Introductory Algebra

Transcription

1 Introductory Algebra Student Media Workbook Manuscript An open source (CC-BY) media work book Adapted by Jesse Frausto and Elaine Pham Santiago Canyon College Continuing Education Division This is an adaptation of various OER textbooks listed on the next page. Sections have been rearranged, transformed, removed and added.

2 ACKNOWLEDGEMENT This workbook is made possible because of the exceptional work done by the following people listed below. We are deeply grateful. Introductory Algebra CK-12 flexbook Available for free download at: Introductory Algebra Student Workbook 6 th edition Scottsdale Community College Development Team Available for free download at: Arithmetic for College Readiness Student Workbook 1 st edition Scottsdale Community College Development Team Available for free download at: Understanding Algebra (with author s permission) James Brennan, Boise State University Available for free download at Beginning Algebra Darlene Diaz Santiago Canyon College Available for free download at html Numerous math video lessons from James Sousa at Tyler Wallace at Salman Khan at Larry Perez at Cover image Fibonacci spirals Photo by Aldo Cavini Benedetti Special thanks to Shannon Carter and Berenice Diaz for their help with typing up the answer key for this book. i

3 Copyright 2017, some rights reserved CC-BY Introductory Algebra is licensed under a Creative Commons Attribution 3.0 Unported License. You are free to share: copy, distribute and transmit the work remix: adapt the work Under the following conditions: Attribution: You must attribute the work in the manner specified by the author or licensor (but not in any way that suggests that they endorse you or your use of the work). With the understanding that Waiver: If you get permission from the copyright holder, any of the above conditions can be waived. Public Domain: Where the work or any of its elements is in the public domain under applicable law, that status is in no way affected by the license. Other Rights: In no way are any of the following rights affected by the license. Your fair dealing or fair use rights, or other applicable copyright exceptions and limitations; The author s moral rights; Rights other persons may have either in the work itself or in how the work is used such as publicity or privacy rights Notice: For any reuse or distribution, you must make clear to others the license term of this work. The best way to do this is with a link to the following web page: This is a human readable summary of the full legal code which can be read at the following URL: ii

4 TABLE OF CONTENTS CHAPTER 1: THE NUMBERS OF ARITHMETIC... 1 SECTION 1.1: THE REAL NUMBER SYSTEM... 4 SECTION 1.2: FACTORS AND DIVISIBILITY... 7 SECTION 1.3: FRACTION SECTION 1.4: DECIMALS SECTION 1.5: INTEGERS SECTION 1.6: ORDER OF OPERATIONS CHAPTER 2: INTRODUCTION TO VARIABLES AND PROPERTIES OF ALGEBRA SECTION 2.1 INTRODUCTION TO VARIABLES SECTION 2.2 PROPERTIES OF ALGEBRA CHAPTER 3: LINEAR EQUATIONS AND INEQUALITIES SECTION 3.1: LINEAR EQUATIONS SECTION 3.2: LINEAR INEQUALITIES SECTION 3.3: LITERAL EQUATIONS CHAPTER 4: LINEAR EQUATION APPLICATIONS SECTION 4.1: INTEGER PROBLEMS SECTION 4.2: MARK-UP AND DISCOUNT PROBLEMS SECTION 4.3: GEOMETRY PROBLEMS SECTION 4.4: VALUE AND INTEREST PROBLEMS SECTION 4.5: UNIFORM MOTION PROBLEMS SECTION 4.6 MIXTURE PROBLEMS CHAPTER 5: GRAPHING LINEAR EQUATIONS SECTION 5.1 GRAPHING AND SLOPE SECTION 5.2 EQUATIONS OF LINES SECTION 5.3 PARALLEL AND PERPENDICULAR LINES CHAPTER 6: SYSTEMS OF TWO LINEAR EQUATIONS IN TWO VARIABLES SECTION 6.1: SYSTEM OF EQUATIONS: GRAPHING SECTION 6.2: SYSTEMS OF EQUATIONS: THE SUBSTITUTION METHOD SECTION 6.3: SYSTEM OF EQUATIONS: THE ADDITION METHOD SECTION 6.4: APPLICATIONS WITH SYSTEMS OF EQUATIONS iii

5 CHAPTER 7: INTRODUCTION TO FUNCTIONS SECTION 7.1: RELATIONS AND FUNCTIONS SECTION 7.2: DOMAIN AND RANGE CHAPTER 8: EXPONENTS AND POLYNOMIALS SECTION 8.1: EXPONENTS RULES AND PROPERTIES SECTION 8.2 SCIENTIFIC NOTATION SECTION 8.3: POLYNOMIALS CHAPTER 9: FACTORING EXPRESSIONS AND SOLVING BY FACTORING SECTION 9.1: GREATEST COMMON FACTOR AND GROUPING SECTION 9.2: FACTORING TRINOMIALS OF THE FORM x 2 + bx + c SECTION 9.3: FACTORING TRINOMIALS OF THE FORM ax 2 + bx + c SECTION 9.4: SPECIAL PRODUCTS SECTION 9.5: FACTORING, A GENERAL STRATEGY SECTION 9.6: SOLVE BY FACTORING CHAPTER 10: RATIONAL EXPRESSIONS SECTION 10.1: REDUCE RATIONAL EXPRESSIONS SECTION 10.2: MULTIPLY AND DIVIDE RATIONAL EXPRESSIONS SECTION 10.3 OBTAIN THE LOWEST COMMON DENOMINATOR SECTION 10.4: ADD AND SUBTRACT RATIONAL EXPRESSIONS CHAPTER 11: RATIONAL EQUATIONS AND APPLICATIONS SECTION 11.1: RATIONAL EQUATIONS SECTION 11.2: WORK-RATE PROBLEMS SECTION 11.3: UNIFORM MOTION PROBLEMS SECTION 11.4: REVENUE PROBLEMS CHAPTER 12: RADICALS SECTION 12.1 INTRODUCTION TO RADICALS SECTION 12.2: ADD AND SUBTRACT RADICALS SECTION 12.3: MULTIPLY AND DIVIDE RADICALS SECTION 12.4: RATIONALIZE DENOMINATORS SECTION 12.5: RADICAL EQUATIONS iv

6 CHAPTER 13: QUADRATIC EQUATIONS AND APPLICATIONS SECTION 13.1: THE SQUARE ROOT PROPERTY SECTION 13.2: COMPLETING THE SQUARE SECTION 13.3: QUADRATIC FORMULA SECTION 13.4: APPLICATIONS WITH QUADRATIC EQUATIONS v

7 Chapter 1 CHAPTER 1: THE NUMBERS OF ARITHMETIC Chapter Objectives By the end of this chapter, students should be able to: The real number system Factors and divisibility Review operations with fractions Review operations with decimals Review operations with integers Review the order of operations CHAPTER 1: THE NUMBERS OF ARITHMETIC... 1 SECTION 1.1: THE REAL NUMBER SYSTEM... 4 A. NATURAL NUMBERS... 4 B. WHOLE NUMBERS... 4 C. INTEGERS... 4 D. RATIONAL NUMBERS... 5 E. IRRATIONAL NUMBERS... 5 F. REAL NUMBERS... 6 SECTION 1.2: FACTORS AND DIVISIBILITY... 7 A. DIVISIBILITY... 7 B. FACTORS... 8 C. GREATEST COMMON FACTOR AND LEAST COMMON MULTIPLE... 9 D. PRIME AND COMPOSITE NUMBER E. PRIME FACTORIZATION, GCF, AND LCM EXERCISE SECTION 1.3: FRACTION A. WHAT IS A FRACTION? B. FRACTIONS IN CONTEXTS C. REPRESENTING UNIT FRACTIONS D. EQUIVALENT FRACTIONS E. WRITING FRACTIONS IN SIMPLEST FORM F. IMPROPER FRACTIONS AND MIXED NUMBERS G. OPERATIONS WITH FRACTIONS EXERCISE SECTION 1.4: DECIMALS A. INTRODUCTION TO DECIMALS B. OPERATIONS WITH DECIMALS C. FRACTION AND DECIMAL CONNECTIONS

8 Chapter 1 EXERCISE SECTION 1.5: INTEGERS A. INTEGERS AND THEIR APPLICATIONS B. PLOTTING INTEGERS ON A NUMBER LINE C. ABSOLUTE VALUE AND NUMBER LINES D. OPPOSITES AND NUMBER LINES E. ORDERING INTEGERS USING NUMBER LINES F. REPRESENTING INTEGERS USING THE CHIP MODEL G. THE LANGUAGE AND NOTATION OF INTEGERS H. ADDING INTEGERS I. SUBTRACING INTEGERS J. CONNECTING ADDITION AND SUBTRACTION K. USING ALGORITHMS TO ADD AND SUBTRACT INTEGERS L. MULTIPLYING AND DIVIDE INTEGERS EXERCISE SECTION 1.6: ORDER OF OPERATIONS A. INTRODUCTION TO EXPONENTS B. THE ORDER OF OPERATIONS WITH ADDITION AND SUBTRACTION C. THE ORDER OF OPERATIONS WITH MULTIPLICATION AND DIVISION D. THE ORDER OF OPERATIONS FOR +,,, E. THE ORDER OF OPERATIONS WITH PARENTHESES F. PEMDAS AND THE ORDER OF OPERATIONS EXERCISE CHAPTER REVIEW

9 INTRODUCTION TO ALGEBRA Chapter 1 Media Lesson Brief Origins of Algebra (Duration 7:17) View the video clip and answer the questions below. 1. Who wrote the 1 st book of Algebra? 2. What is the English translation of the title of the book? 3. Where and where was the book written? 4. What civilization were the stone tablets found exploring some of the fundamental ideas of algebra? When? 5. Who was Diophantus? 6. Who lived in India that also significantly contributed to Algebra? 7. CHAPTER 1: THE NUMBERS OF ARITHMETIC Media Lesson Different number systems (Duration 7:04) View the video and fill in the blanks: 1. As we journey through the rich and vibrant history of mathematics, we can see how ideas and creations grew out of 2. Through time, the mathematical explorations of men and women from around the globe have given us fascinating lenses that 3

10 SECTION 1.1: THE REAL NUMBER SYSTEM Chapter 1 A. NATURAL NUMBERS The real number system evolved over time by expanding the notion of what we mean by the word number. At first, number meant something you could count, like how many sheep a farmer owns. These are called the natural numbers, or sometimes the counting numbers. 1, 2, 3, 4, 5,... The use of three dots at the end of the list is a common mathematical notation to indicate that the list keeps going forever. B. WHOLE NUMBERS At some point, the idea of zero came to be considered as a number. If the farmer does not have any sheep, then the number of sheep that the farmer owns is zero. We call the set of natural numbers plus the number zero the whole numbers. 0, 1, 2, 3, 4, 5,... Natural Numbers together with zero What is Zero? Getting Something from Nothing (Duration 3:52). 1. What are the two roles of 0? 2. Who defined 0 explicitly? 3. 0 and what number made up the binary numerical system formed the foundation for modern computer programing? (Optional) Discovery of Zero ( Duration 5:29) C. INTEGERS Even more abstract than zero is the idea of negative numbers. If, in addition to not having any sheep, the farmer owes someone 3 sheep, you could say that the number of sheep that the farmer owns is negative 3. It took longer for the idea of negative numbers to be accepted, but eventually they came to be seen as something we could call numbers. The expanded set of numbers that we get by including negative versions of the counting numbers is called the integers. Whole numbers plus negatives... 4, 3, 2, 1, 0, 1, 2, 3, 4,... The number zero is considered to be neither negative nor positive. About Negative Numbers How can you have less than zero? Well, do you have a checking account? Having less than zero means that you have to add some to it just to get it up to zero. And if you take more out of it, it will be even further less than zero, meaning that you will have to add even more just to get it up to zero. 4

11 D. RATIONAL NUMBERS Introduction to rational and irrational numbers (Duration 5:54) Chapter 1. Rational numbers are numbers that can be written in the form of aa bb or nnnnnnnnnnnnnnnnnn, where aa and bb dddddddddddddddddddddd are integers (but bb cannot be zero). Rational numbers include what we usually call fractions. Notice that the word rational contains the word ratio, which should remind you of fractions. Some decimals are also rational numbers because some decimals can be converted to fractions. RESTRICTION: The denominator cannot be zero! (But the numerator can) Examples: 3 = 0.75 Rational (terminates) = = 0. 6 Rational (terminates) 5 11 = = 0.45 Rational (terminates) There are numbers that cannot be expressed as a fraction, and these numbers are called irrational because they are not rational. NOTE: The denominator cannot be zero! (But the numerator can). A fraction has the denominator is undefined. For example, 5 0 = undefined. If the numerator is zero, then the whole fraction is just equal to zero. For example, 0 5 = 0 E. IRRATIONAL NUMBERS Irrational numbers are numbers: cannot be expressed as a ratio of integers. as decimals they never repeat or terminate (rational decimals always repeat or terminate) Examples: Irrational (Never repeats or terminates) 2 = ππ Irrational (Never repeats or terminates) Irrational (Never repeats or terminates) Below is an example of irrational numbers on a number line approximately: To get the exact location of on a number line, we can apply the Pythagorean Theorem to a right triangle with the length of each leg equal 1 to find the hypotenuse length of 2 like the diagram on the right. We can use the same method, apply the Pythagorean Theorem to other right triangles to find the exact location of 3, 4, 5 etc. on the number line. 5

12 Making sense of irrational numbers (Duration 4:41) (Skip 1:50 to 3:08) Chapter 1. 1) Which of these numbers is/are rational: A) 3 B) 12.1 C) 5 A. A) only B. A) and B) only C. All of the above D. None of the above 2) What are irrational numbers? A. Numbers which do not make any sense to the common man B. Real numbers which can be expressed as a ratio of two integers in the form p/q and where the denominator is always non-zero C. Numbers which are exactly opposite to rational numbers D. Real numbers which cannot be expressed as a ratio of integers 3) Which of the following statements is true? A. Pi = 22/7 B. Pi = 355/13 C. Pi is the ratio of a circle s circumference to its diameter D. Pi is the ratio of a circle s diameter to its circumference E. Pi is a terminating recurring decimal F. REAL NUMBERS Rational + Irrational numbers All points on the number line When we put the irrational numbers together with the rational numbers, we finally have the complete set of real numbers. Any number that represents an amount of something, such as a weight, a volume, or the distance between two points, will always be a real number. The following diagram illustrates the relationships of the sets that make up the real numbers 6

13 SECTION 1.2: FACTORS AND DIVISIBILITY Chapter 1 A. DIVISIBILITY Divisible: When one number can be divided by another number and the result is an exact whole number that is there is no remainder left. Example: 12 is divisible by 3 because 12 3 = 4 exactly with no remainder. 13 is not divisible by 3 because 13 3 = 4 with remainder 1. Divisibility Rules Media Lesson Divisibility Rules (Duration 9:34). 1. A number is divisible by 2 if. Example: 512: Yes 431: No 2. A number is divisible by 4 if. Example: 3. A number is divisible by 8 if. Example: 4. A number is divisible by 3 if. Example: 5. A number is divisible by 6 if. Example: 6. A number is divisible by 9 if. Example: 7. A number is divisible by 10 if. Example: 8. A number is divisible by 5 if. Example: YOU TRY Determine if the given number is divisible by 2, 3, 4, 5, 6, 8, 9, 10. a) 8064 b) 270 Yes/No Yes/No Yes/No Yes/No By 2 By 6 By 2 By 6 By 3 By 8 By 3 By 8 By 4 By 9 By 4 By 9 By 5 By 10 By 5 By 10 7

14 B. FACTORS Factors of a number are the numbers you multiply together to make that number. Example: 2 x 5 = 10, 2 and 5 are factors. Factor Factor Product 9 x 2 = 18 Chapter 1 Media Lesson Factors (Duration 5:47). Here is another definition of factors. Factors are the numbers that divide evenly into a number. This means a factor divides into another number and there is no remainder. Example: The factor of 15 are 1, 3, 5, 15 since 1 15 = = 15 Determine the factors of each number. 1) 24 2) 54 3) 120 4) 23 FINDING FACTORS USING PERFECT SQUARES Media Lesson Finding all of the Factor of a Number using Perfect Squares (Duration 13:54). Method: To determine all of the factors of a whole number, we will find all the pairs of whole numbers whose product is the number. We will check all the numbers whose square is less than the number we are trying to factor. 8

15 Table of Perfect Squares 2 2 = = = = = = = = = = = = 169 Chapter 1 Directions: Find all factors of the given numbers by finding factor pairs. Use the table of perfect squares to see what the largest number you have to check is. Write your final answer as a list of factors separated by commas. a) 18 Largest number you have to check: List of Factors: b) 90 Largest number you have to check: List of Factors: YOU TRY a) 84 Largest number you have to check: List of Factors: C. GREATEST COMMON FACTOR AND LEAST COMMON MULTIPLE Media Lesson Intro to Greatest Common Factor and Least Common Multiple (Duration: 9:54). a) You and your friends are sending care packages to military service members overseas. Each package will contain brownies and cookies. You have 20 brownies and 12 cookies. Every package made needs to be identical. What is the greatest number of packages you can send that meets this requirement? # of Packages # of Brownies # of Packages # of Cookies 9

16 Chapter 1 b) Judy and Dan are running around a track. Judy can run one lap in 3 minutes while it takes Dan 4 minutes. If they both start at the same time, how many minutes will it take them to meet? Finding the Greatest Common Factor of Two Numbers (Duration 5:12). Common Factors of two numbers are factors that both numbers share. The Greatest Common Factor (GCF) of two numbers is the largest of these common factors. a) Find all factors of 36. Write your final answer as a list of factors separated by commas. List of Factors 36: b) Find all factors of 90. Write your final answer as a list of factors separated by commas. List of Factors of 90: c) List the common factors of 36 and 90: d) Identify the Greatest Common Factor (GCF) of 36 and 90: YOU TRY Finding the GCF of Two Numbers a) Find all factors of 24. Write your final answer as a list of factors separated by commas. List of Factors 24: b) Find all factors of 60. Write your final answer as a list of factors separated by commas. List of Factors of 60: c) List the common factors of 24 and 60: d) Identify the Greatest Common Factor (GCF) of 24 and 60: 10

17 Multiples, Common Multiples, and Least Common Multiple (Duration 3:26) Chapter 1. A multiple of a number is a product of the number with any whole number Common Multiples of two numbers are multiples that both numbers share. The Least Common Multiple (LCM) of two numbers is the least of these common multiples. a) The first six multiples of 8 are: b) The first six multiples of 12 are: c) Some common multiples of 8 and 12 are: d) The Least Common Multiple (LCM) of 8 and 12 is: YOU TRY The method above is called Finding LCM by using list of multiples. a) The first six multiples of 6 are: b) The first six multiples of 4 are: c) Some common multiples of 6 and 4 are: d) The Least Common Multiple (LCM) of 6 and 4 is: D. PRIME AND COMPOSITE NUMBER Prime and composite numbers (Duration 7:41). A prime number is a whole number greater than 1 whose factor pairs are only the number itself and 1. A composite number is a whole number greater than 1 which has at least one factor other than itself and 1. For example: 6 is not prime because it has factors 2 and 3 besides 1 and 6. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23,... To determine if a number is prime or composite, we only need to check to see if the number is divisible by the prime factors whose square is less than the number we are trying to factor 2 2 = = = = = = = = 361 Verify that the following numbers are prime by checking to see if the number is divisible by any prime numbers whose square is less than the number given. a) 89 Largest prime you have to check: 11

18 Chapter 1 b) 163 Largest prime you have to check: YOU TRY Notes: There are an infinite number of prime numbers (the list goes on forever). The numbers 0 and 1 are neither prime nor composite. All even numbers are divisible by 2 and so all even numbers greater than 2 are composite numbers. Verify that the following numbers are prime by checking to see if the number is divisible by any prime numbers whose square is less than the number given. a) 109 Largest prime you have to check: Determine a number is a prime or a composite (Duration 4:39). Determine whether the following numbers are prime, composite, or neither. 24: 2: 1: 17: The Fundamental Theorem of Arithmetic Any integer greater than 1 is either a prime number, or can be written as a unique product of prime numbers (ignoring the order). In other words, all composite numbers are unique products of smaller prime numbers (ignoring the order of the factors). This theorem is saying that prime numbers are building blocks of the integers, just like atoms are the building blocks of matter in chemistry and biology. Modern cryptography, which is absolute essential to computer security, relies on the fundamental theorem of arithmetic. By combining primes and forming very large composite numbers, code makers can create a key that makes a cipher virtually unbreakable. Knowing how to construct numbers has a very practical use in our world today. The Fundamental Theorem Of Arithmetic (Duration 3:51) Optional reading: What are prime numbers, and why are they so vital to modern life? 12

19 Chapter 1 E. PRIME FACTORIZATION, GCF, AND LCM "Prime Factorization" is the process of finding which prime numbers multiply together to make the original number. In other words, prime factorization is the process of breaking down a number into its prime factors. Prime Factorization (Duration 5:34). Find the prime factorizations for the given numbers using factor trees. Write the final result in exponential form and factored form. a) 12 b) 75 c) 155 Factored Form: Factored Form: Factored Form: Exponential Form: Exponential Form: Exponential Form: YOU TRY d) 18 e) 84 f) 266 Factored Form: Factored Form: Factored Form: Exponential Form: Exponential Form: Exponential Form: DETERMINE GCF & LCM USING THE PRIME FACTORIZATION METHOD In this section, we are going to use prime factorization to find a more streamlined approach to finding the GCF and LCM of two numbers. First, let s review the method we used in Section C to find the GCF and LCM. A. To find the GCF of 8 and 12, we would follow the steps below. Step 1: Find all the factors of 8. Factors of 8: 1, 2, 4, 8 Step 2: Find all the factors of 12. Factors of 12: 1, 2, 3, 4, 6, 12 Step 3: The GCF of 8 and 12 is the largest factor they have in common. So the GCF is 4. 13

20 B. To find the LCM of 8 and 12, we would follow the steps below. Chapter 1 Step 1. List some multiples of 8. Step 2. List some multiples of 12. Multiples of 8: 8, 16, 24, 32, 40, 48, Multiples of 12: 12, 24, 36, 48, 60, Step 3. The LCM of 8 and 12 is the smallest multiple they have in common. So the LCM is 24. Prime Factorization, GCF, and LCM (Duration 8:06). 1. Use the prime factorization method to determine the GCF and LCM of 8 and 12. a) Find the prime factorizations of 8 and 12 using factor trees and write the prime factorizations in factored form Factored Form: Factored Form: b) List of common prime factors: (include repeated factors) c) The product of the common prime factors of 8 and 12 is their GCF. Find the GCF. GCF of 8 and 12: d) The LCM of 8 and 12 is their product divided by their GCF. Find the LCM. Show all steps. LCM of 8 and 12: 2. Use the prime factorization method to determine the GCF and LCM of 54 and 90. a) Find the prime factorizations of 54 and 90 using factor trees and write the prime factorizations in factored form Factored Form: Factored Form: b) List of common prime factors: (include repeated factors) c) The product of the common prime factors of 54 and 90 is their GCF. Find the GCF. GCF of 54 and 90: d) The LCM of 54 and 90 is their product divided by their GCF. Find the LCM. Show all steps. LCM of 54 and 90: The method above is call Finding LCM by dividing GCF. 14

21 YOU TRY Chapter 1 Use the prime factorization method to determine the GCF and LCM of 18 and 84. a) Find the prime factorizations of 18 and 84 using factor trees and write the prime factorizations in factored form Factored Form: Factored Form: b) List of common prime factors: (include repeated factors) c) The product of the common prime factors of 18 and 84 is their GCF. Find the GCF. GCF of 18 and 84: d) The LCM of 18 and 84 is their product divided by their GCF. Find the LCM. Show all steps. LCM of 18 and 84: Method 3: Finding LCM using the highest power of prime factors. Finding LCM (Duration 6:25). The method of writing out multiples is only practical for determining the LCM of small integers. As a result, we have a different way to build the LCM of two integers by analyzing prime factors. The LCM will be the product of the prime factors raised to the highest power of the prime factorization of the numbers. 36 = = = = LCM: Determine the LCM of 60 and = 72 = LCM = = 15

22 YOU TRY Chapter 1 Find the LCM 1) 27, = 18 = LCM = = 2) 24, = 32 = LCM = 3) 35, = 25 = LCM = 16

23 EXERCISE Use the divisibility rules to answer the following questions. Explain your reasoning. 1) Is 18 divisible by 3? 2) Is 240 divisible by 5? Chapter 1 3) Is 22 divisible by 2? 4) Is 212 divisible by 4 and 6? 5) Is 44 divisible by 6? 6) Is 456 divisible by 6 and 3? 7) Is 112 divisible by 2 and 3? 8) Is 246 divisible by 2? 9) Is 27 divisible by 9 and 3? 10) Is 393 divisible by 3? 11) Is 219 divisible by 9? 12) Is 7450 divisible by 10? 13) Is 612 divisible by 2 and 3? 14) Is 884 divisible by 4? Find all factors of the given numbers by finding factor pairs. Use the table of perfect squares to see what the largest number you have to check is. Write your final answer as a list of factors separated by commas. Complete the Table of Perfect Squares 2 2 = 5 2 = 8 2 = 11 2 = 3 2 = 6 2 = 9 2 = 12 2 = 4 2 = 7 2 = 10 2 = 13 2 = 15) 12 Largest number you have to check: List of Factors 16) 48 Largest number you have to check: List of Factors 17) 185 Largest number you have to check: Fill in the blanks: List of Factors 18) A number is a whole number greater than 1 whose factor pairs are only the number itself and one. 19) A number is a whole number greater than 1 which has at least one factor other than itself and one 20) The of a number is the number written as a product of only prime factors. 21) Common factors of two or more numbers are factors that both numbers. 22) The of two or more numbers is the largest of the two numbers common factors. 23) A of a number is a product of the number with any whole number. 17

24 24) The is the smallest multiple of two or more numbers. 25) Determine all of the prime numbers less than 50. Chapter 1 Verify that the following numbers are prime by checking to see if the number is divisible by any prime numbers whose square is less than the number given. 26) ) ) 83 29) 39 Determine whether the numbers are prime or composite. If it is composite, show at least one factor pair of the number besides 1 and itself. If it is prime, show the numbers you tested and the results of your division 30) ) 61 32) ) 139 Find the prime factorizations for the given numbers using factor trees. Write the final result in exponential form and factored form. 34) 32 35) ) 72 37) 280 Use two different factor trees to determine the prime factorizations of 90 and 150. Write the final result in exponential form and factored form. 38) Factored form: 90 = Exponential form: 90 = 39) Factored form: 150 = Exponential form: 150 = Find the GCF of the given numbers 40) 8 and 20 41) 30 and ) 16 and 18 43) 22 and 25 44) 12, 8, 24 45) 120, 325, 525 Find the LCM for the given numbers using the listing of multiples method. 46) 4 and 6 47) 10 and 8 48) 15 and 9 49) 12 and 26 18

25 Chapter 1 Find the prime factorizations using factor trees for the following pairs of numbers. Find the GCF Then find LCM by using the dividing the GCF method. 50) 4 and 6 51) 10 and 8 52) 15 and 9 53) 12 and 26 Find the LCM by using the highest power of prime factors method. 54) 125 and ) 156 and ) 126, 266 and 38 57) 36, 18 and 86 58) Penny and Sheldon are assembling hair clips. Penny can assemble a hair clip in 6 minutes and Sheldon can assemble a hair clip in 9 minutes. a) If they start making the hair clips at the same time, what is the least amount of minutes it will take for them finish a hair clip at the same time? b) After this amount of minutes, how many hair clips will Penny have made? c) After this amount of minutes, how many hair clips will Sheldon have made? 59) Kathryn is packing bags of food at the local food pantry. She has 24 jars of tomato sauce and 30 cans of soup. a) If she wants each bag to have the same numbers of tomato sauce and soup, what is the greatest number of bags she can pack? b) How many jars of tomato sauce will each bag have? c) How many cans of soup will each bag have? 60) Paige is buying hot dogs and buns for a family reunion. Each package of hot dogs contains 8 hot dogs. Each package of buns contains 10 buns. a) What is the least total amount of hot dogs and buns she needs to buy in order for the amounts to be equal? b) How many packages of hot dogs will she buy? c) How many packages of buns will she buy? 19

