each street is. If that's the case, you may be able to figure out the perimeter by treating your route as a group of combined shapes. Let's look at this example. We know the lengths of some of the sides, but not all of them. But look closer at the shape. It's actually a combined shape comprised of a group of squares and rectangles. In the top square, we know two sides are 4 miles, so the other side must also be 4 miles. As for this bit, well, we know the entire length of that rectangle is 10 miles. If these two sections are 3 and 4, then the missing part must be 3 miles. And this part is going to be equal to the opposite side of this rectangle, so it's 1 mile. Finally, this part is just 5 minus 2, or 3 miles. So now we know all the sides! It's
334 , or 40 miles. Holy cow! That's way farther than you planned. I think you earned some ice cream. Now, let's say you wake up the next day and all you can do is hobble around your house, leaning on walls for support. But you still want to know how far you're moving, since every step hurts. Since you live in a geometry lesson, your house has some challenging geometric qualities. You can do this! You know a few parts, which is all you need. Just break it up into shapes like this. Let's start with the big one. One side is 8 feet. Then you can see that these two 4foot stretches add up to 8, so that's a square. That means all sides are 8 feet long. As for this rectangle, well, one side is 4, so this side is 4. If this side is 2 and 3, then the missing piece is also 2. Finally, those pesky triangles. But wait, those are special triangles. First, they're right triangles. But more than that, look at the two legs of the top one. They are 3 and 4 feet, which makes it a triangle. So the hypotenuse is 5 feet. If you forget what a triangle is, you could also use the Pythagorean theorem, or a^2 + b^2 = c^2. 3^2 + 4^2 is 25. The square root of 25 is 5. Then there's the final triangle. This is also a triangle, so the missing side is 4 feet. If you put everything together, it's , or 46 feet. Between all that running, walking and geometry, I think you earned extra ice cream today. Lesson Summary In summary, if you know the lengths of all the sides of a shape, simply adding them together will give you the perimeter. With a regular quadrilateral, like a square or a parallelogram, you just need to know one or two sides and you can figure out the rest. With an irregular shape, the perimeter is still just the sum of all the side. If you don't know all the lengths, look to see if you can break it up into a group of combined shapes. Then use the properties of those shapes to determine the missing sides. Area of Triangles and Rectangles Chapter 12 / Lesson 3 How do you find out the area of rectangles and triangles? Learn how in this lesson! We'll look at the formulas, then practice solving problems for each shape.
335 Shapes Rectangles and triangles are all around us... watching us. That's not creepy at all, right? Okay, it's a little creepy. But they're also around us in less ominous, more friendly ways. That field where you play soccer or football? That's a rectangle. As is everything from your cell phone to the state of Colorado. And that rack you use to set up the balls on a pool table? That's a triangle. So is the peanut butter and jelly sandwich you cut diagonally because it's more fun to eat that way. They even come in larger sizes, like the pyramids in Egypt. In this lesson, we're going to learn how to find the area of these shapes. Let's start with rectangles. Rectangles Let's say you're planting a garden. If your garden is 12 feet long and 4 feet wide, how big is it? This garden is a rectangle. And what you want to know is its area. The area of a rectangle is the length times width. So your garden is 12 feet long by 4 feet wide. 12 is our length and 4 is our width. 12 * 4 is 48. So your garden is 48 square feet. Note the 'square.' That means that there are 48 one foot by one foot squares in your garden. That's plenty of space for your vegetables! Let's try another. Let's say you've planted your garden and now you have an abundance of zucchini, so you're making zucchini bread  mmm, tasty. Your bread pan is 9 inches long by 5 inches wide. What is the area? Like your garden, your pan is rectangular. Ah, the circle of life. Or rectangle of life. So we want the length times width, which is 9 * 5. That's 45. So your pan is 45 square inches. Okay, now you have to deal with all your extra peppers, cucumbers, carrots and other vegetables. You had quite a harvest. You decide to can them for the winter. You keep the jars on the shelves in your apocalypse preparedness closet (you know, just in case). If a shelf can hold 6 jars from front to back and 10 jars from left to right, and you have 300 jars (I said it was a big harvest), how many shelves will you need? So our rectangular shelf is 6 jars wide by 10 jars long. 6 * 10 is 60, so one shelf holds 60 jars. That's our area. But wait, we have 300 jars. How many 60jar shelves will we need for 300 jars? Just divide 300 by 60 to get 5. So we'll need 5 shelves.
336 Right Triangles With a belly full of zucchini bread and all that canning behind us, let's talk triangles. Let's say your garden went so well you decide to branch out, so to speak, into growing and selling Christmas trees. That's ambitious of you. You look at your remaining yard. It's 30 feet by 50 feet. So the area of that rectangle is 30 * 50, or 1500 square feet. What if you cut it in half diagonally, like this? Now you can grow trees in one half and still have half a yard for your annual croquet tournaments. How big is the area of the triangle you have for trees? It's half the rectangle, which was 150 square feet. So it's 750 square feet. Do you know what you just did? You just figured out the area formula for triangles. The area of a triangle can be defined as 1/2 * base * height or just 1/2bh. Base and height are just like length and width in a rectangle. In this right triangle, which is a triangle with a right angle, we have half a rectangle. So it makes sense that our formula is the same as a rectangle but cut in half. Oblique Triangles So you plant your trees in your triangular plot and wait. Let's jump forward in time. It's a math lesson. We can do that. Your trees are now grown. Here's one! You notice it forms a triangle. Can you find the area of that triangle? We know the base is 4 feet. But it doesn't form a right triangle. That's okay! We call this an oblique triangle, which is just any triangle that doesn't have a right angle. To determine the area, we just need to know its height, which is this line here. It's the line that's perpendicular from the base. Perpendicular means it forms a right angle, which gets us back into what we had with our right triangle. That's 7 feet. Our formula doesn't change: 1/2 * base * height, or 1/2 * 4 * 7. That's 14. So the area of the triangle is 14 square feet. Staring at your trees makes you hungry, so you decide to eat some pizza. When you look at a slice, you realize that like your fir trees, it's a triangle! Only your pizza slice is much tastier. If the slice is 5 inches along the crust and 7 inches from top to bottom, what's its area? Well, 5 inches is our base and that 7 inches is our height. So this pizza slice's area is 1/2 * 5 * 7. That's 17.5 square inches. Okay, well, subtract a few for the bite you just took while doing the math.
337 Lesson Summary In summary, you've proven to be quite the botanist. And you've also mastered some essential geometry. We learned that the area of a rectangle is length times width. We just need to know those two sides and we can determine the area. With a triangle, it's 1/2 times base times height. In a right triangle, our base and height are the two sides that meet to form a right angle. In an oblique triangle, the height is just a line that's perpendicular from the base. Area of Complex Figures Chapter 12 / Lesson 4 Finding the area of shapes isn't always as simple as it is with squares and rectangles. In this lesson, you'll learn how to find the area of complex figures. This includes finding the area of unusual shapes and finding the area of parts of shapes. Area in the Real World Let's say you need to paint some rooms in your house. How much paint will you need? You need to know the surface area of the walls. If you live in a really boring house, all your walls are neat rectangles and squares. But maybe there are angled ceilings. Or windows  you don't need paint to cover the windows, do you? In fact, determining the area of shapes in the real world can seem more complicated than your standard 'area of a rectangle' geometry problem. So let's look at how you can solve area questions with some unusual shapes. It might just save you an extra trip to the store for more paint. Review of Basic Shapes Before we get into complex figures, let's quickly review the area formulas we'll need. First up is the rectangle. A rectangle is a foursided shape where all the angles are 90 degrees. With rectangles, the sides opposite of each other are equal. Remember that  it's important. The area of a rectangle is width x height. In other words, multiply the long side by the short side and you're done!
338 Next is the square. A square is a type of rectangle where all the sides are equal in length. Therefore, you only need to know one side's length to know all of the sides. That's going to come in handy. Also, the area of a square is s^2, where s is the length of one of the sides. So again, if you know one side of a square, you have all the power. Next is the triangle. The area of a triangle is 1/2 * b * h, or 1/2 * base * height. So you need to know more about triangles to find their areas. But there are a ton of useful triangle facts that are helpful. For example, in a triangle with one right angle, you can use the Pythagorean theorem, a^2 + b^2 = c^2, to determine the length of a missing side. Finally, there are circles. If you've ever stared at an extralarge pizza and wondered how much space it takes up, then you've thought about the area of a circle. The formula is pi * r^2, where r is the radius of a circle. That's 3.14 pi, by the way, not pizza pie. Okay, with those formulas in mind, let's get to the task at hand. Complex Shapes Let's look at a wall you need to paint. Okay, you know the measurements, but how do you account for that space in the corner? Well, what if you draw a line here? A complex shape just became two rectangles, and you're a rectangle ninja. But wait, do you know the measurements for these new shapes? You do! This small rectangle has a width of 3 feet. And if the height above it is 4 feet and the whole wall is 8 feet, then the height of the small rectangle is 4 feet. That rectangle must be 3 times 4, or 12 square feet. As for the big rectangle, again, its height is 8 feet. The width is 10 feet of the whole wall minus the 3 feet of the small rectangle, so it's 7 feet. And 8 times 7 is 56 square feet. Add 56 and 12 and you get 68 square feet. Let's look at another. Here's one from that room with a tiny corner. You avoid it because sometimes there are spiders in that corner, but you don't want to have to tell people you didn't paint the whole wall because of spiders. So what can we do? If we draw a line here, perpendicular to the floor, we have a rectangle and a right triangle. Awesome. That rectangle is just 8 by 10 feet, or 80 square feet. And that triangle? You know it's 6 feet high, because your 6`1`` friend always hits his head when he offers to kill the spiders for you. But how long is it? Well, the whole wall is 17 feet. And the rectangle is 8 feet wide. So it's 17 minus 8, or 9 feet. What's the area of a triangle? 1/2 * b * h. So it's 1/2 * 9 * 6, or 27 square feet. This whole wall, spider territory and all, is 80 plus 27, or 107 square feet.
339 Parts of Shapes We've been avoiding walls with doors and windows, but fear not. They're not so bad. Here's a pretty simple wall with one door. You know the wall is 8 feet high by 12 feet wide. It's a rectangle. So the area is width times height, or 8 times 12, which is 96 square feet. But what about the door? Well, the door is also a rectangle. It's 7 feet by 3 feet, so its area is 21 square feet. To determine the amount of wall space, subtract the door's area from the wall. 96 minus 21 is 75. So this wall is 75 square feet. Next up is this wall, which has two windows and a dog door. Little Bagel the beagle would be very upset if you painted his door shut. Let's start with the overall wall. It's an 11 by 8 rectangle. So, it's 88 square feet. The windows are both 3 by 3 squares, making them 9 square feet each, or 18 total square feet. Now, you had to widen Bagel's door when he put on a little weight. It's a 1 by 1.5 rectangle. That's 1.5 square feet. So there are 19.5 square feet to subtract from the 88, leaving 68.5 square feet. All this talk of painting is making me hungry. Let's go back to that pizza I mentioned earlier. What if you wanted to know how much empty surface space a box has outside of the pizza  you know, space that could be more pizza but sadly isn't? Okay, that pizza in a box is really just a circle inside a square, like this. We know this is a 12inch pizza, which is its diameter. That makes its radius 6. Therefore, the area is pi * r^2, or 36 * pi. That's about 113 square inches of hot, gooey goodness. Now, how big is the box? It's a square with a 13inch side. So the area is 13^2, or 169 square inches. Therefore, the pizzaless space is 169 minus 113, or 56 square inches of cardboard sadness. Lesson Summary In summary, the trick with complex shapes is to break them up into rectangles, triangles and other shapes you know. Use the properties of those simpler shapes to find any missing lengths, solve for the smaller areas and then add them together. With parts of shapes, the same rules apply, though sometimes you just need to subtract one or more of the simpler shapes. Now, enough talking about pizza  I think it's time to go find some. Circles: Area and Circumference
340 Chapter 12 / Lesson 5 Understanding how to calculate the area and circumference of circles plays a vital role in some of our everyday functions. They serve as the foundation for operating with threedimensional figures. Learn more about the area and circumference of circles in this lesson. Circles Are Everywhere When was the last time you jumped on a trampoline? Did you ever wonder how much material was required for the center mat or how much steel was needed to create the frame? If you did, you may not have realized that both the mat and the frame are examples of how area and circumference of circles are used in real life. Let's take a brief look at each of these concepts and examine how they are used in both twodimensional and threedimensional situations. Area As you may recall, area is the amount of space taken up by a twodimensional figure, and it is measured in units squared. Most shapes require a formula to calculate area, and circles are no exception. To calculate the area of a circle, use the formula A = pi * r^2. In the formula, r represents the radius, which is a segment that connects the center of the circle to a point on the edge of the circle. It is half the size of the diameter, and every radius inside of a circle will be the same size. Let's take a brief look at how to calculate area of circles. Take a look at circle 'm'. Here, we see that the diameter is 12 inches long. To calculate the area of circle 'm,' we must cut the diameter in half to determine the radius. When we divide 12 by 2, we see that the radius has a length of 6 inches. From here, we will substitute 6 into the equation to get A = pi * (6)^2. Remember to follow the order of operations  we must square the radius before multiplying by pi. When we do, we are able to determine that the total area of circle 'm' is 36 * pi or inches squared. Now, let's take a look at a realworld example. Joe is purchasing material to build his first trampoline. If he wants the diameter of the mat to be 14 feet long, how much nylon will he have to purchase? Once again, we will cut the diameter in half to determine the radius. When we do, we will see that the radius of the mat will be 7 feet. Substituting this radius into the equation gives us A = pi * (7)^2. Once we complete our calculations using the order
341 of operations, we will find that the total area of the mat is 49 * pi or feet squared. Circumference So, that takes care of calculating the amount of space taken up by a circle, but how can we determine the total distance around a circle? For other shapes, this is referred to as perimeter, but circles don't have actual sides like other shapes. Therefore, the term circumference is used. It is defined as the distance around a circle and is represented by the formula C = 2 * pi * r, where r is the radius. Let's examine a few problems, beginning with Joe's trampoline. In addition to buying nylon for the mat, Joe needs to create a steel frame. With a diameter of 14 feet, how much steel must he purchase to build a frame that will enclose the entire mat? As you recall, with a diameter of 14 feet, the radius will be 7 feet. Let's plug this into the formula for circumference. Once we do, we get 2 * pi * (7). After simplifying, we see that the circumference of the mat is 14 * pi or feet, and that is how much steel Joe will need to purchase for the frame. Here's one more. Clara has started her own hat decorating business. For her current design, she wants to line the brim of the hat with ribbon. If the brim has a diameter of 24 inches, how much ribbon will she need? Since the diameter of the hat is 24 inches, we must divide this number by 2 to determine the radius. When we do, we see that the radius is 12 inches and we are ready to plug this into the formula. Doing so gives us 2 * pi * (12), which will equal 24 * pi or inches. Therefore, she will need inches of ribbon to line her hat. Circles in 3D In addition to these examples, it's important to realize that the formulas for area and circumference of circles create the foundation for calculating the surface area and volume of some threedimensional figures, specifically cylinders, cones and spheres. Recall that surface area is the amount of area taken up by each side of a figure, and volume is the amount of space inside of a figure, or the amount that a threedimensional figure can hold. Let's look at each, beginning with cylinders.
342 The surface area of a cylinder is calculated with the formula SA = (2 * pi * r^2) + (2 * pi * r * h). In the first part of the formula, you can see the area of a circle. It is being multiplied by 2, because there are two circles that form the surface of the cylinder. In the second part of the formula, we see the circumference. It is being multiplied by the height of the cylinder to calculate the area for the rounded side of the figure. Cylinder volume is calculated with the formula V = pi * r^2 * h. The area of a circle shows up here as well. This time, it's being multiplied by the height of the cylinder to determine the amount needed to fill the entire figure. Now let's take a look at the role they play with surface area and volume of cones. Can you spot the area or circumference formula for circles? SA = (pi * r * l) + (pi * r^2) V = (1/3) * pi * r^2 * h In both of these formulas, we can see that the formula for the area of a circle is required. Now, let's take a look at surface area for spheres. What formula do you see? SA = 4 * pi * r^2 Here, once again, we have the area of a circle serving as the foundation for this formula. Without the ability to calculate the area and circumference of a circle, finding surface area and volume of cylinders, cones and spheres would not be possible. Lesson Summary In review, the area of a circle is defined as the amount of space covered by the circle and is calculated using the formula A = pi * r^2. Circumference, the distance around the circle, is calculated using the formula C = 2 * pi * r. For both formulas, r represents the radius of the circle. From jumping on a trampoline to designing hats or baking cakes, area and circumference of circles are present in everyday life, and they form the foundation to calculate volume and surface area of cylinders, cones and spheres. Volume of Prisms and Pyramids
343 Chapter 12 / Lesson 6 In this lesson, we'll learn about prisms and pyramids. We'll look at the different types of prisms and pyramids, as well as practice calculating volume for each shape. Prisms and Pyramids Prisms and pyramids are everywhere. If you've ever seen the cover of Pink Floyd's Dark Side of the Moon album, then you've seen a prism in action. Pyramids are even more pervasive. If you have a dollar bill in your wallet, then you're walking around with a picture of a pyramid. Each of these shapes can be found in different forms. Let's take a closer look at what those are and then work on finding the volume of these shapes. Types of Prisms Let's start with prisms. You've probably seen all different kinds of prisms. There are the glass ones used for tricks with lights, like turning sunlight into rainbows. Tents can also be prisms. If you're a chocolate lover, then you know that Toblerone packages their chocolate in prismshaped boxes. A prism is a threedimensional shape with flat sides and two parallel faces. What does that mean? Well, those examples above are all triangular prisms. Notice that each face is a triangle. Those are the parallel faces. And the sides? Yep, they're flat. They're also parallelograms. One way of thinking about prisms is that if you make a slice anywhere that's parallel to the face, the shape will always be the same. Not all prisms are triangular. There are also square prisms. These include cubes where all six sides are the same, but as long as the faces are squares, it's a square prism. See, prisms are defined by those faces. If the face has five sides? It's a pentagonal prism. Also, a barn. That may be a barn. Volume of Prisms Let's say you need to find the volume of a prism. For example, maybe you're camping and you want to fill your buddy's tent with marshmallows. You need to plan stuff like this.
344 Ok, the volume of a prism is pretty straightforward. Start with the area of the face. If it's a triangle, that's 1/2*b*h. If it's a square, it's s^2, and so on. Then you just multiply it by the height of the prism. So the volume of a prism is the area of the base (B) multiplied by the height between the bases (h), which we can just write as B*h. Now, back to that tent. The front, or face, is a triangle. If the base is 4 feet long and it's 4 feet high, then the area is 1/2*4*4, or 8 square feet. Now, this tent is 7 feet long, so that's the height. The volume is just B*h, or 8*7, which is 56 cubic feet. So you're going to need 56 cubic feet of marshmallows. That's a lot of marshmallows. Let's look at another example. Let's say you've moved up from camping pranks and you're hosting a wine and cheese party. You may have a block of cheese you're trying to slice into cubes. Each cube is one cubic centimeter. How many cubes can you get from this cheese? This is a rectangular prism, so you need to know the area of the rectangle and the height. The face of the cheese is 3 cm long by 7 cm wide. The area of a rectangle is length times width. 3*7=21 square centimeters. The height of this block is 14 cm. So the volume is B*h, or 21*14, which is 294 cubic centimeters. So you'll have 294 tiny cubes of cheese. Oh, and if you're like me, remember to subtract a few that you'll eat while you slice. Types of Pyramids Next, let's look at pyramids. People often confuse prisms and pyramids. But you'll never see a rainbow if you hang a pyramid in your window. When people think of pyramids, they think of the pyramids in Egypt. Why? Because they're totally awesome! A pyramid is a shape with a base connected to an apex. The sides of a pyramid always form triangles. The bases can be any shape with three or more sides. Those ones in Egypt are square pyramids, which means their bases are squares. You can also have triangular pyramids and those with more sides to their bases. Volume of Pyramids Ok, so I bet you've always wondered, just how much stone is in one of those Egyptian pyramids? To find the volume of a pyramid, you need to multiply the area of the base (B) by the height (h), then divide by three, or 1/3*B*h. Why the 1/3? Because
345 base times height would give you the volume of a prism. Since the top of a pyramid is a point, you know it has less volume. It's 1/3 the volume. Ok, the largest pyramid in Egypt is the Great Pyramid of Giza. This is a square pyramid, where each side of the base is about 750 feet long. So the area of the base is 750^2, or 562,500 square feet. The pyramid originally was about 480 feet tall. So the volume is 1/3*B*h, or 1/3*562,500*480. That's 90,000,000 cubic feet. Sounds huge, right? As a side note, the largest pyramid in the world isn't in Egypt. It's in Mexico. It's the Great Pyramid of Cholula, and its volume is over 100 million cubic feet. However, today it's covered in plants and topped with a church. Let's look at a smaller example. Here's a triangular pyramid. It's only 6 inches tall. But hey, it's way easier to build. The base is a triangle. Remember, the area of a triangle is 1/2*b*h. It's important to not get confused with the terms. We're working with two things called base and two things called height here. But let's take it one step at a time and start with the triangle. Its base is 3 inches. And its height is 4 inches. Plug that in to 1/2*b*h and you have 1/2*3*4, or 6 square inches. That's our big B. Remember the volume formula: 1/3*B*h. That'll be 1/3*6*6, which is 12 cubic inches. Tourists may not come to see our humble pyramid, but it is a pyramid. Lesson Summary In summary, prisms can be triangular, square or other shapes. The volume of a prism is B*h, where B is the area of the base and h is the height of the prism. Pyramids can have bases that are triangles, squares, or other shapes, too. The volume of a pyramid is 1/3*B*h, where B is the area of the base and h is the height of the pyramid. Volume of Cylinders, Cones, and Spheres Chapter 12 / Lesson 7 In this lesson, we'll learn about the volume formulas for cylinders, cones and spheres. We'll also practice using the formula in a variety of realworld examples where knowing how to calculate volume is helpful.
346 What Is Volume? Have you ever watched an eating contest? It can be kind of gross. But what always amazes me is when seemingly skinny people win. How can someone fit so much food into such a tiny stomach? It's a question of volume. Volume is the capacity of an object, or how much space it occupies. An eating contest tests the volume of the human stomach. In this lesson, we're not going to test the limits of anyone's stomach, but we are going to learn about the volume formulas for some common shapes. Cylinders A cylinder is basically like a big pile of circles. Imagine you have a poker chip. Then you start winning, and you have a bunch of chips. If you stack them together, you've created a cylinder. This should help you remember the volume formula. The volume of a cylinder is pi*r^2*h, where r is the radius of the circle on the cylinder's end. What looks familiar in there? Pi*r^2  that's the area of a circle. And if you had just one perfectly flat poker chip, its area would be pi*r^2. But since you're a baller, you have a stack. So you take the area formula and you multiply it by the height of your chips. Let's try some examples. Here's a can of a new soda, MegaSurge. Not only is the neon yellow soda packed with 10 times the caffeine of regular soda, it also comes in an oversized can. But how much soda does it hold? You measure the width of the top, which is the diameter of the circle, and it's 6 inches. So the radius, which is half the diameter, is 3 inches. And how tall is it? 9 inches. Let's use our volume formula: pi*r^2*h. Pi*3^2*9. That's about 254 cubic inches. That's a lot of soda! Okay, here's another. After drinking that soda, you decide you need to get back to more healthy foods, so you visit your aunt's farm for some fresh veggies, eggs and other good stuff. She has a grain silo. You want to know how tall it is. She tells you it holds 3,141 cubic feet of grain when full. So you know its volume. And the radius is 5 feet. How tall is it? Let's use our formula! 3141 = pi*5^2*x = 79x. x? 40, so it's about 40 feet tall.
347 Cones As it turns out, your aunt uses that grain to feed her cows. And she uses her cows' milk to make ice cream. Your aunt is kind of awesome. And now you want to know how much ice cream you can pack into your cone. The more ice cream you can pack into the cone, the happier you'll be if the scoop on top should fall off or, you know, get eaten too quickly. The volume of a cone is 1/3*pi*r^2*h, where r is the radius of the circle at the wide end of the cone. Notice how this is the same as the cylinder formula. There's just that extra 1/3. What's that for? Well, one end is a circle and one is a point. So there is 1/3 of the volume of a cylinder. In other words, you'd need 3 coneshaped cones to match the volume of 1 cylindershaped cone. Maybe those people with the cake cones are onto something. But let's talk about the cone you have. It's 5 inches tall and 2 inches across the top. Remember, that 2 inches is the diameter. That means the radius is 1 inch. Okay, the formula: 1/3*pi*r^2*h. That's 1/3*pi*1^2*5, which equals a little over 5 cubic inches. That's not a lot of ice cream. So you convince your aunt to make waffle cones, which are bigger, but how much bigger? You make a waffle cone that's 8 inches tall and 5 inches across the top. So its radius is 2.5 inches. What is the volume it can hold? 1/3*pi*r^2*h. That's 1/3*pi*2.5^2*8, or just over 52 cubic inches. That's about 10 times the ice cream! That's a big waffle cone. Spheres Maybe you should get some exercise after all that ice cream and play some basketball. You decide to impress your friends with your awesome geometry skills by explaining how much air fits in the basketball. But wait, the basketball is a sphere. What's the formula for that? The volume of a sphere is 4/3*pi*r^3, where r is the radius of the sphere. Where does this come from? Well, I could just tell you 'calculus,' but that wouldn't help you remember it. But if you remember that the surface area of a sphere is 4*pi*r^2, you can remember that volume is related to surface area. The other volume formulas we've been discussing involve 3s, so if you add some 3s to the surface area formula, you get 4/3*pi*r^3.
348 That 4/3 is definitely unusual, which is the opposite of spheres. Spheres are everywhere! They're as small as the tiny peas from the garden and as huge as the Death Star in Star Wars. Without spheres, there'd be no baseball, soccer, tennis or golf. Well, there'd be no Earth either, which is kind of important. But back to the basketball. All you need to know is its radius or diameter. You know the diameter is 9 inches, so its radius is 4.5 inches. To the formula! 4/3*pi*r^3. 4/3*pi*4.5^3. That's about 382 cubic inches. Your friends are impressed. In fact, one asks you to help him with his science project. He's building a model of the solar system, and he's trying to understand the difference in sizes between the planets. Your ice cream experiment makes you think that volume may be a good way to help. He knows the diameter of Earth is about 7,900 miles, the diameter of Mars is about 4,200 miles, and the diameter of Jupiter is about 89,000 miles. But what does that mean in volume? Okay, first, let's convert those all to radii. Earth's radius is 3,950 miles. With Mars, it's 2,100 miles. And Jupiter's is 44,500 miles. Let's break out that volume formula: 4/3*pi*r^3. First, Earth: 4/3*pi*3950^3 = 258 billion cubic miles. That's a lot! Next, Mars: 4/3*pi*2100^3 = 39 billion cubic miles. That's a lot smaller. What about Jupiter? 4/3*pi*44,500^3 = 369 trillion cubic miles! That means well over 1,000 Earths could fit inside Jupiter! Your buddy is going to need a bigger model. Lesson Summary We learned about 3 unique shapes. First, cylinders, like that can of MegaSurge or the silo. The formula for the volume of a cylinder is pi*r^2*h. Second, we looked at cones as we feasted on ice cream. The formula for the volume of a cone is 1/3*pi*r^2*h. Finally, we looked at spheres, like a basketball and the planets. The formula for the volume of a sphere is 4/3*pi*r^3. What is Area in Math?  Definition & Formula Chapter 12 / Lesson 8 Area is the size of a twodimensional surface. This lesson will define area, give some
349 of the most common formulas, and give examples of those formulas. A quiz at the end of the lesson will allow you to work out some area problems on your own. Definition of Area The mathematical term 'area' can be defined as the amount of twodimensional space taken up by an object. The use of area has many practical applications in building, farming, architecture, science, and even deciding how much paint you need to paint your bedroom. The area of a shape can be determined by placing the shape over a grid and counting the number of squares that the shape covers, like in this image: The area of many common shapes can be determined using certain accepted formulas. Let's take a look at the most common formulas for finding area. Area Formulas To find the area of a rectangle, you use this formula: Area = length * width The area of a square is found with this formula: Area = s^2, where s = side The formula for the area of a triangle is: Area = (1/2) b * h, where b = base and h = height To find the area of a circle, use this formula: Area = pi * r^2, where r = radius The area of a parallelogram is found using this formula:
350 Area = b * h, where b = base and h = vertical height The formula for the area of a trapezoid is: Area = (1/2) * (a + b) * h, where a =base 1, b = base 2, and h = vertical height An ellipse's area is found this way: Area = pi * a * b, where a = radius of major axis and b = area of minor axis Units of Area Finding the area of a shape always requires the multiplication of two lengths. In a square, it's side multiplied by side. In a circle, it's the radius squared. For an ellipse, it's the radius of the major axis multiplied by the radius of the minor axis. Due to this, the units given to area will always be squared (feet squared, inches squared, etc.). Anything multiplied to itself is squared, whether it is a number or not. Finding the Area Example Problems Let's practice finding the area with some example problems. What is the area of a square with side length of 5 inches? Remember, the formula for finding the area of a square is A = s^2. The sides of this particular square are 5 inches. Plug that into the formula to get A = 5^2 = 25 in^2. What is the area of this parallelogram? Remember, the formula is A = b * h. So, for this example, the area would be A = 3 * 12 = 36 mm^2.
351 Finding the Area of Uncommon Shapes If you are asked to find the area of an uncommon shape, it can be done by breaking the shape into more common shapes, finding the area of those shapes, and then adding the areas together. Let's look at some examples: Find the area of this shape: The first step to solving this problem is to divide the shape into shapes we can find the area of easily. This shape can be divided into a triangle and a square.
352 You can use the information given to determine the lengths you need to calculate the area. Since you know the height from the point of the triangle to the bottom of the square is 10 cm and the height of the square is 8 cm, the height of the triangle must be 2 cm. The base of the triangle is equal to a side of the square, which is 8 cm. You can use these numbers to determine the area. A of square = s^2 = 8^ = 64 cm^2. A of triangle = (1/2) * b * h = (1/2)* 8 * 2 = 8 cm^2. Then you just add the areas together to get the total area of the figure. A = = 72 cm^2. Find the area of the figure shaded in red, given that the dimensions of the rectangle are 11 inches by 7 inches This example is a bit different, since you only want the area of a small portion of the figure. This figure can be broken down into a rectangle and a circle only, this time, the area of the circle needs to be subtracted from the area of the rectangle to get the remaining area. A of rectangle = l * w = 11 * 7 = 77 in^2. A of circle = pi * r^2 = pi * (3.5^2) = in^2. The radius of the circle is determined from the diameter of the circle, which is equal to the width of the rectangle because the circle is as wide as the rectangle. The radius is half of the diameter (1.2 * 7 = 3.5). A = = in^2. Lesson Summary The area of a twodimensional figure is a calculation of the space taken up by the figure. Figures such as squares, triangles, circles, and others have specific formulas that can be used to find their area. The area of other figures can be determined by breaking the figure into parts whose area can be easily determined.
353 Properties of Shapes: Rectangles, Squares and Rhombuses Chapter 13 / Lesson 1 What's the difference between a square and a rectangle? What about a rhombus and a square? In this lesson, we'll look at the properties of these shapes. Studying Shapes Let's talk about shapes. There are all kinds of shapes and they serve all kinds of purposes. If the wheels on your bike were triangles instead of circles, it would be really hard to pedal anywhere. And if bowling balls were cubes instead of spheres, the game would be very different. Here, we're going to focus on a few very important shapes: rectangles, squares and rhombuses. All of them are quadrilaterals. That just means they all have four sides. Remember that 'quad' means 'four.' That means no pentagons or octagons will be discussed here. Sorry, fans of Department of Defense headquarters or, um, stop signs. But even though they all have four sides, they all have their own special properties that make them unique. Rectangles Let's start with rectangles. A rectangle is a foursided shape with all right angles. If you want to know if a shape is a rectangle, you just have two tests. Is it foursided? And are all the angles 90 degrees? If both answers are yes, then you're looking at a rectangle. Rectangles are everywhere. Think about your average room. What's a rectangle in it? Doors, tables, windows, posters on the walls  they're all foursided shapes with all right angles. Even the screen you're looking at right now is probably a rectangle.
354 Rectangles have a few special properties. First, opposite sides are parallel. Second, opposite sides are equal in length. In this rectangle, we know side AB is parallel to side CD. And BC is parallel to AD. Also, if we know AB is 6, then so is CD. If BC is 4, then so is AD. The fun thing about rectangles is that each pair of opposite sides can be a totally different length than the other pair. You can have a super skinny rectangle like this skyscraper or a very even one like this old album cover. Squares That old album cover fits both the definition of a rectangle and the definition of our next shape, the square. Squares are a very special subset of rectangles. A square is a foursided shape with all right angles and sides of equal length. Does that definition look familiar? Here are the steps to define a square: Is it foursided? Are all the angles 90 degrees? If yes, then you have a rectangle. If all the sides are the same length, then it's not only a rectangle, it's also a square. That means that all squares are rectangles. But not all rectangles are squares since a rectangle's pairs of sides can have different lengths. Just like rectangles, squares are everywhere. In addition to the album cover, think about the spaces on a chessboard, stamps, floor tiles and even a snack of crackers and cheese. Because each side of a square has the same length, you don't need to be given much information to solve most problems. For example, if you see this square where you know this side is 5, then you know all the other sides are 5 as well. So the perimeter of this square is 5 *4, or 20. The area of a square is s^2, or one side squared. So the area of this square is 5^2, which is 25. Rhombuses Then there is the rhombus. A rhombus is a little different from a square or rectangle. Here are the questions to ask if you think you're dealing with a rhombus: Is it foursided? Are all the sides equal in length? If both answers are yes, then what you have is a rhombus. Did you notice what was missing? Right angles. A rhombus doesn't need to have right angles. It can, but that's the big difference with a rhombus. I like to think of it like this: The word 'rhombus' is kind of like the word 'rhino.' If a rhino charges at a square and
355 knocks it askew, it's not a square anymore. But it is still a rhombus! Rhinos or no, the definition of a rhombus is a foursided shape with sides of equal length. There are a few notable properties for rhombuses. First, the opposite sides are parallel. That's true for rectangles and squares, too. But in a rhombus, even if the angles aren't 90 degrees, the opposite sides are still parallel to each other. So in this one, AB is parallel to CD. And AD is parallel to BC. Also, the opposite angles are equal. Here, angle A equals angle C. And angle B equals angle D. Plus, here's a fun one: If you draw diagonal lines from the corners, those lines form right angles. Since a square is a rhombus, that's true for squares as well. And no matter how far that rhino pushes the rhombus, those diagonals still form right angles. A square is a rhombus, but a rhombus isn't necessarily a square. And a rectangle can be a rhombus, but if the sides of a rectangle aren't all equal in length, then it's not a rhombus. Lesson Summary In summary, we looked at three different types of quadrilaterals, or foursided shapes. First, there's the rectangle, which is a foursided shape with all right angles. Its opposite sides are parallel and equal in length, but each pair of sides isn't necessarily the same length as the other pair. Second, there's the square, which is a foursided shape with all right angles and sides of equal length. A square is a type of rectangle, just one where all four sides are the same length. Finally, there's the rhombus, which is a foursided shape with sides of equal length. The angles can be 90 degrees, but they don't need to be. So a square is a rhombus, but not every rhombus is a square. Properties of Shapes: Quadrilaterals, Parallelograms, Trapezoids, Polygons Chapter 13 / Lesson 2 Foursided objects are more than just squares and rectangles. In this lesson, we'll discuss quadrilaterals, parallelograms and trapezoids. We'll also discuss polygons, objects that can have more sides than you can count.