26 SECTION 1.3: FRACTION Chapter 1 A. WHAT IS A FRACTION? There are many ways to think of a fraction. A fraction can be thought of as one quantity divided by another written by placing a horizontal bar between the two numbers such as 1 where 1 is called the numerator 2 and 2 is called the denominator. Or, we can think of fractions as a part compared to a whole such as 1 out of 2 cookies or 1 of the cookies. 2 In this lesson, we will look at a few other ways to think of fractions as well. Officially, fractions are any numbers that can be written as nnnnnnnnnnnnnnnnnn but in this course, we will dddddddddddddddddddddd consider fractions where the numerator and denominator are integers. These special fractions where the numerator and denominator are both integers are called rational numbers. Since rational numbers are indeed fractions, we will frequently refer to them as fractions instead of rational numbers. Language of Fractions (Duration 6:18). Each of the phrases below are one way we may indicate a fraction with words. Rewrite the phrases below in fraction form and write the fraction word name. Language Fraction Representation Fraction Word Name 20 divided by 6 8 out of 9 A ratio of 3 to 2 11 per 5 2 for every 7 20

27 B. FRACTIONS IN CONTEXTS Dividend Divisor Quotient 12 3 = 4 Chapter 1 Fractions in Context: Four Models (Duration 8:12).. In the next example, we will look at four different types of fractions in context. 1. Quotient Model (Division): Sharing equally into a number of groups 2. Part-Whole Model: A part in the numerator a whole in the denominator 3. Ratio Part to Part Model: A part in the numerator and a different part in the denominator 4. Rate Model: Different types of units in the numerator and denominator (miles and hours) Represent the following as fractions. Determine whether it is a quotient, part-whole, part to part, or rate model. a. Three cookies are shared among 6 friends. How many cookies does each friend get? b. Four out of 6 people in the coffee shop have brown hair. What fraction of people in the coffee shop have brown hair? c. Tia won 6 games of heads or tails and lost 3 games of heads or tails. What is the ratio of games won to games lost? d. A snail travels 3 miles in 6 hours. What fraction of miles to hours does he travel? What fraction of hours to miles does he travel? YOU TRY e. Jorge bikes 12 miles in 3 hours. What fraction of miles to hours does he travel? f. Callie has 5 pairs of blue socks and 12 pairs of grey socks. What fraction of blue socks to grey socks does she have? g. Callie has 5 pairs of blue socks and 12 pairs of grey socks. What fraction of all of her socks are blue socks? 21

28 Chapter 1 NOTE: Let s revisit the note in the Rational number section. Dividing by zero is undefined. If we consider a fraction is a quotient that is sharing equally into a number of groups. For example, dividing 5 candies among 0 people, how much does each person get? The question does not make sense because we cannot share among 0 people, and we cannot divide by 0. When the numerator is zero, the whole fraction is equal to zero. For example, 0 = 0. When we do 5 not have any candies but want to share them with 5 people. Therefore, each of them will get 0 candy. C. REPRESENTING UNIT FRACTIONS A unit fraction is a fraction with a numerator of 1. In this section, we will develop the idea of unit fractions and use multiple representations of unit fractions. Representing unit fractions (Duration 4.37). a. Plot the following unit fractions on the number line, 1 2, 1 4, 1 Label your points below the 5 number line. Representing unit fractions using area model (Duration 3.38). b. Represent the fractions using the area model. The unit is labeled in the second row of the table Representing unit fractions using the discrete objects (Duration 2.07 c. Represent the unit fractions using the discrete objects. The unit is all of the triangles in the rectangle. Represent 1 of the triangles. 4 22

29 D. EQUIVALENT FRACTIONS Media Lesson Equivalent fractions (Duration 7:37) Chapter 1 Follow the video lesson, take notes and fill in the blanks below. Equivalent fractions are different fractions that are equal or represent the same amount To find equivalent fractions, or the numerator and denominator by the nonzero whole number. We would only the numerator and denominator by a whole number that is a of both the numerator and denominator. Once a fraction does not have any other than 1 between the numerator and denominator, a fraction is simplified or written in lowest terms

30 Chapter Define: Dark blue Rod = 1 1 Determine two equivalent fractions for each fractions for each fractions below Determine the numerator with given denominator Media Lesson Determine the numerator with given denominator (Duration 2:25). Example: Missing numerator Fill in the missing numerator in each fraction below to create equivalent fractions. 1 a) =? b) 14 =? c) 4 =? 7 49 d) =? 5 The denominator indicates how many part is the whole. The numerator indicates how many parts are being considered. 24

31 YOU TRY Chapter 1 Rewrite the given fractions as equivalent fractions given the indicated denominator. 3 a) =? b) 12 =? 5 c) =? d) 6 =? e) 8 =? f) 3 =? 5 15 E. WRITING FRACTIONS IN SIMPLEST FORM What is a simplified Fraction (Duration 11:49 ). The simplest form of a fraction is the equivalent form of the fraction where the numerator and denominator are written as integers without any common factors besides 1. We may also request that a fraction be written in simplest form by using the equivalent language; simplify the fraction, reduce the fraction, or reduce the fraction completely. a) Write the fraction number for each diagram below the figure using one circle as the unit. b) What do the fractions have in common? c) Which fraction do you think is the simplest and why? d) Divide the numerators and denominators of the second and fourth fractions by 2. What do you notice? e) Rewrite the last three fractions below by writing their numerators and denominators in terms of their prime factorizations. Do you see any patterns? f) Simplify your fractions in part e by cancelling out all of the common factors (besides 1) that the numerators and denominators share. 25

32 Simplifying Fractions by Repeated Division and Prime Factorization Simplifying Fractions by Repeated Division and Prime Factorization (Duration 12:40) Chapter 1 We can use two different methods to simplify a fraction: repeated division or prime factorization. 1. Repeated Division: Look for common factors between the numerator and denominator and divide both by the common factor. Continue this process until you are certain the numerator and denominator have no common factors. 2. Prime Factorization: Write the prime factorizations of the numerator and denominator and cancel out any common factors. Simplify the given fractions completely using both the repeated division and prime factorization methods. In each case, state which you think is easier and why. a) b) c) YOU TRY Simplify the given fractions completely using both the repeated division and prime factorization methods. In each case, state which you think is easier and why. a) 6 8 b) c)

33 F. IMPROPER FRACTIONS AND MIXED NUMBERS Media Lesson Improper fractions and mixed numbers (Duration 6:48) Chapter 1.. There are 3 types of fractions. 1. A proper fraction is a fraction with a smaller numerator than denominator. Proper fractions are less than 1. Example: 2. An improper fraction is a fraction, which the numerator is greater or equal to the denominator. Improper fractions are greater than or equal to 1. Example: 3. A mixed number is a number written as the sum of an integer and a proper fraction. Example: To convert from an improper fraction to a mixed number: a. b. c. d. Example: Convert 13 to a mixed number. 5 a. b. c. Shade the circles to represent 13 5 The final answer is : 27

34 Chapter 1 Example: Convert 32 6 to a mixed number To convert a mixed number into an improper fraction Example: Convert 2 3 into an improper fraction. 8 Shade the circles to represent 13 5 Example: Convert 7 5 into an improper fraction. 12 Why is converting a mixed number to an improper fraction important? It is important because you will have to convert mixed numbers to improper fractions in order to multiple or divide mixed numbers. YOU TRY Convert the following fractions to a mixed number if you can. a) 8 3 b) 15 8 c) d) e)

35 G. OPERATIONS WITH FRACTIONS Addition and Subtraction of Fractions Adding fractions (Duration 9:10) Chapter 1. To add fractions with like denominators: 1) Add the numerators, keeping the same denominator. 2) Simplify, if possible = = To add fractions with unlike denominators: 1. Find the least common multiple of the denominators, that number is the least common denominator, LCD The LCD is the least common multiple of the denominators. 2. Create equivalent fractions with the common denominator. 3. Add the numerators, keeping the common denominator the same. 4. Simplify if possible = ERROR! =

36 Add Chapter Adding mixed numbers (Duration 6:35).. To add mixed numbers 1. Obtain a (the least) common denominator. 2. Add 3. If the sum has an improper fraction, convert it to a mixed number and add it to the whole number. 4. Simplify if possible

37 YOU TRY Chapter 1 Add the following fractions and reduce your answer to the lowest term if possible. 1) ) Subtract fractions with like and unlike denominators (Duration 8:29).. To subtract fractions with like denominators: 1) Keep the denominator the same and subtract the numerators. 2) Simplify, if possible = = To subtract fractions with unlike denominators: 1. Find the least common multiple of the denominators, that number is the least common denominator, LCD 2. Create equivalent fractions with the common denominator. 3. Keep the common denominator the same and subtract the numerators 4. Simplify, if possible. Ex: = = Subtract 5 a)

38 Chapter 1 b) c) d) Subtract mixed numbers (Duration 7:15). To subtract mixed numbers 1) Obtain a (the least) common denominator. 2) Check to see is you need to borrow 3) Subtract the fractions and then the whole numbers. 4) Simplify, if needed. Subtract

39 Chapter YOU TRY Subtract and reduce your answer to the lowest term if possible a) b) c) d) Multiplication and Division of Fractions Multiply fractions (Duration 6:41). To multiply fractions 1. Multiply the numerators and multiply the denominators 2. Write the product in simplest form or in lowest terms. A fraction is simplified or in lowest terms when the only common factor between the numerator and denominator is 1. We can simplify before multiplying or after multiplying. Example: Multiply. If applicable, write your answer as both an improper fraction and a mixed number. 1 a) 3 b) c) d)

40 YOU TRY Chapter 1 Multiply. If applicable, write your answer as both an improper fraction and a mixed number. 2 a b c d To divide fractions: 1. Change the second fraction (the divisor) to its reciprocal 2. Simplify or cancel common factors if possible 3. Multiply Divide fractions (Duration 3:06). Example: To divide mixed numbers: 1. Rewrite mixed numbers and whole numbers as improper fractions 2. Change the second fraction (the divisor) to its reciprocal 3. Simplify or cancel common factors if possible 4. Multiply Note: A pair of numbers are reciprocals if their product is one. Each number is considered the reciprocal of the other. For example, the reciprocal of 2 3 is 3 because their product is

41 Divide mixed numbers (Duration 3:47) Chapter 1. Example: YOU TRY Divide. Write your answers in simplest form. a) b) c) d) e) f) g) h)

42 EXERCISE Chapter 1 Each of the phrases below are one way we may indicate a fraction with words. Rewrite the phrases below in fraction form and write the fraction word name. Language Fraction Representation Fraction Word Name 1) A ratio of 5 to 10 2) 9 for every 10 3) 4 out of 7 4) 15 per 2 5) 16 divided by 5 For each problem below, write the fraction that best describes the situation. Be sure to reduce your final result. 6) John had 12 marbles in his collection. Three of the marbles were Comet marbles. What fraction of the marbles were Comet marbles? What fraction were NOT Comet marbles? 7) Jorge s family has visited 38 of the 50 states in America. What fraction of the states have they visited? 8) In a given bag of M & M s, 14 were yellow, 12 were green, and 20 were brown. What fraction were yellow? Green? Brown? 9) Donna is going to swim 28 laps. She has completed 8 laps. What fraction of laps has she completed? What fraction of her swim remains? 10) Last night you ordered a pizza to eat while watching the football game. The pizza had 12 pieces of which you ate 6. Today, two of your friends come over to help you finish the pizza and watch another game. What is the fraction of the LEFTOVER pizza that each of you gets to eat (assuming equally divided). What is the fraction of the ORIGINAL pizza that each of you gets to eat (also assuming equally divided). 11) Represent the unit fraction 1 8 a. Number line using each of the representations below. 36

43 b. Area models. Use the unit labeled in the second row of the table. Chapter 1 c. Discrete objects. 12) Create two fractions equivalent to the given fraction by cutting the given representations into a different number of equal pieces. a) Given fraction 3 4 b) Given fraction is equivalent to the fraction: 3 5 is equivalent to the fraction: Rewrite the given fractions as equivalent fractions given the indicated numerator or denominator 13) Rewrite 2 3 with a denominator of ) Rewrite 6 7 with a numerator of ) Rewrite 5 6 with a numerator of ) Rewrite 4 8 with a denominator of 2. 17) Rewrite with a numerator of 9. 18) Rewrite 5 1 with a denominator of 5. 37

44 Simplify each of the following fractions if possible. 5 19) 20) ) 0 4 Chapter 1 22) ) ) 1 0 Simplify the given fractions completely using both the repeated division and prime factorization methods. In each case, state which method you think is easier and why. Fraction Repeated Division Method Prime Factorization Method 25) 26) ) Complete the table below. Convert. Improper Fraction Mixed Number 28) ) ) ) ) ) Use the diagrams given to represent the values in the addition problem and find the sum. Then perform the operation algebraically. 34) Tom had 1 6 of a carrot cake last night and 2 of a carrot cake today. How much of one whole 6 carrot cake did Tom have? 38

45 Chapter 1 35) Ava walked 3 8 of a mile to the store and then ran another 7 8 of a mile to school. How far did she travel in total? 36) Add or subtract each of the following. Be sure to leave your answer in simplest (reduced) form. If applicable, write your answer as both an improper fraction and a mixed number. 37) ) ) ) ) ) ) 44) ) ) Divide or multiply. Simplify your result if necessary. Write any answers greater than 1 as both an improper fraction and a mixed number. 47) ) ) ) ) 53) 54) ) ) 58) ) )

46 Chapter 1 Solve the following problems. When necessary, write your final answers as both mixed numbers and improper fractions in the simplest term. 60) If Josh ate 1 of a pizza, what fraction of the pizza is left? 4 61) If I drove miles one day and miles the second day and miles the third day, how far did I drive? 62) Melody bought a 2-liter bottle of soda at the store. If she drank 1 of the bottle and her brother drank 8 2 of the bottle, how much of the bottle is left? 7 63) James brought a small bag of carrots for lunch. There are 6 carrots in the bag. Is it possible for him to eat 2 6 of the bag for a morning snack and 5 of the bag at lunch? Why or why not? 6 64) Maureen went on a 3 day, 50 mile biking trip. The first day she biked biked miles. How many miles did she bike on the 3 rd day? miles. The second day she 65) Scott bought a 5 lb bag of cookies at the bakery. He ate 2 5 of a bag and his sister ate 2 9 of a bag. What fraction of the bag did they eat? What fraction of the bag remains? 66) Suppose your school costs for this term were $2500 and financial aid covered 3 4 much did financial aid cover? of that amount. How 4 67) If, on average, about of the human body is water weight how much water weight is present in a 7 person weighing 182 pounds? 68) If, while training for a marathon, you ran 920 miles in 3 1 months, how many miles did you run each 2 month? (Assume you ran the same amount each month) 69) On your first math test, you earned 75 points. On your second math test, you earned 6 as many points 5 as your first test. How many points did you earn on your second math test? 70) You are serving cake at a party at your home. There are 12 people in total and 2 3 cakes. (You ate 4 some before they got there!). If the cakes are shared equally among the 12 guests, what fraction of a cake will each guest receive? 40

47 SECTION 1.4: DECIMALS A. INTRODUCTION TO DECIMALS Introduction to Decimals (Duration 8:03) Chapter 1.. Decimal notation is used to write numbers according to place value in base -10. A decimal point is used to separate whole numbers from numbers from numbers less than 1. It costs $2.89 per gallon of gas. The runner finished in 10.3 seconds. The Dow was up points today. Ten Thousands Thousands Hundreds Tens Ones And Tenths Hundredths Thousandths Ten Thousandths 100,000 1, Read the following number using place values. Write the decimal in using fraction notation , Comparing using >, <, or = Order each list from least to greatest , 3.1, -3,04, 3.11, , , ,

48 Rounding Decimal Numbers Chapter 1 Procedure 1. Identify the round-off place digit. 2. If the digit to the right of the round-off place digit is: Less than 5, do not change the round of place digit. 5 or more, increase the round-off place digit by In either case, drop all digits to the right of the round-off place digit. Round each number to the indicated place value to the hundredths to the tenths to the ten thousandths YOU TRY a) Order the numbers from least to greatest: 2.8, 2.08, 2.88, 2.088, 2.008, 2.808, b) Order from smallest to largest: 4.25, 0.425, 4.05, 4.2, c) Round 3.24 to the nearest tenth. d) Round to the nearest hundredth. e) Round to the nearest tenth. f) Round to the nearest thousandth. 42

49 B. OPERATIONS WITH DECIMALS Chapter 1 Adding and subtracting decimals Add and subtract decimals (Duration 5:35). Procedure 1. Write the numbers vertically and line up the decimal points. Add zeros to the right as needed so each number has the same number of digits to the right of the decimal points. 2. Bring the decimal point down into the sum or difference. 3. Add or subtract as you normally would. Example: YOU TRY Perform the operations indicated below. Be sure to show your work. a) b) c) d) e) f)

50 Multiplying decimals Multiply decimals (Duration 7:54) Chapter 1. Procedure 1. Multiply the numbers just as you would multiply whole numbers 2. Find the sum of the decimal places in the factors. 3. Place the decimal point in the product so that the product has the same number of decimal places as the sum of the decimal places. You may need to write zeros to the left of the number. Example: Multiply 5.2 x x x Multiplying a decimal by a power of ten To multiply a decimal by a power of 10, move the decimal point to the right the same number of places as the number of zeros in the power of 10. It may be necessary to add zeros at the end of the number. Example: Multiply x 100 Why? Multiply ,000 10,000 x YOU TRY Multiply the numbers by the given powers of 10 by moving the decimal point the appropriate number of places a) = b) = c) = d) = 44

51 Multiply the decimals. Chapter 1 e) = f) = g) = h) = Dividing decimals Dividing decimals (Duration 9:11). Dividing a decimal by a whole number: 1. Place the decimal point in the answer directly above the decimal point in the dividend. 2. Divide as if there were no decimal point involve. Example: Dividing a decimal by a decimal 1. If the divisor is a decimal, change it to a whole number by moving the decimal point to the right as many places as necessary. 2. Then move the decimal point in the dividend to the right the same number of places. 3. Place the decimal point in the answer directly above the decimal point in the dividend. 4. Divide until the remainder becomes zero or the remainder repeats itself, or the desired number of decimal places it achieved. 45

52 Chapter 1 Dividing a decimal by a powers of ten Dividing decimal by Powers of Ten (Duration 5:09). Divide the numbers by the given powers of 10 on your calculator then look for patterns to make a general strategy. a) = b) = c) = d) = e) = f) = g) Look for patterns in the examples above and complete the statement below. To divide a decimal number by a power of 10, you move the decimal place. YOU TRY Divide the decimals. a) = b) = c) d) Divide the numbers by the given powers of 10 by moving the decimal point the appropriate places. e) = f) = g) = 46

53 C. FRACTION AND DECIMAL CONNECTIONS Converting Fractions to Decimals Chapter 1 Method 1: To convert the denominator to a power of 10. Then write the number using the correct place value. Converting a Fraction to a Decimal (Power of 10) (Duration 5:50). Ten Thousands Thousands Hundreds Tens Ones And Tenths Hundredths Thousandths Ten Thousandths 100,000 1, ,000 Example: Convert the fraction to the decimal by converting the denominator to a Power of 10. a) b) c) 7 25 d) 6 5 e) f)

54 Method 2: Perform long division. Remember a fraction bar is a division symbol. Chapter 1 a b = a b b Converting a fraction to a decimal (long division) (Duration 6:05). Convert the fractions to decimals. a) 7 25 b) 2 5 c) 7 8 d) 3 11 Converting decimals to fractions Converting decimals to fractions (Duration 5:10). When we rewrite a decimal as a simplified fraction, we will start by writing it as a fraction based on its place value, a power of ten. Observe that 10 s prime factorization is 2 5. So any power of 10 is just a product of 2 s and 5 s. This will make the process of simplification easier because we will only have to check the numerator for factors of 2 s and 5 s. Complete the table below. Show all of your work for simplifying the fraction. Decimal Fraction Simplified Fraction a) 0.8 b) 0.65 c) 0.44 d)

55 YOU TRY Chapter 1 Convert the following fractions to decimals using the indicated method below. Show all of your work. Fraction Powers of 10 method Long division method 7 a) 25 b) 3 5 c) 3 8 d) Convert the following decimals to fractions and simplify your answer. Decimal Fraction Simplified Fraction e) 0.6 f) 0.85 g) h)

56 EXERCISE Perform the following operations with decimals. Show your work. Chapter 1 1) ) ) ) ) ) ) ) ) ) ) ) ) ) Multiply or divide the numbers by the given powers of 10 by moving the decimal point the appropriate number of places. 15) = 16) = 17) = 18) = 19) = 20) = 21) Sylvia just received her monthly water usage data from her local water department. For the past 6 months, her water used (in thousands of gallons) was 19.9, 25.6, 28.8, 22.5, 20.3, and What was her average usage during this time? (Round to the nearest tenth) 22) Glenn normally earns $8.50 per hour in a given 40-hour work-week. If he works overtime, he earns time and a half pay per hour. During the month of October, he worked 40 hours, 50 hours, 45 hours, and 42 hours for the four weeks. How much did he earn total for October? 23) Dave is making a gazebo for his yard. He has a piece of wood that is 13 feet long and he needs to cut it into pieces of length 1.5 inches. How many pieces of this size can he cut from the 13 foot piece of wood? 24) Callie ordered 4 items online. She is charged $2.37 per pound per shipping. The items weighed 3.2 lbs., 4.6 lbs., 9.2 lbs. and 1.5 lbs. How much will be charged for shipping? (Round to the nearest cent). 25) Mary s son wants to go on the Gadget s Go Coaster ride at Disneyland. The height requirement for the ride is 35 inches. He is 29.3 inches tall now. How many inches more does he need in order to get into the ride? 50

57 Complete the table below. Show all of your work for simplifying the fraction. Chapter 1 Decimal Fraction Simplified Fraction 26) ) ) ) Complete the table below. Show all of your work for simplifying the fraction. Fraction Powers of 10 method if possible Long division method 30) ) ) )

58 SECTION 1.5: INTEGERS Chapter 1 INTRODUCTION Now that we have discussed the Base-10 number system including whole numbers and place value, we can extend our knowledge of numbers to include integers. The first known reference to the idea of integers occurred in Chinese texts in approximately 200 BC. There is also evidence that the same Indian mathematicians who developed the Hindu-Arabic Numeral System also began to investigate the concept of integers in the 7th century. However, integers did not appear in European writings until the 15th century. After conflicting debate and opinions on the concept of integers, they were accepted as part of our number system and fully integrated into the field of mathematics by the 19th century. A. INTEGERS AND THEIR APPLICATIONS Integers and their Applications (Duration 3:24). Definition: The integers are all positive whole numbers and their opposites and zero.... 4, 3, 2, 1, 0, 1, 2, 3, 4 The numbers to the left of 0 are negative numbers and the numbers to the right of 0 are positive numbers. We denote a negative number by placing a symbol in front of it. For positive numbers, we either leave out a sign altogether or place a + symbol in front of it. Determine the signed number that best describes the statements below. Circle the word that indicates the sign of the number. a) Tom gambled in Vegas and lost $52 Statement Signed Number b) Larry added 25 songs to his playlist. c) The airplane descended 500 feet to avoid turbulence. YOU TRY Determine the signed number that best describes the statements below. Circle the word that indicates the sign of the number. Statement Signed Number a) A balloon dropped 59 feet. 52

59 b) The altitude of a plane is 7500 feet Chapter 1 c) A submarine is 10,000 feet below sea level B. PLOTTING INTEGERS ON A NUMBER LINE Number lines are very useful tools for visualizing and comparing integers. We separate or partition a number line with tick marks into segments of equal length so the distance between any two consecutive major tick marks on a number line are equal. Plotting Integers on a Number Line (Duration 6:05). Plot the negative numbers that correspond to the given situations. Use a to mark the correct quantity. Also label all the surrounding tick marks and scale the tick marks appropriately. a) The temperature in Greenland yesterday was 5 What does 0 represent in this context? b) The altitude of the plane decreased by 60 feet. What does 0 represent in this context? YOU TRY Plot the negative numbers that correspond to the given situations. Use a to mark the correct quantity. Also label all the surrounding tick marks and scale the tick marks appropriately. Akara snorkeled 30 feet below the surface of the water. What does 0 represent in this context? What does 0 represent in this context? 53

60 C. ABSOLUTE VALUE AND NUMBER LINES Chapter 1 Absolute Value and Number Lines (Duration 4:50). Definition: The absolute value of a number is the positive distance of the number from zero. Notation: Absolute value is written by placing a straight vertical bar on both sides of the number. 50 = 50 RRRRRRRR tthee aaaaaaaaaaaaaaaa vvvvvvvvvv oooo 50 eeeeeeeeeeee = 50 RRRRRRRR tthee aaaaaaaaaaaaaaaa vvvvvvvvvv oooo 50 eeeeeeeeeeee 50 Answer the questions below based on the given example. The submarine dove 15 meters below the surface of the water. a) What integer best represents the submarine s location relative to the surface of the water? b) What word indicates the sign of this number? c) What does 0 represent in this context? d) Plot your number from pppppppp aa and 0 on the number line below. e) Draw a line segment that represents this value s distance from zero on the number line below. f) Write the symbolic form of the absolute value representation. When we want to talk about how large a number is without regard as to whether it is positive or negative, we use the absolute value function to represent the distance from that number to the origin (zero) on the number line. That distance is always given as a non-negative number. In short: If a number is positive (or zero), the absolute value function does nothing to it: 4 = 4 If a number is negative, the absolute value function makes it positive: - 4 = 4 54

61 YOU TRY Chapter 1 Answer the questions below based on the given example. The temperature dropped 8 Celsius overnight. a) What integer best represents the change in temperature? b) What word indicates the sign of this number? c) What does 0 represent in this context? d) Plot your number from pppppppp aa and 0 on the number line below. e) Draw a line segment that represents this value s distance from zero on the number line below. f) Write the symbolic form of the absolute value representation. Find the absolute value of: 1) 5 = 2) 5 = 3) 120 = D. OPPOSITES AND NUMBER LINES Opposites and number lines (Duration 4:48). Definition: The opposite of a nonzero number is the number that has the same absolute value of the number, but does not equal the number. Another useful way of thinking of opposites is to place a negative sign in front of the number. The opposite of 4 is (4) = 4 The opposite of 4 is ( 4) = 4 Answer the questions below to use number lines to find the opposite of a number. a) Plot the number 5 on the number line below. b) Draw an arrow that shows the reflection of 5 about the reflection line to find 5 ss opposite c) The opposite of 5, or (5) is d) Draw an arrow that shows the reflection of 5 about the reflection line to find 5 ss opposite e) The opposite of 5, or ( 5) is f) Based on the pattern above, what do you think ( ( 5)) equals? 55

62 YOU TRY Chapter 1 Answer the questions below to use number lines to find the opposite of a number. a) Plot the number 4 on the number line below. b) Draw an arrow that shows the reflection of 4 about the reflection line to find 4 ss opposite c) The opposite of 4 is E. ORDERING INTEGERS USING NUMBER LINES Ordering Integers Using Number Lines (Duration 5:12). Fact: If two numbers are not equal, one must be less than the other. One number is less than another if it falls to the left of the other on the number line. Equivalently, if two numbers are not equal, one must be greater than the other. One number is greater than another if it falls to the right of the other on the number line. Notation: We use inequality notation to express this relationship. 2 < 5, read 2 is less than 5 6 > 3, read 6 is greater than 3 Although we typically read the < sign as less than and the > sign as greater than because of the equivalency noted above, we can also read them as follows: 2 < 5, is equivalent to 5 is greater than 2 6 > 3, is equivalent to 3 is less than 6 Plot the given numbers on the number line. Determine which number is greater and insert the correct inequality symbol in the space provided. a) Plot 5 and 3 on the number line below. Write the number that is further to the right: Insert the correct inequality symbol in the space provided: b) Plot 4 and 7 on the number line below. Write the number that is further to the right: Insert the correct inequality symbol in the space provided:

63 YOU TRY Chapter 1 Plot the given numbers on the number line. Determine which number is greater and insert the correct inequality symbol in the space provided. Plot 8 and 2 on the number line below. Write the number that is further to the right: Insert the correct inequality symbol in the space provided: F. REPRESENTING INTEGERS USING THE CHIP MODEL Representing integers using manipulatives (Duration 3:11). Observe the two images below. Although they both have a total of 5 chips, the chips on the left are marked with " + " signs and the chips on the right are marked with " " signs. This is how we indicate the sign each chip represents Determine the value indicated by the sets of integer chips below. a) b) Number: c) Number: d) Number: Number: YOU TRY Determine the value indicated by the sets of integer chips below. a) b) Number: Number: 57

64 G. THE LANGUAGE AND NOTATION OF INTEGERS Chapter 1 The Language and Notation of Integers (Duration 6:28). The + symbol: 1. In the past, you have probably used the symbol + to represent addition. Now it can also represent a positive number such as + 4 read positive Let s agree to say the word plus when we mean addition and positive when we refer to a number s sign. The symbol: 1. In the past, you have probably used the symbol to represent subtraction. Now it can also mean a negative number such as 4 read negative 4 or the opposite of 4 2. Let s agree to say the word minus when we mean subtraction and negative when we refer a number s sign. Write the given numbers or mathematical expressions using correct language using the words opposite of, negative, positive, plus or minus. Number or Expression Written in Words a) 6 b) ( 6) c) d) 3 ( 4) YOU TRY Write the given numbers or mathematical expressions using correct language using the words opposite of, negative, positive, plus or minus. Number or Expression Written in Words a) 3 b) ( 7) c) 4 + ( 2) d) 1 ( 5) 58

65 H. ADDING INTEGERS Chapter 1 Adding integers using manipulatives (Duration 8:24). We call the numbers we are adding in an addition problem the addends. We call the simplified result the sum. a) Using integer chips, represent positive 5 and positive 3. Find their sum by combining them into one group. Addend Addend Sum = b) Using integer chips, represent negative 5 and negative 3. Find their sum by combining them into one group. Addend Addend Sum ( 5) + ( 3) = c) Using integer chips, represent positive 5 and negative 3. Find their sum by combining them into one group. Addend Addend Sum 5 + ( 3) = d) Using integer chips, represent negative 5 and positive 3. Find their sum by combining them into one group. Addend Addend Sum ( 5) + 3 = 59

66 Results: Addends have the same sign = 8 ( 5) + ( 3) = 8 Addends have different signs 5 + ( 3) = 2 ( 5) + 3 = 2 Chapter 1 Summary of the Addition of Integers When adding two numbers with the same sign, 1. Add the absolute values of the numbers 2. Keep the common sign of the numbers When adding two numbers with different signs, 1. Find the absolute value of the numbers 2. Subtract the smaller absolute value from the larger absolute value 3. Keep the original sign of the number with the larger absolute value. YOU TRY a) Using integer chips, represent negative 6 and negative 4. Find their sum by combining them into one group. Addend Addend Sum ( 6) + ( 4) = b) Using integer chips, represent negative 6 and positive 4. Find their sum by combining them into one group. Addend Addend Sum ( 6) + 4 = Adding Integers Using a Number Line (Duration 3:05). Use a number line to represent and find the following sums. a) = 60

67 b) ( 5) + ( 3) = Chapter 1 c) 5 + ( 3) = d)( 5) + 3 = YOU TRY Use a number line to represent and find the following sums. a) 7 + ( 3) = b) ( 7) + 3 = I. SUBTRACING INTEGERS The Language of Subtraction (Duration 4:38). Minuend Subtrahend Difference 9 6 = 3 Symbolic Minus Language Subtracted from Language 5 3 Less than Language Decreased by Language 5 ( 3) 61

68 YOU TRY Symbolic Minus Language Subtracted from Language Less than Language Chapter 1 Decreased by Language 6 ( 5) Subtracting integers with manipulatives Take away method (Duration 4:02). Using integer chips and the take away method, represent the following numbers and their difference. a) 5 3 Minuend Subtrahend Take Away Simplified Difference b) ( 5) ( 3) 5 3 = Minuend Subtrahend Take Away Simplified Difference ( 5) ( 3) = Subtracting integers with manipulatives Comparison method (Duration 4:19). Using integer chips and the comparison method, represent the following numbers and their difference. a) 3 5 Minuend Subtrahend Comparison Simplified Difference 3 5 = 62

69 b) 5 ( 3) Minuend Subtrahend Comparison Simplified Difference Chapter 1 5 ( 3) = c) ( 5) 3 Minuend Subtrahend Comparison Simplified Difference ( 5) 3 = YOU TRY Using integer chips and the method indicated, represent the following numbers and their difference. a) ( 6) ( 2) Minuend Subtrahend Take Away Simplified Difference ( 6) ( 2) = b) 3 4 Minuend Subtrahend Comparison Simplified Difference 3 4 = J. CONNECTING ADDITION AND SUBTRACTION You may have noticed that we did not write a set of rules for integer subtraction like we did with integer addition. The reason is that the set of rules for subtraction is more complicated than the set of rules for addition and, in general, wouldn t simplify our understanding. However, there is a nice connection between integer addition and subtraction that you may have noticed. We will use this connection to rewrite integer subtraction as integer addition. Fact: Subtracting an integer from a number is the same as adding the integer s opposite to the number. Rewriting Subtraction as Addition (Duration 4:38). Rewrite the subtraction problems as equivalent addition problems and use a number line to compute the result. a) 4 7 Rewrite as addition: 63

70 b) 6 ( 2) Rewrite as addition: Chapter 1 YOU TRY Rewrite the subtraction problems as equivalent addition problems and use a number line to compute the result. a) 4 6 Rewrite as addition: b) 3 ( 4) Rewrite as addition: K. USING ALGORITHMS TO ADD AND SUBTRACT INTEGERS Using Algorithms to Add and Subtract Integers (Duration 6:30). Thus far, we have only added and subtracted single digit integers. Now we will use base blocks and the ideas developed in this lesson to add and subtract larger numbers. We will follow the protocol below. 1. If given a subtraction problem, rewrite it as an addition problem. 2. Use the rules for addition to add the signed numbers as summarized below. 3. Use regrouping or decomposing from Lesson 1 to carry in addition when necessary or borrow in subtraction when necessary. 4. Write the associated standard algorithm that represents this process. Summary of the Addition of Integers When adding two numbers with the same sign, 1. Add the absolute values of the numbers 2. Keep the common sign of the numbers When adding two numbers with different signs, 1. Find the absolute value of the numbers 2. Subtract the smaller absolute value from the larger absolute value 3. Keep the original sign of the number with the larger absolute value. 64

71 Use the Standard Algorithms to solve the addition and subtraction problems below. a) b) Chapter 1 c) Use your results from above and your knowledge of integer addition and subtraction to find the following. ( 275) + ( 308) = 275 ( 308) = ( 275) = ( 275) ( 308) = YOU TRY ( 308) = ( 275) 308 = Use the Standard Algorithms to solve the addition and subtraction problems below. a) b) c) Use your results from above and your knowledge of integer addition and subtraction to find the following. ( 137) + ( 324) = 137 ( 324) = ( 324) = ( 137) 324 = L. MULTIPLYING AND DIVIDE INTEGERS To multiply or divide two signed numbers 1. Multiply or divide the absolute values. 2. If both signs are the same, the sign of the result is always positive. 3. If the signs are different, the sign of the result is negative. Remember the following rules: (+)(+)=(+) (+)(-)=(-) (-)(+)=(-) (-)(-)=(+) Using algorithms to multiply integers (Duration 2:05). a) Use the Standard Algorithm to find b) Use your results from above and your knowledge of integer multiplication to find the following = = 14( 23) = ( 14)( 23) = 65

72 YOU TRY Chapter 1 a) Use the Standard Algorithm to find b) Use your results from above and your knowledge of integer multiplication to find the following = 25( 32) = = ( 25)( 32) = Using algorithms to divide integers (Duration 3:02). a) Find using the Standard Algorithm. b) Use your results from above and your knowledge of integer division to find the following. c) ( 4) = = ( 564) ( 4) = 4 YOU TRY a) Find using the Standard Algorithm. b) Use your results from above and your knowledge of integer division to find the following ( 3) = = 3 = 66

73 EXERCISE 1) Determine the signed number that best describes the statements below. Chapter 1 Statement Signed Number a) The boiling point of water is 212 o F b) Carlos snorkeled 40 feet below the surface of the water c) Jack lost 32 pounds. d) Jill gained 5 pounds. e) The company suffered a net loss of twelve million dollars. f) The elevation of Death Valley is about 280 feet below sea level g) The elevation of Longs Peak is about 14,000 feet above sea level 2) Plot the numbers 4 and 1 on the number line below. 3) Plot the numbers 4 and 1 on the number line below. 4) Plot the numbers 20, 5, and 30 on the number line below. 5) Label the following number line so that it includes 0 and the integers from 3 to 7. 6) Label the following number line so that it includes 0 and the integers from 8,000 to 12,000. 7) Plot the numbers that correspond to the given situations. Use a to mark the correct quantity. Also label all the surrounding tick marks. Make sure to include 0 on your number line and scale the tick marks appropriately. a. Shelby lost 8 pounds 67

74 b. Juan snorkeled 25 feet below the surface of the water Chapter 1 8) Consider the number line shown below. Elevation (in meters) relative to sea level a. What does 3 represent in this situation? b. What does 2 represent in this situation? c. What does 0 represent in this situation? 9) Use the number line to plot the given number and use the reflection line to find the opposite. a. Plot the number 2. Make sure to scale the tick marks on your number line appropriately. The opposite of 2 is b. Plot the number 30. Make sure to scale the tick marks on your number line appropriately. The opposite of 30 is 10) Plot the number 8. Make sure to scale the tick marks on your number line appropriately. The opposite of 8 is 11) Insert the correct <, >, = symbol in the space provided. a. 3 9 f = k. 5 5 b. 5 1 c. 0 8 d e. 8 2 g h i. 8 5 j. 4 0 l m n. 1,213 1,123 o. 4,651 4,650 68

75 12) Write TRUE or FALSE in the space provided. Chapter 1 a) If two numbers are positive, the one that is closest to zero is greater. b) If two numbers are negative, the one that is closest to zero is greater. c) If one number is positive and one number is negative, the positive number is greater. 13) Camden, SC had a record low temperature of -19 F on Jan 21, 1985, and Monahans, TX had a record low temperature of -23 F on Feb 8, (Data Source Wikipedia: a. Plot these numbers on the number line below, and label all the surrounding tick marks. Make sure to include 0 on your number line and scale the tick marks appropriately. b. Write an inequality statement that compares the two numbers. c. Which of the two temperatures was colder? 14) Liquid hydrogen evaporates at about 400. Liquid nitrogen evaporates at about 300. a. Plot these numbers on the number line below, and label all the surrounding tick marks. Make sure to include 0 on your number line and scale the tick marks appropriately. b. Write an inequality statement that compares the two numbers. c. Which liquid has the lower evaporating temperature? 15) Determine the value indicated by the sets of integer chips below. Chip Representation Number a) b) c) d) 69

76 16) Write the following numbers from least to greatest. Chapter 1 Ordering from least to greatest: 17) Write + or in the blank next to each of the following words. negative opposite plus positive minus 18) Write the given numbers or mathematical expressions using correct language using the words opposite of, negative, positive, plus, or minus. Number or Expression a. 5 b. ( 5) c. +5 d. 5 3 e. (+2) f g h. 4 + ( 9) i. (5 1) Written in Words 19) Complete the table. Write the symbolic operations in words using the indicated language below. Symbolic Minus Language Subtracted from Language Less than Language Decreased by Language ( 4) 70

77 Chapter 1 20) Using integer chips, represent the expressions and their combined amount. Use the table to show how you did this using + for positive chips and for negative chips and find the sum. Addend Addend Sum a b. 4 + ( 2) c (-3) d e. 6 + ( 4) f g ) Using integer chips, represent the following numbers and their difference. Use the table to show how you did this using + for positive chips and for negative chips. Minuend Subtrahend Circle Subtrahend Taken Away from Minuend Simplified Difference a. 5 3 b. 5 ( 3) c. 2 6 d. 6 2 e. 5 ( 4) 22) Perform the indicated operations. a b. 3 + ( 1) c d. 5 ( 5) e. 5 ( 5) f. 5 5 g. 5 5 h. 2 ( 5) i. 2 ( 5) j. 5 + ( 5) k. 3 + ( 3) l m. 2 5 n. 2 5 o. 5 2 p. 8 + ( 9) q r. 41 ( 41) s t. 5 ( 2) u. 5 ( 2) v. 35 ( 22) w x ( 200) 71

78 23) Perform the indicated operations. a. 6 2 b. 4( 2) c. 3 1 Chapter 1 d e. 2( 5) f. ( 24) ( 4) g h. ( 9) ( 9) i. 32 ( 4) j. 5( 6) k. 8 ( 4) l. ( 12) ( 2) m. 5 x 3 n. 6 x 3 o. 8 ( 4) p. 10 ( 5) q r. 7 4 Represent the application problem in symbolic form and evaluate. Then write your answer as a complete sentence. (Note: Make sure to use an addition statement even though a subtraction statement may apply as well). 24) Sara hiked down a mountain for 3 hours. Each hour, her elevation decreased by 30 meters. Compute her change in elevation in meters relative to her starting point. 25) Tom gained 10 pounds and then lost 12 pounds. What is his total change in weight relative to his original weight? 26) Joanne lost 3 pounds per month for 6 months. Find Joanne s total change in weight relative to her original weight. 27) Carlos lowers the temperature of his freezer by 7 degrees. It was originally set to 4 degrees Celsius. What is the new temperature of the freezer in degrees Celsius? 28) Leslie bought coffee 8 days this month and charged it to her checking account. She spent 6 dollars each time she visited the store. Determine the change in dollars in her checking account. 29) Malala's pool was filled 9 inches below the top of the pool. She drained the pool 5 inches. What is the water level relative to the top of the pool? 30) A total of 10 friends have a debt of 50 dollars. If they share the debt equally, what number represents the change in dollars for each friend? 31) Allie had 5 dollars in her debit account. She returned an internet purchase and they removed a charge of 10 dollars from her debit account. 32) The temperature in Minneapolis changed by 32 degrees in 8 days. If the temperature changed by the same amount each day, what was the change in temperature per day? 33) Kayla camped at 9 miles relative to sea level. She then hiked 4 miles upwards. What is her current altitude relative to sea level? 34) Tally bought 50 packages of printer paper for her business. Each package contained 300 sheets of paper. How many sheets of paper is this in total? 72

79 35) Sheldon has 140 dollars in his checking account and Penny has 150 dollars in her checking account. How much did they have all together? Chapter 1 36) Ken had 15 dollars in his checking account and wrote a check for 21 dollars. What is the balance in his checking account in dollars? 73

80 SECTION 1.6: ORDER OF OPERATIONS A. INTRODUCTION TO EXPONENTS Introduction to Exponents (Duration 5:54) Chapter 1 Solve the problem below. Use the rectangle below to represent the problem visually. a) Don makes a rectangular 20 square foot cake for the state fair. After he wins his award, he wants to share it with the crowd. First he cuts the cake into 2 pieces. Then he cuts the 2 pieces into 2 pieces each. Then he cuts all of these pieces into two pieces. He continues to do this a total of 5 times. How many pieces of cake does he have to share? b) Write a mathematical expression that represents the total number of pieces in which Don cut the cake. Terminology We will use exponential expressions to represent problems such as the last one. Exponents represent repeated multiplication just like multiplication represents repeated addition as shown below. MMMMMMMMMMMMMMMMMMMMMMMMMMMM: 5 2 = = 10 EEEEEEEEEEEEEEEEEE: 2 5 = = 32 Language and Notation of Exponents (Duration 7:32) In the exponential expression, is called the base 5 is called the exponent We will say 2 5, as 2 raised to the fifth power or 2 to the fifth Since exponents represent repeated multiplication, and we call the numbers we multiply factors, we will also use this more meaningful language when discussing exponents. 2 5 mmmmmmmmmm 5 ffffffffffffff oooo 2 or We also have special names for bases raised to the second or third power. a) For 3 2, we say 3 squared or 3 to the second power b) For 4 3, we say 4 cubed or 4 to the third power 74

81 Represent the given exponential expressions in the four ways indicated. a) 6 2 b) 6 2 Chapter 1 Expanded Form Expanded Form Word Name Word Name Factor Language Factor Language Math Equation Math Equation c) ( 6) 2 d) ( 5) 3 Expanded Form Expanded Form Word Name Word Name Factor Language Factor Language Math Equation Math Equation YOU TRY Represent the given exponential expressions in the four ways indicated. a) 7 2 b) ( 7) 2 Expanded Form Expanded Form Word Name Word Name Factor Language Factor Language Math Equation Math Equation 75

82 B. THE ORDER OF OPERATIONS WITH ADDITION AND SUBTRACTION The Order of Operations with Addition and Subtraction (Duration 4:29) Chapter 1 Solve the problem below. Be sure to indicate every step in the process of your solution. a) Suppose on the first day of the month you start with $150 in your bank account. You make a debit transaction on the second day for $60 and then make a deposit on the third day for $20. What is the balance in your account on the third day? b) What string of operations (written horizontally) can be used to determine the amount in your account? Rule 1: When we need to add or subtract 2 or more times in one problem, we will perform the operations from left to right The Order of Operations with Addition and Subtraction (cont.) (Duration 5:35) Use a highlighter to highlight the operations in the problem. Determine the number of operations to be performed in the problem. Then compute the results by using the convention of performing the operations from left to right. a) # of operations b) 12 ( 5) ( 1) # of operations YOU TRY Use a highlighter to highlight the operations in the problem. Determine the number of operations to be performed in the problem. Then compute the results by using the convention of performing the operations from left to right. a) ( 2) # of operations b) 8 + ( 5) 6 ( 2) + 9 # of operations 76

83 C. THE ORDER OF OPERATIONS WITH MULTIPLICATION AND DIVISION Chapter 1 The Order of Operations with Multiplication and Division (Duration 4:22) Solve the problem below. Be sure to indicate every step in the process of your solution. a) Suppose you and your three siblings inherit $40,000. You divide it amongst yourselves equally. You then invest your portion and make 5 times the amount of your portion. How much money do you have? Be sure to indicate every step in your process. b) What string of operations (written horizontally) can be used to determine the result? Rule 2: When we need to multiply or divide 2 or more times in one problem, we will perform the operations from left to right. The Order of Operations with Multiplication and Division (Cont.) (Duration 3:27) Use a highlighter to highlight the operations in the problem. Determine the number of operations to be performed in the problem. Then compute the results by using the convention of performing the operations from left to right. a) 6 4 ( 2) 2 # of operations b) ( 3) # of operations YOU TRY Use a highlighter to highlight the operations in the problem. Determine the number of operations to be performed in the problem. Then compute the results by using the convention of performing the operations from left to right. a) 8( 2) ( 4) ( 2) # of operations b) ( 1)(2) # of operations 77

84 D. THE ORDER OF OPERATIONS FOR +,,, Order of Operations for +,,, (Duration 5:21) Chapter 1 Solve the two problems below. Be sure to indicate every step in your process a) Bill went to the store and bought 3 six-packs of soda and an additional 2 cans. How many cans did he buy in total? What string of operations (written horizontally) can be used to represent this problem? b) Amber went to the store and bought 3 six-packs of cola and an additional 2 six-packs of diet cola. How many cans did she buy in total? What string of operations (written horizontally) can be used to represent this problem? Rule 3: Unless otherwise indicated by parentheses, we perform multiplication and division before addition and subtraction. We continue to perform the operations from left to right. Order of Operations for +,,, (cont. ) (Duration 4:04) Use a highlighter to highlight the operations in the problem. Determine the number of operations to be performed in the problem. Perform the operations in the appropriate order.. a) ( 3) # of operations b) ( 3) # operations YOU TRY Use a highlighter to highlight the operations in the problem. Determine the number of operations to be performed in the problem. Perform the operations in the appropriate order. Show all intermediary steps. a) ( 3) # of operations b) ( 3 )( 4) # operations 78

85 E. THE ORDER OF OPERATIONS WITH PARENTHESES Chapter 1 There are cases when we want to perform addition and subtraction before multiplication and division in the order of operations. So we need a method of indicating we want to make such a modification. In the next media problem, we will discuss how to show this change. Grouping symbols ( ), { }, [ ] If there are several parentheses in a problem, we will start with the inner most parenthesis and work our way out as we apply order of operations to the expression. To avoid confusion with multiple parentheses, we use different types of grouping symbols such as { } and [ ] and ( ). These grouping symbols all mean the same thing and imply the expression inside must be evaluated first. Rule 4: If we want to change the order in which we perform operations in an arithmetic expression, we can use parentheses to indicate that we will perform the operation(s) inside most parentheses first. Order of operations with grouping symbols ( ), { }, [ ] (4:42) Example: Different types of parentheses: Always do first! (4 + 2) [ 5 2 (2 + 3) ] YOU TRY 7 { [20 (4 + 6) ] } Simplify the expression completely: 2{8 2 7[32 4( ) ] ( 1) } 79

86 Chapter 1 Grouping symbols Fraction bar There are several types of grouping symbols that can be used besides parentheses. One type is a fraction bar. If we have a fraction, the entire numerator and the entire denominator must be evaluated prior to reducing the fraction. In these cases, we can simplify the numerator and denominator simultaneously. Order of operations with fraction bar (Duration 5:00) When simplifying fractions, always simplify and first, then. Example: Top: 4 2 ( ) 5 + 3(5 4) Bottom: Top: (4 + 5)(2 9) 2 3 ( ) Bottom: YOU TRY Simplify the expression completely: 24 ( 8)

87 Chapter 1 Grouping symbols absolute value Another type of grouping symbol that also has an operation is an absolute value. When there is absolute value, we evaluate the expression inside the absolute value first, just as if it were a normal parenthesis. Then take the absolute value. Order of operations with absolute value (Duration 4:52) Absolute Value Just like, Make positive Example: 2 4 (5 + 4) ( ) YOU TRY Simplify the expression completely: ( 8) ( 5) 2 81

88 Chapter 1 F. PEMDAS AND THE ORDER OF OPERATIONS Finally, we will consider problems that may contain any combination of parentheses, exponents, multiplication, division, addition and subtraction. PEMDAS and the Order of Operations Duration ( Rule 5: Exponents are performed before the operations of addition, subtraction, multiplication and division. P E M D A S Simplify items inside Parentheses ( ), brackets [ ] or other grouping symbols first. Simplify items that are raised to powers (Exponents) Perform Multiplication and Division next (as they appear from Left to Right) Perform Addition and Subtraction on what is left. (as they appear from Left to Right) Use a highlighter to highlight the operations in the problem. Determine the number of operations to be performed in the problem. Then compute the results by using the correct order of operations. Check your results on your calculator. a) (8 3) 2 4 b) c) ( 3) 2 4( 3) + 2 YOU TRY Parentheses Exponents Multiplication or Division (whichever comes 1st) Addition or Subtraction (whichever comes 1st) Use a highlighter to highlight the operations in the problem. Determine the number of operations to be performed in the problem. Then compute the results by using the correct order of operations. Check your results on your calculator. a) 7 (2 3) 2 # of operations b) ( 4) 2 + 5( 4) 6 # operations 82

89 EXERCISE Represent the given exponential expressions in the four ways indicated. 1) 2 5 Expanded Form Word Name Factor Language Math Equation Chapter 1 2) 5 2 3) 5 2 4) ( 5) 2 5) 3 2 6) 2 3 7) 2 3 Expanded Form Word Name Factor Language Math Equation Expanded Form Word Name Factor Language Math Equation Expanded Form Word Name Factor Language Math Equation Expanded Form Word Name Factor Language Math Equation Expanded Form Word Name Factor Language Math Equation Expanded Form Word Name Factor Language Math Equation 8) ( 2) 3 Expanded Form Word Name Factor Language Math Equation 83

90 9) ( 5) 4 Expanded Form Word Name Factor Language Math Equation Chapter 1 10) ) ( 3) 5 12) 5 4 Expanded Form Word Name Factor Language Math Equation Expanded Form Word Name Factor Language Math Equation Expanded Form Word Name Factor Language Math Equation Simplify the expressions completely. 13) 6 4( 4) 14) 3 + (8) 4 15) ) [ 9 (2 5)] ( 6) 17) 6 + ( 3 3)2 3 18) ) [ 1 ( 5)] ) ) [ ( 6)] ( 5 + 3) 22) ( 1) 3 + ( 6) [ 1 ( 3)] 23) ( 6 6)3 24) 5( 5 + 6) ) 3 {3 [ 3(2 + 4) ( 2)]} 26) ( ) ( 4) 27) ( 7 5) [ 2 2 ( 6)] 28) 10 6 ( 2)

91 CHAPTER REVIEW KEY TERMS AND CONCEPTS Chapter 1 Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson. Natural Numbers Whole Numbers Integers Rational Numbers Irrational numbers Real Numbers Absolute Value Prime numbers Composite numbers The Fundamental Theorem of Arithmetic Divisible Factor Product Greatest common factor (GCF) Common multiples 85

92 Chapter 1 Least common multiple (LCM) Prime numbers Composite numbers Absolute Value Opposite Numerator Denominator Equivalent fraction Simplest form Dividend Divisor Quotient Prime factorization Proper fraction Improper fraction Mixed number Least (lowest) common denominator Reciprocal 86

93 Chapter 2 CHAPTER 2: INTRODUCTION TO VARIABLES AND PROPERTIES OF ALGEBRA Chapter Objectives By the end of this chapter, students should be able to: Introduction to variables Different meanings of variables Evaluate algebraic expressions Writing algebraic expressions Properties of Algebra Commutative Associative Identity Inverse Distributive Contents CHAPTER 2: INTRODUCTION TO VARIABLES AND PROPERTIES OF ALGEBRA SECTION 2.1 INTRODUCTION TO VARIABLES A. WHAT IS A VARIABLE? B. MEANING OF A VARIABLE IN MATHEMATICS C. VARIABLE EXPRESSIONS D. WRITING ALGEBRAIC EXPRESSIONS E. EVALUATE ALGEBRAIC EXPRESSIONS F. LIKE TERMS AND COMBINE LIKE TERMS EXERCISE SECTION 2.2 PROPERTIES OF ALGEBRA EXERCISE CHAPTER REVIEW

94 SECTION 2.1 INTRODUCTION TO VARIABLES Chapter 2 A. WHAT IS A VARIABLE? When someone is having trouble with algebra, they may say, I don t speak math! While this may seem weird to you, it is a true statement. Math, like English, French, Spanish, or Arabic, is a like a language that you must learn in order to be successful. In order to understand math, you must practice the language. Action words, like run, jump, or drive, are called verbs. In mathematics, operations are like verbs because they involve doing something. Some operations are familiar, such as addition, multiplication, subtraction, or division. Operations can also be much more complex like a raising to an exponent or taking a square root. Variable is one of the most important concepts of mathematics, without variables we would not get far in this study. Variable is a symbol, usually an English letter, written to replace an unknown or changing quantity. Definitions A variable, usually represented by a letter or symbol, can be defined as: A quantity that may change within the context of a mathematical problem. A placeholder for a specific value. An algebraic expression is a mathematical statement that can contain numbers, variables, and operations (addition, subtraction, multiplication, division, etc ). Introduction to variables and variable expressions (Duration 7:54) Definition: A is a that represents an. a) How many hours will you study tomorrow? b) How much will you pay to have your car repaired? A consists of,, and like,, and. Often and are used. The difference is a contain a variable while the. 88

95 Mathematical Expressions Writing variable expressions a) It costs $9 per adult and $5 per child to go to the movies. Equations Chapter 2 Variable expression for the cost of a group go to the movies: b) It costs $30 per day to rent a car plus $0.10 per mile. Variable expression for the total rental cost: Evaluating variable expressions c) Evaluate 5xx + 7 when xx = 6 d) Evaluate 4mm 3nn when mm = 9 and nn = 7 e) Evaluate pp 2 3qq + 7 when pp = 11, qq = 8 f) Evaluate 36 dd and ff = 1 +7cc 9ff when dd = 9, ee = 3, Why all the letters in algebra? (Duration 3:03) What question students ask a lot when they start Algebra? What are other things besides letters that can be used as a variable? 89

96 B. MEANING OF A VARIABLE IN MATHEMATICS Meaning of a variable (Duration 4:13) Chapter 2 Definitions of a variable 1) Define variable as a changing value A variable is a letter that represents or that may within the context of a mathematical problem. Example: 2) Define variable as a placeholder A variable is a letter that represents or that will remain based on the confines of the mathematical problem. Example: Example: Determine if the description describes a changing value (CV) or a placeholder (P) then determine a possible variable. Changing Value (CV) or Scenario Variable Placeholder (P) The elevation of Mount Humphrey s The water level of a pool as it is being filled The number of calories consumed throughout the day The monthly car payment with a fixed interest loan The amount of gas consumed by your car as you drive The cost of a new textbook for a specific class from the bookstore at the beginning of the semester YOU TRY Scenario The number of cars in the parking lot at this moment The number of cars pass by an intersection throughout the day The altitude of an airplane during a trip The temperature in the Dead Valley on at midnight on January 1 st in 1905 The balance of your checking account today Changing Value (CV) or Placeholder (P) Variable 90

97 Why aren't we using the multiplication sign? (Duration 3:08) Chapter 2 Why the multiplication sign x is not being used much when we get to algebra? Why is 'x' the symbol for an unknown? (Duration 3:57) Why is it that xx is the unknown? C. VARIABLE EXPRESSIONS In mathematics, especially in algebra, we look for patterns in the numbers that we see. Using mathematical verbs and variables studied in previous lessons, expressions can be written to describe a pattern. Definition An algebraic expression is a mathematical phrase combining numbers and/or variables using mathematical operations. An algebraic expression consists of coefficients, variables, and terms. Given an algebraic expression, a coefficient is the number in front of the variable. variable is a letter representing any number. term is a product of a coefficient and variable(s). constant: a number with no variable attached. A term whose value never changes. For example, tt, 22xx, 33ssss, 77xx 22, 55aabb 33 cc are all examples of terms because each is a product of a coefficient and variable(s). 5555yy 22 tttttttt 2222 tttttttt 33 cccccccccccccccc tttttttt Algebraic expression vocabulary (Duration 5:52) Terms:. Constant Term:. Example 1: Consider the algebraic expression 4xx 5 + 3xx 4 22xx 2 xx + 17 a) List the terms: b) Identify the constant term: Factors:. Coefficient:. 91