356 All About Shapes Triangles? They're ok, but you hear about them everywhere. Squares, too. What about the more unusual shapes? Those ones with the funny names and weird properties. Let's learn a bit about them. Quadrilateral First up: the quadrilateral. It's a mouthful, I know. But what do you see in that word? Quad. Where have you seen that? Quadruplets? How many babies is that? It's four! Or, about three more than a normal person can handle. A Quadrilateral is just a foursided shape. Squares and rectangles are quadrilaterals. So are more unusual shapes, like this boomerang. What do we know about quadrilaterals? If they have four sides, they also have four vertices, or corners. And while those vertices can be any angle, they need to add up to 360 degrees. Why? Think about a square. All the angles are 90 degrees. 90 times 4 is 360. Now, what about that boomerang? If that first angle is 130 degrees and the two small corners are 20 degrees, when we make those last two lines meet, the angle has to be 190 degrees = 360. No matter how you change the angles, they always add up to 360. The Perimeter of Any Quadrilateral is just the sum of the four sides. In this quadrilateral, the sides, in inches, are 4, 4, 5 and 2. Therefore, the perimeter is , or 15 inches. Is every square a quadrilateral? They always have exactly four sides, so yes. But is every quadrilateral a square? Well, no. Some are rectangles. And some are other things, like parallelograms. Parallelograms Do you remember what parallel means? Two parallel lines are two lines that will never meet. They're like the lines painted on the sides of an endless straight road.
357 A Parallelogram is a foursided shape with opposite sides that are parallel. Because of the parallel lines, the opposites sides are equal in length. There are some unique properties of the angles inside parallelograms. First, the opposite angles are equal. Also, the adjacent angles are supplementary, which means they add up to 180 degrees. The perimeter of this parallelogram, where the sides are 4, 4, 5 and 5, is 18. When we have a specific type of quadrilateral like this, we can determine its area. The Area of a Parallelogram is the base times the height. So in this example, where the base is 7 inches and the height is  whoa, wait, that's not the height! Remember, the height is the distance from the base to the top, so while that side is 4 inches, the height is actually 3 inches. So the area is 7 times 3, or 21 square inches. This is the same area formula as with squares and rectangles. And if you remember, squares and rectangles also have sets of parallel sides, so they're both parallelograms! Trapezoids But what if only one set of sides of a quadrilateral is parallel? Then, it can't be a parallelogram. But it is a trapezoid. A Trapezoid is a foursided shape with at least one set of parallel sides. It can have two and be a parallelogram. But if two sides aren't parallel, then it's just the lowly trapezoid. That's a weird name, and it can be kind of a weird shape. You know what it reminds me of? A table. And guess what? That's where the name comes from! Like a table, two sides, or the tabletop and the base, or the floor, need to be parallel. If not, your poor meatball may roll off the table and right out the door. So, in a trapezoid, the parallel sides are called the bases. And just like with a table, the other sides are called legs. The perimeter is the same as with parallelograms. The perimeter of this trapezoid? Let's see. The bases are 8 inches and 6 inches. The legs are 5 inches each = 24 inches. Area is a little more complicated. You can't just multiply base times height, because the bases are different. So the area of a trapezoid is the average of the bases times the height. Or a + b / 2 * h. Let's find the area of this one. The bases are 8 inches and 6 inches. What's the height? The legs are each 5 inches. Remember, the height is the distance from one base to the
358 other. The height of this trapezoid is 4 inches. Let's plug those numbers into our formula = / 2 = 7. 7 * 4 = 28. That's 28 square inches. Polygons We've been talking about foursided objects, but what's the term that also includes shapes with different numbers of sides? Whoa. No. That's polygamy, or marrying multiple people. I meant polygons. Totally different. But notice that they both begin with poly. The prefix poly comes from the Greek word for many. And gon means angles. So a polygon is a closed, twodimensional shape with many sides and angles. Polygons are polygamous in that they are shapes that join multiple sides and angles. There's really no more to them. They can have three sides or eight, like an octagon, or a million, like the megagon. They don't need to have parallel lines or right angles. If you think about that, it means that a triangle is a polygon. So are squares, rectangles and, yes, quadrilaterals, parallelograms and trapezoids. They all are closed shapes with many sides, so they're all polygons! The perimeter of a polygon is still just the sum of the sides. It's 3 for this one and 5 for this one and 8 for this one. Since polygons can get awfully complicated, determining their areas is more advanced math than we're going to get into here. But no matter how big or small that stop sign is, you still need to obey it! Lesson Summary In summary, a Quadrilateral is any foursided shape. A Parallelogram is a foursided shape with opposite sides that are both parallel and equal in length. The area of a parallelogram is the base times the height. A Trapezoid is a foursided shape with at least one set of parallel sides. The area of a trapezoid is the average of the bases times the height. Finally, a Polygon is a closed, twodimensional shape with many sides. Everything from a triangle to an octagon to a megagon is a type of polygon. For all of these shapes, the perimeter is the sum of the sides.
359 How to Identify Similar Triangles Chapter 13 / Lesson 3 Similar triangles have the same characteristics as similar figures but can be identified much more easily. Learn the shortcuts for identifying similar triangles here and test your ability with a quiz. The Language of Similarity Similar figures are figures that have the same shape but are different sizes. They have congruent corresponding angles and proportional corresponding sides. Similar triangles are a type of similar figure, and determining their similarity is much easier thanks to the triangle similarity theorems. These theorems, which are Angle  Angle (AA), Side  Angle  Side (SAS) and Side  Side  Side (SSS), make it possible to determine triangle similarity with minimal calculations. Before we go any further, let's review key terms that will help these theorems make sense. When parts of a figure are corresponding, this means that they are in the same location in each figure. Sides are proportional when the ratios between the corresponding sides are congruent. So, if you create ratios or fractions comparing all of the corresponding sides, each will have the same value and reduce to the same number. When discussing congruent or similar figures, the included angle is the angle formed by the congruent or proportional sides. In this triangle, since side AB is congruent to side DE and side BC is congruent to side EF, then angles B and E are the included angles. Triangle Similarity Theorems Okay. Now that we're refreshed on vocabulary, let's further examine each of the similarity theorems. We'll begin with Angle  Angle (AA). For two triangles to be similar by Angle  Angle (AA), two angles of one triangle are congruent to two angles of another triangle. Look at triangle JKL. Angle J is 52 degrees and angle K is 60 degrees. If we subtract 52 and 60 from 180 (the total number of degrees that all angles in a triangle must add up to), we will see that angle L is 68 degrees. Now, look at triangle MNO. If angle M is congruent to angle J, and angle N is congruent to angle K, what can we say about angles L and O? We can say that they
360 are congruent as well. With three pairs of congruent angles, the triangles are the same shape but different sizes, meaning that their sides are proportional and the triangles are similar. For Side  Angle  Side (SAS), an angle of one triangle must be congruent to the corresponding angle of another triangle, and the lengths of the sides including these angles are in proportion. With congruent included angles, the proportional sides can't fluctuate, and the third side in both triangles must be a specific length. So, with two sides already proportional, the lengths of the third sides must also be proportional, proving triangle similarity. The last theorem is Side  Side  Side (SSS), which means that the three sets of corresponding sides of two triangles are in proportion. If the ratios of all corresponding sides are equal, then the sides are similar and so are the triangles. When determining which theorem proves similarity, don't overthink it; just look at the letters in each theorem. For triangles to be similar by Angle  Angle (AA), the measures of two angles in each triangle will be provided. If similar by Side  Angle  Side (SAS), then you will have the measures of two sides and the included angles of both triangles. For Side  Side  Side (SSS), you will have all three side lengths for both triangles. Let's practice. Are They Similar? Is triangle ABC similar to triangle DEF? Let's examine the given information. We have the lengths for two sides in both triangles and the measures of the included angles. This sounds like Side  Angle  Side (SAS). But, before concluding similarity by this theorem, we must check for congruent angles and proportional sides. Angle B and angle E both measure 63 degrees, so the included angles are congruent. To set up the side proportions, always compare the two smallest sides together and the two largest sides together, going in the same order between triangles. For this example, our ratios are 3/6 and 5/8. If we convert to decimals, 3/6 =.5 and 5/8 =.625. Since these ratios are not equal, triangle ABC is not similar to triangle DEF. For our next example, determine if triangle RST is similar to triangle WXY. Since we were given the lengths of all three sides in both triangles, Side  Side  Side (SSS) is the only theorem that can prove similarity. Let's set up our proportions. The two smaller sides have a ratio of 6/3, the largest sides have a ratio of 10/5 and the remaining sides have a ratio of 8/4. From simplifying, all of the ratios equal two.
361 Therefore, triangle RST is similar to triangle WXY by the Side  Side  Side (SSS) similarity theorem. Let's do one more. Is triangle CRE similar to triangle PHB? Based on the given information, these triangles can only be congruent by Angle  Angle (AA). But, it appears that only one pair of angles are congruent. So you may think the triangles aren't similar. Before coming to a conclusion, let's calculate the measure of angle E. Subtracting , we find that angle E measures 45 degrees. With this information, we now see that angle R and angle H are congruent, as well as angle E and angle B. Therefore, triangle CRE is similar to triangle PHB by the Angle  Angle (AA) similarity theorem. Lesson Summary Similar triangles possess the same characteristics as other similar figures: congruent corresponding angles and proportional corresponding sides. The triangle similarity theorems, which are Angle  Angle (AA), Side  Angle  Side (SAS) and Side  Side  Side (SSS), serve as shortcuts for identifying similar triangles. When assessing similarity, always begin by examining the provided information, which will help you figure out which theorem to use when determining if triangles are similar or not. Applications of Similar Triangles Chapter 13 / Lesson 4 Similar triangles are used to solve problems in everyday situations. Learn how to solve with similar triangles here, and then test your understanding with a quiz. Using Similar Triangles Sarah is standing outside next to a flagpole. The sun casts a 4 ft. shadow of Sarah and a 7 ft. shadow of the flagpole. If Sarah is 5 ft. tall, how tall is the flagpole? Without a ladder and measuring stick, you may think that solving this problem is impossible. However, finding the solution may be easier than you think. For this problem, and others like it, solving becomes a matter of similar triangles. In another lesson, we learned that similar triangles have congruent corresponding angles and proportional corresponding sides and can be quickly identified by the AngleAngle (AA), SideAngleSide (SAS), and SideSideSide (SSS) similarity
362 theorems. To solve with similar triangles, we will use their side lengths to set up proportions, meaning that we will create fractions for the corresponding sides and set them equal to each other. Let's begin our practice with some basic examples. Solving with Similar Triangles Take a look at triangle ABE and triangle ACD. From the picture, we see that Angle B is congruent to Angle C and Angle E is congruent to Angle D. Therefore, these triangles are similar by AA. Let's set up our proportions. By going in the same order for each fraction, our beginning proportion is 8/5 = x/6. To get x by itself, we must crossmultiply, which will give us 5x = 48. Next, we will divide both sides by 5 to conclude that x = 9.6. Here's another one: This time, we have triangle MOQ and triangle NOP. Once again, based on the congruent marks in the picture, we know that these two triangles are similar by AA. So, let's start the problem. In the figure, we were given the length of side MO, meaning that we also need the length of side QO for the proportion. But, we were not given this information, so we must calculate it. To get the length of QO, we must add QP + PO since these two pieces form segment QO. By doing this, we add to see that QO = 15. Now, we are ready for our proportion. By going in the same order, our proportion is 7/15 = x/18. Then, when we crossmultiply, we see that 15x = 126. Once we divide both sides by 15, we can conclude that x = 8.4. Applications of Similar Triangles Now that we've covered some of the basics, let's do some realworld examples, starting with Sarah and the flagpole. Recall that Sarah is 5 ft. tall and has a 4 ft. shadow, and we are looking for the height of the flagpole, which has a 7 ft. shadow. Do you see the similar triangles? Sarah and the flagpole are the vertical legs and the shadows are the horizontal legs. Connecting the end of the shadows to the top of Sarah and the flagpole complete the triangles. Since the sun is casting both shadows, and since Sarah is standing right beside the flagpole, the top angles in both triangles are congruent. Additionally, since Sarah and the flagpole are perpendicular with the
363 ground, each triangle has a right angle. Therefore, these triangles are similar by the AA similarity theorem. Now we can write a proportion. Using x to represent the height of the flagpole, we begin with 5/4 = x/7. After crossmultiplying, we have 4x = 35, and then, by dividing both sides by 4, we see that x = Therefore, the flagpole has a height of 8.75 ft. Let's try another: Julie places a mirror on the ground 30 ft. away from the base of her office building and walks backwards until she can see the top of the building in the mirror. If Julie is 5.5 ft. tall and is standing 5ft. away from the mirror, how tall is the building? For this scenario, Julie and the building serve as the vertical legs of the triangles and the distances from the mirror to Julie and the building are the horizontal legs. The triangles are completed by connecting the mirror to the top of both Julie and the building. Since Julie can see the top of her building through the mirror, the two angles formed at the mirror are congruent. Also, Julie and her office building form right angles with the ground, showing us that these two triangles are also similar by AA. Now, it's time for the proportion. By letting x represent the height of the building, our beginning proportion is x/30 = 5.5/5. From crossmultiplying, we get 5x = 165. Then, by dividing both sides by 5, we can conclude that x = 33. Therefore, the height of Julie's office building is 33 ft. Lesson Summary In conclusion, similar triangles can be applied to solve everyday, realworld problems. Since the triangles are similar, solving with them requires us to set up proportions that compare their corresponding sides. After going in the same order to set up the proportions, crossmultiplying is necessary to complete and solve the problem. Properties of Congruent and Similar Shapes Chapter 13 / Lesson 5 In this lesson, we'll look at triangles, rectangles and other shapes that share properties. This includes both congruent and similar shapes. We'll also practice identifying the
364 missing properties of these shapes. Comparing Shapes Triangles, rectangles, parallelograms... geometric figures come in all kinds of shapes. This diversity of figures is all around us and is very important. For example, making stop signs octagons and yield signs triangles helps us to differentiate them from a distance. Figures of the same shape also come in all kinds of sizes. The debit card in your wallet and the billboard on the interstate are both rectangles, but they're definitely not the same size. If they were, you'd either never be able to read that billboard, or your wallet would need to be a really inconvenient size. When we study figures, comparing their shapes, sizes and angles, we can learn interesting things about them. So let's get to it! Congruent Shapes Sometimes the easiest shapes to compare are those that are identical, or congruent. Congruent shapes are figures with the same size and shape. You could also think of a pair of cars, where each is the same make and model. They're alike in every way. Well, until one gets awesomely tricked out. For a more geometrybased example of congruency, look at these two rectangles. We can see that both figures have the same lengths and widths. Next, look at these hexagons. They aren't turned the same way, but they are congruent. The sides and angles all match. If we took one, turned it, and put it on top of the other, you'd see that they match perfectly. That's what being congruent means. Sometimes, you'll be given special clues to indicate congruency. Let's look at two congruent triangles. When two shapes, sides, or angles are congruent, we'll use this symbol. We know angle A is congruent to angle D because of these symbols. Likewise, angle B is congruent to angle E and angle C is congruent to angle F. We also have these hash marks to indicate that line AB is congruent to line DE, line BC is congruent to line EF and line AC is congruent to line DF.
365 Practice with Congruent Shapes When you have congruent shapes, you can identify missing information about one of them. Consider these two triangles. We know they're congruent, which enables us to figure out angle F and angle D. We just need to figure out how triangle ABC lines up to triangle DEF. If AB is congruent to DE and AC is congruent to DF, then angle A is going to be congruent to angle D. So angle D is 55 degrees. We could use the same logic to determine that angle F is 35 degrees. Or we could just know that the sum of the interior angles of a triangle is 180, and subtract 55 and 90 from 180 to get 35. Either way, we now know all the angles in triangle DEF. Sometimes you have even less information to work with. But you can still figure out quite a bit. Consider these triangles. All we're given is the statement that triangle MNO is congruent to triangle PQR. We also know the measures of angles O and Q. This is actually everything we need to know to figure out everything about these two triangles. Use the order of the vertices to guide you. M corresponds to P, N to Q and O to R. So angle M is congruent to angle P, N to Q and O to R. That means angle R is 50 degrees and angle N is 100 degrees. Since we need the angles to add up to 180, angles M and P must each be 30 degrees. Similar Shapes Similar shapes are much like congruent shapes. The key difference is that similar shapes don't need to be the same size. So similar shapes are figures with the same shape but not always the same size. Remember those two cars we looked at? They're similar. But so are one car and a Matchbox version. That Matchbox car's the same shape, just much smaller. Here are two similar rectangles. Notice that the 2/5 is equal to 4/10. If we knew the rectangles were similar, but we didn't know the length of the orange one, we could set up the equation 2/5 = 4/x and solve for x. Here are two similar triangles. We'd identify them as similar using this symbol. We'd say triangle ABC is similar to triangle DEF. We're given the lengths of the sides, so we can see that AB/DE = BC/EF = AC/DF. In similar shapes, the corresponding angles are congruent. That means that angle A is congruent to angle D, angle B is congruent to angle E and angle C is congruent to angle F.
366 Practice with Similar Shapes Let's try practicing with a few similar shapes. Here are two similar rectangles. Can you figure out x? You just need to set up a simple equation: 3/6 = 7/x. Cross multiply: 3x = 42. x = 14. We did it! Here's a pair of triangles. This time, there are two variables: x and y. Let's start with x. 8/6 = 12/x. 8x = 72. x = 9. Ok, now y. 8/6 = 16/y. 8y = 96. y = 12. That's it! The properties of similar shapes aren't limited to rectangles and triangles. They work for more complicated shapes, too. Let's say you want to build a scale model replica of the Millennium Falcon from Star Wars in your garage. Why? Why not? The original ship is about 115 feet long and 85 feet wide. Your garage? It's only 24 feet by 20 feet. If you want to make it as big as possible, then you'll make your ship 24 feet long. How wide will it be? 115/24 = 85/x. 115x = x = 18. So your ship will be 24 feet by 18 feet. Granted, that leaves you no room to walk around it or fit it through the door. But that's ok. It probably won't fly. Or will it? Lesson Summary In summary, congruent shapes are figures with the same size and shape. The lengths of the sides and the measures of the angles are identical. They're exact copies, even if one is oriented differently. Similar shapes are figures with the same shape but not always the same size. Because the shapes are proportional to each other, the angles will remain congruent. And you can always find the length of the sides by setting up simple equations. Parallel, Perpendicular and Transverse Lines Chapter 13 / Lesson 6 What are the different types of lines? Where are they visible in the real world and how can you recognize them? Find out here and test your knowledge with a quiz.
367 Lines, Lines Everywhere Take a look at your surroundings. Are you sitting at a desk? Are you close to a window with blinds? If you look out that window, can you see the next street or a highway? If you answered yes to any of these questions, then you are surrounded by lines, which are everywhere! In this lesson, we are going to take a closer look at parallel lines, perpendicular lines and transverse lines. Each of these types of lines are classified as coplanar lines, meaning that they are located on the same plane, which is a flat, twodimensional surface. Let's examine and practice with each one. Parallel Lines Parallel lines are defined as coplanar lines that do not intersect. They have the same slope and, just as the definition states, will never, ever meet at any point. Think about it: since slope is referred to as rise over run, having the same slope means that two lines will rise and run at the exact same rate, ensuring that they will never intersect each other. Let's take a look at reallife examples of parallel lines. First, we have a window with blinds. Here, you can see that each blind is moving in the same direction and never touches another blind. Next, we have a parking lot. Notice that all of the lines are going in the same direction. Perpendicular Lines Perpendicular lines are coplanar lines that intersect and form a 90degree angle. So, any time you have perpendicular lines, you will also have right angles and vice versa. The slopes of perpendicular lines are opposite reciprocals of each other. Being opposite means that one slope will be positive and the other will be negative. Being reciprocals means that one slope will be the upside down or flipped version of the other. Perpendicular lines are also visible in the real world. Take a look at a desk. Can you see how the top of it lays flat on all the legs? This means that the top of the desk is perpendicular to the legs and forms ninetydegree angles, which keeps things from sliding off of it.
368 Parallel, Perpendicular or Neither? Now, let's practice what we've learned so far. If line g = 3x + 7 and line h = 3x  2, are these lines parallel, perpendicular or neither? Let's begin by looking at their slopes, which are the numbers in front of the x variables. Line g has a slope of three and line h has a slope of negative three. Their slopes are the same number, but one is positive and the other is negative. so they are not exactly the same. For this reason, we know that line g is not parallel to line h. Also, though their slopes are opposites, they are not reciprocals of each other. Therefore, we can also conclude that these two lines are not perpendicular. For our next example, line j = 4/3x + 2 and line k = 3/4x + 5. Are these two lines parallel, perpendicular or neither? By looking in front of the x variables, we see that line j has a slope of fourthirds, and line k has a slope of negative threefourths. These slopes are not congruent, so the lines cannot be parallel. However, one slope is positive and the other slope is negative. Additionally, these slopes are reciprocals or flipped fractions of each other. Therefore, we can conclude that the lines are perpendicular. Transverse Lines A transversal is a line that intersects two or more coplanar lines at different points. For example, in this figure, line t is a transversal because it intersects both line a and line b. Transversals have an important role in geometry because they are needed to form alternate interior angles, alternate exterior angles, consecutive interior angles and corresponding angles. In the real world, transversals are highly visible on street maps. Take a look at this one. Here, you can see that Elm Street is a transversal to Asbury Street, W. Taylor Street and Villa Avenue. Now, let's practice identifying a transversal. Take a look at the following scenario. Which line is the transversal? Line a does not intersect any other line. Line b intersects line c. Line c intersects line b and line d, and line d intersects line c. Therefore, since a transversal must intersect at least two lines, we can conclude that line c is the transversal.
369 Lesson Summary In review, remember that all of the lines we discussed are coplanar. Parallel lines have congruent slopes, perpendicular lines have opposite reciprocal slopes, and to be a transversal, a line must intersect at least two other lines at different points. From office furniture to highway road maps, lines are everywhere. Whether parallel, perpendicular or transverse, lines provide structure for our everyday lives. Types of Angles: Vertical, Corresponding, Alternate Interior & Others Chapter 13 / Lesson 7 In addition to basic right, acute, or obtuse angles, there are many other types of angles or angle relationships. In this lesson, we will learn to identify these angle relationships and discuss their measurements. Introduction Have you ever looked at a sliced pizza and noticed that the beginning of each pizza slice was the same size? Did you ever take the time to wonder why that was? Or have you ever examined the lines in a parking lot? While you may have noticed all the lines that form the parking spaces, did you ever think about the angles that were formed? Well, if you haven't before, I'm sure you're thinking of them now. In every pizza and in every parking lot, there are many different angles and angle relationships. In this lesson, we are going to learn about these relationships, ways to identify relationships and examine the measures of these angles. Types of Angles The first angle relationship that we will discuss is vertical angles. They are defined as a pair of nonadjacent angles formed by only two intersecting lines. They are known as 'Kissing Vs' and always have congruent measures. In the figure, angles 1 and 3 are vertical, as well as angles 2 and 4. The second relationship is corresponding angles. They are considered to be in the same location at each point of intersection. For example, take a look at angles 1 and 3.
370 They are both in the upper left corner. Another pair of corresponding angles is angles 6 and 8, which are both in the lower right corner. Next we have alternate interior angles. Located between the two intersected lines, these angles are on opposite sides of the transversal. Angles 2 and 7, as well as angles 3 and 6 are examples of alternate interior angles. Similarly, we also have alternate exterior angles that are located outside of the two intersected lines and on opposite sides of the transversal. An example of this relationship would be angles 1 and 8, as well as angles 4 and 5. The last angle relationship is consecutive interior angles. These angles are located on the same side of the transversal and inside of the two lines. In the diagram, angles 2 and 3 are consecutive interior angles, and so are angles 6 and 7. With the exception of vertical angles, all of these relationships can only be formed when two lines are intersected by a transversal. Determining Angle Relationships With so many similarities, you may be wondering how to determine the relationship between angles formed by transversals and intersecting lines. All you have to do is ask yourself these three basic questions: 1. Are the angles in the same location at both points of intersection? This means that we want to know if they are both in the upper left, upper right, lower left, or lower right corners of the intersections. If you determine that they are in the same location, then these angles must be corresponding angles, and you are finished. If they are not corresponding angles, move on to questions 2 and Are the angles on the same side or opposite sides of the transversal? If they are on the same side, then the angles are considered consecutive. If they are on opposite sides, then the angles are considered alternate. 3. Are the angles inside or outside of the two intersected lines? If they are inside the two lines, then they will be classified as interior. If located outside the two lines, then the angles are considered to be exterior. From here, we will combine our answers for questions 2 and 3 to determine the relationship between the angles. Before we get into our examples, let's go ahead and discuss the relationship between the measures of these angles.
371 Angle Measures As mentioned earlier, a pair of vertical angles will always be congruent. No matter what the diagram looks like, each pair of vertical angles will always have the same measurement. But the story is a little different when observing corresponding, alternate interior, alternate exterior or consecutive interior angles. We cannot make any assumptions about their values, unless we have one specific condition: parallel lines. When the two lines intersected by the transversal are parallel, corresponding angles are congruent, alternate interior angles are congruent, alternate exterior angles are congruent, and consecutive interior angles become supplementary, which means they have a sum of 180 degrees. Putting It All Together Now that we are aware of these relationships and their measures, let's tie all of this information together by examining a couple of basic problems. In the diagram, please notice that line a is parallel to line b. Line c is the transversal. For our first example, the measure of angle 1 = 6x  3 and the measure of angle 8 = 4x What is the value of x? Before we can solve this problem, we must determine the relationship between these two angles. Let's answer those three questions from earlier. For question 1, we notice that these angles are not in the same location. Angle 1 is in the upper left corner of the top intersection, and angle 8 is in the lower right corner of the bottom intersection, so they cannot be corresponding angles. Let's move to question 2. We see that these two angles are on opposite sides of the transversal, so we can classify them as alternate angles. In response to question 3, both angles are outside of the two lines, and must be exterior angles. When we combine our answers to these questions, we can conclude that they are alternate exterior angles. Since the two lines are parallel, we know that their measures must be equal. So, to solve for x, we will set 6x  3 = 4x From here, we will subtract 4x from both sides and then add 3 to both sides to get 2x = 36. Once we divide both sides by 2, we will find that the value of x = 18. For our next example, let angle 6 = 15y  11 and angle 4 = 20y What is the measure of angle 7?
372 Before we can find the measure of the angle, we must first solve for y. To do this, let's begin by determining the relationship between the two angles. We will revisit the three questions from earlier. In response to question 1, we see that angle 4 and angle 6 are not in the same location at each intersection, so they are not corresponding angles. To answer question 2, they are on the same side of the transversal, which makes them consecutive angles. And to address question 3, they are both within the two lines, which makes them interior. Therefore, we see that angles 4 and 6 are consecutive interior angles. Since the two lines are parallel, we know that we should add these angles together and set them equal to 180. When we do, we get 15y y + 16 = 180. Once we combine our like terms, we have 35y + 5 = 180. From here, we will subtract 5 from both sides to get 35y = 175, and then we will divide both sides by 35 to get y = 5. Though this is a great start to our problem, we are not yet finished. We were asked to determine the measure of angle 7, even though we were only given information about angles 6 and 4. Let's find the connection between the angles. In examining the picture, we notice that the vertex of angle 7 meets the vertex of angle 6. These two angles are nonadjacent and are formed by only two intersecting lines. By definition, these are vertical angles, and their measures must be congruent. Therefore, if we find the measure of angle 6, we will also know the measure of angle 7. Since we know that y has a value of 5, we will substitute this into the equation for angle 6. We will see that angle 6 = 15(5) After we simplify, we see that angle 6 measures 64 degrees. As we stated before, angle 6 and angle 7 are congruent because they are vertical angles. Therefore, we can conclude that angle 7 is also 64 degrees. Lesson Summary In review, vertical angles are angles formed by the intersection of two lines while alternate interior angles, alternate exterior angles, corresponding angles and consecutive interior angles are formed by the intersection of two lines and a transversal. Determining if angles are vertical requires simple observation, while determining other angle relationships can be done by asking yourself three questions: 1. Are the angles in the same location? 2. Are they on the same side or opposite sides of the transversal?
373 3. Are they inside or outside of the two lines? When referencing angle measures, we know that the measures of vertical angles are always congruent; however, the relationships of measures for the other angles are determined by whether the two intersected lines are parallel or not. Angles and Triangles: Practice Problems Chapter 13 / Lesson 8 Want more practice solving with angle pairs? How about more review for solving angles in triangles? Look no further. Get more practice here, and test your ability with a quiz. Solving Angles In another lesson, we learned about the different types of angles: consecutive interior, alternate interior, alternate exterior and corresponding. We discovered that when two lines are parallel, all of the angle pairs are congruent, except for consecutive interior angles, which are supplementary. We also learned about vertical angles, which are always congruent. Let's do some practice with these angles. In the figure, let Angle 5 = 30y + 31, and let Angle 9 = 22y What is the value of y? Angle 5 and Angle 9 don't match any of the angle pairs, so let's find the connection between their measures. We notice that Angle 5 corresponds to Angle 1, and Angle 1 corresponds to Angle 9. Knowing that all corresponding angles are congruent, Angle 5 = Angle 1, and Angle 1 = Angle 9. So, by the transitive property of equality, we can conclude that Angle 5 = Angle 9. By substituting the equations, we have 30y + 31 = 22y From here, we can subtract 31 from both sides to get 30y = 22y + 24, and then subtract 22y from each side to get 8y = 24. To finish, we will divide both sides by 8 to determine that y = 3. Let's do another. This time, let Angle 4 = 14x  23, and let Angle 14 = 4x + 5. Find the measure of Angle 15. Once again, these angles are not a special angle pair; so let's find the connection. Angle 4 corresponds to Angle 12, and Angle 12 is consecutive interior to Angle 14. Therefore, Angle 4 = Angle 12, and Angle 12 + Angle 14 = 180. With this knowledge, we can replace Angle 12 with Angle 4 to get Angle 4 + Angle 14 = 180. With the
374 equations, we have 14x x + 5 = 180. Combining like terms gives us 18x  18 = 180, and then, by adding 18 to both sides, we get 18x = 198. Last, we will divide both sides by 18 to conclude that x = 11. Now we can find the value of Angle 15, which is vertical to and congruent with Angle 14. Substituting 11 into the equation, we see that Angle 14 = 4(11) + 5, which equals 49. Therefore, we can also conclude that Angle 15 = 49 degrees. Solving Triangles When working with triangles, remember that the sum of all three angles in every triangle is 180 degrees. Let's get started. In this first triangle, let's solve for x. For each angle, we either have a measure or an equation. For that reason, let's add all of the angles together to equal 180 degrees. Doing so, we have x = 180, and by combining like terms, we have 10x + 80 = 180. Next, let's subtract 80 from both sides to get 10x = 100, and then let's divide each side by 10 to finish with x = 10. Here's another. This is triangle JKL. What is the measure of Angle L? By having information for all three angles, we will add them together to equal 180. Remember that the square in the angle tells us that the angle measures 90 degrees. So we can begin with 10y y + 35 = 180. Combining like terms gives us 25y = 180. From here, we will subtract 130 from both sides, leaving 25y = 50. Then, let's divide both sides by 25 to see that y = 2. Now, by substitution, we see that Angle L = 15(2) + 35, which equals 65 degrees. Let's do one more. In triangle DEF, Angle D is two times a number, Angle E is forty more than five times the number, and Angle F is five more than two times the number. What is the measure of Angle E? Since we don't have the exact equation for each angle, we have to use the descriptions to create them. All of the descriptions reference some unknown number. Not knowing what this number is, we will call it x in each equation. Therefore, Angle D = 2x, Angle E = 5x + 40, and Angle F = 2x + 5. To solve, we begin with 2x + 5x x + 5 = 180. Combining like terms gives us 9x + 45 = 180, and then subtracting 45 from both sides leaves us with 9x = 135. From here, we will divide both sides by 9 to get x = 15. To complete the problem, we will
375 substitute 15 into the equation for Angle E to see that Angle E = 5(15) + 40, which equals 115 degrees. Lesson Summary In review, when solving with angles and lines, always begin by determining the special angle pair or the connection between the angles you were given. This will send you in the right direction for determining whether you should set the angles equal to each other or add them together to equal 180 degrees. But remember, to solve these problems in this manner, the two lines must be parallel. When it comes to triangles, remember that all of the angles in every triangle must add together to equal 180 degrees. So, when you have information for each angle in a triangle, this is the best way to start and solve the problem. The Pythagorean Theorem: Practice and Application Chapter 13 / Lesson 9 The Pythagorean theorem is one of the most famous geometric theorems. Written by the Greek mathematician Pythagoras, this theorem makes it possible to find a missing side length of a right triangle. Learn more about the famous theorem here and test your understanding with a quiz. Introduction Jessica has a square garden. To make watering her plants easier, she would like to create a diagonal pathway that connects the opposite corners. If one side of her garden is 13 feet long, what will be the length of her diagonal pathway? This scenario presents a perfect example of geometry in real life. Since the garden is in the shape of a square, we will be able to use the Pythagorean theorem to figure out how long the diagonal pathway will be. Let's take a closer look at this theorem and how it's used so that we can calculate the solution to this problem. The Theorem and Its Purpose Written by the Greek mathematician Pythagoras, the Pythagorean theorem states that in right triangles, the sum of the squares of the two legs is equal to the square of
376 the hypotenuse. In clearer terms, if you square both of the legs and then add these numbers together, you will end up with the same value as the hypotenuse squared. As a formula, this theorem is written as a^2 + b^2 = c^2, where a and b represent the two legs of the triangle, and c represents the hypotenuse. When labeling any right triangle with these variables, it is important to know that a and b can be written on either leg. However, c must be written on the hypotenuse, which is the longest side and opposite the right angle. The purpose of the Pythagorean theorem is to make it possible to find the length of any missing side of a right triangle. For the theorem to be usable, the value of the other two sides must be known. Let's take a look at a few basic problems. Pythagorean Theorem Basics In our first triangle, one leg has a length of 4 centimeters, and the second leg has a length of 7 centimeters. Let's find the length of the hypotenuse. Since I have the value for the two legs, which are represented by a and b, we will let a = 4 and b = 7. We are now ready to plug this information into the formula. Once we do, we get 4^2 + 7^2 = c^2. After simplifying, we end up with = c^2, and when we combine our like terms, we have 65 = c^2. For our last step, we will take the square root of both sides to get a final answer of c = Therefore, we can see that the length of the diagonal is 8.06 centimeters. In our next triangle, the measure of one leg is 12 inches and the measure of the hypotenuse is 20 inches. For this triangle, we have the value of the hypotenuse, but must find the length of the missing leg. Therefore, we will let a = 12 and c = 20. By substituting this information into the formula, we have 12^2 + b^2 = 20^2. When we simplify, we get b^2 = 400. At this point, we should subtract 144 from both sides to get b^2 = 256. Once we take the square root of both sides, we see that b, the missing leg in our triangle, has a value of 16 inches. Did anything about the numbers in that last example stand out to you? Notice that all of the side lengths are whole numbers, not fractions or decimals. This means that the side lengths for this triangle are a Pythagorean triple, which is defined as a set of nonzero whole numbers that satisfy the Pythagorean theorem.
377 A few more examples of Pythagorean triples are 3, 4, 5, and 28, 45, 53. If you plug these numbers into the Pythagorean theorem, you will see that 3^2 + 4^2 = 5^2, and 28^2 + 45^2 = 53^2. Pythagorean Theorem in Everyday Life Now that we have an understanding of the Pythagorean theorem, let's apply it to a few everyday examples, beginning with Jessica's garden from the beginning of the lesson. Let's review the problem again. Jessica has a square garden. To make watering her plants easier, she would like to create a diagonal pathway that connects opposite corners. If one side of her garden is 13 feet long, what will be the length of her diagonal pathway? For this situation, we should definitely use the Pythagorean theorem. We know that in a square, all four angles measure 90 degrees and all four sides are congruent. So if one side has a length of 13 feet, then all the other sides have a length of 13 feet. By taking half of this square, the diagonal becomes the hypotenuse and we have a right triangle with both legs equal to 13 feet. We are now ready to plug in our information to find the length of the hypotenuse. When we do, we will see that 13^2 + 13^2 = c^2. By simplifying and combining like terms, we end up with 338 = c^2, and c = as our final answer, which will be the length of Jessica's diagonal pathway. Let's do another. Elizabeth needs to repair her roof. She props a 21foot ladder against the side of her 20foot home. If the top of the ladder reaches the top of her house, how far away from her home is the base of the ladder? In this problem, we are still working with a right triangle. The ladder is the hypotenuse, while the height of the home and the ground between the home and the ladder are the legs of this right triangle. Using the given information, our beginning equation is 20^2 + b^2 = 21^2. Simplifying brings us to b^2 = 441. From here, we subtract 400 from both sides to get b^2 = 41, from which we see that b = Therefore, we can conclude that the base of the ladder is 6.40 feet away from the home.