98 Chapter 2 Example 2: Complete the table below. List the Factors Identify the Coefficient 4mm xx 1 2 bbh 2rr 5 Example 3: Consider the algebraic expression 5yy 4 8yy 3 + yy 2 yy 7 4 a) How many terms are there? b) Identify the constant term. c) What is the coefficient of the first term? d) What is the coefficient of the second term? e) What is the coefficient of the third term? f) List the factors of the fourth term. YOU TRY Consider the algebraic expression 2mm 3 + mm 2 2mm 8 a) How many terms are there? b) Identify the constant term. c) What is the coefficient of the first term? d) What is the coefficient of the second term? e) List the factors of the third term. D. WRITING ALGEBRAIC EXPRESSIONS Write algebraic expressions (Duration 6:18) Example 1: Juan is 6 inches taller than Niko. Let NN represent Niko s height in inches. Write an algebraic expression to represent Juan s height. Niko s height: Juan s height: Example 2: Juan is 6 inches taller than Niko. Let JJ represent Juan s height in inches. Write an algebraic expression to represent Niko s height. Niko s height: Juan s height: 92

99 Chapter 2 Example 3: Suppose sales tax in your town is currently 9.8%. Write an algebraic expression representing the sales tax for an item that costs DD dollars. Cost: Sales tax: Example 4: You started this year with $362 saved and you continue to save an additional $30 per month. Write an algebraic expression to represent the total amount saved after mm months. Number of months: Total amount saved: Example 5: Movie tickets cost $8 for adults and $5.50 for children. Write an algebraic expression to represent the total cost for AA adults and CC children to go to a movie. Number of adults: Number of children: Total Cost: YOU TRY Complete the following problems. Show all steps as in the media examples. a) There are about 80 calories in one chocolate chip cookie. If we let nn be the number of chocolate chip cookies eaten, write an algebraic expression for the number of calories consumed. Number of cookies: Number of calories consumed: b) Brendan recently hired a contractor to do some necessary repair work. The contractor gave a quote of $450 for materials and supplies plus $38 an hour for labor. Write an algebraic expression to represent the total cost for the repairs if the contractor works for hh hours. Number of hours worked: Total cost: c) A concession stand charges $3.50 for a slice of pizza and $1.50 for a soda. Write an algebraic expression to represent the total cost for PP slices of pizza and SS sodas. Number of slices of pizza: Number of sodas: Total cost: 93

100 The story of xx (Duration 7:09) Chapter 2 Example 1: Tell the story of xx in each of the following expressions. a) xx 5 b) 5 xx c) 2xx d) xx 2 Example 2: Tell the story of xx in each of the following expressions. a) 2xx + 4 b) 2(xx + 4) c) 5(xx 3) 2 2 Example 3: Write an algebraic expression that summarizes the stories below. a) Step 1: Add 3 to xx Step 2: Divide by 2 b) Step 1: Divide xx by 2 Step 2: Add 3 Example 4: Write an algebraic expression that summarizes the stories below. Step 1: Subtract xx from 7 Step 2: Raise to the third power Step 3: Multiply by 3 Step 4: Add 1 94

101 Chapter 2 Below is a table of common English words converted into a mathematical expression. You can use this table to assist in translating expressions. Operation Words Example Translation Added to 4 added to nn nn + 4 More than 2 more than yy yy + 2 Addition Subtraction Multiplication Division Division Power Equals The sum of The sum of rr and ss rr + ss Increased by mm increased by 6 mm + 6 The total of The total of 8 and xx 8 + xx Plus cc plus 2 cc + 2 Minus xx minus 1 xx 1 Less than 5 less than yy yy 5 Less 4 less rr 4 rr Subtracted from 3 subtracted from tt tt 3 Decreased by mm decreased by 10 mm 10 The difference between The difference between xx and yy xx yy Times 12 times xx 12 xx Of One-third of vv 1 3 vv The product of The product of nn and kk nnnn oooo nn kk Multiplied by yy multiplied by 3 3yy Twice Twice dd 2dd oooo 2 dd Divided by nn divided by 4 The quotient of The ratio of Per The quotient of tt and xx The ratio of xx to pp 2 per bb The square of The square of yy yy 2 The cube of The cube of kk kk 3 Is Are Equal Gives Is equal to Is equivalent to Yields Results in was nn 4 tt xx xx pp 2 bb 95

102 YOU TRY Chapter 2 Complete the following problems. a) Tell the story of xx in the expression xx 3 5 b) Write an algebraic expression that summarizes the story below: Step 1: Multiply xx by 2 Step 2: Add 5 Step 3: Raise to the second power. The beauty of algebra (Duration 10:06) What algebraic expression did Sal start with when he discussed why the abstraction of mathematics is so fundamental? E. EVALUATE ALGEBRAIC EXPRESSIONS Evaluate algebraic expressions (Duration 9:33) To evaluate a algebraic or variable expression: PEMDAS P: E: MD: AS: 96

103 Chapter 2 Example 1: Find the value of each expression when w = 2. Simplify your answers. a) ww 6 b) 6 ww c) 5ww 3 d) w 3 e) 3ww 2 f) (3ww) 2 g) 4 5ww h) 5ww 4 i) 3 ww Example 2: Evaluate aaaa + cc given aa = 5, bb = 7, and cc = 3 Example 3: Evaluate aa 2 bb 2 given aa = 5 and bb = 3 Example 4: A local window washing company charges $11.92 for each window plus a reservation fee of $7. a) Write an algebraic expression to represent the total cost from the window washing company for washing ww windows. b) Use this expression to determine the total cost for washing 17 windows. YOU TRY Given aa = 5, bb = 1, cc = 2, evaluate the expressions below. Show all steps. Evaluate bb 2 4aaaa Evaluate 2aa 5bb + 7cc 97

104 F. LIKE TERMS AND COMBINE LIKE TERMS Definition Chapter 2 Two terms are like terms if the base variable(s) and exponent on each variable are identical. For example, 3xx 2 yy and 7xx 2 yy are like terms because they both contain the same base variables, xx and yy, and the exponents on xx (the xx is squared on both terms) and yy are the same. Combining like terms: If two terms are like terms, we add (or subtract) the coefficients, then keep the variables (and exponents on the corresponding variable) the same. Like terms and combine like terms (Duration 6:30) Example 1: Identify the like terms in each of the following expressions 3aa 6aa + 10aa aa 5xx 10yy + 6zz 3xx 7nn + 3nn 2 2nn 3 + 8nn 2 + nn nn 3 Example 2: Combine the like terms 3aa 6aa + 10aa aa 5xx 10yy + 6zz 3xx 7nn + 3nn 2 2nn 3 + 8nn 2 + nn nn 3 YOU TRY Combine the like terms. a) 3xx 4xx + xx 8xx b) 5 + 2aa 2 4aa + aa

105 EXERCISE Tell the story of xx in each of the following expressions. Chapter 2 1) xx 11 2) xx + 5 3) 5xx 4) xx 5 5) xx 3 6) 2 xx 7) 2xx 3 8) 8xx 2 9) (2xx) 2 10) 7 2xx 11) 5(7 xx) 3 12) 3xx Write an algebraic expression that summarizes the stories below. 13) Step 1: Add 8 to xx Step 2: Raise to the third power 14) Step 1: Divide xx by 8 Step 2: Subtract 5 15) Step 1: Subtract 3 from xx Step 2: Multiply by 7 16) Step 1: Multiply xx by 10 Step 2: Raise to the 3rd power Step 3: Multiply by 2 17) Step 1: Add 5 to xx Step 2: Divide by 2 Step 3: Raise to the second power Step 4: Add 8 19) Step 1: Subtract xx from 2 Step 2: Multiply by 8 Step 3: Raise to the third power Step 4: Add 1 Step 5: Divide by 3 18) Step 1: Raise xx to the second power Step 2: Multiply by 5 Step 3: Subtract from 9 20) Step 1: Multiply xx by 4 Step 2: Add 9 Step 3: Divide by 2 Step 4: Raise to the fifth power Find the value of each expression when bb = 88. Simplify your answers. 21) bb 11 22) bb ) 5bb 24) bb 2 25) bb 3 26) 2 bb Evaluate each of the following given qq = ) 2qq 3 28) 8qq 2 29) (2qq) 2 30) 4 7qq 31) 7 2qq 32) 2 qq 99

106 Evaluate the following expressions for the given values. Simplify your answers. Chapter 2 33) bb for aa = 6, bb = 4 2aa 35) 3 5 aaaa for aa = 8, bb = xx 8 34) for xx = 3 5+xx 36) 3xx 2 + 2xx 1 for xx = 1 37) xx 2 yy 2 for xx = 3, yy = 2 38) 2xx 7yy for xx = 5, yy = 3 39) Shea bought cc candy bars for $1.50 each. Write an algebraic expression for the total amount Shea spent. 40) Suppose sales tax in your town is currently 9%. Write an algebraic expression representing the sales tax for an item that costs dd dollars. 41) Ben bought mm movie tickets for $8.50 each and pp bags of popcorn for $3.50 each. a) Write an algebraic expression for the total amount Ben spent. b) Use this expression to determine the amount Ben will spend if he buys 6 movie tickets and 4 bags of popcorn. 42) Noelle is 5 inches shorter than Amy. Amy is AA inches tall. Write an algebraic expression for Noelle's height. 43) Jamaal studied h hours for a big test. Karla studied one fourth as long. Write an algebraic expression for the length of time that Karla studied. 44) A caterer charges a delivery fee of $45 plus $6.50 per guest. a) Write an algebraic expression to represent the total catering cost if gg guests attend the reception. b) Use the expression to determine the amount of a company luncheon of 50 guests. 45) Tickets to the museum cost $18 for adults and $12.50 for children. a) Write an algebraic expression to represent the cost for aa adults and cc children to visit the museum. b) Use this expression to determine the cost for 4 adults and 6 children to attend the museum. 46) Consider the algebraic expression 5nn 8 nn 5 + nn 2 + nn 8 2 a) How many terms are there? b) Identify the constant term. c) What is the coefficient of the first term? d) What is the coefficient of the second term? e) What is the coefficient of the third term? f) What is the coefficient of the fourth term? Combine the like terms. 47) 3dd 5dd + dd 7dd 48) 3xx 2 + 3xx 3 9xx 2 + xx xx 3 49) aa 2bb + 4aa + bb ( 2bb) 50) 3xx 7yy + 9xx 5yy 7 100

107 SECTION 2.2 PROPERTIES OF ALGEBRA Properties of real numbers are the basic rules when we work with expressions in Algebra. Addition Multiplication aa + 00 = aa aa 11 = aa Chapter 2 Identity Hint: the number stays true to its identity. Any number plus zero is the same number = 7 xx + 0 = xx + 0 = Any number multiplies by one is the same number. 7 1 = 7 xx 1 = xx 1 = Inverse Hint: Inverse reverse undo Additive Inverse aa + ( aa) = 00 Any number plus its opposite equals zero. 2 + ( 2) = 0 xx + ( xx) = 0 Multiplicative inverse aa 11 aa = 11 if aa is not zero Any number multiplies by its reciprocal is one = 1 xx 1 xx = 1 Commutative Hint: Commute move switch places aa + bb = bb + aa You can add in any order = xx + 5 = 5 + xx + = + aa bb = bb aa You can multiply in any order. 2 3 = 3 2 xx 2 = 2 xx = Associative Hint: associate different groups parenthesis (aa + bb) + cc = aa + (bb + cc) When you add, you can group in any combination. (3 + 5) + 2 = 3 + (5 + 2) (xx + yy) + zz = xx + (yy + zz) (aa bb) cc = aa (bb cc) When you multiply, you can group in any combination. (6 2) 7 = 6 (2 7) (xx yy) zz = xx (yy zz) Distributive Hint: Distribute give out to Multiplying a number by a group of numbers added together or subtracted each other is the same as doing each multiplication separately. 5( ) = (xx + yy) = 2xx + 2yy 101

108 Properties of Real Numbers (Duration 9:59) Chapter 2 The commutative properties The associative properties Identity properties Inverse properties 102

109 Distributive property Chapter 2 NOTE: Subtraction and division do not have the associative and communicative properties. 103

110 EXERCISE Identify the property that justifies each problem below. 1) ww + 0 = ww Name: 2) xx + xx = 0 Name: 3) 3(uu + vv) = 3uu + 3vv Name: 4) 5ww(6zz) = (6zz)5ww Name: 5) (5) 1 5 xx = 1xx Name: 6) 2 (4aa)bb = (2 4)aaaa Name: 7) (5xx + 5) + 4 = 5xx + (5 + 4) Name: 8) 2xx(3yy + 7zz) = 2xx(7zz + 3yy) Name: Chapter 2 Find the additive inverse and the multiplicative inverse for each problem below Additive inverse Multiplicative inverse 9) 6 10) ) ) mm (given mm 0) Complete the following table by performing the indicated operations by computing the result in the parentheses first as the order of operations requires. Problem 1 Problem 2 Are the Results the Same? (5 + 7) (7 + 3) 13) Addition (10 5) 4 10 (5 4) 14) Subtraction 15) Multiplication (2 3) 4 2 (3 4) 16) Division (600 30) (30 5) 104

111 Chapter 2 Use the commutative or associative properties to perform the operations in the order you find most simple. Show all your work. 17) ( ) ) 5 (6 8) 19) (15 + 4) ) Apply distributive property and combine like terms if possible. 21) 8nn(5 mm) 22) 7kk( 6xx + 6) 23) 6(1 + 6xx) 24) 8xx(5 + 10nn) 25) ( 5 + 9aa) 26) (nn + 1)2 27) 4(xx + 7) + 8(xx + 4) 28) 8(nn + 6) 8(nn + 8) 29) 8xx + 9( 9xx + 9) 30) 4vv 7(1 8vv) 31) 10(xx 2) 3 32) 7(7 + 3vv) + 10(3 10vv) 33) (xx 2 8) (2xx 2 7) 34) 9(bb + 10) + 5bb 35) 2nn( 10nn + 5) 7(6 10nn 3mm) 105

112 CHAPTER REVIEW KEY TERMS AND CONCEPTS Chapter 2 Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson. Variable Algebraic expression Coefficient Term Constant Like terms Identity property of addition Identity property of multiplication Additive Inverse property Multiplicative Inverse property Associative property of addition Associative property of multiplication Commutative property of addition Commutative property of multiplication Distributive property 106

113 Chapter 3 CHAPTER 3: LINEAR EQUATIONS AND INEQUALITIES Chapter Objectives By the end of this chapter, the student should be able to Solve linear equations (simple, dual-side variables, infinitely many solutions or no solution, rational coefficients) Solve linear inequalities Solve literal equations with several variables for one of the variables Contents CHAPTER 3: LINEAR EQUATIONS AND INEQUALITIES SECTION 3.1: LINEAR EQUATIONS A. VERIFYING SOLUTIONS B. ONE-STEP EQUATIONS C. TWO-STEP EQUATIONS D. GENERAL EQUATIONS E. SOLVING EQUATIONS WITH FRACTIONS EXERCISES SECTION 3.2: LINEAR INEQUALITIES A. GRAPHING LINEAR INEQUALITIES B. SOLVING LINEAR INEQUALITIES C. TRIPARTITE INEQUALITIES EXERCISES SECTION 3.3: LITERAL EQUATIONS A. SOLVING FOR A VARIABLE EXERCISES CHAPTER REVIEW

114 SECTION 3.1: LINEAR EQUATIONS Chapter 3 A. VERIFYING SOLUTIONS A linear equation is made up of two expressions that are equal to each other. A linear equation may have one or two variables in it, where each variable is raised to the power of 1. No variable in a linear equation can have a power greater than 1. Linear equation: 2yy = 3xx + 1 (each variable in the equation is raised to the power of 1) Not a linear equation: yy 2 = 3xx + 1 (y is raised to the power of 2, therefore this is not linear) The solution to an equation is the value, or values, that make the equation true. Given a solution, we plug the value(s) into the respective variable(s) and then simplify both sides. The equation is true if both sides of the equation equal each other. Is it a solution? (Duration 5:00) A solution to an equation is the for the that makes the equation. To test a possible solution, the with the. Example. Is aa = 3 the solution to 4aa 18 = 2aa? Explain your answer. YOU TRY a) Verify that xx = 3 is a solution to the algebraic equation 5xx 2 = 8xx + 7. b) Is mm = 1 a solution to the algebraic equation mm + 9 = 3mm + 5? c) Is aa = 5 a solution to the algebraic equation 4(aa + 1) = 6(1 aa)? B. ONE-STEP EQUATIONS The Addition Property of Equality If aa = bb, then for any number cc, aa + cc = bb + cc That is, if we are given an equation, then we are allowed to add the same number to both sides of the equation to get an equivalent statement. 108

115 Addition Principle (Duration 5:00) Chapter 3 To clear a negative we it to. Example (follow the structure in the video and fill in the diagram below) xx 9 = 4 The Multiplication Property of Equality If aa = bb, then for any number cc, aa cc = bb cc That is, if we are given an equation, then we are allowed to multiply by the same number on both sides of the equation to get an equivalent statement. We use these two properties to help us solve an equation. To solve an equation means to undo all the operations of the equation, leaving the variable by itself on one side. This is known as isolating the variable. Multiplication (Division) Principle (Duration 5:00) To clear multiplication we both sides by the. Example (follow the structure in the video and fill in the diagram below) 8xx =

116 Chapter 3 NOTE: When using the Multiplication Property of Equality on an equation like xx = 4 It is easier to think of the negative in front of the variable as a 1 being multiplied by xx, that is 1 xx = 4 We then multiply both sides by 1 to isolate the variable. ( 1) 1 xx = 4 ( 1) 1 xx = 4 xx = 4 When using the Multiplication Property of Equality on an equation where the coefficient is a number other than 1 3xx = 3 We take the coefficient s reciprocal then multiply both sides of the equation by that reciprocal. This will isolate the variable, that is xx = xx = 3 3 xx = 1 YOU TRY Solve. a) xx + 7 = 18 b) rr 4 = 5 c) 4 + bb = 45 d) 3 = 19 + mm e) 3yy = 42 f) 5 = xx C. TWO-STEP EQUATIONS Steps to solve a linear two-step equation. 1. Apply the Addition Property of Equality. 2. Apply the Multiplication Property of Equality to isolate the variable. 3. Check by substituting your answer into the original equation. Basic Two Step (Duration 4:59) 110

117 Chapter 3 Simplifying we use order of operations and we before we. Solving we work in reverse so we will first and then second. Example (follow the structure in the video and fill in the diagram below) 9 = 5 2xx YOU TRY Solve for the variable in each of the following equations. Check your answers. a) Solve: 2bb 4 = 12 Check: b) Solve: 4 + 3rr = 5 Check: c) Solve: 3 = 19 2mm Check: d) Solve: 11 yy = 32 Check: D. GENERAL EQUATIONS We will now look at some more general linear equations, that is, equations that require more than two steps to solve. These equations may have more than one of the same variable on each side of the equal sign and/or may contain parentheses General Equations (Duration 5:00) xx 5 = 4xx + 7 3(4nn 2) = 5(nn + 3) Move variables to one side by. 111

118 Sometimes we may have to first. Simplify by and Chapter 3 on each side. Example (follow the structure in the video and fill in the diagram below) 2xx + 7 = 5xx 3 Use the following steps to solve a general equation. 1. Simplify each side of the equation. Remove parentheses if necessary. Combine like terms. 2. Add terms on each side of the equation so that all terms containing the variable are on one side of the equal sign and all constant terms are on the other side. 3. Simplify each side of the equation by combining like terms. 4. Apply the Multiplication Property of Equality to isolate the variable. 5. Check by substituting the solution into the original equation. YOU TRY Solve for the variable in each of the following equations. Check your answers. a) Solve: xx 5 = 4xx + 7 Check: b) Solve: 3(4nn 2) = 5(nn + 3) Check: 112

119 c) Solve: 4 (2yy 1) = 2(5yy + 9) + yy Check: Chapter 3 E. SOLVING EQUATIONS WITH FRACTIONS When solving linear equations with fractions, it is vital to remember the Multiplication Property of Equality. Previously, we ve only dealt with coefficients that were integers. Now we will be looking at coefficients that are rational numbers. 5xx 6 = 5 We can manipulate the left side of this equation as such 5 xx = 5 6 Looking at it this way, we can then use the Multiplication Property of Equality and multiply both sides of the equation by the coefficient s reciprocal xx = xx = 6 xx = 6 Another way to solve this type of equation is to clear the fractions in the equation by multiplying by the LCD. Distributing with Fractions (Duration 5:00) Important: Always first and second. Solve the equation below by multiplying the equation by the LCD. 2 3 (xx + 4) = xx

120 Chapter 3 YOU TRY a) Solve: xx 6 = 5 Check: b) Solve: 3 4 aa = 8 Check: c) Solve: 0 = 5 4 xx 6 5 Check: 114

121 Chapter 3 EXERCISES Solve for the variable in each of the following equations. Reduce, simplify, and check your answers. Show all steps, and box your answer. 1) 8xx 2 = 30 2) 5 xx = 3 3) 1 2 xx 4 = 8 4) 2 xx + 3 = ) 4xx 8 = xx + 7 6) 3 4 xx 1 2 = 9 8 xx ) 6xx 4( 2xx + 8) = 10 8) 2(4xx 2) = 2(xx 8) 9) (2xx 7) (4xx + 8) = 4(xx + 6) 10) 2(4xx + 3) = 8xx ) 5(xx + 6) xx = 4(xx + 7) ) mm = ) xx = 4 3 xx xx

122 SECTION 3.2: LINEAR INEQUALITIES Chapter 3 A. GRAPHING LINEAR INEQUALITIES An algebraic inequality is a mathematical sentence connecting an expression to a value, variable, or another expression with an inequality sign. Below is a table of inequalities we will be using Symbol In Words Examples < less than 1 < 2 1 is less than 2 > greater than 4 > 3 4 is greater than 3 less than or equal to is less than 5 greater than or equal to is equal to 1 not equal is not equal to 4 A solution to an inequality is a value that makes the inequality true. For example, a solution to the inequality xx < 1 may be 0 since 0 is indeed less than 1. However, 2 cannot possibly be a solution since 2 is not less than 1. NOTE: The inequality symbols < and > can be quite easy to interpret, however, the inequalities symbols and on the other hand, can be tricky. For example, xx 1 is read as xx is less than or equal to 1. The keyword here is the word or. The word or tells us that our solution can be less than 1 or equal to 1. So 0 is a solution to this inequality since 0 is less than 1. As it turns out, 1 is also a solution to this inequality. The solution 1 is not less than 1 but it is equivalent to 1, thus 1 is a solution. Notice that this reasoning does not work with strict inequalities. To graph an inequality, let us look at xx < 1. We first draw a number line and mark the number in our inequality on the line. 1 We then draw an open circle or closed circle (depending on the inequality symbol) on the number line, above the number we marked. The final step is to draw a line in the direction of the solutions. 1 1 Remember: We use an open circle with the symbols < and >, and a closed circle with the symbols or 116

123 Interval Notation (Duration 3:04) Chapter 3 Interval notation is used to a graph with numbers. Interval notation will always be read to. (, ) We use parentheses for less/greater than, and for less/greater than or equal to. The symbols and will always use parentheses. Example, graph the interval (, 1) on the number line below. YOU TRY a) Determine whether the number 4 is a solution to the following inequalities. xx > 1 xx < 1 xx 9 xx > 4 xx 4 b) Graph the following inequalities in the box below. Write the solution using interval notation. Inequality Graph Interval Notation xx > 2 xx 2 xx < 2 xx 2 B. SOLVING LINEAR INEQUALITIES A linear inequality has the form aaaa + bb < cc where aa, bb, and cc are real numbers. This definition is the same for,, or >. 117

124 Chapter 3 To solve linear inequalities we use the following properties (in the following properties we use the < symbol. Keep in mind that these properties work with the other inequality symbols too): The Addition Property of Inequalities For real numbers aa, bb, and cc, if aa < bb, then aa + cc < bb + cc. The Multiplication Property of Inequalities For real numbers aa, bb, and cc > 00, if aa < bb, then aa cc < bb cc If cc < 00, then aa cc > bb cc When we are multiplying or dividing by a negative number, we reverse the sign of the inequality. Steps to solve a general equation. 1. Simplify each side of the inequality. Remove parentheses if necessary. Collect like terms. 2. Add terms on each side of the inequality so that all terms containing the variable are on one side and all constant terms are on the other side. 3. Simplify each side of the inequality by combining like terms. 4. Apply the Multiplication Property of Inequalities to isolate the variable. 5. Check by substituting the solution (endpoint and a value from the solution set) into the original inequality. Solving (Duration 5:00) Solving inequalities is just like. The only exception is if you or by a, you must. Example. Solve the inequality below using the video as a guide. Write the solution in interval notation. 7 5xx 17 YOU TRY Solve the inequality, check your answer, and graph the solution on a number line. Give the solution in interval notation. a) 3xx > xx

125 b) 3 5aa 2(aa + 5) Chapter 3 c) 5(xx + 2) 3(xx + 4) C. TRIPARTITE INEQUALITIES We have been dealing with inequalities where the variable (or expression containing the variable) is on the left or the right side of a number. We will now look at a special type of inequality called a tripartite inequality, where the expression containing the variable is between two numbers, for example 1 < xx < 1 The solution for this inequality in interval notation is ( 1, 1). Graphing the solution we get -1 1 Below is another example. 1 xx < 1 The solution for this inequality in interval notation is [ 1, 1). Graphing the solution we get -1 1 Tripartite (Duration 5:00) A tripartite inequality is a part inequality. We use a tripartite inequality when our variable is two numbers When solving these type of inequalities we will. When graphing, we will graph the inequality the numbers. Example. Solve in the inequality below. Write the solution in interval notation. 5 < 5 4xx

126 Chapter 3 YOU TRY a) Which of the following values are in the solution set for 3 nn < 5? nn = 5 nn = 3 nn = 0 nn = 4.9 nn = 5 nn = 12 b) Write a compound inequality to represent the following situation. Clearly indicate what the variable represents. A number is greater than or equal to 5 but les than 8 120

127 EXERCISES 1) Which of the following values are in the solution set for xx < 3? Chapter 3 xx = 0 xx = 1 xx = 5 xx = 3 xx = 5 xx = 5 3 2) Which of the following values are in the solution set for xx 1? xx = 0 xx = 1 xx = 5 xx = 3 xx = 5 xx = 5 3 3) Which of the following values are in the interval [ 2, )? xx = 0 xx = 1 xx = 5 xx = 3 xx = 5 xx = 5 3 4) Which of the following values are in the interval (, 1)? xx = 0 xx = 1 xx = 5 xx = 3 xx = 5 xx = 5 3 5) Which of the following values are in the interval ( 1, 5]? xx = 0 xx = 1 xx = 5 xx = 3 xx = 5 xx = 5 3 6) Which of the following values are in the interval 5 < xx 3? xx = 0 xx = 1 xx = 5 xx = 3 xx = 5 xx = 5 3 For questions 7-14, solve the inequality, check your answer, and graph the solution on a number line. Give the solution in interval notation. 7) 7 4xx 5 8) 4xx 2xx ) 14mm + 8 > 6mm 8 10) 5( 2aa 8) 9aa ) 6xx + 13 < 5(2xx 3) 12) xx 7 13) 5 xx ) 4 < 8 3mm 11 15) Translate the statement into a compound inequality. A number nn is greater than 0 and less than or equal to 8 121

128 SECTION 3.3: LITERAL EQUATIONS Chapter 3 A. SOLVING FOR A VARIABLE In this section will be constructing linear equations from sentences. Use the steps below as a guide when approaching each problem. Steps for Writing and Solving Equations 1) Read and understand the problem. Underline the givens and circle the goal. 2) Form a strategy to solve the problem. 3) Choose a variable to represent the unknown quantity. 4) Read every word in the problem, and translate the given information into an algebraic equation. 5) Solve the equation. 6) Write your answer in a complete sentence. Example: The cost of leasing a new Ford mustang is $2,311 for a down payment and processing fee plus $276 per month. For how many months can you lease this car with $10,000? Solution: Step 1. The cost of leasing a new Ford mustang is $2,311 for a down payment and processing fee plus $276 per month. For how many months can you lease this car with $10,000? Step 2. Since $2,311 is a down payment, this number must be constant, in other words, this number does not change no matter how many months go by. The $276 does change as the months go by. In the first month we pay $276 on top of the down payment. In the second month we pay $276 + $276 plus the down payment, and so on. Take note that the down payment is a one time payment not a monthly payment like the $276. Step 3. We will let mm be our variable to represent the number of months we can lease the car. Step 4. By the information that is given, the linear equation is 276mm = 10,000. Step 5. Solving for mm we get 27.8 months. So how many months can we lease the car? If we say we can lease the car for 28 months (rounding up) then we would go over the $10,000 limit. So we must round down and say 27 months. Step 6. We can lease this car for 27 months. YOU TRY a) You have just bought a new Sony 55 3D television set for $1,600. The value of the television set decreases by $250 per year. How long before the television set is worth half of its original value? 122

129 Chapter 3 EXERCISES For each of the following, underline the Givens and circle the Goal of the problem. Form a Strategy, Solve, and Check. Show all work, and write your answers in complete sentences. 1) John is a door to door vacuum salesman. His weekly salary, SS, is $200 plus $50 for each vacuum he sells. This can be written as SS = vv, where vv is the number of vacuums sold. If John earns $1000 for a week s work, how many vacuums did he sell? 2) Paul is planning to sell bottled water at the local Lollapalooza. He buys 2 crates of water (2000 bottles) for $360 and plans on selling the bottles for $1.50 each. Paul s profits, PP in dollars, from selling bb bottles of water is given by the formula PP = 1.5bb 360. How many bottles does Paul need to sell in order to break even? 3) A new Sony 55 3D television costs $2,499. You are going to pay $600 as a down payment, and pay the rest in equal monthly installments for one year. Write an equation to represent this situation, and use it to determine how much you should pay each month. Clearly indicate what the variable in your equation represents. Solve the equation, and write your answer in a complete sentence. 4) Your yard is a mess, and you decide to hire a landscaper. The Greenhouse charges a $20 consultation fee plus $11 per hour for the actual work. Garden Pros does not charge a consulting fee, but charges $15 per hour for the actual work. Write an equation that will help you determine the number of hours at which the two companies charge the same. Clearly indicate what the variable represents. Solve the equation, and write your answer in a complete sentence. 5) Let pp represent the marked price of an item at Toys R Us. Emma s aunt gave her a $50 gift card to Toys R Us for her birthday. If sales tax is currently 9%, set up an equation to express how much she can spend using her gift card. Solve the equation, and interpret your answer in a complete sentence. 6) Carlos recently hired a roofer to do some necessary work. On the final bill, Carlos was charged a total of $1105, where $435 was listed for parts and the rest for labor. If the hourly rate for labor was $67, how many hours of labor was needed to complete the job? 123

130 CHAPTER REVIEW KEY TERMS AND CONCEPTS Chapter 3 Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson. Linear Equation Solution to a Linear Equation Addition Property Of Equality Multiplication Property of Equality Algebraic Inequality < > Addition Property of Inequalities Multiplication Property of Inequalities 124

131 CHAPTER 4: LINEAR EQUATION APPLICATIONS Chapter Objectives Chapter 4 By the end of this chapter, students should be able to Translate sentences into equation Model and solve - Discount and mark-up problems - Geometry problem with perimeter and triangles - Investment problems - Mixture problems - Uniform motion problems SECTION 4.1: INTEGER PROBLEMS A. NUMBER PROBLEMS B. CONSECUTIVE INTEGERS EXERCISE SECTION 4.2: MARK-UP AND DISCOUNT PROBLEMS A. MARK-UP PROBLEMS B. DISCOUNT PROBLEMS EXCERSISE SECTION 4.3: GEOMETRY PROBLEMS A. TRIANGLES B. PERIMETER EXCERSISE SECTION 4.4: VALUE AND INTEREST PROBLEMS A. VALUE PROBLEMS WITH COINS B. SIMPLE INTEREST PROBLEMS C. VALUE/INTEREST PROBLEMS WITH 1 VARIABLE EXCERSISE SECTION 4.5: UNIFORM MOTION PROBLEMS A. DISTANCE = RATE X TIME B. OPPOSITE DIRECTIONS C. CATCH-UP D. TOTAL TIME EXCERSISE SECTION 4.6 MIXTURE PROBLEMS EXCERSISE CHAPTER REVIEW

132 Chapter 4 Word problems can be tricky. The goal is becoming proficient in translating an English sentence into a mathematical sentence. In this chapter, we focus on word problems modeled by a linear equation and solve. Below is a table of common English words converted into a mathematical expression. You can use this table to assist in translating expressions and equations. Operation Words Example Translation Added to 4 added to nn nn + 4 More than 2 more than yy yy + 2 Addition Subtraction Multiplication Division Power Equals The sum of The sum of rr and ss rr + ss Increased by mm increased by 6 mm + 6 The total of The total of 8 and xx 8 + xx Plus cc plus 2 cc + 2 Minus xx minus 1 xx 1 Less than 5 less than yy yy 5 Less 4 less rr 4 rr Subtracted from 3 subtracted from tt tt 3 Decreased by mm decreased by 10 mm 10 The difference between The difference between xx and yy xx yy Times 12 times xx 12 xx Of One-third of vv 1 3 vv The product of The product of nn and kk nnnn oooo nn kk Multiplied by yy multiplied by 3 3yy Twice Twice dd 2dd oooo 2 dd Divided by nn divided by 4 The quotient of The ratio of Per The quotient of tt and xx The ratio of xx to pp 2 per bb The square of The square of yy yy 2 The cube of The cube of kk kk 3 Is Are Equal Gives Is equal to Is equivalent to Yields Results in was nn 4 tt xx xx pp 2 bb 126

133 SECTION 4.1: INTEGER PROBLEMS A. NUMBER PROBLEMS Find number problems (Duration 4:47) Chapter 4 Translate: Is/Were/Was/Will Be: More than: Subtracted from/less than: Example: a) Five less than three times a number is nineteen. What is the number? b) Seven more than twice a number is six less than three times the same number. What is the number? YOU TRY a) If 28 less than five times a number is 232, what is the number? b) Fifteen more than three times a number is the same as ten less than six times the number. What is the number? B. CONSECUTIVE INTEGERS Another type of number problem involves consecutive integers. Definition Consecutive integers are integers that come one after the other, such as 3, 4, 5, and so on (or equivalently, 3, 2, 1, ). If we are trying to find several consecutive integers, it important to identify the first integer and then assign all the following integers. E.g., if xx is the first integer, then xx + 11 will be the next, and xx + 22 will be the following, and so on. If we are trying to find several even or odd consecutive integers, it important to identify the first integer and then assign all the following even or odd integers. E.g., if xx is the first integer, then xx + 22 will be the next, and xx + 44 will be the following, and so on. 127

134 Chapter 4 Find consecutive numbers problems (Duration 4:59) Consecutive Numbers: First: Second: Third: Example: a) Find three consecutive numbers whose sum is 543 b) Find four consecutive integers whose sum is 222 YOU TRY a) The sum of three consecutive positive integers is 93. What are the positive integers? b) The sum of three consecutive even positive integers is 246. What are the numbers? c) Find three consecutive odd positive integers so that the sum of twice the first integer, the second integer, and three times the third integer is

135 EXERCISE 1) When five is added to three more than a certain number, the result is 19. What is the number? Chapter 4 2) If five is subtracted from three times a certain number, the result is 10. What is the number? 3) When 18 is subtracted from six times a certain number, the result is 42. What is the number? 4) A certain number added twice to itself equals 96. What is the number? 5) A number plus itself, plus twice itself, plus 4 times itself, is equal to 104. What is the number? 6) Sixty more than nine times a number is the same as two less than ten times the number. What is the number? 7) Eleven less than seven times a number is five more than six times the number. Find the number. 8) Fourteen less than eight times a number is three more than four times the number. What is the number? 9) The sum of three consecutive integers is 108. What are the integers? 10) The sum of three consecutive integers is 126. What are the integers 11) Find three consecutive integers such that the sum of the first, twice the second, and three times the third is ) The sum of two consecutive even integers is 106. What are the integers? 13) The sum of three consecutive odd integers is 189. What are the integers? 14) The sum of three consecutive odd integers is 255. What are the integers? 15) Find three consecutive odd integers such that the sum of the first, two times the second, and three times the third is

136 SECTION 4.2: MARK-UP AND DISCOUNT PROBLEMS Chapter 4 When paying for our meal at a restaurant, we do not pay just the price of the food. We also pay a percentage for sales tax. Imagine our food cost $65, but the sales tax is 8%. Then we would pay the original $65 plus 8% of that $65. In mathematical terms, the final price would be ffffffff cccccccc + ssssssssss tttttt = tttttttttt bbbbbbll $65 = $70.20 Likewise, if we were buying an item on sale, we would not pay the original price of the item, but a new lower price based on a percentage of the original. Suppose we want to buy a $38 sweater, and it s on sale for 15% off. We would take 15% of the $38 off from the original $38. In mathematical terms, the final price would be cccccccc oooo ssssssssssssss dddddddddddddddd = ffffffffff pppppppppp $38 = $32.30 Some of us may not be familiar with retail, but a business must first acquire merchandise before the business sells it to the public. They add, to their original cost, a percentage of that cost to make a profit when they sell an item. Consider a business owner that purchases a vase for $350. The owner marks up the price by 25% to sell the vase at a higher price and make a profit. In mathematical terms, the final price to the consumer would be Notice how these scenarios have the same format: cccccccc tttt tthee oooooooooo + mmmmmmmm-uuuu = ffffffffff pppppppppp $350 = $ OOOOOOOOOOOOOOOO PPPPPPPPPP ± PPPPPPPPPPPPPP OOOOOOOOOOOOOOOO PPPPPPPPPP = FFFFFFFFFF PPPPPPPPPP The plus or minus is determined by whether we are increasing or decreasing the original price. In these previous three scenarios, the discount problem was the only time we were decreasing the original price. We will be dealing with two prices. We first need to determine which price came first in the timeline because that price is the original. A. MARK-UP PROBLEMS Mark-up formula Given the original cost of an item CC, the mark-up rate rr, the selling price of the item including the mark-up is given by SSSSSSSSSSSSSS pppppppppp = OOOOOOOOOOOOOOOO cccccccc + (MMMMMMMM-uuuu rrrrrrrr)(oooooooooooooooo cccccccc) SS = CC + rrrr where SS is the selling price. Note, the mark-up rate is a percentage and should be converted to a decimal when using the formula. 130

137 Mark-up problems (Duration 3:51) Chapter 4 Mark-up Formula S C r Example: a) The cost to a distributor for a product is $34. The distributor sells the product for $58 to his customers. What is the mark-up rate? Be sure to round your answer to the nearest whole percent, e.g would be 44%. YOU TRY a) A retailer acquired a laptop for $2,015 and sold it for $3, What was the percent markup? 131

138 B. DISCOUNT PROBLEMS Discount formula Chapter 4 Given the original cost of an item RR, the discount rate rr, the sale price of the is given by SSSSSSSS pppppppppp = RRRRRRRRRRRRRR pppppppppp (MMMMMMMM-uuuu rrrrrrrr)(rrrrrrrrrrrrrr pppppppppp) SS = RR rrrr where SS is the sale price. Note, the discount rate is a percentage and should be converted to a decimal when using the formula. Discount problems (Duration 3:28) Discount Formula S R r Example: a) The sale price of a product is $220, which is 40% off the regular price. Find the regular price. Be sure to round your answer to the nearest cent. YOU TRY a) Sue bought a sweater for $ after a 15% discount. How much was it before the discount? 132

139 EXCERSISE Chapter 4 1) A jewelry storeowner bought a diamond in St. Thomas. He then sold it in the United States for $5,500. If he sold it at an 87% mark-up, what was his original cost for the diamond in St. Thomas? Round your answer to the nearest cent. 2) Sari bought a $125 pair of shoes on sale for $ What was the percent discount? 3) Mickey wants to buy her first new car, but she only has $35,500 to spend. She has her eye on a $34,000 hybrid. However, taxes, license and registration tack on an extra 9.5% of the purchase price. What would her actual cost be? Can she afford this car? 4) Rusty took his family out to eat at a nice restaurant. The bill was $ after a sales tax of 8.75%. How much was the food before taxes? 5) As a sales representative for Nerds-Are-Us, Mal gets an 18% employee discount. If she purchased a tablet for $705.20, what was the price before the discount? 6) Jay found a rare book at an estate sale and purchased it for $350. He then sold it online for $770. What was his percent markup? 7) Maria found a computer for $4,500. If the sales tax is 6.5%, what did she pay for the computer? 8) Bella s clothing store pay $7.25 per pair of earrings. The store marks-up 175%. How much is the selling price for a pair of earrings. Round your answer to the nearest cent. 9) The Fancy steakhouse marks-up their steaks 325%. Each piece of premium steak cost $12.00 at wholesale price. What is the selling price of a piece of steak? 133

140 Chapter 4 SECTION 4.3: GEOMETRY PROBLEMS Another example of translating English sentences to mathematical sentences comes from geometry. We will discuss triangles and perimeter problems. A. TRIANGLES Sum of Angles in a Triangle Given a triangle with three angles. The sum of the three angles is I.e., if the angles in a triangle are aa, bb, and cc, then aa + bb + cc = 180 The sum of the three interior angles of a triangle of any size is always 180 degrees. Sum of angles in a triangle (Duration 4:59) Angles of a triangle add to Example: a) Two angles of a triangle are the same measure. The third angle is 30 degrees less than the first. Find the three angles. b) The second angle of a triangle measures twice the first. The third angle is 30 degrees more than the second. Find the three angles. YOU TRY a) The second angle of a triangle is double the first. The third angle is 40 less than the first. Find the three angles. 134

141 B. PERIMETER Another geometry problem involves perimeter or the distance around an object. Chapter 4 Perimeter of a rectangle The formula for the perimeter of a rectangle is given by PP = where ww is the width and ll is the length of the rectangle. Perimeter problems (Duration 5:00) Formula for Perimeter of a rectangle: Width is the side. Example: a) A rectangle is three times as long as it is wide. If the perimeter is 62 cm, what is the length? b) The width of a rectangle is 6 cm less than the length. If the perimeter is 52 cm, what is the width? YOU TRY The perimeter of a rectangle is 44 cm. The length is 5 less than double the width. Find the dimensions. 135

142 EXCERSISE 1) The second angle of a triangle is the same size as the first angle. The third angle is 12 degrees larger than the first angle. How large are the angles? Chapter 4 2) Two angles of a triangle are the same size. The third angle is 12 degrees smaller than the first angle. Find the measure the angles. 3) Two angles of a triangle are the same size. The third angle is 3 times as large as the first. How large are the angles? 4) The third angle of a triangle is the same size as the first. The second angle is 4 times the third. Find the measure of the angles. 5) The second angle of a triangle is 3 times as large as the first angle. The third angle is 30 degrees more than the first angle. Find the measure of the angles. 6) The second angle of a triangle is twice as large as the first. The measure of the third angle is 20 degrees greater than the first. How large are the angles? 7) The second angle of a triangle is three times as large as the first. The measure of the third angle is 40 degrees greater than that of the first angle. How large are the three angles? 8) The second angle of a triangle is five times as large as the first. The measure of the third angle is 12 degrees greater than that of the first angle. How large are the angles? 9) The second angle of a triangle is three times the first, and the third is 12 degrees less than twice the first. Find the measures of the angles. 10) The second angle of a triangle is four times the first and the third is 5 degrees more than twice the first. Find the measures of the angles. 11) The perimeter of a rectangle is 150cm. The length is 15cm greater than the width. Find the dimensions. 12) The perimeter of a rectangle is 304 cm. The length is 40 cm longer than the width. Find the length and width. 13) The perimeter of a rectangle is 152 meters. The width is 22 meters less than the length. Find the length and width. 14) The perimeter of a rectangle is 280 meters. The width is 26 meters less than the length. Find the length and width. 15) The perimeter of a college basketball court is 96 meters and the length is 14 meters more than the width. What are the dimensions? 136

143 SECTION 4.4: VALUE AND INTEREST PROBLEMS A common type of word problems is value and interest problems. Chapter 4 A. VALUE PROBLEMS WITH COINS We will use a common formula to model the word problems called AAAAAA: Value problems (Duration 5:00) AAAAAAAAAAAA VVVVVVVVVV = TTTTTTTTTT Value table: The equation always comes from the Example: a) Brian has twice as many dimes as quarters. If the value of the coin is $4.95, how many of each does he have? b) A child has three more nickels than dimes in her piggy-bank. If she has $1.95 in the bank, how many of each does she have? YOU TRY a) In a child s piggy bank are a mix of 11 quarters and dimes that has a total value of $1.85. How many coins of each does the child have? b) A man has a collection of stamps made up of 5-cent stamps and 8-cent stamps. There are three times as many8-cent stamps as 5-cent stamps. The total value of all the stamps is $3.48. How many of each stamp does the man have? 137

144 B. SIMPLE INTEREST PROBLEMS Simple Interest (Duration 6:17) Chapter 4 Interest represents a change in money. If you have a saving account, the. If you have a loan, the. Simple Interest Formula II = PP = rr = tt = Example: a) If you invest $3,500 in saving account that pays 4% simple interest, how much interest will you earn after 3 years? What will the new balance be? b) You borrows $6,000 from a loan shark. If you will owe $7,200 in 18 months, what would be the simple interest rate? YOU TRY a) Jesse invests $500 in a simple interest account at 5%. How much interest will he earn after 4 years? b) Karla earned $180 of interest at 4% after she deposited $1500 in a saving account. How long did she leave her money in there? 138

145 Chapter 4 C. VALUE/INTEREST PROBLEMS WITH 1 VARIABLE For interest problems below, we can use a similar formula II = PPPPPP, but in this section, we will assume that the time is one year, ie., tt = 1 to get the formula below. PPPPPPPPPPPPPPPPPP rrrrrrrr tttttttt = IIIIIIIIIIIIIIII PPPPPPPPPPPPPPPPPP rrrrrrrr 1 = IIIIIIIIIIIIIIII PPPPPPPPPPPPPPPPPP rrrrrrrr = IIIIIIIIIIIIIIII Interest problems with 1 variable (Duration 5:00) Interest table: The equation always comes from the Example: a) Sophia invested $1900 in one account and b) Carlos invested $2500 in one account and $1500 in another account that paid 3% higher $1000 in another which paid 4% lower interest rate. After one year she had earned interest. At the end of a year, he had earned $113 in interest. At what rates did she invest? $345 in interest. At what rates did he invest? YOU TRY a) A woman invests $4,000 in two accounts: one at 6% interest, the other at 9% interest for one year. At the end of the year, she had earned $270 in interest. How much did she invest in each account? b) John invests $5,000 in one account and $8,000 in an account paying 4% more in interest. He earned $1,230 in interest after one year. At what rates did he invest? 139

146 EXCERSISE Chapter 4 1) A collection of dimes and quarters is worth $ There are 103 coins in all. How many of each is there? 2) A collection of half dollars and nickels is worth $ There are 34 coins in all. How many are there? 3) A purse contains $3.90 made up of dimes and quarters. If there are 21 coins in all, how many dimes and how many quarters were there? 4) A boy has $2.25 in nickels and dimes. If there are twice as many dimes as nickels, how many of each kind has he? 5) $3.75 is made up of quarters and half dollars. If the number of quarters exceeds the number of half dollars by 3, how many coins of each denomination are there? 6) A collection of 27 coins consisting of nickels and dimes amounts to $2.25. How many coins of each kind are there? 7) $3.25 in dimes and nickels were distributed among 45 boys. If each received one coin, how many received dimes and how many received nickels? 8) A coin purse contains 18 coins in nickels and dimes. The coins have a total value of $1.15. Find the number of nickels and dimes in the coin purse. 9) The total value of dimes and quarters in a bank is $6.05. There are six more quarters than dimes. Find the number of each type of coin in the bank. 10) A child s piggy bank contains 44 coins in quarters and dimes. The coins have a total value of $8.60. Find the number of quarters in the bank. 11) A coin bag contains nickels and dimes. The number of dimes is 10 less than twice the number of nickels. The total value of all the coins is $2.75. Find the number of each type of coin in the bank. 12) A total of 26 bills are in a cash box. Some of the bills are one dollar bills, and the rest are five dollar bills. The total amount of cash in the box is $50. Find the number of each type of bill in the cash box. 13) A bank teller cashed a check for $200 using twenty-dollar bills and ten-dollar bills. In all, 12 bills were handed to the customer. Find the number of twenty-dollar bills and the number of ten-dollar bills. 14) Find the simple interest rate for $2,000 at 6% for 2 years. 15) What is the balance of an account after 10 years with an initial $3,000 deposit at 3% interest? 16) Jerry got a car $22,000 car loan with the simple interest rate at 4.25% for 60 months. How much would be he pay to the bank in total when the loan term is over? 140

147 Chapter 4 17) Lupita deposited $800 into a saving account 4 years ago. She has left her money alone since then. She has $896 in her account today. What is the interested rate of the account? 18) Elena deposited $2,000 in her saving account at 5%. How long would it take for her for her to earn $1,000 interest? 19) You want to borrow $10,000 to buy a car. The dealership has a promotion at 1.99% simple rate. How much do you have to pay in interest after 60 months? 20) Jessica makes $395 on interest in her saving account after 5 years at 2%. How much did she deposited into her account initially? 21) How long would it take Jose to earn $660 in interested after he deposited $1000 into an account with a 5.5% rate? 22) Jose invested $12,500 with Bank USA and $14,500 with at Bank of Pacifica that offered 1% higher. He earned $3,385 in total interest after one year. At what rate did each bank offer? 23) Angela invested $2,500 in the Wealthy Future mutual fund plan and $6,500 in the Blue Chip plan that yield 2% more. The total interest after one year was $1,030. At what rate did each plan offer? 24) A college fund is invested with two different banks, $4,100 with Bank A and $5,900 at Bank B, which offered at 1.5% higher rate. The combined annual interest was $1, What was the interest rate at each bank? 25) Jason earned $256 interest last year on his investments. If $1,600 was invested at a certain rate of return and $2,400 was invested in a fund with a rate that was double the rate of the first fund, find the two rates of interest. 26) Natalia invested $3,500 in the AB High yield fund and $5,000 in the Emerging Market Core fund. The AB high yield rate is 2.5% higher than the Emerging Market rate. Natalia earned the total interest after one year is $385. What was interest rate for each fund? 27) A person borrowed $6,500 loan at QuickCash and $8500 loan at CashNow. QuickCash charge 3% less than CashNow. After one year, he paid a total interest of $1,455. What was the simple interest rate each company charge him? 28) Kenny took out two different loans to buy a used car for his teenage daughter. He got $3000 at Prosperity Bank and $3000 with Trusty Bank, which charged him 1.5% higher. He paid $300 in simple interest to both banks after a year. At what rate did each bank charge him? 29) Yang Yang owned two credit card balances of $7,500 and $3,500. The card with the $3,500 balance charged 1.4% more than the other. At the end of one year, he paid is $797. What was the simple interest of each card? 30) Samantha earned $1,480 in interest last year on her investments. If $5,000 was invested in Aggressive portfolio with a 4% higher rate of return than $11,000 was invested in Moderate portfolio. Find the return rate of each portfolio. 141

148 Chapter 4 SECTION 4.5: UNIFORM MOTION PROBLEMS Another common application of linear equations is uniform motion problems. When solving uniform motion problems, we use the relationship the distance formula dddddddddddddddd = rrrrrrrr tttttttt or rrrrrrrr tttttttt = dddddddddddddddd (ssssssssss) A. DISTANCE = RATE X TIME Distance problems Find d (Duration 1:39) Example: If you drive 65 miles per hours for 4 hours, how far have you traveled? Distance problems Find t (Duration 2:28) Example: Google maps shows your trip will require you to drive 377 miles. If you average 58 miles per hours, how long will the drive take you? Distance problems Find r (Duration 2:51) Example: If you drive for 7 hours and travel 385 miles, what is your rate? 142

149 B. OPPOSITE DIRECTIONS Distance opposite directions (Duration 5:00) Chapter 4 Distance table: The equation always comes from the Example: a) Brian and Jennifer both leave the convention at the same time traveling in opposite directions. Brian drove 35 mph and Jennifer drove 50 mph. After how much time were they 340 miles apart? b) Maria and Tristan are 126 miles apart biking towards each other. If Maria bikes 6 mph faster than Tristan and they meet after 3 hours, how fast did they each ride? YOU TRY a) Two joggers start from opposite ends of an 8- mile course running towards each other. One jogger is running at a rate of 4 miles per hour, and the other is running at a rate of 6 miles per hour. After how long will the joggers meet? b) Bob and Fred start from the same point and walk in opposite directions. Bob walks 2 miles per hour faster than Fred. After 3 hours, they are 30 miles apart. How fast did each walk? c) Two campers left their campsite by canoe and paddled downstream at an average speed of 12 mph. They turned around and paddled back upstream at an average rate of 4 mph. The total trip took 1 hour. After how much time did the campers turn around downstream? 143

150 C. CATCH-UP Distance Catch up (Duration 4:50) Chapter 4 A head start: the head start to his/her Catch up: Example: a) Raquel left the party traveling 5mph. 4 hours later Nick left to catch up with her, traveling 7mph. How long will it take him to catch up? b) Trey left on a trip traveling 20 mph. Julian left 2 hours later, traveling in the same direction at 30 mph. After how many hours does Julian pass Trey? YOU TRY a) Mike leaves his house traveling 2 miles per hour. Joy leaves 6 hours later to catch up with him traveling 8 miles per hour. How long will it take her to catch up with him? 144

151 D. TOTAL TIME Distance Total time (Duration 5:00) Chapter 4 Consider: Total time of 8 When we have total time, for the first box we use and for the second we use. Example: a) Lupe rode into the forest at 10 mph, turned b) Ian went on a 230-mile trip. He started driving around and returned by the same route 45 mph. However, due to construction on the traveling 15 mph. If her trip took 5 hours, how second leg of the trip, he had to slow down to long did she travel at each rate? 25 mph. If the trip took 6 hours, how long did he drive at each speed? YOU TRY a) On a 130-mile trip, a car traveled at an average speed of 55 mph and then reduced its speed to 40 mph for the remainder of the trip. The trip took 2.5 hours. For how long did the car travel 40 mph? 145

152 EXCERSISE Chapter 4 1) A train traveled for 2 hours at a speed of 45 mph and 5 hours at 95 mph. What is the total miles that the train traveled? 2) The distance from LAX to San Francisco is 385 miles. If Gabriela drove at the average speed of 55mph, how long did it take her to arrive? 3) Sofia and her friends drove from Orange County to Las Vegas to celebrate her 21 st birthday. It took them 4 hours to get there. The distance from Orange County to Vegas is 262 miles. What was their average speed? 4) It takes 6 hours to fly from John Wayne airport to JFK airport, New York that is 2454 air miles. What is the average speed of the plane? 5) Ana rode from Orange to Yosemite National Park with her bike group a total of 450 miles. It took them 45 hours in total. What was their average speed? 6) A is 60 miles from B. An automobile at A starts for B at the rate of 20 miles an hour while an automobile at B starts for A at the rate of 25 miles an hour. How long will it be before the automobiles meet? 7) Two trains travel toward each other from points which are 195 miles apart. They travel at rate of 25 and 40 miles an hour, respectively. If they start traveling at the same time, how long before the trains will meet? 8) A passenger and a freight train start toward each other at the same time from two points 300 miles apart. If the rate of the passenger train exceeds the rate of the freight train by 15 miles per hour, and they meet after 4 hours, what are the rates of the passenger and train? 9) A man having ten hours at his disposal made an excursion, riding out at the rate of 10 miles an hour and returning on foot at the rate of 3 miles an hour. Find the distance he rode. 10) A boy rides away from home in an automobile at the rate of 28 miles an hour and walks back at the rate of 4 miles an hour. The round trip requires 2 hours. How far does he ride in the automobile? 11) A family drove to a resort at an average speed of 30 mph and later returned over the same road at an average speed of 50 mph. Find the distance to the resort if the total driving time was 8 hours. 12) Annie, who travels 4 miles an hour starts from a certain place 2 hours in advance of Brandie, who travels 5 miles an hour in the same direction. How many hours must Brandie travel to overtake Annie? 146