378 Lesson Summary In summary, the Pythagorean theorem, which can only be used with right triangles, is a very important theorem that allows us to find the length of a missing side in a triangle. The theorem is interpreted as the formula a^2 + b^2 = c^2. The legs are represented by a and b and can be labeled interchangeably, but the hypotenuse must be labeled c. In order for this theorem to be used, two sides of the right triangle must be known. Applying Scale Factors to Perimeter, Area, and Volume of Similar Figures Chapter 13 / Lesson 10 How do shapes change sizes yet retain their proportions and similarities to other shapes? In this lesson, we'll look at what a scale factor is and how to apply it. We'll consider scale factors with regards to three different aspects of similar shapes: perimeter, area and volume. Issues of Scale When I was a kid, I had all kinds of action figures  HeMan, Star Wars, Ghostbusters. I liked to play with them together, but that didn't always make sense. One set of action figures in particular, Thundercats, never fit. The Thundercats action figures were way bigger than all the other action figures. So a Thundercat could never pilot a Star Wars speeder bike. And my Han Solo figure was way too small for the Thundercats' vehicles. The scales just weren't the same. This is an example of why scale factors matter. What Is a Scale Factor? A scale factor is simply a number that multiplies the dimensions of a shape. This can make a shape larger. Larger shapes will have a scale factor greater than one. So, if the scale factor is three, then the dimensions of the new shape will be three times larger than that of the original. Let's say you own a doughnut shop and you want a giant strawberryfrosted doughnut on top of your shop. You might use a scale factor of 25. Every tasty inch of a regular
379 doughnut would be 25 inches on the model. So, a 5inch doughnut would be 125 inches, or almost 10.5 feet tall. A scale factor can also make a shape smaller. Smaller shapes will have a scale factor of less than one. You've probably seen this with Matchbox cars, which are often shrunkdown versions of real cars. This also works with dollhouses. For example, a model of a house may have a scale factor of 1/50. That means that 1 inch on the model is equal to 50 inches on the actual house. So, a 4foottall dresser would be about 1 inch tall in the model. If the scale factor is one, then the two shapes are the same, or congruent. The scale factor of this car to this car is one. That's not very interesting, so we don't usually talk about scale factors of one. If you're into variables, the equation for a scale factor can be written as y = Cx, where x is the original dimension, y is the new dimension, and C is the scale factor, or the amount by which the original is multiplied. Perimeter Examples Let's talk about how scale factors influence shapes in geometry, starting with perimeter. Here's a rectangle. Its width is 3 and its length is 5. What's the perimeter? , or 16. Now let's apply a scale factor of 4 for the new rectangle. It will have dimensions that are 4 times that of the original. Instead of a width of 3, it will be 3 x 4, or 12. And instead of a length of 5, it'll be 5 x 4, or 20. This new rectangle is similar to the original, which means it has the same shape, but not necessarily the same size. What impact did this have on the perimeter? The new perimeter is , or to 64? That's 4 times the original. So the change in perimeter is equal to the scale factor. Area Examples Things are a little different with area. Let's look at those same two rectangles. What's the area of the first? The area of a rectangle is length times width, so it's 5 x 3, which is 15. What about the second one? 20 x 12, which is 240. Does 15 x 4 = 240? No. What is the relationship between 15 and 240? If you divide 240 by 15, you get 16.
380 And what was our scale factor? is 4^2. So the change in area is equal to the scale factor squared. Let's look at another example. Here's a triangle with a base of 5 and a height of 4. The area of a triangle is ½ times base times height. So this triangle's area is ½ x 5 x 4, which is 10. Let's make a new triangle using a scale factor of 3. This new triangle has a base of 15 and a height of 12. Its area is ½ x 15 x 12, or 90. Okay, remember that our scale factor was 3. And 3^2? That's 9. So the change in area should be 9 times the original. Does 10 x 9 = 90? Yes! So it works. When you're working with scale factors, square the scale factor to determine the area of the new figure. If you think about it, it makes sense why area would be the scale factor squared. Area involves two dimensions multiplied together. With scale factor, all you're really doing is multiplying the scale factor times itself. Volume Examples Next, there's volume. Like going from perimeter to area, going from area to volume means adding a layer. In this case, it's a third dimension. Instead of squaring the scale factor, guess what? We're going to cube it! So the change in volume is equal to the scale factor cubed. Cubing a number is raising it to the third power. So, if you remember that volume involves three dimensions, you can remember to cube the scale factor. Let's try this out. Here's a rectangular prism. Let's make this interesting. Let's say it's a box of cookies. It's 4 inches by 2 inches by 3 inches. The volume of a rectangular prism will be length times width times height. So its volume is 4 x 2 x 3, or 24 cubic inches. That's not going to hold a lot of cookies, even if they're small. So let's scale it! Let's use a scale factor of 3. Here's our new box. It's 12 by 6 by 9. The volume of this box will then be 12 x 6 x 9, which is 648 cubic inches. Now that box will hold a lot of cookies. We just need some milk. Oh, but what about the change in volume? I said it's the scale factor cubed. But what's 3 cubed? It's 27. And what happens if we multiply the original volume, 24, times 27? Yep. It's 648. But what about that milk? Let's do one more volume example. Here's a kids' size glass of milk. The volume of a cylinder is pi times r^2 times h, where r is the radius of the
381 circle on the top and h is the height. This glass has a radius of 1 inch and a height of 4 inches. So its volume is pi x 1^2 x 4, or about 12.6 cubic inches. We have a huge box of cookies, so we need a bigger glass of milk. What scale factor should we use? Since scale factors are cubed with volume, remember that even a small change will have significant ramifications. Let's try a scale factor of 2. That will make our radius 2 inches and our height 8 inches. That doesn't seem unreasonable, right? Now, we don't need to do the volume formula. We can just cube the scale factor. 2^3 = 8. If our original volume was 12.6, then our new volume is 12.6 x 8, or 101 cubic inches. For those of you who don't think of your milk in cubic inches, that's about 1.75 quarts of milk. Holy cow! Lesson Summary In summary, a scale factor is simply a number that multiplies the dimensions of a shape. It's a way of describing the relationship between similar shapes, whether it's a house and a dollhousesize replica or a pair of triangles. The perimeter of a scaled object will be equal to the scale factor. If the scale factor is three, then the perimeter of the new object will be three times the original perimeter. The area of a scaled object will be equal to the scale factor squared. Again, if the scale factor is three, the area of the new object will be nine times, or three squared, the area of the original object. Finally, the volume of a scaled object will be equal to the scale factor cubed. So if the scale factor is three, the volume of the new object will be three cubed, or 27 times, the volume of the original object.
382 What Is a Number Line? Chapter 14 / Lesson 1 A number line is a visual representation of all real numbers. In this lesson, we'll learn how to identify points on a number line. We'll also practice addition and subtraction, letting the number line do all the hard work. Learning about Numbers Let's say you have eight apples. Then I come along and take five of your apples. That's not cool. Maybe you were going to make a pie. Apple theft is a real problem! It's also the way most people begin to learn about numbers. But it's not the only way. When you're ready to move beyond fruit transactions, you're ready for number lines. What Is a Number Line? A number line is a picture representation of real numbers evenly laid out on a straight line. It looks like this. As a reminder, real numbers include just about everything: whole numbers like 7, 15 and 5,832,622; rational numbers like 1/2 and 6.8; and irrational numbers like pi. You could plot any of these on a number line. What can't you put on a number line? Imaginary numbers, like the square root of 1 and also infinity. You can make your number line as long as you want; you could never find infinity on it. Same goes for numbers you make up, like seventytwelve. A number line has a few basic parts. In the middle, there is the origin, which is 0. How can you remember this? Well, in the beginning, there was nothing. I read that somewhere. Also, the origin of most things is nothing. That includes the score at the start of a game or my bank account at the start of a month. To the right are positive numbers. To the left are negative numbers. Both theoretically go on forever in either direction (though they never reach infinity). It's like the
383 interstate going east and west forever, which is how it seems when you're in the middle of Kansas! Since your screen or paper isn't as big as Kansas, we put arrows on the ends to indicate the line keeps going. When you draw a number line, you can plot numbers on it however it suits your purpose. Maybe you just want integers between 5 and 5. Maybe you want fractions between 1 and 1. Maybe you want odd numbers. Or maybe you're like Scrooge McDuck calculating his wealth and you want increments in the millions. It's up to you! Identifying Points Sometimes you just need to find a point on a number line. Let's say you were asked to find 4.5 on this line. Where is it? Well, it's halfway between 4 and 5, right here. Or maybe you'll be asked to identify a point on a line. For example, what is this point? Okay, notice that there are two marks between each whole number, so each one represents 1/3. This one is 2 marks past 1, so it's 1 and 2/3. This is just like finding points on a ruler; you just need to figure out what each hash mark represents. Addition and Subtraction Numbers lines are particularly useful with addition and subtraction problems. Remember that apple piracy I mentioned? That's a great way to learn subtraction if you happen to have a bunch of apples handy. But you can do that problem with a number line. I said you had eight apples. So find the 8 on the line. Then I took five. Since this is subtraction, we go to the left. Why? Because when you subtract, your numbers left. Okay, count back 5. We're at 3. So you have three apples left! That's enough for a tart, right? Now, what if I gave you back four apples? Okay, this is addition, so move to the right. Why? Because getting more numbers is just right. Okay, count one, two, three, four. And we're at 7. I think we're back in pie territory.
384 Negative Numbers Number lines are a great way to work with negative numbers. You can't have 4 apples. I may be an apple thief, but I'm not that good. But on a number line, you can solve problems involving negatives. What's ? Start at 8 and move 5 which way? It's addition, so go right. One, two, three, four, five: = 3. What about 29? Start at 2 and move 9 to the left  for subtraction. One, two, three, four, five, six, seven, eight, nine. We're at 7, so 29 = 7. Sometimes you'll need to bounce back and forth between the positive and negative sides of the origin. But the principle of what you're doing doesn't change. What is ? Whoa. This would be a mess with apples. But we can do it with a number line. Start at 9, then move 12  where? Your numbers left, so go left. Now we're at 3. If we add 4, we go which way? Getting more is right, so go right. Now we're at positive 1. Then we go left 2, and we're at 1. So, = 1. With a number line, that wasn't so bad, was it? How about one more? Start at 7, then go 6 to the right: we're at 1. Now 2 to the left: 3. Now 5 to the right: positive 2. That's it! Just let the number line guide you. Lesson Summary In summary, a number line is a picture representation of real numbers evenly laid out on a straight line. You can draw the numbers however you'd like. Remember that the middle is the origin, or 0. When you add, you go to the right. When you subtract, you go to the left. Pay attention to what each hash mark represents, and that's all there is to it. What Are the Different Parts of a Graph? Chapter 14 / Lesson 2 Being able to read a graph isn't just vital for an algebra class. Graphs and charts are used everywhere! We'll take a crash course on the basic x/y plane used in algebra and the fundamental vocab you need.
385 What is a Grid? You've probably used grids before in real life. For example, a chessboard is a grid. Along the bottom, there are columns labeled A through H, and along the side, there are rows labeled 1 through 8. If you want to talk about a specific square on the board, you can refer to it by its number and letter. For example, the pawn here is at B3. The type of grid used in math problems is called the Cartesian plane or xy plane. A Cartesian plane is a little different from a chessboard, but you can think about it the same way. Using a Cartesian plane can help you solve all kinds of math problems and understand the relationships between things in the real world. In this lesson, you'll learn about the parts of a Cartesian plane, so you'll be all ready to use it when you need to. X & Y Axes On a chessboard, the horizontal location of a piece is labeled with a letter and the vertical location is labeled with a number. On the Cartesian plane, it's a little bit different: both the horizontal and vertical position are labeled with numbers. On a Cartesian plane, you can also have locations that are described with negative numbers. The numbers that describe a particular location on the Cartesian plane are called coordinates. The xcoordinate of an object on the Cartesian plane tells you how left or right it is from the center of the graph. Every graph has a line called the xaxis, which marks the horizontal location of points on the graph. This is basically a number line, and if you've used number lines before, it should look very familiar. If you're moving to the left along the xaxis, the number will be negative, and if you're moving to the right, it will be positive. For example, the xcoordinate of this point is 5, because it's 5 units to the right of the center of the graph. The xcoordinate of this point is 5, because it's 5 units to the left of center. Using xcoordinates lets you distinguish between points: for example, Point A in this graph has an xcoordinate of 3, and Point B has an xcoordinate of 5. But how would you distinguish between Point B and Point C? They're obviously not in the same place, but they both have the same xcoordinate! That's where the ycoordinates come in. The line that shows the vertical position of points on the graph is called the yaxis. The ycoordinate of an object tells you how
386 far up or down it is on the yaxis, relative to the center of the graph. You can remember which axis is x and which is y by looking at the letters: the letter y has a long tail that goes down, so the yaxis is the axis that goes up and down. Looking at the ycoordinates lets us describe the differences between these two points. Point B has an xcoordinate of 5 and a ycoordinate of 2. Point C has an xcoordinate of 5 and a ycoordinate of 4. Points & Coordinate Pairs It's a little bit clunky to walk around saying things like 'This is the point with an xcoordinate of 5 and a ycoordinate of 2,' so we use a shorthand to talk about it. Instead of all that stuff, you can write the numbers as an ordered pair. An ordered pair is defined as a set of coordinates shown as two numbers inside parentheses. The xcoordinate always comes first, and the ycoordinate always comes second. So the coordinates of Point B on the graph would be (5, 2). Here are some examples of ordered pairs. A point that's on one of the two axes has a 0 for the other coordinate. For example, this point is labeled (5, 0), because it's 5 units to the right of the middle of the graph, and 0 units up or down. This point is labeled (0, 3), because it's 3 units up from the middle of the graph, and 0 units to the right or left. The point at the very middle of the graph is called the origin, and its coordinates are (0, 0), because it's 0 units away from the center of the graph in both directions. If we draw a line on the graph, sometimes the line crosses either the xaxis or the yaxis. The point where a line crosses an axis is called an intercept. If it's the xaxis, it's the xintercept; if it's the yaxis, it's the yintercept. A line can have either an xintercept, a yintercept, or both. Quadrants If you take a step back and look at this graph, you'll see that the x and yaxes divide it into four parts. Each of these parts is called a quadrant. To make it easy to talk about quadrants, each one gets a number, starting from the first quadrant at the top right. In the first quadrant, both x and y are positive. The top left is the second quadrant. Here, x is negative because we're left of the origin, but y is positive. The bottom left is the
387 third quadrant, where both x and y are negative, and the bottom right is the fourth quadrant, where x is positive but y is negative. Lesson Summary In this lesson, you learned about the Cartesian plane, which is a type of grid that mathematicians use to show the location of lines and points. As you start doing more advanced algebra and geometry, this will help you solve all kinds of problems. The Cartesian plane has an xaxis and a yaxis that cross each other at the origin. The xaxis is horizontal and measures distance to the left and right of the origin. The yaxis is vertical and measures distance up and down. It's basically like two number lines, with one going the usual way and another one flipped on its side to measure distance up and down. Points on the Cartesian plane are labeled as ordered pairs, in the format (x, y), where x represents the xcoordinate and y represents the ycoordinate. If a line crosses one of the axes, the point where it crosses is called the xintercept or yintercept. The Cartesian plane is divided into four quadrants, starting with the first quadrant at the top right, and moving counterclockwise. That's a lot of terminology to learn all at once, so don't sweat it if you don't get it all immediately. As you start using Cartesian planes in actual problems, you'll pick it up really fast, and, eventually, you won't even have to think about it. Plotting Points on the Coordinate Plane Chapter 14 / Lesson 3 If you'll be working with a graph, otherwise known as the coordinate plane, it's essential to understand how it works. This includes learning the parts of a graph, identifying points and plotting points. Descartes and Cartesian Coordinates Once upon a time, I lived in rural Maine. I sometimes had to ask locals for directions. In rural Maine, the windy, confusing roads aren't always marked well, if it all. And people had a habit of saying things like, 'Turn left where the Paxtons' barn used to be,' or 'Turn right after the field where they were gonna build that corner store.' Not helpful.
388 Way back in the 17th century, French philosopher and mathematician Rene Descartes had a similar problem. You might remember him as the guy who said, 'I think, therefore I am.' He was never lost in Maine, but he was trying to make geometry easier. There's a story, and it may or may not be true, that Descartes was sick in bed one day when he noticed a fly on the ceiling. He realized he could describe its position based on the grid of ceiling tiles. Maybe it was five tiles in from the east wall and four tiles from the south wall. And, if that story's true, the concept of the Cartesian coordinate plane was born. Named for Descartes (though less Frenchsounding), Cartesian coordinates are a method of identifying a point on a graph relative to a set of lines. You might hear this referred to with or without the word 'Cartesian' on it. Don't worry, it's still the same thing. Parts of a Graph Before we get to identifying and plotting points, let's look at the parts of a graph. The most common version of the coordinate plane looks like this. Basically, you have a grid. On that grid, there are two lines. The xaxis is the horizontal line on a graph. Remember, x is a cross, so it goes 'across' the graph. The yaxis is the vertical line on a graph. You can see that that the letter Y is reaching for the sky  so 'Y to the sky.' The axes meet at the origin. Everything begins at the origin, even graphs. Both axes are at 0 here. The xaxis is positive to the right and negative to the left. That's exactly like a number line. In fact, that's really all an xaxis is. But when it gets together with its friend the yaxis, it becomes a graph. The yaxis is positive going up and negative going down. As you can see, the axes break up the graph. The four sections of the graph are called quadrants. They're numbered with Roman numerals. To remember which quadrant is which, think positive. Quadrant I is here, where both the axes are positive. Then it just goes counterclockwise through Quadrant II, Quadrant III and Quadrant IV. Why counterclockwise? Blame the Babylonians. No, really. The ancient Babylonians measured angles in that direction, and today, mathematicians think that that was passed on to graphs. Think of it this way: You have to go back in time to find the answer, which is counterclockwise on a clock.