153 Chapter 4 13) A motorboat leaves a harbor and travels at an average speed of 8 mph toward a small island. Two hours later a cabin cruiser leaves the same harbor and travels at an average speed of 16 mph toward the same island. In how many hours after the cabin cruiser leaves will the cabin cruiser be alongside the motorboat? 14) A car traveling at 48 mph overtakes a cyclist who, riding at 12 mph, has had a 3-hour head start. How far from the starting point does the car overtake the cyclist? 15) Two men are traveling in opposite directions at the rate of 20 and 30 miles per hour at the same time and from the same place. In how many hours will they be 300 miles apart? 16) A motor boat leaves a harbor and travels at an average speed of 18 mph to an island. The average speed on the return trip was 12 mph. How far was the island from the harbor if the total trip took 5 hours? 17) A jet plane traveling at 570 mph overtakes a propeller-driven plane that has had a 2-hour head start. Thepropeller-drivenplaneistravelingat190mph. How far from the starting point does the jet overtake the propeller-driven plane? 18) As part of flight training, a student pilot was required to fly to an airport and then return. The average speed on the way to the airport was 100 mph, and the average speed returning was 150 mph. Find the distance between the two airports if the total flight time was 5 hours. 19) A car traveling at 56 mph overtakes a cyclist who, riding at 14 mph, has had a 3-hour head start. How far from the starting point does the car overtake the cyclist? 20) A bus traveling at a rate of 60 mph overtakes a car traveling at a rate of 45 mph. If the car had a 1- hour head start, how far from the starting point does the bus overtake the car? 21) A truck leaves a depot at 11a.m. and travels at a speed of 45 mph. At noon, a van leaves the same place and travels the same route at a speed of 65 mph. At what time does the van overtake the truck? 22) Three campers left their campsite by canoe and paddled downstream at an average rate of 10 mph. They then turned around and paddled back upstream at an average rate of 5 mph to return to their campsite. How long did it take the campers to canoe downstream if the total trip took 1 hour? 147

154 Chapter 4 SECTION 4.6 MIXTURE PROBLEMS The last common type of word problems in this chapter is mixture problems. Mixture problems are problems where we mix two different categories resulting in one new category. E.g., we can mix cashews with almonds and create a mix of nuts, or we can mix alcohol with water and create a new solution, etc. We will use a common formula to model the word problems called AAAAAA: AAAAAAAAAAAA CCCCCCCCCCCCCCCCCCCCCCCCCC oooo CCCCCCCC = TTTTTTTTTT oooo FFiiiiiiii Mixture problems Known starting amount (Duration 5:00) Mixture table: The equation always comes from the Example: a) A storeowner wants to mix chocolate and nuts b) You need a 55% alcohol solution. On hand, you to make a new candy. How many pounds of have 600 ml of 10% alcohol mixture. How chocolate costing $8.50/lb. should be mixed much of the 95% mixture should you add to with 25 pounds of nuts that cost $2.50/ lb. to obtain your desired solution? make a mixture worth $4.33/lb.? YOU TRY a) A chemist has 70 ml of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane? b) A café sells a two-origin coffee for $2.50. The café has a single-origin coffee that costs $3.00 for 40 ml. How much of a second coffee that costs $1.50 should the café mix with the first to create the two-origin coffee? 148

155 EXCERSISE 1) How much antifreeze should be added to 5 quarts of a 30% mixture of antifreeze to make a solution that is 50% antifreeze? 2) A tank contains 8000 liters of a solution that is 40% acid. How much water should be added to make a solution that is 30% acid? Note that water has 0% of acid. Chapter 4 3) Of 12 pounds of salt water, 10% is salt. In another mixture, 3% is salt. How many pounds of the second should be added to the first to get a mixture of 5% salt? 4) How many pounds of a 4% solution of borax must be added to 24 pounds of a 12% solution of borax to obtain a 10% solution of borax? 5) A 100 lb. bag of animal feed is 40% oats. How many pounds of oats must be added to this feed to produce a mixture which is 50% oats? 6) A 20-ounce alloy of platinum that costs $220 per ounce is mixed with an alloy that costs $400 per ounce. How many ounces of the $400 alloy should be used to make an alloy that costs $300 per ounce? 7) How many pounds of tea that cost $4.20 per pound must be mixed with 12 pounds of tea that cost $2.25 per pound to make a mixture that costs $3.40 per pound? 8) How many liters of a solvent that costs $80 per liter must be mixed with 6 liters of a solvent that costs $25 per liter to make a solvent that costs $36 per liter? 9) How many kilograms of hard candy that cost $7.50 per kilogram must be mixed with 24 kilograms of jelly beans that cost $3.25 per kilogram to make a mixture that sells for $4.50 per kilogram? 10) How many kilograms of soil supplement that costs $7 per kilogram must be mixed with 20 kilograms of aluminum nitrate that costs $3.50 per kilogram to make a fertilizer that costs $4.50 per kilogram? 11) How many pounds of Lima beans that costs 90 per pound must be mixed with 16 pounds of corn that costs 50 cents per pound to make a mixture of vegetables that costs 65 cents per pound? 12) How many liters of a blue dye that costs $1.60 per liter must be mixed with 18 liters of anil concentration that costs $2.50 per liter to make a mixture that costs $1.90 per liter? 13) How many ounces of dried apricots must be added to 18 ounces of a snack mix that contains 20% dried apricots to make a mixture that is 25% dried apricots? 14) How many pounds of coffee that is 40% Java beans must be mixed with 80 pounds of coffee that is 30% Java beans to make a coffee blend that is 32% Java beans? 15) The manager of a garden shop mixes grass seed that is 60% rye grass with 70 pounds of grass seeds that is 80% rye grass to make a mixture that is 74% rye grass. How much of the 60% mixture is used? 149

156 CHAPTER REVIEW KEY TERMS AND CONCEPTS Chapter 4 Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson. Consecutive integers Mark-up formula Discount formula Sum of Angles in a Triangle Perimeter of a rectangle formula AVT formula Simple Interest formula Distance formula Mixture formula 150

157 Chapter 5 CHAPTER 5: GRAPHING LINEAR EQUATIONS Chapter Objectives By the end of this chapter, students should be able to: Find the slope of a line from two points or a graph Find the equation of a line from its graph, the standard form, two given points Obtain equations of parallel and perpendicular lines Contents CHAPTER 5: GRAPHING LINEAR EQUATIONS SECTION 5.1 GRAPHING AND SLOPE A. POINTS AND LINES B. OBTAINING THE SLOPE OF A LINE FROM ITS GRAPH C. OBTAINING THE SLOPE OF A LINE FROM TWO POINTS EXERCISES SECTION 5.2 EQUATIONS OF LINES A. THE SLOPE-INTERCEPT FORMULA B. LINES IN SLOPE-INTERCEPT FORM C. GRAPHING LINES D. VERTICAL AND HORIZONTAL LINES E. POINT-SLOPE FORMULA F. OBTAINING A LINE GIVEN TWO POINTS EXERCISES SECTION 5.3 PARALLEL AND PERPENDICULAR LINES A. THE SLOPE OF PARALLEL AND PERPENDICULAR LINES B. OBTAIN EQUATIONS FOR PARALLEL AND PERPENDICULAR LINES EXERCISES CHAPTER REVIEW

158 SECTION 5.1 GRAPHING AND SLOPE Chapter 5 A. POINTS AND LINES In this chapter, we will begin looking at the relationship between two variables. Typically one variable is considered to be the INPUT, and the other is called the OUTPUT. The input is the value that is considered first, and the output is the value that corresponds to or is matched with the input. We write the input and its corresponding output as "(iiiiiiiiii, oooooooooooo). This is known as an ordered pair. For example, Input Output Ordered Pairs 4 3 (4, 3) 5 8 (5, 8) In an ordered pair, order matters. Let us take a look at the ordered pair (4,3). Since 4 appears first in this ordered pair, we know that 4 is the input. Likewise, since 3 appears second, we know that 3 is the output that belongs to 4. We can also refer to these numbers as coordinates. To plot ordered pairs we use the Cartesian plane. The Cartesian plane is made up of a horizontal real number line (which we call the xx-axis) and a vertical real number line (which we call the yy-axis). The vertical and horizontal lines intersect at the point (0,0), which is called the origin. The Cartesian plane is divided into four quadrants. Quadrant II Quadrant I Quadrant Quadrant III -2 IV yy -axis xx -axis To plot the ordered pair (4,3) we will look at the first coordinate, 4. We start at the origin and move to the right (the positive direction) by four units. Looking at the second coordinate, 3, we will then go up (in the positive direction) by three units. This is the point (4,3). 5 3 (4,3)

159 Chapter 5 A line is made up of an infinite number of points. To draw a line, however, we only need two points. What a line represents are the solutions to a linear equation. An example of a linear equation is yy = 2xx + 1 where xx is the input, and yy is the output. If we want to graph a linear equation, then we will need to make a table of inputs and outputs. Let us graph the linear equation above. For the table we are creating, we are allowed to pick any inputs we want. One person can pick the input 1 and another can pick the input 1,000. There is no wrong input you can choose for a linear equation, but we would like to keep things as simple as possible. Let us choose the following. Input (xx value) Output (yy value) 0? 1? 2? To find the corresponding outputs to the inputs we have chosen, we plug in one xx value into the linear equation and solve for yy. Let us find all the outputs: Filling in our chart For xx = 0: yy = 2(0) + 1 yy = 1 For xx = 1: yy = 2(1) + 1 yy = yy = 3 For xx = 2: yy = 2( 2) + 1 yy = yy = 3 Input (xx value) Output (yy value) Plotting these ordered pairs allows us to draw the line for the linear equation yy = 2xx

160 Chapter 5 Two important points worth mentioning are the xx and yy intercepts of the line. The xx-intercept of a line is the point (xx, 0), that is, the point where the line crosses the xx-axis. The yy-intercept of a line is the point (0, yy), that is, the point where the line crosses the yy-axis. Below are some examples of xx and yy intercepts. The cross is indicated by an x. x-intercept y-intercept Points and lines (Duration 2:57) stop at 2:57 The positive numbers on the xx-axis are located in what direction? The negative numbers on the xx-axis are located in what direction? The positive numbers on the yy-axis are located in what direction? The negative numbers on the yy-axis are located in what direction? We give to points on the xxxx-plane using these two number lines. First we give direction to the point going to, then we give direction to the point going up. Example: Graph the points. ( 2, 3), (4, 1), ( 2, 4), (0, 3) and ( 1,0) YOU TRY 154

161 Plot and label the points. Chapter 5 a) ( 4, 2) b) (3, 8) c) (0, 5) d) ( 6, 4) e) (5, 0) f) (2, 8) g) (0, 0) B. OBTAINING THE SLOPE OF A LINE FROM ITS GRAPH The slope of a line is the measure of the line s steepness. We denote the slope of a line with the symbol mm. To find the slope of a line from its graph we look at the change in yy over the change in xx, that is, mm = cchaaaaaaaa iiii yy cchaaaaaaaa iiii xx = rrrrrrrr rrrrrr In order to determine the rise and run of a graph, let us look at an example. Let us graph the linear equation yy = xx + 1 rise To find the rise we start at a well-defined point. In our graph above we started at ( 2, 1). Then locate a second well-defined point, in our case above we let that second point be (2, 3). Now, starting at our initial point we rise up four units until we get to the exact same level as the second point. This is shown as a dotted vertical line above. Next, we move towards the second point which is four units to the right. This is shown as a dotted horizontal line above. Since we rose up by four units, we say that the rise is 4. run Since we ran to the right by four units, we say that the run is 4. Thus mm = rrrrrrrr rrrrrr = 4 4 = 1 So mm =

162 NOTE: If the slope is positive, then the slope will be rising from left to right. If the slope is negative, then the slope will be declining from left to right. Chapter 5 m is positive m is negative We will now look at two special lines: the vertical line and the horizontal line. A vertical line has the form xx = cc, where cc is a constant number. Here is an example of the vertical line xx = (2, 4) (2, -4) If we were to pick the two well-defined points to be (2, 4) and (2, 4), then the rise would have a value of 8. However, the run will have a value of 0 since we do not move to the right or left. Thus mm = rrrrrrrr rrrrrr = 8 = dddddddd nnnnnn eeeeeeeeee 0 Since we can t divide by 0, the slope of the line does not exist. A horizontal line has the form yy = cc, where cc is a constant number. Here is an example of the horizontal line yy = 2. (-3, 4) (3, 4) If we were to pick the two well-defined points to be ( 3, 4) and (3, 4), then the rise would have a value of 0 since we do not move up or down. The run, however, will have a value of

163 Thus mm = rrrrrrrr rrrrrr = 0 6 = 0 Since 0 divided by anything is 0, our slope does exist and is 0. Chapter 5 To summarize: The slope of a vertical line does not exist The slope of a horizontal line does exist and has a value of 0. Slope from two points (Duration 5:00) If we select points on a line we should be able to determine the. For example, if we are given the coordinates (3, 3) and (6, 5), we should be able to determine the. The slope of the two given coordinates is, therefore the yy-intercept is equal to. We these two pieces of information, the linear equation is. YOU TRY a) Find the slope of the line below. b) Find the slope of the line below. (-3, -7) (3, 5) (-3, 11) (3, -7)

164 Chapter 5 C. OBTAINING THE SLOPE OF A LINE FROM TWO POINTS In the previous chapter we found the slope of a line by its graph. Another way to find the slope of a line (if we weren t given its graph) is to look at any two points belonging to that line. Let us look at a modified definition of slope. mm = rrrrrrrr cchaaaaaaaa iiii yy = rrrrrr cchaaaaaaaa iiii xx = yy 2 yy 1 xx 2 xx 1 The last expression is what we are interested in. If we are given two points (xx 1, yy 1 ) and (xx 2, yy 2 ), then we just need to take the difference of the two yy values and divide them by the difference of their respective xx values. For example, if we have the points ( 1, 1) and (1, 4), then So mm = 3 2. Slope from two points (Duration 5:00) mm = yy 2 yy 1 xx 2 xx 1 = ( 1) = 3 2 Slope is calculated by. When we say rise over run we think of the rise as the change in. We think of the run as the change in. Follow the video and find the slope between ( 2, 5) and ( 17, 4). YOU TRY Find the slope between the given two points a) ( 4, 3) and (2, 9) b) ( 4, 1) and ( 4, 5) c) (4, 6) and (2, 1) d) (3, 1) and ( 2, 1) EXERCISES For problems 1-4 find the slope of the line. 158

165 1) 2) Chapter ) 4) For problems 5-16 find the slope of the line through each ordered-pair. 5) ( 16, 14), (11, 14) 6) ( 4, 14), (16, 8) 7) (12, 19), (6, 14) 8) ( 5, 7), (18, 14) 9) (1, 2), ( 6, 14) 10) (13, 2), (7, 7) 11) ( 16, 2), (15, 10) 12) (8, 11), ( 3, 13) 13) (11, 2), (1, 17) 14) ( 2, 10), ( 2, 15) 15) ( 18, 5), (14, 3) 16) (19, 15), (5, 11) For problems find the value of xx or yy so that the line through the points has the given slope. 17) (2, 6) aaaaaa (xx, 2); mm = ) ( 3, 2) aaaaaa (xx, 6); mm = ) (xx, 5) aaaaaa (8, 0); mm = ) (8, yy) aaaaaa ( 2, 4); mm = ) (2, 5) aaaaaa (3, yy); mm = 6 22) (6, 2) aaaaaa (xx, 6); mm =

166 SECTION 5.2 EQUATIONS OF LINES A. THE SLOPE-INTERCEPT FORMULA The slope-intercept form of a linear equation is given by Chapter 5 yy = mmmm + bb Where mm is the slope and bb is the yy intercept (recall that the yy-intercept is a point, so we really have (0, bb)). When finding the equation of a line we would like the final result to be in slope-intercept form. This not only makes it easier to solve for yy but it also gives us two important pieces of information: the slope and the yy-intercept of the line. Slope intercept equation (Duration 5:00) Give the equation of the line with a slope of 3 and a yy-intercept of 2. 4 YOU TRY a) Find the equation of the line with slope 2 and yy intercept (0, 3). b) Find the equation of the line B. LINES IN SLOPE-INTERCEPT FORM Just by looking at a linear equation that is in slope-intercept form gives us the slope and yy-intercept. With these two pieces of information we can readily graph its line. There will be times when the linear equation is not in slope-intercept form, and in those cases it could benefit us to manipulate the equation to where it is in slope-intercept form. Put in intercept form (Duration 4:08) 160

167 We can put a linear equation in slope-intercept form to help us identify the and. If the equation is not in this form, then we identify these key points of information. To put an equation in intercept form we. Follow the video and give the slope and yy-intercept of the graph Chapter 5 yy + 4 = 2 (xx 4) 3 YOU TRY a) Write the equation 3xx 9yy = 6 in slope-intercept form. Find the slope and the yy-intercept of the line. C. GRAPHING LINES Now that we know what information is given to us when a linear equation is in slope-intercept form and how to manipulate a linear equation to get it into slope-intercept form, let us use this form to start graphing lines. Graph (Duration 4:57) We can graph an equation by identifying the and. Once we have identified this key information we will start the graph at the and use the to change to the next point. Remember slope is over. Follow the video and put the linear equation 3xx 2yy = 2 in slope intercept form. Then graph your answer. 161

168 YOU TRY Chapter 5 a) Graph yy = 1 xx 2 by using the slope and yy-intercept. 3 b) Graph the equation xx + 2yy = 10 using the slope and yy-intercept. D. VERTICAL AND HORIZONTAL LINES Using the slope-intercept form when dealing with horizontal and vertical lines can help us understand them a little better. Below is an example of a horizontal line that is in slope-intercept form yy = 0xx + bb = bb Recall that a horizontal line has a slope of 0, hence the reason mm = 0. If we let xx = 1, then yy = bb. If we let xx = 3, then yy = bb. No matter what value we choose for xx, the yy value will always be the same. The equation for a horizontal line is therefore yy = bb or its yy-coordinate of the graph. Unfortunately it is impossible for us to represent a vertical line in slope-intercept form because a vertical line s slope does not exist. This tells us that there is no yy in its equation. To represent a vertical line as an equation, we will simply make xx equal to its xx-coordinate of the graph. The equation for a vertical line is therefore xx = cc, where cc is the intercept. 162

169 Chapter 5 Vertical and Horizontal (Duration 2:17) Vertical lines, when graphed, will always go through the. Vertical lines are always equals the. Horizontal lines, when graphed, will always go through the. Horizontal lines are always equals the. 1. Label the axes and graph yy=-2 2. Find the equation of the graph below YOU TRY a) Graph yy = 4. b) Graph xx = 4. E. POINT-SLOPE FORMULA The slope-intercept formula gives us two pieces of information that makes graphing relatively easy: the slope of the line and its yy-intercept. Unfortunately we may not be given the yy-intercept all the time. In these cases we can use the point-slope formula below 163

170 yy yy 1 = mm(xx xx 1 ) Chapter 5 where mm is the slope and (xx 1, yy 1 ) is a point on the line. Point slope (Duration 5:00) Give the equation of the line that passes through ( 3, 5) and has the slope of 2 5. YOU TRY a) Using the point-slope formula, write the equation of the line passing through the point (1, 2) with a slope of 1. Write your final answer is slope-intercept form. 2 b) Using the point-slope formula, write the equation of the line passing through the point ( 2, 4) with a slope of 2. Write your final answer is slope-intercept form. 5 F. OBTAINING A LINE GIVEN TWO POINTS Let us look at the slope-intercept form again. In the previous section we used the point-slope formula to write an equation of a line given the slope and a point (not necessarily the yy-intercept). Now we will look at another scenario: given only two points (xx 1, yy 1 ) and (xx 2, yy 2 ). If we are given two points (and no slope) then we can construct a linear equation in the following way: 1. We can use the two given points to find the slope using the slope formula. 2. Once we obtain the slope, we can then use the point-slope formula together with the slope and any of the two points that we were given to create the linear equation. 3. Once we have the equation we can then put it in slope-intercept form to help us graph. Given two points (Duration 5:00) One important fact is that to find the equation of a line we must have the. 164

171 Recall that the formula for slope is Chapter 5 mm = Find the equation of the line through (1, 4) and (3, 5) and give the answer in slope-intercept form. YOU TRY a) Find the equation of the line passing through the points (1, 2) and ( 1, 3) and write your final answer in slope-intercept form. b) Find the equation of the line passing through the points (3, 2) and (1, 5) and write your final answer in slope-intercept form. 165

172 Chapter 5 EXERCISES For problems 1-3 write the equation of the line in slope-intercept form given the slope and the yyintercept. 1) mm = 1 3, yy-intercept = 1 2) mm = 1 yy-intercept = -2 3) mm = 2 5, yy-intercept = 5 For problems 4-6 write the equation of the line in slope-intercept form given the graph. 4) 5) 6) For problems 7-15 write the equation of the line in slope-intercept form given the equation. 7) 2xx + yy = 1 8) xx = 8 9) yy 4 = 4(xx 1) 10) yy + 1 = 1 2 (xx 4) 11) 6xx 11yy = 70 12) xx 7yy = 42 13) yy 3 = 2 3 (xx + 3) 14) 0 = xx 4 15) xx 10yy = 3 For problems sketch the graph of each line. 16) yy = 6 5 xx 5 17) xx yy + 3 = 0 18) 3yy = 5xx ) yy = 3 2 xx 1 20) 4xx + 5 = 5yy 21) 3yy = xx For problems write the equation of the line in slope-intercept form given a point passing through the line and its slope. 22) (2, 2); mm = ) ( 4, 1); mm = ) (0 5); mm = ) ( 1, 4); mm = ) (2, 1); mm = ) (4, 3); mm = 2 28) (0, 2); mm = ) (2, 3); mm = uuuuuuuuuuuuuuuuuu 30) (2, 2); mm = 0 For problems write the equation of the line in slope-intercept form given two points on the line. 31) (5, 1) aaaaaa ( 3, 0) 32) (3, 5) aaaaaa ( 5, 3) 33) (1, 3) aaaaaa ( 3, 3) 34) ( 4, 1) aaaaaa (4, 4) 35) ( 5, 1) aaaaaa ( 1, 2) 36) (4, 1) aaaaaa (1, 4) 166

173 SECTION 5.3 PARALLEL AND PERPENDICULAR LINES A. THE SLOPE OF PARALLEL AND PERPENDICULAR LINES Parallel Lines Chapter 5 Lines ll 1 and ll 2 are said to be parallel to each other if their slopes are the same, that is, mm 1 = mm 2 where mm 1 is the slope of ll 1 and mm 2 is the slope of ll 2 Perpendicular Lines Lines ll 1 and ll 2 are said to be perpendicular to each other if they have negative reciprocal slopes. Slopes of parallel and perpendicular lines (Duration 5:23) Parallel lines are two or more lines in a plane that never. Perpendicular lines are two or more lines that intersect at a angle. Follow the video and find the slope of a line perpendicular to the line yy = 3xx + 2. YOU TRY a) Find the slope of a line parallel to 2yy 2xx = 10. b) Find the slope of a line that is perpendicular to 2yy 2xx =

174 Chapter 5 B. OBTAIN EQUATIONS FOR PARALLEL AND PERPENDICULAR LINES Once we have the slope for a line that is perpendicular or parallel to another line, it is possible to find the equation for the perpendicular or parallel line given a point that is on one of these lines. Equations for parallel and perpendicular lines (Duration 5:00) Parallel lines have the same and perpendicular lines have slopes. Once we know the slope and a point we can use the formula. Follow the video and find the equation of the line parallel to the line 2xx 5yy = 3 that goes through the point (5, 3). YOU TRY a) Find the equation of a line passing through (1, 2) and parallel to 2xx 3yy = 6. b) Find the equation of the line passing through (6, 9) and perpendicular to yy = 2xx + 1. Write your final answer in slope-intercept form. 168

175 EXERCISES For problems 1-3 find the slope of a line parallel to the given line. Chapter 5 1) yy = 4xx 5 2) 7xx + yy = 2 3) yy = 10 3 xx 5 For problems 4-6 find the slope of a line perpendicular to the given line. 4) xx = 3 5) yy = 4 6) 8xx 3yy = 9 For problems 7-16 find the equation of the line passing through the point and given the line to be parallel or perpendicular. Write your final answer in slope-intercept form. 7) (5, 2); parallel to yy = 7 5 xx + 4 8) (1, 2); perpendicular to xx + 2yy = 2 9) ( 1, 3); parallel to yy = 3xx ) (1, 3); perpendicular to xx + yy = 1 11) (1, 4); parallel to yy = 7 5 xx ) ( 3, 5); perpendicular to 3xx + 7yy = 0 13) (1, 1); parallel to yy = 3xx 1 14) ( 2, 5); perpendicular to yy 2xx = 0 15) (2, 5); parallel to xx = 0 16) (1, 1); perpendicular to xx + yy = 1 169

176 CHAPTER REVIEW KEY TERMS AND CONCEPTS Chapter 5 Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson. Input Output Cartesian plane Origin Ordered pair xx-intercept yy-intercept Slope of a vertical line Slope of a horizontal line Slope formula Slope-intercept form Point-slope form Slope of parallel lines Slope of perpendicular lines 170

177 CHAPTER 6: SYSTEMS OF TWO LINEAR EQUATIONS IN TWO VARIABLES Chapter 6 Contents CHAPTER 6: SYSTEMS OF TWO LINEAR EQUATIONS IN TWO VARIABLES SECTION 6.1: SYSTEM OF EQUATIONS: GRAPHING A. VERIFYING SOLUTIONS B. SOLVE A SYSTEM BY GRAPHING EXERCISES SECTION 6.2: SYSTEMS OF EQUATIONS: THE SUBSTITUTION METHOD A. THE SUBSTITUTION METHOD B. SUBSTITUTION: SPECIAL CASES EXERCISES SECTION 6.3: SYSTEM OF EQUATIONS: THE ADDITION METHOD A. THE ADDITION METHOD B. THE ADDITION METHOD WITH MULTIPLICATION C. MULTIPLYING TWO EQUATIONS D. ADDITION: SPECIAL CASES EXERCISES SECTION 6.4: APPLICATIONS WITH SYSTEMS OF EQUATIONS A. VALUE & INTEREST PROBLEMS B. MIXTURE PROBLEMS C. UNIFORM MOTION WITH UNKNOWN RATES EXERCISES CHAPTER REVEW

178 SECTION 6.1: SYSTEM OF EQUATIONS: GRAPHING Chapter 6 A. VERIFYING SOLUTIONS In chapter 2 we solved single variable linear equations. For example, to solve for xx given the linear equation xx + 3 = 4, we must isolate the variable xx. This is done by moving any term with an xx to the left of the equal sign and the rest of the terms to the right of the equal sign. Doing this gives us xx = 1. What if we wanted to solve for two variables? For example, if we wanted to solve for xx and yy in the equation xx + yy = 3 To solve for xx all we would need to do is isolate it. Doing so gives us xx = yy + 3 The problem here is that even when xx is by itself, we still don t know what xx is equal to. Having yy on the other side of the equal sign puts us in a tough spot since we have no idea what yy is. It turns out that in order to solve a linear equation of two variables, we will need another equation. In other words, to solve for two variables xx and yy, we will need two equations. When solving for more than one equation and one variable, we call the set of equations a system of equations. When dealing with a system of equations, we are looking for the values that make both equations true. If only one equation is true, then we have the wrong answer and must try again. A system of two linear equations in two variables is of the form aaaa + bbbb = cc dddd + eeee = ff Where aa, bb, cc, dd, ee, and ff are coefficients and xx and yy are variables. Given an ordered pair (xx, yy), we can check to see if this is a solution to a system by plugging the ordered pair into both equations and verifying that both are true. For example, to verify that the point (4, 1) is a solution to the system 1 xx + yy = xx + yy = 2 4 we will plug the point (4, 1) into the first equation. 1 (4) + (1) = = 3 3 = 3 Seeing that this equation is true, let s verify the next one. 3 (4) + (1) = = 2 2 = 2 Since both equations are true, we say the point (4, 1) is a solution to the system. Verifying solutions (Duration 2:18 ) To solve a system of equations we want to find the value of and the value of that satisfies equations. The point of intersection is the point that lies on lines 172

179 Example: a) Given the graph, identify the solution to the system of equations. Verify the solution. 7 Chapter YOU TRY a) Is the ordered-pair (2, 1) the solution to the system 3xx yy = 5 xx + yy = 3 B. SOLVE A SYSTEM BY GRAPHING One way to solve a system of linear equations is by graphing each linear equation on the same xxxx-plane. When this is done, one of three cases will arise: Case 1: Two Intersecting Lines If the two lines intersect at a single point, then there is one solution for the system: the point of intersection. Case 2: Parallel Lines If the two lines are parallel to each other (not touching), then there is no solution for the system. Case 3: Same Lines If one line is on top of the other line (equivalent lines), then there are infinitely many solutions for the system. Solve by graphing (Duration 3:44) 173