389 Identifying Points Okay, enough with time travel. Let's get to identifying some points. If you look at this grid, it looks just like those ceiling tiles in Descartes' bedroom. Well, I think it does. I've never been to his house. If you want to identify a point, like this one, start at the origin. Always start at the origin. First, we count along the xaxis. Why? Because x comes before y. This point is 1, 2, 3 to the right. Now we count up. It's 1, 2 up. We would write this point as (3, 2). That right there? That's called an ordered pair. An ordered pair is just a pair of numbers in order, indicating a point on a graph. X is first, then y. Here's another one. This one is in Quadrant II. Remember, Quadrant I is all positive, then we go counterclockwise to Quadrant II. This one is 1, 2, then 1, 2, 3, 4. So it's (2, 4). Let's try Quadrant III. 1, 2, 3, 4, then 1, 2, 3. So it's (4, 3). Why so negative, Quadrant III? Okay, let's complete the circle with a point in Quadrant IV. Remember, we're now in positive territory with the xaxis. 1, 2, 3, 4, 5. Then 1, 2. So it's (5, 2). Plotting Points Plotting points is the exact same thing as identifying points, only in reverse. Where would (2, 4) be? Remember, x before y. 1, 2. Then 1, 2, 3, 4. Right here. What quadrant is that? Quadrant II. What about (3, 3)? 1, 2, 3. And 1, 2, 3. That's in Quadrant III. Here's another one: (2, 5). That's all positive, so it'll be in Quadrant I. We go 1, 2. Then 1, 2, 3, 4, 5. And we did it! Lesson Summary In summary, a coordinate plane, sometimes called a Cartesian coordinate plane, is just a graph. It has a horizontal xaxis, a vertical yaxis and an origin at (0, 0). There are four quadrants on a graph, which go I, II, III, IV. We label points using an ordered pair, which always begin with the x coordinate. To identify or plot a point, begin by counting left or right on the xaxis, then go up or down on the yaxis.
390 Graph Functions by Plotting Points Chapter 15 / Lesson 1 Let's say you have a function. What does it look like? How do you graph it? In this lesson, we'll learn how to graph functions by determining and plotting points. Functions Imagine it's a dark and stormy night. You're lost in the woods. You sense there's something chasing you. You start to run, but you trip over a fallen branch. You turn over and look behind you. And there, in a flash of lightning, is this: f(x). Wait, hold on. You don't have to be afraid of functions. I know  they look scary. Where'd that f come from and what does it mean? And why is it holding that x prisoner? Functions have intimidated math students for as long as there have been bad stories that start with it being a dark and stormy night. But in this lesson, we're going to see that functions are just like any other equation. And, in fact, it's very simple to make a graph of a function. First, let's get a few things straight. A function is a rule that uses an input to produce an output. Let's say you see f(x) = 2x. That just means that for each value of our input, x, our function, f, produces an output of 2x. So if we input 5, our output is 5*2, or 10. If we input 7, our output is 7*2, or 14. So f(x) = 2x is actually similar to y = 2x. Again, if x = 5, y = 10. In this equation, y is a function of x. The value of x determines the value of y. Now we can get into graphs. The graph of a function is a visual representation of all of the points on the plane of (x, f(x)). Note that f(x) is taking the place of y in that ordered pair. This graph we're making is just a collection of points. You know that famous painting by Seurat? The one that looks like a normal picture, but when you look closer, you see that it's really a collection of countless tiny dots? A line on the graph is the same thing. It's just an infinite number of dots.
391 Find Ordered Pairs To graph a function, we take that idea and follow a simple threestep plan. Let's try this out with this function: f(x) = 3x  4. Okay, let's start. Step one: Find ordered pairs. Your ordered pairs are your dots. If you could find an infinite number of them, you'd have your graph. But we don't have time for that here, though that could be a fun project. We're just going to fill in this chart. In the x column, let's pick some numbers. It doesn't matter too much what we pick, but we want to make sure that if this line curves, we check enough numbers to notice that. Let's use 1, 0, 1, 2. To find the ordered pairs, we plug our x values into the function. Start with 1. Just plug that into our function: f(1) = 3(1)  4. So our f(x) value here is 7. Let's add that to our chart. That means our first ordered pair is (1, 7). 0 is next. f(0) = 3(0)  4. That's just 4. Next is 1. f(1) = 3(1)  4. That's 1. Finally, there's 2. f(2) = 3(2)  4. That'll be 2. Plot the Points Okay, we have our points. Now comes Step two: Plot the points. Let's move over to the coordinate plane and plot these ordered pairs. Here's (1, 7) and then here's (0, 4). If you know you have a straight line, you really only need two points. But we're being cautious and plotting four points. So here's (1, 1), and here's (2, 2). Connect the Dots That leaves our final step: Step three: Connect the dots. You probably practiced this in preschool. Did you know then that connecting the dots would be a practical skill in advanced math? Anyway, before we connect the dots, let's make sure we know what they're doing.
392 Some functions have curves. You'd be able to tell because a straight line wouldn't go through all of our points. But if we draw a line here, it does. So we have our line. And that's the graph of f(x) = 3x  4. Practice Problems Now that we know the three steps, let's get some practice. First, let's try f(x) = 3x + 6. Remember step one? Find ordered pairs. Let's make a chart. Okay, now pick some values for x. How about we try even numbers? 2, 0, 2 and 4. Next, plug them into our function. f'(2) = 3(2) + 6. That'll be 12. f(0) = 3(0) + 6. That's just 6. f(2) = 3(2) + 6. That's going to be 0. Finally, f(4) = 3(4) + 6. That's 6. I'm noticing a pattern here. Okay, step two: plot the points. Here's (2, 12). Here's (0, 6). Here's (2, 0). And here's (4, 6). Finally, step three: connect the dots. This one is definitely a straight line. And here it is! This is graph of f(x) = 3x + 6. Let's do a trickier one: f(x) = x^ Okay, step one: find ordered pairs. Since x is squared, let's make sure we use both positive and negative numbers. Let's use 4, 2, 2 and 4. So what's f(4)? f(4) = (4)^ That's 18. Next is 2. f(2) = (2)^ That's 6. Then 2. f(2) = 2^ That's also 6. And since we're squaring x, we know that 4 will be the same as 4: 18. Now step two: plot the points. Okay, here's (4, 18). Here's (2, 6). Here's (2, 6) and here's (4, 18). Finally, we connect the dots. Now, you can probably tell that this isn't going to be a straight line. This line curves.
393 We can see it curves up here and also up here, but how do we know where it starts? That will be the smallest possible value of y. Since we're squaring x, the smallest y value will be when x is 0. Why? As we saw, when we square a negative number, we get a positive number. Therefore, every negative x value will give us a larger y value than 0. So let's check what y is when x = 0. f(0) = 0^ It's 2! So that's the bottom of our graph, (0, 2). If we connect the dots, we get a parabola! This example is a good reminder that you can always try more points if you're not sure what the graph will be. Just like that painting, sometimes you need a lot of dots to be sure. Hopefully, you don't need as many as Seurat, though. Lesson Summary In summary, when we want to graph a function, we want to make a visual representation of all of the points on the plane of (x, f(x)). We follow three simple steps. Step one: find ordered pairs. Find as many as you want, though I recommend at least four or five. Step two: plot the points. See what they look like on the graph. Then step three: connect the dots. Channel your inner preschooler and graph your function! Identify Where a Function is Linear, Increasing or Decreasing, Positive or Negative Chapter 15 / Lesson 2 Functions do all kinds of fun things. In this lesson, learn how to identify traits of functions such as linear or nonlinear, increasing or decreasing and positive or negative. Opposites Opposites  they're everywhere: yin and yang; cats and dogs; Republicans and Democrats; bacon and foods that just aren't bacon. The idea of opposites also comes into play with functions. In this lesson, we're going to look at a few different kinds of opposites that matter for differentiating functions. Feel free to pet a cat or dog as you watch, or munch on bacon, just don't pet your cat with bacon. They don't like that.
394 Linear or Nonlinear First up, let's talk about linear or nonlinear functions. A linear function is a function that represents a straight line. As you might expect, a nonlinear function is a function that represents a line that isn't straight. That's surprising, I know. But that's really all it is. There are many ways of thinking about linear functions, but usually the simplest is to just remember that linear means line and nonlinear means, well, not a line. If you're asked to identify a function as linear or nonlinear based on a graph, you're really just looking for a straight line. This one? Linear. This one? Nonlinear. This one? Linear. This one? Nonlinear. This one? Chicken. If you just have the function and no graph, you can make a table. In fact, sometimes you'll be given a table of x and y values and asked if the function is linear or nonlinear. Here's one: x y
395 In a linear function, the y values will follow a constant rate of change as the x values. Here, notice that the x values are increasing by 2 each time. The y values are increasing by 5 each time. So this is linear. What about this one? x y Here, the x values are going up by 2 again, but each time the x values go up by 2, the y values go up by different amounts. So they're not constant, and this function is not linear. Increasing or Decreasing Next, let's look at increasing or decreasing. Maybe your waistline is increasing as the bacon on your plate is decreasing. To be increasing, a function's y value is increasing as its x value increases. In other words, if when x1 < x2, then f(x1) < f(x2), the function is increasing. To be decreasing, the opposite is true  a function's y value is decreasing as its x value increases. In other words, if when x1 < x2, then f(x1) > f(x2), the function is decreasing. An increasing function looks like this. Here, when x is 0, y is 1. When x is 5, y is about 1. As x goes up, so does y. That's increasing. Decreasing looks like this: Here, the y values are getting smaller as the x values increase. When you have a graph like this, just think of increasing and decreasing as going up or down from left to right. If a line rises, it's increasing. If it falls, it's decreasing. You could also think of slope. A positive slope is increasing, while a negative slope is decreasing.
396 In a nonlinear function like this: It's both increasing and decreasing. This one is increasing until x = 0 and decreasing when x is greater than 0. If you were asked when this function is increasing, you'd say when x < 0. Positive or Negative Finally, let's look at positive or negative. As in, my dog has a positive outlook about everything, especially if it involves going for walks or smelling other dogs. Meanwhile, my cats have a negative outlook about things, especially things involving my dog. A function is positive when the y values are greater than 0 and negative when the y values are less than zero. Here's the graph of a function. Where is it positive? When x < 2. And it's negative when x > 2. Here's another. This one is positive when x > 3 or x < 3. It's negative when x is < 3 and > 3. Of course, you can do this without the graph. Let's consider f(x) = x^2 + 3x  2. Is it positive or negative when x = 1? Just plug in 1 for x. So f(x) = (1)^2 + 3(1)  2. That's , or is negative, so f(x) is negative when x = 1. What about when x = 3? So f(x) = (3)^2 + 3(3)  2, which is , or is positive, so f(x) is positive when x = 3. Lesson Summary In summary, we learned about opposites. There are linear and nonlinear functions. Linear functions represent straight lines, while nonlinear functions are lines that aren't straight. There are increasing and decreasing functions. In increasing functions, the y values increase as the x values increase. In decreasing functions, the y values decrease as the x values increase. Finally, there's positive and negative. This just means is y positive or negative for a given x value?
397 As for cats and dogs, well, I guess we didn't learn much about them, but that's OK. Linear Equations: Intercepts, Standard Form and Graphing Chapter 15 / Lesson 3 Do you know what to do if an equation doesn't look like y=mx+b?! If not, then this video is for you. Chances are the equation is in standard form, so we'll learn how to use standard form equations, how to graph them and why they can be helpful. How to Convert to Standard Form This video is on the different forms of a linear equation, specifically slopeintercept form and standard form, and how we can graph lines given to us in either of those forms. Slopeintercept form (y = mx+b) is the one that most people are familiar with. It's the most common one you see but that doesn't mean it's the only way to represent a linear equation. For example, if I had the line y = 3x+4, given to me in slopeintercept form, by using inverse operations and taking a positive 3x and undoing it with a 3x on both sides, I end up with the equation 3x+y = 4. These two equations are equivalent. They mean the same thing. They are the same thing, they're just written differently. What I end up with in this second one is what's called standard form. Essentially, because the xs and ys are on the same side of the equation. The generic standard form equation is Ax+By = C. Converting an equation from slopeintercept to standard form This is probably the second most common form of a linear equation that you see, but unlike slopeintercept form, the As and Bs do not necessarily give us any useful information like the m and b does in the slopeintercept form. That doesn't mean that there aren't still some advantages to standard form over slopeintercept form.
398 'SlopeIntercept' Way to Graph a Linear Equation What we're going to find out is that the orientation of the variables in standard form makes finding the x and yintercepts pretty quick and easy, which will allow us to kind of use a shortcut when it comes to graphing. Here we have a question that asks us to graph the line 3x+2y = 6. So let's do it the way we know how, which is by using slopeintercept form to use the m value and b value to graph our line. But because this isn't given to us in slopeintercept form, it requires us to first put it in slopeintercept form by using inverse operations to get the y by itself. This means we have to first undo the 3x with a positive 3x to both sides; we have to undo a times by 2 with a divide by 2 to both sides. Now we have the equation y = (3/2)x+3. This equation is equivalent to the one we started with, just written in a different way. Now that it's written in slopeintercept form, I know that I can use my m (my slope) and my b (my yintercept) to graph it. I begin at 3 on the yaxis, and then I go up 3 and over 2 to find my next point using the slope. You could continue going up 3 and over 2 as many times as you wanted, but you notice that all your dots are in the same straight line. You can connect that line and you have your graph. Which isn't too bad, but these steps in the beginning that required us to first get the y by itself are unnecessary and sometimes they can be a little complicated, especially with the fractions. We'd rather know a way to do it without having to solve for y. 'Standard Form' Way to Graph a Linear Equation Graph for the linear equation y = (3/2)x + 3
399 So let's take a look at the exact same problem: graph 3x+2y = 6. But this time, try to do it without having to do all those beginning steps where we get y by itself using inverse operations and having to deal with fractions and all that messy stuff. We know that at the xintercept, y is 0, and at the yintercept, x is 0. So because we know this, it turns out that the x and yintercepts are really easy to find. Check it out; if I know that the xintercept at y is 0, I can simply substitute 0 in for y into my equation, which gives me 3x+(2*0) = 6. Well, 2*0 just turns into 0, so this term just cancels out and all that we're left with are the xs, and it's a very simple, quick and easy division to both sides, because division undoes the multiplication of the 3, and we find that x = 2, which is my xintercept. So I have the coordinates (2,0). I can find the yintercept in exactly the same way. This time, plugging in 0 for x, gives me the equation 2y = 6 because the xs disappear. Again, I simply undo the times by 2 and divide by 2 and I find that y = 3, which gives me the point (0,3). And I have two points and we're done. I can put those two points on my graph  (2,0), (0,3)  and as long as you have two points, you connect them with your line, and we end up with the exact same line that we did before, but this time we didn't have to solve for y and we didn't have to deal with fractions. It was actually a little bit easier. Lesson Summary So to review, we've talked about slopeintercept form and also standard form. We should also probably quickly discuss the pros and cons to each. Slopeintercept form is a little more intuitive because it gives us more information straight from the rule; it tells us the m and the b that are obvious right from the rule that I can translate quickly into information on the graph. But standard form is nice if we're trying to find intercepts because it makes it easy to substitute in 0 and solve for the remaining value. So if you're given an equation in standard form and you're asked to graph, there's no reason you really have to change it into slopeintercept form first, and we can use this shortcut of finding the intercepts to do it without having to do all the work of solving for y.
400 How to Find and Apply The Slope of a Line Chapter 15 / Lesson 4 In this lesson, we'll discover all about the slopes of lines. We'll learn about different types of slopes and how to find the slopes of lines. We'll use two methods: the slope formula and the slopeintercept form. What Is a Slope? How often do you think about slopes? Maybe not that often. But they're pretty important. I used to live in San Francisco, where I'd ride my bike around. Slopes are very important there. They're kind of unavoidable. The greater, or steeper, the slope, the less likely I wanted to take that route. I wanted to find the flattest route, or the route with the smallest slopes. When you're working with lines, that's all slopes are  they're like different types of hills. Fortunately, you don't have to try to pedal your bike up the slopes of lines. Officially, slope is defined as the steepness of a line. We use the letter m to signify a line's slope. Why m for slope? There isn't even an m in the word slope. Some people say it's because the French verb for 'to climb' is 'monter,' but that may be a myth. I like to think that it's because the letter M is a slopefilled adventure. Notice how it has slopes built right in! But I made that up. Finding Slopes So how do you find a line's slope? You might hear slope referred to as 'rise over run.' That's a way of saying that a slope is a line's vertical change divided by its horizontal change. Or, how much it rises over how much it runs. The formula for finding a slope is m = (change)y/(change)x, which equals (ysub2  ysub1)/(xsub2  xsub1). Those ysub1s and xsub1s just symbolize points on the line. To find a slope, you need two points on the line: (xsub1, ysub1) and (xsub2, ysub2). Since we're talking about straight lines, the slope is constant, so it doesn't matter where those points are. Knowing how the line changes vertically (or its rise) over how it changes horizontally (or its run) is all we need. Sometimes, you'll be given two points on a line and asked to find the slope. If so, just plug those points into the formula. Just remember: rise over run means the y value goes on top, so the yaxis is the vertical one.