180 Chapter 6 When we talk about solving a system of equations what we re looking for is the combination of xx and yy that simultaneously make both equations. Another way of saying that is it s the point that lies on lines at the same time. Example: a) Solve the following system by graphing. 2xx + yy = 5 xx 3yy = 8 YOU TRY a) Solve the system by graphing. If possible, write the solution as an ordered pair. 6xx 3yy = 9 2xx + 2yy = 6 b) Solve the system by graphing. If possible, write the solution as an ordered pair. 3 xx + yy = xx + yy =

181 EXERCISES Solve each system by graphing. When possible, write the solution as an ordered pair. 1) 2) 3) Chapter 6 yy = 3 yy = xx 4 yy = 1 3 xx + 2 yy = 5 3 xx 4 xx + 3yy = 9 5xx + 3yy = 3 4) 5) 6) 2xx + 3yy = 6 2xx + yy = 2 2xx + yy = 2 xx + 3yy = 9 2xx yy = 1 0 = 2xx yy 3 7) 8) 9) yy + 7xx = 4 yy 3 + 7xx = 0 yy = 5 4 xx 2 yy = 1 4 xx + 2 yy = 2xx + 2 yy = xx 4 10) 11) 12) yy = 1 2 xx + 4 yy = 1 2 xx + 1 6xx + yy = 3 xx + yy = 2 xx + 2yy = 6 5xx 4yy = 16 13) 14) 15) 2yy + xx = 4 2 = xx yy 16 = xx 4yy 2xx = 4 4yy 5xx + 1 = yy yy + xx = 3 175

182 SECTION 6.2: SYSTEMS OF EQUATIONS: THE SUBSTITUTION METHOD Chapter 6 A. THE SUBSTITUTION METHOD In the previous section we saw that one way to solve a system of linear equations is to graph each equation on the same xxxx-plane. If the graph is not accurate, then it can be difficult to see the solution. In this section we will look at another method to solving a system of linear equations: the substitution method. Let us look at an example. 6xx 3yy = 9 (aa) 2xx + 2yy = 6 (bb) To the right of each equation is a label. This will make keeping track of our work much easier. The idea behind the substitution method is to solve one equation for one variable, then substitute this value into the second equation. Once this substitution is made, the second equation becomes a one variable equation. Let us walk through the example above. Step 1: Solving for a variable (any variable you want) Looking at the first equation (aa), let us solve for yy. 6xx 3yy = 9 3yy = 6xx 9 yy = 2xx + 3 So yy = 2xx + 3. We don t know what yy is since there is an xx in it, but that s okay. Step 2: Substitution Now let us take a look at the second equation (bb). Since we solved for yy in the first equation, we know what yy is equal to. We will use this value and plug it into the second equation (the variable we are plugging into the equation is in red) 2xx + 2yy = 6 2xx + 2(2xx + 3) = 6 Step 3: Simplify and solve Notice how the second equation changes with the substitution we performed above! Our two variable equation 2xx + 2yy = 6 just became the one variable equation 2xx + 2(2xx + 3) = 6. And we know how to solve one variable equations. 2xx + 2(2xx + 3) = 6 2xx + 4xx + 6 = 6 6xx + 6 = 6 6xx = 12 xx = 2 Hence xx = 2 Step 4: Finding the second value We are almost done. We have what xx is equal to and now we just need what yy is equal to. Looking back at Step 1 we actually have an idea of what yy is. Now that we have a value for xx, let us plug this value (shown in blue) into the Step 1 equation and simplify yy = 2xx + 3 yy = 2( 2) + 3 yy = yy = 1 Therefore, our solution to this system is xx = 2 and yy = 1. Since the solution represents the point of intersection, let us write our solution as a point: ( 2, 1). 176

183 Chapter 6 If there exists at least one solution to the system, then the system is called a consistent system. A consistent system with one unique solution (as shown above) is said to have an independent solution. A consistent system with more than one solution is said to have a dependent solution. If no solution exists to the system, then the system is called an inconsistent system. Substitute Expression (Duration 4:45) Just as we can replace a variable with a number, we can also replace it with an. Whenever we substitute it is important to remember. Example: a) Solve the system by substitution. yy = 5xx 3 xx 5yy = 11 b) Solve the system by substitution. 2xx 6yy = 24 xx = 5yy 22 Solve for a variable (Duration 5:00) To use substitution we may have to a lone variable. If there are several lone variables. Example: a) Solve the system by substitution. 6xx + 4yy = 14 xx 2yy = 13 b) Solve the system by substitution. 5xx + yy = 17 7xx + 8yy = 5 177

184 Chapter 6 YOU TRY a) Solve the system by substitution. 2xx 3yy = 7 yy = 3xx 7 b) Solve the system by substitution. 3xx + 2yy = 1 xx 5yy = 6 B. SUBSTITUTION: SPECIAL CASES There are two special cases that arise in a system of linear equations that we should look at: when we have parallel lines and when we have duplicate lines. Once a system is graphed it can be readily seen what case we have, however, when solving a system algebraically it can be difficult to tell what case we are dealing with. Infinitely many solutions (equivalent lines) Let us look at an example. We are still using the substitution method, so the steps we will use to arrive at our solution will mimic the ones used in part A of this section. Solve the system by substitution. yy + 4 = 3xx (aa) 2yy 6xx = 8 (bb) Step 1: Solving for a variable (any variable you want) Looking at the first equation (aa), let us solve for yy. yy + 4 = 3xx yy = 3xx 4 So yy = 3xx 4. Step 2: Substitution 2yy 6xx = 8 2(3xx 4) 6xx = 8 Step 3: Simplify and solve 6xx 8 6xx = 8 8 = 8 What just happened? The variable xx just got cancelled out and now we are left with no variable. However, we are left with a true statement: 8 = 8. This means that we have an infinitely many solutions. Graphically we have a line on top of another line (equivalent lines). 178

185 No solution (parallel lines) Let us look at another example. Solve the system by substitution. 6xx 3yy = 9 (aa) 2xx + yy = 5 (bb) Chapter 6 Step 1: Solving for a variable (any variable you want) Looking at the first equation (aa), let us solve for yy. 6xx 3yy = 9 3yy = 6xx 9 yy = 2xx + 3 So yy = 2xx + 3. Step 2: Substitution 2xx + yy = 5 2xx + (2xx + 3) = 5 Step 3: Simplify and solve 2xx + 2xx + 3 = 5 3 = 5 The variable xx got cancelled out again, however, we are now left with a false statement: 3 = 5. This means that we have no solution. Graphically we have two parallel lines Special cases (Duration 4:38) If the variables subtract out to zero then it means either there is or. Example: a) Solve the system by substitution. xx + 4yy = xx = 12yy b) Solve the system by substitution. 5xx + yy = 3 8 3yy = 15xx YOU TRY Solve the system by substitution 179

186 5xx 6yy = 14 2xx + 4yy = 12 Chapter 6 180

187 Chapter 6 EXERCISES Solve each system by substitution. Determine if each system is consistent, independent or dependent, or inconsistent. 1) 2) 3) yy = 2xx 9 yy = 2xx 1 yy = 3xx + 2 yy = 3xx + 8 yy = 6xx 6 3xx 3yy = 24 4) 5) 6) yy = 5 3xx + 4yy = 17 yy = 8xx + 19 xx + 6yy = 16 xx 5yy = 7 2xx + 7yy = 20 7) 8) 9) 6xx + yy = 20 3xx 3yy = 18 yy = xx + 5 yy = 2xx 4 yy = 3xx + 13 yy = 2xx 22 10) 11) 12) yy = 7xx 24 yy = 3xx xx 4yy = 8 yy = 6xx + 2 yy = xx + 4 3xx 4yy = 19 13) 14) 15) xx 2yy = 13 4xx + 2yy = 18 6xx + 4yy = 16 2xx + yy = 3 5xx 5yy = 20 2xx + yy = 7 16) 17) 18) xx + 5yy = 15 3xx + 2yy = 6 6xx + 6yy = 12 8xx 3yy = 16 2xx + yy = 7 5xx + 3yy =

188 SECTION 6.3: SYSTEM OF EQUATIONS: THE ADDITION METHOD A. THE ADDITION METHOD A third method to solving a system of linear equations is the addition method (also called the elimination method). Let us look at an example to see how the addition method is carried out. Given the system below, let us solve it by the addition method. Chapter 6 3xx 4yy = 8 5xx + 4yy = 24 Step 1: Lining up the variables To begin the process of the addition method we want to start by lining up the variables. Notice in the system above how the first term in each linear equation have the same variable, xx. Likewise how the second term in each linear equation have the same variable, yy. Finally, on the right side of the equal sign, we have the numbers with no variables. Step 2: Adding both equations Now that the linear equations are lined up nicely, let us add them together 3xx 4yy = 8 + 5xx + 4yy = 24 8xx + = 16 xx = 2 Notice that adding both equations together allows us to eliminate one variable and solve for the other simultaneously. So xx = 2 Step 3: Finding the second value Now that we have one value let us solve for the other. To get the second value all we need to do is plug the value we found into one of the initial linear equations. It doesn t matter which one we choose, so let us pick the first one. 3xx 4yy = 8 3( 2) 4yy = 8 6 4yy = 8 4yy = 14 yy = 14 4 yy = 7 2 Our solution is therefore xx = 2 and yy = 7. Since our solution represents a point of intersection, let 2 us write our solution as an ordered pair: 2, 7 2. Addition method (Duration 4:18) If there is no lone variable, it may be better to use. This method works by adding the and sides of the equations together. 182

189 Example: a) Solve the system using the addition method. 8xx 3yy = 12 2xx + 3yy = 6 Chapter 6 b) Solve the system using the addition method. 5xx + 9yy = 29 5xx 6yy = 11 YOU TRY a) Solve the system by using the addition method. 4xx + 2yy = 0 4xx 9yy = 28 B. THE ADDITION METHOD WITH MULTIPLICATION Sometimes the systems we are dealing with don t add together nicely, meaning that one of the variables is not readily able to cancel out. In cases like these we may multiply one of the equations by some constant, allowing us to create a scenario where we can cancel out one of the variables. For example, 6xx + 5yy = 22 2xx + 3yy = 2 (aa) (bb) If we were to add both of these equations together we would get 4xx + 8yy = 24.This result is pointless for us. All we did was end up with a third equation with both variables still in it. To eliminate a variable, we must first decide what variable we want to get rid of first. It doesn t matter which variable we choose to eliminate nor does it matter which equation we choose to modify, we just need to pick one. Let us choose to eliminate the variable xx. To start the elimination process we will multiply equation (bb) by 3. Equation (bb) turns into 3 (2xx + 3yy) = 2 3 6xx + 9yy = 6 Rewriting the system with the new (bb) equation 6xx + 5yy = 22 (aa) 6xx + 9yy = 6 (bb) 183

190 Chapter 6 Notice how the coefficients belonging to the xx variable in each linear equation is now the exact opposite of each other. This gives us the scenario we had in the previous section. Now all we need to do is follow our previous steps. Step 1: Lining up the variables This is done for us already. Step 2: Adding both equations 6xx + 5yy = xx + 9yy = 6 0xx + 14yy = 28 yy = 2 Step 3: Finding the second value Our solution is therefore ( 2, 2). 6xx + 5yy = 22 6xx + 5(2) = 22 6xx + 10 = 22 6xx = 12 xx = 2 Addition method with multiplication (Duration 4:58) Addition only works if one of the variables have. To get opposites we can multiply of an equation to get the values we want. Be sure when multiplying to have a and in front of a variable. Example: a) Solve the system using the addition method. 2xx 4yy = 4 4xx + 5yy = 21 b) Solve the system using the addition method. 5xx 3yy = 3 7xx + 12yy = 12 YOU TRY Solve the system by using the addition method. 184

191 xx 5yy = 28 xx + 4yy = 17 Chapter 6 C. MULTIPLYING TWO EQUATIONS In the previous section we multiplied one equation by some constant to change it to our liking. However, there are systems that require us to multiply both equations by some constant. Let us solve the system below by adding. 3xx + 6yy = 9 (aa) 2xx + 9yy = 26 (bb) To eliminate xx we will multiply both equations by a constant to obtain the LCM (3, 2) = 6 with opposite signs: We ll multiply equation (aa) by 2 and multiply equation (bb) by 3 This gives us 2 (3xx + 6yy) = (2xx + 9yy) = xx + 12yy = 18 6xx 27yy = 78 Notice how the coefficients of xx are the same but with opposite signs. This gives us a scenario we are familiar with. Let us solve the system. Step 1: Lining up the variables This is done for us already. Step 2: Adding both equations 6xx + 12yy = xx 27yy = 78 0xx 15yy = 60 yy = 4 Step 3: Finding the second value 6xx + 12yy = 18 6xx + 12( 4) = 18 6xx 48 = 18 6xx = 30 xx = 5 Our solution is therefore (5, 4). Multiplying two equations (Duration 4:58) 185

192 Sometimes we may have to multiply by something to get opposites. Chapter 6 The opposites we look for is the of both coefficients. Example: a) Solve the system using the addition method. 6xx + 4yy = 26 4xx 7yy = 13 b) Solve the system using the addition method. 3xx + 7yy = 2 10xx + 5yy = 30 YOU TRY Solve the system by using the addition method. 8xx 8yy = 8 10xx + 9yy = 1 D. ADDITION: SPECIAL CASES Like the substitution method, the addition method will also bring about true or false statements. These are the special cases that we have talked about before, but it is important to go over them once more using the addition method. Infinitely many solutions (equivalent lines) Solve the system by using the addition method 2xx 5yy = 3 6xx + 15yy = 9 Let s get rid of the xx variable. We ll multiply the top equation by

193 3 (2xx 5yy) = 3 3 6xx 15yy = 9 Chapter 6 Now that the xx variables match and are opposite each other, let us add them together. 6xx 15yy = 9 + 6xx + 15yy = 9 0xx + 0yy = 0 So we have 0 = 0. Since this is a true statement this implies that we have infinitely many solutions. No solution (parallel lines) Solve the system by using the addition method. 4xx 6yy = 8 6xx 9yy = 15 Let s get rid of the xx variable. We ll multiply the top equation by 6 and the bottom by 4. 6 (4xx 6yy) = (6xx 9yy) = 15 4 This gives us 24xx + 36yy = xx 36yy = 60 0xx + 0yy = 12 So we have 0 = 0. Since this is a false statement this implies that we have no solution. Special cases (Duration 4:21) If the variables subtract out to zero then it means either there is or. Example: a) Solve the system using the addition method. 2xx 4yy = 16 3xx 6yy = 20 b) Solve the system using the addition method. 10xx + 4yy = 6 25xx 10yy =

194 Chapter 6 YOU TRY a) Solve the system by using the addition method. xx 2yy = 7 xx + 2yy = 7 188

195 Chapter 6 EXERCISES Solve each system by addition. Determine if each system is consistent, independent or dependent, or inconsistent. 1) 2) 3) 9xx + 5yy = 22 9xx 5yy = 13 4xx 6yy = 10 4xx 6yy = 14 2xx yy = 5 5xx + 2yy = 28 4) 5) 6) 2xx + 4yy = 24 4xx 12yy = 8 5xx + 10yy = 20 6xx 5yy = 3 9xx 2yy = 18 5xx 7yy = 10 7) 8) 9) 7xx + 5yy = 8 3xx 3yy = 12 9yy = 7 xx 18yy + 4xx = 26 7xx + yy = 10 9xx yy = 22 10) 11) 12) 5xx 5yy = 15 5xx 5yy = 15 10xx 5yy = 0 10xx 10yy = 30 xx + 3yy = 1 10xx + 6yy = 10 13) 14) 15) 6xx + 4yy = 4 3xx yy = 26 5xx + 4yy = 4 7xx 10yy = 10 4xx 5yy = 12 10xx + 6yy = 30 16) 17) 7xx + 10yy = 13 4xx + 9yy = yy = 12xx xx yy = 0 189

196 SECTION 6.4: APPLICATIONS WITH SYSTEMS OF EQUATIONS Chapter 6 A. VALUE & INTEREST PROBLEMS Here we will look at applications that we ve gone over before, but using a different approach to solve. Now that we know how to solve a system of linear equations, we will solve the applications in this sections using two variables (instead of one like in previous sections). Example. There were 41 tickets sold for an event. Tickets for children cost $1.50 and tickets for adults cost $2.00. Total receipts for the event were $ How many of each type of ticket were sold? To solve this problem we ll make a table to help organize our information. Amount Value (in $) Total Value Adult tickets aa $2 2aa Children tickets cc $ cc Total 41 $73.50 We need to find out how many children tickets were sold and how many adult tickets were sold. We will let cc represent children tickets and let aa represent adult the tickets. It is important to understand that each equation we make represents something from the problem. Our first equation will represent the amount of tickets. The total amount of tickets sold is 41, which includes children cc, and adults aa. This leads us to conclude that our first equation is cc + aa = 41 We will make the next equation represent the total value of the purchase. If one child ticket was sold, then the purchase will be $1.50. If two were sold the purchase would be $ If three were sold the purchase would be $ Since we have no idea how many children tickets were sold, we will multiply by cc. Likewise for adults. Remember, that the total purchase for children and adult tickets is given to be $ Our second equation is then 1.5cc + 2aa = Our system is cc + aa = cc + 2aa = To solve this system let s multiply the first equation by 2. Now adding this to the second equation 2 (cc + aa) = cc 2aa = 82 2cc 2aa = cc + 2aa = cc = 8.50 cc = 17 So the number of children tickets sold is cc = 17. To find the number of adult tickets, let us plug in the value we just got into the first equation. cc + aa = aa = 41 aa = 24 The number of adult tickets sold is aa = 24. In conclusion, the number of adult tickets sold is 24 and the number of children tickets sold is

197 Word Problem 1 (Duration 4:28) Chapter 6 Example: a) If 105 people attended a concert and tickets for adults costs $2.50 while tickets for children cost $1.75 and total receipts for the concert were $228, how many children and how many adults went to the concert? YOU TRY Aaron invests $9,700 in two different accounts. The first account paid 7%, the second account paid 11% in interest. At the end of the first year he had earned $863 in interest. How much was in each account? 191

198 Chapter 6 B. MIXTURE PROBLEMS Example: A farmer has two types of milk, one that is 24% butterfat and another which is 18% butterfat. How much of each should he use to end up with 42 gallons of 20% butterfat? Organizing the information in a table: Amount Concentration Total Butterfat 2222% butterfat xx xx 1111% butterfat yy yy 2222%butterfat (42) Using our chart to set up the equation as before, we get Multiplying the top equation by 0.18 we get Adding the equation together gives us xx + yy = xx yy = (xx + yy) = xx 0.18yy = xx 0.18yy = xx = xx = 0.84 xx = 14 The amount of 24% butterfat is therefore 14 gallons. This implies that the amount of 18% butterfat is 28 gallons. Word problem 2 (Duration 3:23) Example: a) A chemist needs to create 100mL of 38% acid solution. On hand she has a 20% acid solution and a 50% acid solution. How many ml of each solution should she use? 192

199 YOU TRY Chapter 6 A solution of pure antifreeze is mixed with water to make a 65% antifreeze solution. How much of each should be used to make 70 liters? C. UNIFORM MOTION WITH UNKNOWN RATES Example: Turkey the Pigeon travels the same distance of 72 miles in 4 hours against the wind as it does traveling 3 hours with the wind in the local skies. What is the rate of Turkey the Pigeon in still air and the rate of the wind? First we ll make a table to organize the given information and then create an equation. Let rr represent the rate of Turkey the Pigeon and ww represent the rate of the wind. Rate Time distance With the wind rr + ww 3 3(rr + ww) Against the wind rr ww 4 4(rr ww) Turkey travels a distance of 72 miles with the wind and against it. The equations are then 3(rr + ww) = 72 4(rr ww) = 72 To make things easier for us let us divide the first equation by 3 and the second by 4. Doing so gives us rr + ww = 24 rr ww = 18 Adding both equations together we get 2rr = 42 which implies that rr = 21. Hence ww =

200 Word problem 3 (Duration 5:43) Chapter 6 Example: a) An airplane travels 1,200 miles in 4 hours with the wind. The same trip takes 5 hours against the wind. What is the speed of the plane in still air and what is the wind speed? YOU TRY A boat travels upstream for 156 miles in 3 hours and returns in 2 hours traveling downstream in a local stream of water. What is the rate of the boat in still water and the rate of the current? 194

201 EXERCISES Chapter 6 1) There were 429 people at a play. Admission was $1 each for adults and 75 cents each for children. The receipts were $ How many adults and how many children attended? 2) There were 200 tickets sold for a women s basketball game. Tickets for students were 50 cents each and for adults 75 cents each. The total amount of money collected was $ How many of each type of ticket was sold. 3) A total of $27,000 is invested, part of it at 12% and the rest at 13%. The total interest after one year is $3385. How much was invested at each rate? 4) Millicent earned $435 last year in interest. If $3,000 was invested at a certain rate of return and $4,500 was invested in a fund with a rate that was 2% lower, find the two rates of interest. 5) A syrup manufacturer has some pure maple syrup and some which is 85% maple syrup. How many liters of each should be mixed to make 150 liters which is 96% maple syrup. 6) A goldsmith combined an alloy that costs $4.30 per ounce with an alloy that costs $1.80 per ounce. How many ounces of each were used to make a mixture of 200 ounces costing $2.50 per ounce? 7) A boat travels upstream for 336 miles in 4 hours and returns in 3 hours traveling downstream in a local stream of water. What is the rate of the boat in still water and the rate of the current. 195

202 CHAPTER REVEW KEY TERMS AND CONCEPTS Chapter 6 Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson. System of equations System of equations with one solution System of equations with no solution System of equations with infinitely many solutions Consistent system Inconsistent system 196

203 CHAPTER 7: INTRODUCTION TO FUNCTIONS Chapter objectives By the end of this chapter, students will learn: Determine whether a relation is a function Use function notation Use the vertical test Find the domain and range Contents CHAPTER 7: INTRODUCTION TO FUNCTIONS SECTION 7.1: RELATIONS AND FUNCTIONS A. RELATIONS AND FUNCTIONS B. VERTICAL LINE TEST C. FUNCTION NOTATION EXERCISES SECTION 7.2: DOMAIN AND RANGE A. DOMAIN AND RANGE EXERCISES CHAPTER REVIEW

204 SECTION 7.1: RELATIONS AND FUNCTIONS A. RELATIONS AND FUNCTIONS A relation is any set of ordered pairs. We don t have to talk about numbers when we think about relations. For example, we could be talking about names and favorite colors: Names (Input) Stella Bruno Victoria Favorite Color (Output) Blue Brown Green and Pink Let us express each name and respective color as an ordered pair: (SSSSSSSSSSSS, BBBBBBBB) (BBBBBBBBBB, BBBBBBBBBB) (VVVVVVVVVVVVVVVV, GGGGGGGGGG) (VVVVVVVVVVVVVVVV, PPPPPPPP) The first coordinate (called the input) represents the name of the person and the second coordinate (called the output) represents what color they like. The name VVVVVVVVVVVVVVVV is written twice since she likes two colors. A relation just compares two sets of information. Here it was names and favorite colors, but it could have easily been anything else. A function is a relation in which each input value is paired with exactly one unique output value. The example above is a relation but not a function. Why? Stella and Bruno like one color but Victoria likes two colors. In other words, one of the inputs (Victoria) has two outputs (Green and Pink). A function may only have one unique output. An example of a function can be a person and their respective birthday month. Names (Input) Luna Dante Clara Birthday (Output) June December February Let us express each name and birthday month as an ordered pair: (LLLLLLLL, JJJJJJJJ) (DDDDDDDDDD, DDDDDDDDDDDDDDDD) (CCCCCCCCCC, FFFFFFFFFFFFFFFF) We can see that every person has only one birthday, hence this relationship is a function. Now, it is possible that two people have the same birthday. An example is given below Names Luna Dante Clara Spike Birthday June December February June Writing this information in ordered pairs: (LLLLLLLL, JJJJJJJJ) (DDDDDDDDDD, DDDDDDDDDDDDDDDD) (CCCCCCCCCC, FFFFFFFFFFFFFFFF) (SSSSSSSSSS, JJJJJJJJ) 198

205 We see that Luna and Spike both have a birthday in June. However, the inputs (names) still have one unique output (birthday). Luna belongs to June, Dante belongs to December, and so on. Not one input goes to two different outputs like in the first example with Victoria. Hence, this is a function. In general, we say the output depends on the input. We call the output variable the dependent variable and the input variable the independent variable. If the relation is a function, then we say that the output is a function of the input. Introduction to relations (Duration 11:01) A is any set of ordered pairs. A is a relation in which input value is paired with output value. One way to represent the relationship between the input and output variables in a relation or function is by means of a table of values. Example 1: Which of the following tables represent functions? a) b) Input Output Input Output Yes No Yes No c) Input Yes Output No Relations and functions can also be represented as a set of points or ordered pairs. Example 2: Which of the following sets of ordered pairs represent a function? A = {(0, 2), (1, 4), ( 3, 3), (5, 0)} B = {( 4, 0), (2, 3), (2, 5)} C = {( 5, 1), (2, 1), ( 3, 1), (0, 1)} D = {(3, 4), (3, 2), (0, 1), (2, 1)} E = {(1, 3)} Input Input Output Output 199

206 Example 3: On the graphs below, plot the points for A, B, C, and D from example 2, then circle the problem points: B. VERTICAL LINE TEST One way to tell if our set of information represents a function is to look at its graph. If we can draw vertical lines and the lines hit the graph only once, then we have a function. If they hit more than once, then we do not have a fuction. If all lines intersect the graph of a relation in, the relation is also a. output for input. If any line intersects the graph of a relation at, the relation fails the test and is a function. exists for (or all) inputs value (s). Example 4: Use the Vertical Line Test to determine which of the following graphs are functions. Behavior of graphs Increasing Decreasing Constant 200

207 Dependent and independent variables In general, we say that the output depends on the input. Output variable = Input Variable = If the relation is a function, then we say that the output the input. YOU TRY Is it a function? Circle Yes or No for each of the following a) Yes or No b) Yes or No c) Yes or No Input Output {(2, 3), ( 5,2), ( 3,1)} Use the Vertical Line Test to determine if the graph is a function. d) e)

208 C. FUNCTION NOTATION If a relation is a function, we say that the output is a function of the input. In function notation this is written as: ff(iiiiiiiiii) = oooooooooooo For example, if yy is a function of xx, then we can write which is read ff of xx is yy. Why is it necessary to use function notation? The necessity stems from using multiple equations. Function notation allows one to easily decipher between the equations. Suppose Joseph, Lacy, Kevin, and Alfred all went to the theme park together and chose to pay $2.00 for each ride. Each person would have the same equation CCCCCCCC = 2 rrrrrrrrrr or in short CC = 2rr. Without asking each friend, we could not tell which equation belonged to whom. By substituting function notation for the dependent variable, it is easy to tell which function belongs to whom. By using function notation, it will be much easier to graph multiple lines. We can use function notation to write functions to represent the total each friend spent at the park. Joseph s total: JJ(rr) = 2rr Lacy s total: LL(rr) = 2rr Kevin s total: KK(rr) = 2rr Alfred s total: AA(rr) = 2rr Function notation allows you to easily see the input value for the independent variable inside the parentheses. NOTE: It is important to point out that ff(xx) ff (xx) Function Notation (Duration 7:48) If a relation is a function, we say that. FFFFFFFFFFFFFFFF NNNNNNNNNNNNNNNN: ff ( ) = Example: If yy is a function of xx, then we can write. 202

209 Example 1: The function VV(mm) represents value of an investment (in thousands of dollars) after mm months. Explain the meaning of VV(36) = 17.4 Ordered pairs Example 2: (iiiiiiiiii, oooooooooooo) ff(iiiiiiiiii) = oooooooooooo Ordered Pair Function Notation (iiiiiiiiii, oooooooooooo) ff(iiiiiiiiii) = (oooooooooooo) (2, 3) ff(2) = 3 ( 4, 6) (, ) ff( ) = ff(5) = 1 Example 3: Consider the function: ff = {(2, 4), (5, 7), (8, 0), (11, 23)} ff(5) = ff( ) = 0 Table of values Example 4: The function BB(tt) is defined by the table below tt BB(tt) BB(12) = BB(tt) = 18 when tt = Graph gg(2) = Ordered pair: gg( ) = 2 Ordered pair: 203

210 gg(0) = Ordered pair: gg( ) = 1 Ordered pair: YOU TRY a) Complete the table below. Ordered Pair Function Notation (8, 1) ff ( ) = (, ) ff (0) = 11 b) The function k(x) is defined by the following table xx kk(xx) kk(2) = Ordered Pair: kk(xx) = 1 when xx = Ordered Pair: c) At an ice cream factory, the total cost production is a function of the number of gallons of ice cream produced. The function CC(gg), gives the cost, in dollars, to produce g gallons of ice cream. Explain the meaning of CC(580) = 126 in terms of ice cream production. 204