401 Types of Slopes Before we try out the slope formula, let's look at what a slope really is. Here's a line with a slope of 1/2. That means it goes up 1 space for every 2 it goes over. That's its rise over run: 1 over 2. Here's one with a slope of 2. If you remember your fractions, you know that 2 is just 2/1, so this line goes up 2 for every 1 it runs. Those slopes were positive. What about negatives? Rise over run doesn't change. We just go in the other direction. Here's a line with a slope of 2/3. It still rises 2 for every 3 it runs. It just runs the other way. Just remember that slopes work like so many things that normally go left to right  everything from words on a page (at least in English) to the time bar on this video. So, positive slopes rise from left to right. Meanwhile, negative slopes rise from right to left, or backwards. Oh, and there are two other kinds of slopes. Check out this line. If you were riding on this hill, well, that'd be all right. Horizontal lines have a slope of 0. Why? Well, it's not rising, so the numerator on the fraction is 0. 0 divided by anything is 0. Then there's this one. I think I remember a street in San Francisco that looked like this. As for the slope, it will have a 0 on the bottom of the fraction. What happens when you divide by 0? Your calculator gets angry. Vertical lines have undefined slopes. Practice Using Points Let's try finding a few slopes using the slope formula. Let's say we're told a line goes through (1, 2) and (3, 3). What is the slope? Remember the formula? m = (ysub2  ysub1)/(xsub2  xsub1). What are our y values? 2 and 3. And our x values? 1 and 3. Let's plug them in. We get m = (32)/(31). That's 1/2. So our slope is 1/2. In the graph, you can see it rises 1 and runs 2. Rise and run  it kind of sounds like a zombie movie, doesn't it? And like in a zombie movie, you wouldn't run until they rise, right? It's always rise first, then run. Or, you know, stay and fight. Me? I'm running. Ok, here's another. We have a line passing through (1, 3) and (4, 2). Don't lose track of those negatives. Let's plug it into our formula. m = (23)/(4  (1)). That's
402 5/3, or 5/3. So it rises 5 and runs 3. Note that despite all the negatives, this is still a positive slope. Let's try one more of these. We have two points: (1, 2) and (3, 1). Using our formula, we get m = (1  (2))/(31). That's 3/4, or 3/4. A negative slope! And it looks like this. Yep, that's negative. I should note that it doesn't matter what you make ysub1 and ysub2. If we try that last one, reversing the order of the points, we get m = (21)/(1  (3)), which is 3/4. The same thing! So don't worry about order. Just focus on rise over run. Slopes from Equations What do you do if you're not given two points? What if, instead, you have an equation of a line? Maybe you need to find the slope of 2x + 3y = 12. Well, you could try to find two points on the line, then use the slope formula. But there's an easier way. You just need to move things around so your line is in slopeintercept form. Slopeintercept form looks like this: y = mx + b. As before, m is the slope. b is the yintercept. That's why it's called slopeintercept form. Why is b the yintercept? Well, remember that the yintercept is where x = 0. If we plug 0 in for x in y = mx + b, we get the yintercept. So in 2x + 3y = 12, just get y alone on one side of the equation. First, subtract 2x from both sides to get 3y = 2x Then divide both sides by 3. We get y = 2/3x + 4. Remember, the slope is the coefficient of x. So the slope of this line is 2/3. If you come across an equation that already looks like y = mx + b, like y = 3/4x  5, then your work is done for you! Whatever is in front of that x is your slope. Practice Using Equations Let's practice converting a few equations. Here's one: y  4x = 2. Remember, we need to make it look like y = mx + b. So just add 4x to both sides to get y = 4x + 2. That's it! Our slope is 4, right here! What about this one? x + 4y = 12. Ok, let's first subtract x from both sides. 4y = x Note that I could've done x, but I want my x listed first to make sure I'm matching the slopeintercept form. Next, let's get that y alone by dividing both sides by 4. We get y = 1/4x Our slope? Right here. 1/4.
403 Ok, how about one more? 5x  4y = 0. Wait. Do you see what's different? How can our b be 0? Remember, the b in y = mx + b is the yintercept. The yintercept is the place where x = 0. So this line crosses the x and yintercepts at the same time, at (0, 0). But what about the slope? Let's subtract 5x from both sides to get 4y = 5x + 0 (we don't really need that +0, but let's keep it as we practice the slopeintercept form). Next, divide by 4. We get y = 5/4x + 0. And our slope? 5/4. And guess what? Now we even know two points on this line: (0, 0) and (4, 5). Lesson Summary In summary, slope is just the steepness of a line, and it's represented by m. A positive slope goes up from left to right, while a negative slope goes up from right to left. The slope of a horizontal line is 0, while a vertical line has an undefined slope. We can find the slope of a line when we're given two points using the slope formula: m = (change)y/(change)x, or (ysub2  ysub1)/(xsub2  xsub1). That's just rise over run. If we're given an equation, just rearrange it to the slopeintercept form, which is y = mx + b, where m is the slope and b is the yintercept. And if you're ever in San Francisco, watch out for those vertical streets. They're dangerous! How to Find and Apply the Intercepts of a Line Chapter 15 / Lesson 5 How can you take an equation of a line and find out where it will cross the x or yaxis? In this lesson, we'll define xintercepts and yintercepts and learn how to find them. Intercepts Sometimes, when I'm driving, I'm not sure how to get to where I'm going. I may be going down a road with a vague idea that it will connect with another road. Yet, I'm not 100% sure of that, and even if it does, I don't know where that will happen. What I want to know is where my road will intercept with another. That's the concept we're going to learn about here. But we're going to look at intercepts in terms of graphs, not me getting lost.
404 X And YIntercepts But just like roads in the real world, lines on graphs intercept with the graph's axes. There are two kinds of intercepts: xintercepts and yintercepts. The xintercept of a line is the place where the line crosses the xaxis. That means it's the place where y = 0. The yintercept of a line is the place where a line crosses the yaxis. That means it's the place where x = 0. In both cases, there's a simple way to figure out the intercepts. Just take the equation of the line and make one of the variables zero. Then solve for the other variable. Whatever that variable equals, that's where your line crosses the axis. The hardest part of intercepts is remembering which variable to make zero. Again, with xintercepts, y = 0, and with yintercepts, x = 0. Just keep your focus on the goal. If we want to know the xintercept, we'll want to know what x is. Think about that for a second. You need y to be 0 for the xintercept, because the xaxis is where y always is zero. Practice Problems This may make more sense when looking at some lines. Let's start with this one: y = 3x + 6. What's the xintercept? Remember, we want to know what x is. So make y = 0. That's 0 = 3x + 6. We get 3x = 6. Divide by 3 and you get x = 2. So this line crosses the xaxis at x = 2, which we write as (2, 0). What about the yintercept? Here, we want to know what y is, so make x = 0. So y = 3x + 6 becomes y = 3(0), which disappears, + 6. That's just y = 6. So it's going to cross the yaxis at y = 6, or (0, 6). Let's put those dots on a graph. Then you can connect the dots and we have our line! Let's try another line: y = 2. Wait, what? Is that a line? It is. But how can you find the xintercept? If you make y = 0, you'd get 0 = 2, which is impossible. That's because the line of y = 2 looks like this. It's a horizontal line. It intercepts the yaxis at 2, but it never intercepts the xaxis. It's parallel to the xaxis. Ok, that was a tricky one. Let's try this one: 2x + 3y = 12. What's the xintercept? Just make y = 0 and solve for x! We start with 2x + 3(0) = 12. Ok, 2x = 12. x = 6. So this line crosses the xaxis at x = 6, which is also (6, 0).
405 What about the yintercept? Just make x = 0! 2(0) + 3y = 12. That's 3y = 12. Divide by 3 and you get y = 4. So it crosses the yaxis at (0, 4). If we put those two points on a graph, then connect the dots, that's our line! What a nice line we've made. But wait. How can we be sure we made the right line? Let's test it! It looks like this line also goes through (3, 2). Will those values for x and y work in our equation? Remember, it's 2x + 3y = 12. So 2(3) + 3(2) = 12. That's = 12. It works! If only finding my way when I'm driving was this easy! Lesson Summary In summary, the xintercept is the spot where a line crosses the xaxis. To find it, just make y = 0, then solve for x. The yintercept is the spot where a line crosses the yaxis. To find this one, just make x = 0, then solve for y. Graphing Undefined Slope, Zero Slope and More Chapter 15 / Lesson 6 There are two special cases when it comes to slopes on the xy plane: horizontal and vertical lines. Without any more information, these examples can be pretty confusing. But with a little instruction, they end up being some of the easiest lines to graph! Slope Basics & Review In this lesson, you'll take a look at some oddball slopes. But first, let's review the basics. Slope is the amount of vertical change per unit of horizontal change. In other words, as a line moves one unit to the right, how many units does it go up or down? You learned in another lesson that if a line goes up from left to right, its slope is positive, and if it goes down from left to right, its slope is negative. You also know how to identify whether the slope is positive or negative when the line is written in y = mx + b form. M represents the slope of the line, so if m is positive, the slope is positive, and the line will be slanting upwards. If m is negative, the slope is negative, and the line will be slanting downwards. But what if you get one of these?
406 In this lesson, you'll learn how to deal with both of those cases. They might look tricky when you first start out, but they're not actually that bad once you get to know them  in fact, they're some of the easiest slopes to handle! Undefined Slope First, we'll start with this one.
407 If the line is vertical, it means that the slope is undefined: it has no value that we can express in numbers. That's a pretty crazy concept, so let's take a look at what's going on here. You know that as a line gets steeper and steeper, the slope gets bigger and bigger if it's positive, or smaller and smaller if it's negative. In both cases, the absolute value of the slope increases as it gets steeper. The bigger a number gets, the steeper the slope is. But the problem with numbers is that they can just keep getting bigger and bigger. There's no such thing as the biggest number. So, no matter how steep the slope is, there will always be a slope that's even steeper. You can just add or subtract 1. To get a completely vertical slope, you'd have to have a number that was simultaneously the biggest and the smallest number in existence, but that's not possible. Neither of those numbers exist, and they certainly can't be the same number. That's why the slope is undefined. Mathematically, the slope isn't actually a real number; hence we call it undefined. So if m is undefined, how do we write this line as an equation? If you look more closely at the graph, you'll see that the xvalue is exactly the same for the entire line. So to represent this line numerically, we use the equation x =. The line here is the line x = 2. The xvalue is the same for every value of y: it's always 2.
408 Zero Slope Now let's look at a similar problem: what about a horizontal line? As a line gets flatter and flatter, the absolute value of the slope gets smaller and smaller. But unlike the increasing slopes, there is a limit to how much a number can decrease. There is such a thing as a number with the smallest absolute value: 0. That's why slope of a horizontal line is 0. You can also think of this mathematically. Slope represents how many units the line goes up for every unit it moves to the right. This line doesn't go up at all, so it goes up 0 units for every 1 unit of movement to the right. That makes the slope 0/1, or 0. If we plug this into the y = mx + b form, we get y = 0x + b. Since any number multiplied by 0 is 0, we represent this line with the equation y =. Another way of looking at this is to notice that the value of y is always the same. Lesson Summary In this lesson, you learned about the slopes of horizontal and vertical lines. Vertical lines have an undefined slope, and they're written in the form x =. Horizontal lines have a slope of 0, and they're written in the form y =. These slopes look a little complicated when you first approach them, but once you understand the concept behind them, they're really not that hard. They rely on all the same concepts that you learned about for normal slopes; they're just the very extremes of what can possibly happen. Now try out some for yourself on the quiz questions! Equation of a Line Using PointSlope Formula Chapter 15 / Lesson 7 It's time for a road trip to Las Vegas, and after four hours of driving at 60 mph... Are we there yet? Learn the pointslope form of the equation of a line to help answer this ageold question.
409 Road Trip to Las Vegas Graph of a location as a function of time A good buddy of mine is getting married tonight in Las Vegas, so I'm going to go on a little road trip. At 2 p.m., I'm on the road to Vegas and I'm going about 60 mph. At 6 p.m., I'm still on the road to Vegas and I'm still going 60 mph. I'm out of gas, but that's another problem. Whether or not I'm in Vegas, and make it there in time for this 10 p.m. wedding, is going to depend on where I start. If I drive for 4 hours from Los Angeles, I might be in Las Vegas. If I'm trying to make it there from Miami or Seattle, I'm probably not there, especially not if I'm only going 60 mph. Let's take a look at a graph of my location as a function of time. I know that my velocity is my change in location divided by my change in time. On this particular graph, it's going to be the slope of any line that is connecting my location at one point in time to my location at another point in time. The slope of the line connecting these two points will be 60, but I don't exactly where these two points are. I don't know if I'm starting in Los Angeles or Phoenix or Seattle. So I don't know if I can possibly make it to this wedding by 10 p.m., because I don't know where I've started. PointSlope Form Let's look at this mathematically. Let's use what's called pointslope form of an equation. This is going to relate time with my location. To use pointslope form, I need to know a point, such as my starting city, and the slope, such as my speed. For example, let's say that at 2 p.m., I'm at mile marker 40. That's my starting point. My speed will be 60 mph. I don't want to get pulled over on the way to this wedding. If I
410 ask you intuitively where I will be after 2 hours, at 4 p.m., you might tell me that I will be at mile marker 160, if I've been going 60 mph. The location now minus the old location equals 120 miles How did you do that? You said that the change in time, which is the amount of time I'm driving, is 2 hours. I'm going 60 mph in those 2 hours. That means that I'm going to travel a total of 120 miles. I'm not going to be at mile marker 120, though, because I didn't start out at mile marker 0. I'm going an additional 120 miles from where I am now. So the change in location is going to be 120 miles. That's location now minus my old location. So I'm going to take 120 and add it to my current location, which is mile marker 40. In 2 hours, going 60 mph, I will go 120 miles further than I am now. I will get all the way to mile marker 160. The Formula We can write this very mathematically as y=y sub 1 + m(x  x sub 1). Here, our initial point, our starting location, is (x sub 1, y sub 1). That's the current time and the current location: x sub 1 is the time and y sub 1 is my location. m is my slope, delta y / delta x. That's how steep this line is, the change in y divided by the change in x. What I have here is my starting location plus my miles per hour times the amount of time, and that gives my location in the future. Example #1 Let's do an example. Let's say that at 1 p.m., I start out at mile marker 10, and I'm going to go 72 mph. Let's use pointslope form to relate the time with my location. Pointslope form isy=y sub 1 + m(x  x sub 1). Let's start filling things in. m is the slope, so that's 72, because I'm going 72 mph. x is x; that's my time. x sub 1 and y sub 1 are my current location. So x sub 1 is the current time (this point here), which is just
411 1 for 1 p.m. y sub 1 is the location at 1 p.m., which is just mile marker 10. I can plug in those two numbers to get y= (x  1). If I want to know where I am at any given point in time, I can plug in that value for x and solve to find my location, y. So, at 4 p.m., x=4 and y is then (41), which is 226 miles. That's my new mile marker. Example of pointslope formula Example #2 All right, let's go back to this wedding problem. Let's say I'm not driving. After going 72 mph, I got too many speeding tickets, so I'm letting someone else do the driving. The wedding is at 10 p.m., and it's at mile marker 100 in Las Vegas. Currently, it's 2 p.m. and we're at mile marker 5. I leave it up to my buddy and I fall asleep in the back seat. At 6 p.m., I wake up and we're at mile marker 25. Is there any way that we're going to get to Vegas by 10 p.m.? Here I don't know how fast we're going, so I can't use pointslope formula. But I do have two points, and I know that we're going a constant speed. I can find out what our velocity has been between 2 p.m. and now and use that to find my speed for my pointslope formula. Let's first find the slope. The two points are (2, 5), because at 2 p.m. I was at mile marker 5, and (6, 25) because at 6 p.m. I was at mile marker 25. I can find the slope of this line: delta y / delta x, which is the change in location divided by the change in time. My location has gone from 5 to 25, which is a difference of 20, and time has gone from 2 to 6, which is 4 hours. So, I've gone 20 miles in 4 hours. That gives me 5 mph. I really need to talk to this guy. We're on the freeway here. Why are we going 5 mph? Is there really bad traffic or something? That's neither here nor there. I know my slope now, and I have a point. In fact, I have two points. And I can use either one in pointslope form to find out an equation for where I will be at what time.
412 This can be simplified to y = 5x  5. So pointslope form says that y=y sub 1 + m(x  x sub 1). I'm going to use my current location as my point, so y sub 1 is 25, for mile marker 25, x sub 1 is 6, because it's 6 p.m., and m is my speed, which is a lousy 5 mph. Pointslope form gives me y=25 + 5(x  6). I can simplify this and I end up with y=5x  5. Where will I be at 10 p.m.? At 10 p.m., x will be 10. Plug in x=10, and I get 505, which is y=45. I will be at mile marker 45. That's nowhere close to Vegas. Vegas is at mile marker 100. When will I get to Las Vegas? This means that I need to find what value of x will give me a value of y that's 100. Let's solve 100=5x  5. I'm going to add 5 to both sides, then divide by 5, and find that x=21 hours after I started. That means I won't get there until 9 a.m. tomorrow. Well, so much for celebrating that wedding. I probably won't even make the after party. Lesson Summary Let's recap. To find an equation for the line that goes through point (x sub 1, y sub 1) with some slope m, we use the pointslope formula: y=y sub 1 + m(x  x sub 1). We can use this, for example, to find out where we're going to be at any given point in time on our road trip to Las Vegas. If, instead of having a point and a slope, we're given two points, we first can calculate the slope and then use pointslope formula. We use this in the case where, for example, I fall asleep on the road to Vegas and I only know the time and our location.
413 How to Use The Distance Formula Chapter 15 / Lesson 8 You can't always rely on your smartphone to tell you how far you need to travel to get from point A to point B. The distance formula will tell you the distance between any two points on a graph. In this lesson, you'll learn where it comes from and how to use it! Introduction Now, I like technology as much as the next guy and have been known to use my smartphone to tell me how to get to the grocery store just four blocks away, but even so, every once in a while, I enjoy going camping and getting out of cell phone range for a few days to get away from it all. I actually recently went hiking up and over the Continental Divide close to Boulder, Colorado with a good friend of mine. We were planning our route for the day, and we picked a spot on our map that looked like a cool place to spend the night. Thing was, it looked kind of far away, and we weren't sure if we'd be able to make it in one day. On top of that, our phones weren't working and couldn't tell us how far away it was. On the map, it was diagonal from us, which meant we couldn't just count over how many boxes over it was. That meant it was time for some good, oldfashioned Pythagorean Theorem: a^2 + b^2 = c^2. By counting straight over and then straight up we had a right triangle, and the distance between us and the campsite was just the side length c. A few quick calculations later, and we found that we were about 8.5 miles away  not close, but doable! Now, at this point in the lesson, you might be wondering, 'Isn't this supposed to be about the distance formula?' Why are we using the Pythagorean Theorem? Well, the reason I show you this is because the distance formula is really just a condensed version of the Pythagorean Theorem. It takes all the steps we just did and combines them into one formula. Because it turns so many steps into one line of math, it's actually a pretty messy formula and can be easy to make mistakes on. So, before I show you the formula, I wanted you to see where it comes from. Think of the distance formula as just a shortcut. If you ever forget the shortcut or feel like the shortcut isn't for you, you can always just use the Pythagorean Theorem  like the way I showed you.
414 The Distance Formula So here it is, the distance formula: d = sqrt((x  x1)^2 + (y2  y1)^2). Now, although it seems like one big mess of letters and math, just think of it as c = sqrt(a^2 + b^2). The way we find a in the distance formula is just doing (x2  x1), and the way we find b is just (y2  y1). This is basically what we just did previously by counting on the map. So, if we were to find the distance to the campsite with the formula this time, it would look like this. (x2  x1) would just be (18), and (y2  y1) would be (72). Those two little problems give us 7 and 5, and then squaring both those numbers gives us 49 and 25. Look familiar? Adding those two together gives us 74, and then taking the square root finally leaves us with 8.6, just like before. The only difference between this way and the Pythagorean Theorem was that we had a 7 for a few steps doing it this way, whereas it was a +7 before. That will happen a lot of the time, but it doesn't matter, because when we multiply it by itself (when we square it), it turns positive anyway; the negative goes away. And that's it! That's the distance formula. Let's just try a few quick examples with the formula to give you a little practice. Example 1 You'll usually see the problems stated like this: Find the distance between the points (5, 9) and (2, 3). Now, we could just graph these two points, go straight over and straight up to make a triangle, and then use a^2 + b^2 to find c^2, like the Pythagorean Theorem, but let's get some practice using the distance formula instead. We'll start by identifying x1, x2, y1 and y2. x1 is the first x value and x2 the second; same thing for the ys. Now we plug these values into their places in the formula, giving us this: d = sqrt((5  (2))^2 + (93)^2). Doing the subtraction leaves us with 7 and 6. Then squaring these numbers turns them into 49 and 36. Adding them together makes 85, and then taking the square root gives us our answer as around 9.2. Example 2 Not too bad, right? Let's do one more for good measure. Find the distance between (2, 1) and (2, 4). Again, start by identifying x1, x2, y1 and y2. Then, substitute them into
415 the formula: d = sqrt((2  (2))^2 + ((4)  (1))^2). This time, we've got to be extra careful with our negatives. 2  (2) is like 2 + 2, so that gives us 4, and 4  (1) is like , which is 3. That leaves us here, with 4 squared and 3 squared on the inside of the square root. These turn into 16 and 9, which add together to 25, and the square root of 25 is just plain old 5. Lesson Summary And that's the distance formula! Let's quickly review what we've learned. The distance formula is a condensed version of the Pythagorean Theorem (a^2 + b^2 = c^2) and looks like this: d = sqrt((x2  x1)^2 + (y2  y1)^2). x1, x2, y1 and y2 are just the x and y coordinates of these two points. And finally, be careful when you have to plug negatives into the distance formula. Sometimes two negatives make a positive. How to Use The Midpoint Formula Chapter 15 / Lesson 9 The formula for the midpoint of a line segment will tell you how to find the middle of any line segment on the x, y plane. Learn about this formula and see how it is used to find the midpoint of a line segment. Line Segments and Midpoints This lesson is on the midpoint formula. Now, in order to understand what the formula is, it's probably best to know what a midpoint itself is. But, in order to know what a midpoint is, we have to know what a line segment is. Luckily, neither of these things is too complicated, so it should take only a minute to go over them. First off, a line segment: as the name implies, it's pretty much a line, but it's only one piece of a line. A real line goes on forever in both directions. But, a line segment has two endpoints; it kind of stops on either end. Because they don't go on forever, every line segment has a spot that is right in the middle  the midpoint. Whether the line segment is long, or short, or horizontal or any other way, the midpoint is the point right in the middle. Knowing exactly where the midpoint is can be handy for a number of reasons, but it isn't always obvious how to find it.
416 When the line segment is either perfectly horizontal or perfectly vertical, the only things you need to know how to do is count and then divide by 2. But, when the line segment is diagonal (like this), knowing exactly where the middle is isn't quite so straightforward. The Midpoint Formula So what do you do? Well that's what the midpoint formula is made for. As long as we have the endpoints, we can simply substitute the values in, and the midpoint coordinates pop out! But, formulas like this are easy to forget, so let's quickly take a look at where this comes from. That way, in case you do forget, you can still figure it out the long way. All the formula really does is find the middle, or the average, of the xs and then the middle, or average, of the ys, separately. In our previous example, the line segment started at x = 4, and ended up at x = 10. You could just count your way to the middle, or find the average by adding them together and then dividing by 2. Either way, we find that the midpoint must be at where x = 7. The ys, on the other hand, went from 1 to 5. That makes the ycoordinate the number right in the middle, which is 3. And sure enough, there's our midpoint, (7, 3)! Example 1 Throwing some negative numbers into the mix can make it a little trickier, so let's try a few more for some quick practice. How about this: find the midpoint between the two points (5, 3) (1, 3). We could use the formula by calling the first x, x1, the second one, x2. Do the same thing with the ys, and substitute the values in like so. The xs are pretty straightforward, but notice how the ys have canceled out and turned into a 0? Nothing wrong with that! Zero divided by anything is just zero, so that makes our midpoint (3, 0). Example 2 Last example: find the midpoint between (1, 6) and (2, 2). When we substitute into our formula, we notice that this example isn't going to give us as nice of an answer as the rest of them did. That's because we end up with an odd number when we add the xs together. And sure enough, halfway between 1 and 2 is 1.5!
417 Lesson Summary To quickly review what we've learned: The midpoint formula helps us find the point that is in the exact middle of any line segment. If we know the endpoints of the line segment, then the midpoint has the coordinates given by this formula. All this is really is the average, or the middle, of the xcoordinates, and the average of the ys as well. What is a Parabola? Chapter 15 / Lesson 10 A parabola is the U shape that we get when we graph a quadratic equation. We actually see parabolas all over the place in real life. In this lesson, learn where, and the correct vocab to use when talking about them. Parabolas in Everyday Life These vertices are also the maximum heights of the parabolas A big part of a college algebra class is getting introduced to the different types of relationships we see in math. The most basic is a linear function, which only has plain xs (such as y = 2x + 4). But once you get past those, the next step is to a quadratic function, which has xˆ2s (such as y = x^2 + 4). There's a lot to learn about quadratics, but the best place to start is with their graphs.
418 Anytime you graph a quadratic equation you end up with what is called a parabola. Parabolas have been behind the scenes of sports, celebrations, and wars for ages. When the first javelin was thrown in the Greek Olympics, or when the first firework was launched in China, or even when the first cannon was fired in the Civil War, they all flew through the air in the shape of a parabola. Today, parabolas are still around in things just like this, but they've also made their way into more modern inventions, like video games. Back around 2007, I actually had an idea for a video game that would use parabolas. I thought it might be fun to just shoot things across the screen. So, what if these birds had their eggs stolen by some pigs, and the birds got really angry and wanted to get back at the pigs to get their eggs back. Maybe we could make these birds shoot across the screen with slingshots, and see if we could attack the pigs and get our eggs back. So, here we've got a likely scenario. Let's try to launch our bird that's really angry at that mean old pig, and let it fly. Eh, we missed. Let's try again. I want to aim up a little bit more this time, let's try again. And, yeah! We got him. The game draws in those little dots to help you aim your shots, but the path they sketch out is actually a perfect parabola. Notice that depending on the angle we launch the bird at, we get a slightly different shape. But, even if we shot the bird almost straight up, or even really close to the ground, it would still be a parabola because there are lots of different kinds. The line in the middle of a parabola represents the axis of symmetry; the arrows point to the roots Defining Spots of a Parabola Depending on how we shoot the bird, each parabola would have a different maximum height, which is our first vocabulary word. The maximum is the highest
419 yvalue that the parabola reaches. In this case, that represents the height that the bird gets. The name of the actual point on the parabola where it gets to the maximum is our second vocab word; it's called the vertex. You might say that the vertex is in the middle of the parabola. That's because the parabolas are symmetrical, they're the same on either side. This means our third vocab word is the line that goes straight down through the middle of the parabola to divide it in half. It's called, the axis of symmetry. Just like any other graph, parabolas' intercepts where the curve intersects either the x or the yaxis. In parabola word problems, the xintercepts will often be the place where your flying object hits the ground, just like here. These xintercepts of quadratic equations (and also bigger functions) can also be called roots. The vertices for concave up parabolas indicate minimum heights For our last few vocabulary terms, we'll need to abandon our video game analogy, or maybe, imagine a new version on some crazy backwards planet where gravity is upside down. This is because parabolas can be concave down like the examples we've been talking about, or concave up, which means the whole shape is just flipped upside down. All the vocab we've talked about is exactly the same for concave up parabolas except one. Now, instead of our vertices being a maximum, they indicate the minimum that the parabola will reach. Lesson Summary To review, parabolas are the shape that graphs of quadratic equations take. They look kind of like a big letter U, and happen anytime something is launched into the air. They can be concave up or concave down, have vertices where a maximum or minimum happens, intercepts where they cross one of the two axes and an axis of symmetry that divides them in half.
420 Parabolas in Standard, Intercept, and Vertex Form Chapter 15 / Lesson 11 By rearranging a quadratic equation, you can end up with an infinite number of ways to express the same thing. Learn about the three main forms of a quadratic and the pros and cons of each. Kinds of Parabolas Any time you throw something into the air, it's going to follow a parabolic path. From throwing your wrapper into the trash to throwing a 50yard touchdown to launching a bird that seems a little bit angry, we see parabolas all over the place. It makes sense then that we want to be able to graph them because the graphs can help us answer questions, like will my bird hit its target? But as is often the case in math, there is more than one way to go about it. In this case, by simply rearranging the parts of the quadratic equation, we can end up with an infinite number of ways to express the same thing. While most of the ways to write the quadratic equation are redundant and useless, there are three forms that actually have unique uses. These three main forms that we graph parabolas from are called standard form, intercept form and vertex form. Each form will give you slightly different information and have its own unique advantages and disadvantages. In this video, we'll go through both for all the different forms. The a value tells if the parabola is concave up or concave down
421 Standard Form Let's begin with standard form, y = ax^2 + bx + c. There it is in general form, and here are a few specific examples of what one might look like: y = x^2 + x + 1 and y = 4x^25x + 9. To be completely honest, the main reason this one makes the cut as a useful form is because it's the easiest and most basic to write. While the other forms will require some fancy rearranging with algebra tricks, like factoring or completing the square, most quadratics will be in standard form straight from the beginning. This means that you can dive right into the problem from the getgo, while the other forms will often make you do work before you can even begin. Once we get past that, though, standard form doesn't have too much to offer. Perhaps, its most useful trait is that the a value tells you whether the parabola is concave up (positive a value) or concave down (negative a value), but it turns out that all the forms are going to have this ability. The second trait of standard form has to do with the yintercept of the parabola. Since the yintercept is where x=0, substituting this in shows us that the a and b terms drop out, leaving us with only the c value. Therefore, the c value is always the yintercept. This is kind of cool, but substituting x=0 into the other forms to find the yintercept is pretty easy too. The last thing you can do with standard form is calculate the axis of symmetry with the formula x = b / 2a. Once again, while this is kind of cool, finding the axis of symmetry is possible and actually easier with the other forms. Intercept Form The axis of symmetry lies directly between the two roots The next form we'll go over is intercept form, y = a(x  p)(x  q). This is the general form, and here are a few specific examples: y = (x  1)(x + 5) and y = 3(x + 5)(x + 9).
422 While it is true that every once in awhile you'll be given a problem that's already in intercept form, it will often be the case that you'll have to first factor the standardform equation to make it look like intercept form. Although this can sometimes be a headache, there are advantages to doing the work. The a value will, again, tell you whether the parabola is concave up or down, and if you want to find the yintercept, you can simply substitute in x=0 and quickly evaluate a(p)(q). Where intercept form gets its name and passes standard form in usefulness, is in its ability to not just tell you where the yintercept is but also where the xintercepts are. Because the xintercepts are where y=0, substituting in either p or q will give you a zero in your product, turning the entire equation into zero. Therefore, p and q are the two xintercepts, or roots, of your quadratic. Be careful with the signs on your roots, though. Because the general equation has a p and q, an (x  5) would actually mean a root at x=5, while an (x + 5) would mean a root at x= 5. Lastly, because parabolas are symmetrical, the axis of symmetry must lie directly in between the two roots. This means you can find it on your graph by working your way into the middle or algebraically by calculating the average between the two points: x = (p + q)/2. Vertex Form The h and k values represent the vertex of the parabola And finally we come to vertex form: y = a(x  h)^2 + k. This is the general form, and these are some specific examples: y = 9(x + 5)^21 and y = (x  3)^21. This time, getting your quadratic into this form requires you to complete the square, which is possibly the hardest algebraic trick of them all. But if you can, you are going to be rewarded for your hard work. First off, the a value still tells us whether it's concave up or down, and the yintercept is still easily found by substituting in x= 0 and evaluating. But now, just like intercept form gave us the intercepts, vertex form
423 will give us the vertex of our parabola straight from the equation: h is going to become the xcoordinate, and k will become the ycoordinate, of our vertex. Now, we can easily tell where the axis of symmetry is simply by remembering that it goes right through the middle of the graph where the vertex is. Therefore, the axis of symmetry is just the line x = h. Lesson Summary To review, depending on how you organize it, a quadratic equation can be written in three different forms: standard, intercept and vertex. No matter the form, a positive a value indicates a concaveup parabola, while a negative a value means concave down. Standard form is the most basic and easy to come up with but has limited helpfulness. Intercept form is what you get if you're willing to factor the quadratic first, and in addition to all that standard form tells you, it also gives you the xintercepts (or roots) of the parabola. Vertex form is what you get if you complete the square on the standardform equation first, and in addition to all that standard form tells you, it also gives you the coordinates of the vertex. How to Graph Cubics, Quartics, Quintics and Beyond Chapter 15 / Lesson 12 Finding the graph of a polynomial isn't too hard. If you can graph, you already know the major steps! With a couple rules and helpful memorizing tools, you will be graphing polynomials in no time! Review Basic Graphs In this lesson, we're looking at how to graph several equations that contain different exponential values. Let's start by reviewing the graphs of y = x^2, y = x^4 and y = x^6. We notice that the graphs with positive leading coefficients and even exponents have an end behavior  that is, the ends of their graphs go up to the left and up to the right. Next, let's look at what happens when our graphs have negative coefficients in front. In this screen, we see the graphs of y = x^2, y = x^4 and y = x^6. If the leading coefficient is negative, their end behavior is opposite, so it will go down to the left and down to the right.
424 Let's continue our review with odd exponents. Here you see the graphs of y = x^3, y = x^5 and y = x^7. We notice that the graphs with positive leading coefficients and odd exponents have an end behavior going down to the left and up to the right. As we add a negative sign in front of our leading coefficient, we see that our graphs change. If the leading coefficient is negative, their end behavior goes up to the left and down to the right. Local Maximum and Minimums The previous graphs were all simple forms of a more complex graphing process. Most of the graphs we will look at have a whole bunch of hills and valleys called local maximums and local minimums. To graph an equation that is more complex, we will need to know where the hills and valleys occur. A relative local maximum is the highest bump in a graph; we can think of this as a hill. A relative local minimum is the lowest bump on the graph; we can think of this as a valley. Let's see what these type of graphs look like. Let's look at the graph of y = x^3 + 3x^216x As you can see, the highest bump in the graph is our local max, or hill. It isn't our maximum point because the graph goes up to infinity on the end of our graph here. So it is our local max. We can also notice local minimum, or the valley. The minimum is not an absolute minimum because the graph goes down to negative infinity here. This equation can also give us tips about how many bumps we could find in the graph. To calculate how many bumps any equation will have, we take the degree of the equation and subtract one. The degree of the equation is the largest exponential value found in the equation. Sample test question on graphing polynomials Let's look at the equation y = x^3 + 3x^216x The degree of this equation is 3. To find out how many bumps we can find, we take the degree of the equation and
425 subtract one: 31 = 2. So at most we would find two bumps in our graph. Now we're ready to start the fun part  graphing without a calculator using what we know about graphing polynomials. On a multiplechoice test, we aren't going to literally graph the equation; we are going to pick the answer that shows the graph. With what we have learned, this should be fairly easy. Sample Test Question Let's look at a sample test question. Which graph represents the equation y = x^310x^211x + 180? Looking at these four multiple choice options, let's use what we know about graphing polynomials to select the correct answer. To begin, let's graph what we know about this equation on a blank Cartesian plane. We know that our graph will go down to the left and up to the right because the leading coefficient, 1, is positive and its exponential power of 3 is odd. Answer choice B can be eliminated because it goes up to the left and down to the right. Evaluate the equation for x = 0 in order to find the yintercept We can also determine that there will be two bumps on the graph because the degree power of the equation is 3. To calculate the number of bumps, remember we subtract the degree power of the equation minus 1. So, 31 = 2 bumps. So the maximum number of bumps we will find in this equation will be two. So right away we can cancel out any graph that has more than two bumps. We can cancel out graph A because it has more than two bumps. We are now left with two answer choices: C and D. The next way that we can determine which graph matches the equation is to find the yintercept. To find the yintercept, we will need to evaluate the equation for x = 0. Starting with our equation, y = x^310x^211x + 180, we will plug in a zero for every x value we see. Now our equation looks like y = (0)^310(0)^211(0)
426 To solve this equation, we will first do our exponents. Zero raised to any exponent is zero. Our equation now looks like y = 010(0)  11(0) Next, we need to do all of the multiplication. Again, any number times zero equals zero. Our equation simplifies to y = 180. So our yintercept for this equation is (0, 180). It looks like the graphs of answer choices C and D both intersect the yaxis at (0, 180). The yintercept does not help us in this case pick our correct answer. Let's graph some other points that we can find on a blank Cartesian plane to see which graph best matches the correct answer. If our equations were easy to factor, we could find the xintercepts. Unfortunately, we aren't that lucky with this equation. So we are going to plot some points to see how it looks in the middle, or between the ends of the graph. Easy points that I would try with virtually any equations would be when x = 1 and 1. We already know that when x = 0 that was our yintercept. After working out our equation, the y point on the graph will be 180. First let's solve for x = 1. To do so, we will evaluate our equation by plugging in 1 for all of the x values. Our equation was y = x^310x^211x When we plug in our 1 values for x, it becomes y = (1)^310(1)^211(1) To work this equation, we will need to compute all of the exponents, making our equation y = 110(1)  11(1) Next, we simply need to multiply our equation, so now we get y = After adding, our solution will be y = 180. So this ordered pair is (1, 180). We can see that this point on the graph would be beside our yintercept. Now we can see that we have two pairs of coordinates in our table: (1,180) and (0,180). Next, let's evaluate the equation for x = 1. When we let x = 1, we plug in 1 in for all x values. Our equation now looks like y = (1)^310(1)^211(1) We need to solve this equation the same way  by first completing the exponents and then multiplying.
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