211 EXERCISES Which of the following relations are functions? 1) Input Output ) Input Output ) Input Output Which of the following relations are functions? 4) {(1, 8), (5, 2), (1, 6), (7, 3)} 5) {(2, 4, ), (6, 4), (0, 0), (5, 0)} 6) {(1, 1), (2, 2), (3, 3), (4, 4)} Which of the following graphs are functions? 7) 8) The function rr(xx) is defined by the following table of values. Use the table to fill in the blanks. xx rr(xx) ) rr(9)= 10) rr(3) = 11) rr( ) = 1 12) rr( ) = 3 Consider the function gg = {(22, 55), (00, 66), (55, 88), ( 33, 77)}, fill in the blanks. 13) gg(0)= 14) gg(5) = 15) gg( ) = 7 16) gg( ) = 5 205

212 SECTION 7.2: DOMAIN AND RANGE A. DOMAIN AND RANGE The domain of a function is the set of all possible values for the input variable. The range of a function is the set of all possible values for the output variable. For example, given the set BB = {(2, 4), (5, 7), (8, 0), (11, 23)}, let us find the domain and range. The domain of BB is the set of all possible values for the input, in other words the first coordinate of each ordered pair: {2, 5, 8, 11}. The range of BB is the set of all possible values for the output, in other words the second coordinate of each ordered pair: { 4, 0, 7, 23}. NOTE: It is standard to write the domain and range from smallest to largest. If a number is repeated, then we write the number once and ignore the rest. Introduction to domain and range (Duration 10:39) The of a function is the set of all possible values for the ( ) variable. The of a function is the set of all possible values for the ( ) variable. Example 1: Consider the function below xx kk(xx) Input values Domain: { } Output values: Example 2: Consider the function: BB = {(2, 4), (5, 7), (8, 0), (11, 23)} Input values Domain: { } Output values: Range: { } Range: { } Example 3: Consider the graph of ff(xx) shown below Domain: xx Range: ff(xx) 206

213 Example 4: Determine the Domain and Range of each of the following graphs: Domain Domain Domain Range Range Range YOU TRY a) Consider the function below xx kk(xx) Input values Domain: Output values: Range: b) Consider the graph of ff(xx) shown below Domain: xx Range: ff(xx) 207

214 EXERCISES 1) The function rr(xx) is defined by the following table of values. Use the table below to fill in the blanks. xx rr(xx) a) The domain of rr(xx) is b) The range of rr(xx) is 2) Consider the function gg(xx) = {(2, 5), (0, 6), (5, 8), ( 3, 7)}, fill in the blanks. a) The domain of gg is b) The range of gg is 3) Given ff(4) = 8, ff(3) = 11, ff(0) = 6, fill in the blanks. a) The domain of ff is b) The range of ff is c) Write the function ff as a set of ordered pairs. 4) Use the graph below to fill in the blanks: a) Domain: b) Range: c) gg( 3) = d) gg(0) = e) gg(rr) = 4 when rr = f) gg(rr) = 0 when rr = 5) Use the graph below to fill in the blanks: a) Domain: b) Range: c) pp( 1) = d) pp(0) = e) pp(tt) = 5 when tt = f) pp(tt) = 3 when tt = 208

215 CHAPTER REVIEW KEY TERMS AND CONCEPTS Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson. Relation Function Vertical line test Dependent variable Independent variable Function notation Compare: Find ff(4) Find xx when ff(xx) = 4 Domain Range Domain Range 209

216 210

217 Chapter 8 CHAPTER 8: EXPONENTS AND POLYNOMIALS Chapter Objectives By the end of this chapter, students should be able to: Simplify exponential expressions with positive and/or negative exponents Multiply or divide expressions in scientific notation Evaluate polynomials for specific values Apply arithmetic operations to polynomials Apply special-product formulas to multiply polynomials Divide a polynomial by a monomial or by applying long division CHAPTER 8: EXPONENTS AND POLYNOMIALS SECTION 8.1: EXPONENTS RULES AND PROPERTIES A. PRODUCT RULE OF EXPONENTS B. QUOTIENT RULE OF EXPONENTS C. POWER RULE OF EXPONENTS D. ZERO AS AN EXPONENT E. NEGATIVE EXPONENTS F. PROPERTIES OF EXPONENTS EXERCISE SECTION 8.2 SCIENTIFIC NOTATION A. INTRODUCTION TO SCIENTIFIC NOTATION B. CONVERT NUMBERS TO SCIENTIFIC NOTATION C. CONVERT NUMBERS FROM SCIENTIFIC NOTATION TO STANDARD NOTATION D. MULTIPLY AND DIVIDE NUMBERS IN SCIENTIFIC NOTATION E. SCIENTIFIC NOTATION APPLICATIONS EXERCISE SECTION 8.3: POLYNOMIALS A. INTRODUCTION TO POLYNOMIALS B. EVALUATING POLYNOMIAL EXPRESSIONS C. ADD AND SUBTRACT POLYNOMIALS D. MULTIPLY POLYNOMIAL EXPRESSIONS E. SPECIAL PRODUCTS F. POLYNOMIAL DIVISION EXERCISE CHAPTER REVIEW

218 SECTION 8.1: EXPONENTS RULES AND PROPERTIES A. PRODUCT RULE OF EXPONENTS Chapter 8 Product rule of exponents (Duration 2:57) aa 3 aa 2 = (aa aa aa)(aa aa) = aa 5 Product rule: aa mm aa nn = aa mm+nn! Example 1: (2x 3 )(4x 2 )( 3x) = Example 2: (5a 3 b 7 )(2a 9 b 2 c 4 ) = Warning! The rule can only apply when you have the same base. YOU TRY Simplify: a) b) xx 1 xx 3 xx 2 c) (2xx 3 yy 5 zz)(5xxyy 2 zz 3 ) B. QUOTIENT RULE OF EXPONENTS Quotient rule of exponents (Duration 3:12) Quotient Rule: aamm = aamm nn aann aa 5 aa aa aa aa aa = = aa 2 aa3 aa aa aa Example 1: aa7 bb 2 aa 3 bb = Example 2: 8mm7 nn 4 6mm 5 nn = YOU TRY Simplify a) b) 5aa3 bb 5 cc 2 2aabb 3 cc c) 3xx 5 xx 3 yy 212

219 C. POWER RULE OF EXPONENTS Power rule of exponents (Duration 5:00) Chapter 8 (ab) 3 = = Power of a product: (aaaa) mm = aa mm bb mm aa bb 3 = = Power of a Quotient: aa bb mm = aamm bbmm, if b is not 0. (aa 2 ) 3 = = Power of a Power: (aa mm ) nn = aa mm nn Example 1: (5aa 4 bb) 3 Example 2: 5mm3 9nn 4 2 Warning! It is important to be careful to only use the power of a product rule with multiplication inside parenthesis. This property is not allowed for addition or subtraction, i.e., YOU TRY (aa + bb) mm aa mm + bb mm (aa bb) mm aa mm bb mm Simplify: a) xx3 yy 2 5 b) c) (xx 3 yyzz 2 ) 4 d) (4xx 2 yy 5 ) 3 e) aa3 bb cc 8 dd 5 2 f) 4xxxx 8zz 2 213

220 D. ZERO AS AN EXPONENT Zero as Exponent (Duration 3:51) Chapter 8 aa 3 aa 3= Zero Power Rule: aa 00 = 11 Example 1: (5xx 3 yyzz 5 ) 0 Example 2: (3xx 2 yy 0 )(5xx 0 yy 4 ) YOU TRY Simplify the expressions completely a) (3x 2 ) 0 b) 2mm0 nn 6 3nn 5 E. NEGATIVE EXPONENTS Negative Exponents (Duration 4:44) aa 3 aa 5 = = Negative Exponent Rule: aa mm = 11 When a and b are not 0. aa mm 1 = aamm aa mm aa mm bb = bb mm aa = bbmm aa mm Example 1: 7xx yyzz 4 Example 2: 2 5aa 4 Warning! It is important to note a negative exponent does not imply the expression is negative, only the reciprocal of the base. Hence, negative exponents imply reciprocals. YOU TRY a) xx b) aa 3 bb 2 cc 2dd 1 ee 4 214

221 Chapter 8 F. PROPERTIES OF EXPONENTS Putting all the rules together, we can simplify more complex expression containing exponents. Here we apply all the rules of exponents to simplify expressions. Exponent Rules Product aa mm aa nn = aa mm+nn Power of a Product (aaaa) mm = aa mm bb mm Negative Power aa mm = 11 aa mm Quotient aa mm = aamm nn aann Power of a Quotient aa bb mm = aamm bb mm Reciprocal of Negative Power 11 = aamm aa mm Power of Power (aa mm ) nn = aa mm nn Zero Power aa 00 = 11 Negative Power of a Quotient aa mm bb = bb mm aa = bbmm aa mm Properties of Exponents (Duration 5:00) Example 1: (4x 5 y 2 z) 2 (2xx 4 yy 2 zz 3 ) 4 Example 2: 2x2 y 3 4 x 4 y 6 2 (x 6 y 4 ) 2 YOU TRY Simplify and write your final answers in positive exponents. 4xx 5 yy 3 3xx 3 yy 2 a) 6xx 5 yy 3 b) 3aabb3 2 aabb 3 2aa 4 bb 0 215

222 EXERCISE Simplify. Be sure to follow the simplifying rules and write answers with positive exponents. Chapter 8 1) ) ) 3mm 4mmmm 4) 2mm 4 nn 2 4nnmm 2 5) (3 3 ) 4 6) (4 4 ) 2 7) (2uu 3 vv 2 ) 2 8) (2aa 4 ) 4 9) 10) xx 2 yy 4 xxyy 2 11) (xxxx) 3 12) ) ) 3nnmm2 3nn 15) 4xx 3 yy 4 3xxyy 3 16) xx 2 yy 4 4xxxx 17) 3xx 4xx 2 18) (uu 2 vv 2 2uu 4 ) 3 19) (xx 3 yy 4 2xx 2 yy 3 ) 2 20) 2xx(xx 4 yy 4 ) 4 21) 2xx 7 yy 5 3xx 3 yy 4xx 2 yy 3 22) (2xx)3 xx ) 2yy17 (2xx 2 yy 4 ) ) 2mmnn4 2mm 4 nn 4 3 mmnn 4 25) 28) 2xxyy 5 2xx 2 yy 3 2xxyy 4 yy 3 26) 2xx2 yy 2 zz 6 2zzxx 2 yy 2 (xx 2 zz 3 ) 2 27) 2bbaa 7 2bb 4 bbaa 2 3aa 3 bb 4 29) 2aa2 bb 2 aa 7 (bbaa 4 ) 2 30) 2yy (xx 0 yy 2 ) 4 yyxx 2 yy 4 2 2yy 4 31) 2aa 2 bb 2 aa 7 (bbaa 4 ) 2 32) nn3 nn 4 2 2mmmm 33) 2yy 3 xx 2 2 2xx 2 yy 4 xx 2 34) 2qq 3 pp 3 rr 4 2pp 3 (qqqqpp 3 ) 2 35) 2xx 4 yy 2 (2xxyy 3 ) 4 36) 2xx 3 yy 2 3xx 3 yy 3 3xx 0 37) uuvv 1 2uu 0 vv 4 2uuuu 38) 2aa2 bb 3 aa ) 2xxyy 2 4xx 3 yy 4 4xx 4 yy 4 4xx 40) 2bb 4 cc 2 2bb 3 cc 2 4 aa 2 bb 4 216

223 SECTION 8.2 SCIENTIFIC NOTATION A. INTRODUCTION TO SCIENTIFIC NOTATION One application of exponent properties is scientific notation. Scientific notation is used to represent really large or really small numbers, like the numbers that are too large or small to display on the calculator. Chapter 8 For example, the distance light travels per year in miles is a very large number (5,879,000,000,000) and the mass of a single hydrogen atom in grams is a very small number ( ). Basic operations, such as multiplication and division, with these numbers, would be quite cumbersome. However, the exponent properties allow us for simpler calculations. Introduction of scientific notation (Watch from 0:00 9:00) 10 0 = 10 1 = 10 2 = 10 3 = = Avogadro number: 602,200,000,000,000,000,000,000 = Definition of scientific notation (Duration 4:59) Standard Form (Standard Notation): Scientific Notation: b: b positive: b negative: Example: Convert to Scientific Notation a) 48,100,000,000 = b) = 217

224 Chapter 8 Definition Scientific notation is a notation for representing extremely large or small numbers in form of aa xx 10 bb where 1 < a < 10 and b is number of decimal places from the right or left we moved to obtain a. A few notes regarding scientific notation: b is the way we convert between scientific and standard notation. b represents the number of times we multiply by 10. (Recall, multiplying by 10 moves the decimal point of a number one place value.) We decide which direction to move the decimal (left or right) by remembering that in standard notation, positive exponents are numbers greater than ten and negative exponents are numbers less than one (but larger than zero). Case 1. If we move the decimal to the left with a number in standard notation, then b will be positive. Case 2. If we move the decimal to the right with a number in standard notation, then b will be negative. B. CONVERT NUMBERS TO SCIENTIFIC NOTATION Convert standard notation to scientific notation (Duration 1:40) Example: Convert to scientific notation = = YOU TRY Convert the following number to scientific notation a) 14,200 b) c) How long is a Light-Year? The light-year is a measure of distance, not time. It is the total distance that a beam of light, moving in a straight line, travels in one year is almost 6 trillion (6,000,000,000,000) miles. Express a light year in scientific notation. (Source: NASA Glenn Educational Programs Office 12/aerores.htm) C. CONVERT NUMBERS FROM SCIENTIFIC NOTATION TO STANDARD NOTATION To convert a number from scientific notation of the form aa xx 10 bb to standard notation, we can follow these rules of thumb. If b is positive, this means the original number was greater than 10, we move the decimal to the right b times. If b is negative, this means the original number was less than 1 (but greater than zero), we move the decimal to the left b times. 218

225 Convert scientific notation to standard notation (Duration 2:22) Chapter 8 Example: Rewrite in standard notation (decimal notation) a) b) YOU TRY Covert the following scientific notation to standard notation a) b) D. MULTIPLY AND DIVIDE NUMBERS IN SCIENTIFIC NOTATION Converting numbers between standard notation and scientific notation is important in understanding scientific notation and its purpose. Next, we multiply and divide numbers in scientific notation using the exponent properties. If the immediate result is not written in scientific notation, we will complete an additional step in writing the answer in scientific notation. Steps for multiplying and dividing numbers in scientific notation Step 1. Rewrite the factors as multiplying or dividing a-values and then multiplying or dividing 10 b values. Step 2. Multiply or divide the a values and apply the product or quotient rule of exponents to add or subtract the exponents, b, on the base 10s, respectively. Step 3. Be sure the result is in scientific notation. If not, then rewrite in scientific notation. Multiply and divide scientific notation (Duration 2:47) Multiply/ Divide the Use on the 10s Example: a) ( )( ) b) Multiply scientific notations with simplifying final answer step (Duration 3:47) Example: a) ( )( ) b) ( )( ) 219

226 Divide scientific notations with simplifying final answer step (Duration 3:44) Chapter a) b) YOU TRY Multiply or divide a) (2.1 xx 10 7 )(3.7 xx 10 5 ) b) 4.96 xx xx 10 3 c) (4.7 xx 10 3 )(6.1 xx 10 9 ) d) ( )( ) e) f) xx xx 10 7 E. SCIENTIFIC NOTATION APPLICATIONS Scientific notation application example 1 (Duration 2:36) Example 1: There were approximately 50,000 finishers of the 2015 New York City Marathon. Each finisher ran a distance of 26.1 miles. If you add together the total number miles ran by all the runners, how many times around the earth would the marathon runners have ran? Assume the circumference of the earth to be approximately 2.5 x 10 4 miles. Total distance = 220

227 Scientific notation application example 2 (Duration 3:24) Chapter 8 Example 2: If a computer can conduct 400 trillion operations per second, how long would it take the computer to perform 500 million operations? 400 trillion = 500 million = Number of Operations: Rate of Operations: YOU TRY a) It takes approximately 3.7 x 10 4 hours for the light on Proxima Centauri, the next closet star to our sun, to reach us from there. The speed of light is 6.71 x 10 8 miles per hours. What is the distance from there to earth? Given distance = rate x time. Express your answer in scientific notation By ESO/Pale Red Dot - CC BY 4.0, a) If the North Pole and the South Pole ice were to melt, the north polar ice would make essentially no contribution since it is float ice. However, the south polar ice would make a considerable contribution since it overlays the Antarctic land mass and is not float ice. If Antarctic ice melted, it would become approximately 1.5 x 10 9 gallons of water. If it takes roughly, 6 x 10 6 gallons of water to fill 1 foot of the earth, estimate how many feet the earth s oceans would rise? Express your answer in the standard form. (Source: NASA Glenn Educational Programs Office 221

228 Chapter 8 EXERCISE Write each number in scientific notation 1) 885 2) ) ) ) ) 15,000 Write each number in standard notation. 7) ) ) ) ) ) Simplify. Write each answer in scientific notation. 13) ( )( ) 14) ( )( ) 15) ( )( ) 16) ( )( ) 17) ( )( ) 18) ( )( ) ) 20) 21) ) ) 24) ) ) ) Scientific Notation Applications (Source: NASA Glenn Educational Programs Office 28) The mass of the sun is 1.98 x 1,033 grams. If a single proton has a mass of 1.6 x grams, how many protons are in the sun? 29) Pluto is located at a distance of 5.9 x centimeters from Earth. At the speed of light (2.99 x cm/sec), approximately how many hours does it take a light signal (or radio message) to travel to Pluto and return? Write your answer standard form. 30) The planet Osiris was discovered by astronomers in 1999 and is at a distance of 150 light-years (1 light-year = 9.2 x kilometers). a) How many kilometers is Osiris from earth? Express your answer in scientific notation. b) If an interstellar probe were sent to investigate this world up close, traveling at a maximum speed of 700 km/sec or 7 x 10 2 km/sec, how many seconds would it take to reach Osiris? c) There is about 3.15 x 10 6 seconds in a year. How many years would it take to reach Osiris? 222

229 Chapter 8 SECTION 8.3: POLYNOMIALS A. INTRODUCTION TO POLYNOMIALS Algebraic Expression Vocabulary (Duration 5:52) Definitions Terms: Parts of an algebraic expression separated by addition or subtraction (+ or ) symbols. Constant Term: A number with no variable factors. A term whose value never changes. Factors: Numbers or variable that are multiplied together Coefficient: The number that multiplies the variable. Example 1: Consider the algebraic expression 4xx 5 + 3xx 4 22xx 2 xx + 17 a. List the terms: b. Identify the constant term. Example 2: Complete the table below List of Factors 4mm xx 1 2 bbh 2rr 5 Identify the Coefficient Example 3: Consider the algebraic expression 5yy 4 8yy 3 + yy 2 yy 7 4 a. How many terms are there? b. Identify the constant term. c. What is the coefficient of the first term? d. What is the coefficient of the second term e. What is the coefficient of the third term? f. List the factors of the fourth term. YOU TRY Example 3: Consider the algebraic expression 3xx 5 + 4xx 4 2xx + 8 a. How many terms are there? b. Identify the constant term. c. What is the coefficient of the first term? d. What is the coefficient of the second term e. What is the coefficient of the third term? f. List the factors of the third term. 223

230 Introduction to polynomials (Duration 7:12) Chapter 8 Definitions Polynomial: An algebraic expression composed of the sum of terms containing a single variable raised to a non-negative integer exponent. Monomial: A polynomial consisting of one term, example: Binomial: A polynomial consisting of two terms, example: Trinomial: A polynomial consisting of three terms, example: Leading Term: The term that contains the highest power of the variable in a polynomial, example: Leading Coefficient: The coefficient of the leading term, example: Constant Term: A number with no variable factors. A term whose value never changes. Example: Degree: The highest exponent in a polynomial, example: Example 1: Complete the table below Polynomial Name Leading Coefficient Constant Term Degree 24aa 6 + aa mm 3 + mm 2 2mm 8 5xx 2 + xx 3 7 2xx + 4 4xx 3 YOU TRY Complete the table below Polynomial Name Leading Coefficient Constant Term Degree nn 2 2nn + 8 7yy 2 6xx 7 224

231 Introduction to polynomials 2 (Duration 2:58) Chapter 8 Given: 9yy + 7yy 3 5 4yy 2 1 st term: Degree: Coefficient: 2 nd term: Degree: Coefficient: 3 rd term: Degree: Coefficient: 4 th term: Degree: Coefficient: Leading coefficient: Degree of leading term: Degree of polynomial: Write the polynomial in descending order: (Or write the polynomial in the standard form) Standard form of a polynomial The standard form of a polynomial is where the polynomial is written with descending exponents. For example: Rewrite the polynomial in standard form and identify the coefficients, variable terms, and degree of the polynomial 12xx 2 + xx 3 xx + 2 The standard form of the above polynomial is xx 3 12xx 2 xx + 2. The coefficients are 1; 12; 1, and 2; the variable terms are xx 3, 12xx 2, xx. The degree of the polynomial is 3 because that is the highest degree of all terms. YOU TRY Write the following polynomials in the descending order or in standard form: a) 3xx 9xx 3 + 2xx 6 + 7xx xx 4 b) 5mm 2 5mm mm 3 2mm 7 B. EVALUATING POLYNOMIAL EXPRESSIONS Evaluating algebraic expressions (Duration 7:48) To evaluate an algebraic or variable expression, the value of the variables into the expression. Then evaluate using the order of operations. Example 1: If we are given 5xx 12 and xx = 17 we can evaluate. 5xx 12 = 5 ( ) 12 = Example 2: Let xx = 3, yy = 7, zz = 2 Evaluate xx 3yy + 7 Evaluate 2xx 2 + 5yy zz 3 225

232 Chapter 8 Example 3: Let = 3. Evaluate 9 yy 8yy + 2 Example 4: Let xx = 3, yy = 5. Evaluate 4xx 3yy 2 Example 5: Let = 2. Evaluate 3xx 2 xx 2 + 2xx + 9 xx 2 yy 2 Example 6: Let xx = 2, yy = 3. Evaluate xx 2 2yy 3 YOU TRY a) Evaluate 2xx 2 4xx + 6 when xx = 4 b) Evaluate xx 2 + 2xx + 6 when xx = 3 C. ADD AND SUBTRACT POLYNOMIALS Combining like terms review Combine like terms 1 (Duration 4:36) Definition Like terms: Two or more terms are like terms if they have the same variable or variables with the same exponents. Which of these terms are like terms? 2xx 3, 2xx, 2yy, 7xx 3, 49, 0xx 2, yy 2 Like terms: Like terms: To combine like terms, we. The variable factors. Example: Simplify each polynomials, if possible. a) 4xx 3 7xx 3 b) 2yy 2 + 4yy yy yy 5 + 2yy 226

233 Combine like terms 2 (Duration 2:15) Chapter 8 Combine like terms a) xx 2 yy + 3xxyy 2 + 4xx 2 yy b) 7mm 4 + 2mm + 9 YOU TRY Combine like terms a) 5xx 2 + 2xx 5xx 2 3xx + 1 b) 3xxyy 2 2xx yy 5xxyy 2 3 c) 3xx 2 yyyy + 9xx 2 5xxyy 2 zz 3yy 2 + 5xx 2 d) 3xx 2 3xx + 5yy 2 aaxx xx 10yy 2 Add and subtract polynomials Add and subtract polynomials (Duration 3:53) To add polynomials: To subtract polynomials: a) (5xx 2 7xx + qq) + (2xx 2 + 5xx 14) b) (3xx 3 4xx + 7) (8xx 3 + 9xx 2) Add and subtract polynomials (Duration 5:04) c) (2xx 5 6xx 3 12xx 2 4) (11xx 5 + 8xx + 2xx 2 + 6) d) ( 9yy 3 6yy 2 11xx + 2) ( 9yy 4 8yy 3 + 4xx 2 + 2xx) 227

234 YOU TRY Chapter 8 Perform the operation below. a) (4xx 3 2xx + 8) + (3xx 3 9xx 2 11) b) (5xx 2 2xx + 7) (3xx 2 + 6xx 4) c) (2xx 2 4xx + 3) + (5xx 2 6xx + 1) (xx 2 9xx + 8) D. MULTIPLY POLYNOMIAL EXPRESSIONS 1. Distributive property review Distribute property review (Duration 6:08) Distributive Property aa(bb + cc) = aaaa + aaaa aa = 2 bb = 3 cc = 4 Example: Use the distributive property to expand each of the following expressions a) 5(2xx + 4) b) 3(xx 2 2xx + 7) c) (5xx 4 8) d) 2 5 xx YOU TRY Use the distributive property to expand each of the following expressions. a) 4( 5xx 2 + 9xx 3) b) 7( 2mm 2 + mm 2) 2. Multiply a polynomial by a monomial Multiply a polynomial by a monomial (Duration 2:46) To multiply a monomial by a polynomial: Example 1: 5xx 2 (6xx 2 2xx + 5) Example 2: 3xx 4 (6xx 3 + 2xx 7) 228

235 YOU TRY Chapter 8 Multiply a) 4xx 3 (5xx 2 2xx + 5) b) 2aa 3 bb(3aabb 2 4aa) 3. Multiplying with binomials Multiply binomials (Duration 4:27) To multiply a binomial by a binomial: This process is often called, which stands for Example: a) (4xx 2)(5xx + 1) b) (3xx 7)(2xx 8) YOU TRY Multiply a) (3xx + 5)(xx + 13) b) (4xx + 7yy)(3xx 2yy) 4. Multiply with trinomials Multiply with trinomials (Duration 5:00) Multiplying trinomials is just like, we just have to. Example: a) (2xx 4)(3xx 2 5xx + 1) b) (2xx 2 6xx + 1)(4xx 2 2xx 6) 229

236 YOU TRY Chapter 8 Multiply a) (2xx 5)(4xx 2 7xx + 3) b) (5xx 2 + xx 10)(3xx 2 10xx 6) E. SPECIAL PRODUCTS There are a few shortcuts that we can take when multiplying polynomials. If we can recognize when to use them, we should so that we can obtain the results even quicker. In future chapters, we will need to be efficient in these techniques since multiplying polynomials will only be one of the steps in the problem. These two formulas are important to commit to memory. The more familiar we are with them, the next two chapters will be so much easier. 1. Difference of two squares Difference of two squares (Duration 2:33) Sum and difference (aa + bb)(aa bb) = = Sum and difference shortcut: (aa + bb)(aa bb) = Example: a) (xx + 5)(xx 5) b) (6xx 2)(6xx + 2) YOU TRY Simplify: a) (3xx + 7)(3xx 7) b) (8 xx 2 )(8 + xx 2 ) 230

237 Chapter 8 2. Perfect square trinomials Another shortcut used to multiply binomials is called perfect square trinomials. These are easy to recognize because this product is the square of a binomial. Let s take a look at an example. Perfect Square (Duration 3:40) Perfect square (aa + bb) 2 = Perfect square shortcut: (aa + bb) 2 = Example: a) (xx 4) 2 b) (2xx + 7) 2 YOU TRY Simplify: a) (xx 5) 2 b) (2xx + 9) 2 c) (3xx 7yy) 2 d) (6 2mm) 2 F. POLYNOMIAL DIVISION Dividing polynomials is a process very similar to long division of whole numbers. Before we look at long division with polynomials, we will first master dividing a polynomial by a monomial. 1. Polynomial division with monomials Dividing polynomials by monomials - Separated fractions method (Duration 8:14) We divide a polynomial by a monomial by rewriting the expression as separated fractions rather than one aa+bb fraction. We use the fact: = aa + bb cc cc cc Example: 6w 8 a) b) 3xx 6 30ω

238 Chapter 8 c) 6xx 3 +2xx 2 4 4xx d) 20aa2 +35aa 4 5aa 2 YOU TRY Simplify 9xx 5 +6xx 4 18xx 3 24xx 2 a) 3xx 2 b) 8xx3 +4xx 2 2xx+6 4xx 2 Long division review (Duration 3:55) Long division review Long division steps: This method may seem elementary, but it isn t the arithmetic we want to review, it is the method. We use the same method as we did in arithmetic, but now with polynomials. Dividing polynomials by monomials Long division method (Duration 5:00) Example: a) 5xx 5 +18xx 9xx 3 3xx 2 232

239 b) 15aa6 25aa 5 +5aa 4 5aa 4 Chapter 8 YOU TRY Divide using the long division method a) 8xx xx 4 + 4xx 3 4xx 3 b) nn4 nn 3 + nn 2 nn c) 12xx 4 24xx 3 + 3xx 2 6xx 233

240 2. Polynomial division with polynomials Divide a polynomial by a polynomial (Duration 5:00) Chapter 8 Polynomial division with polynomials On division step, only focus on the Example 1: Divide xx3 2xx 2 15xx+30 xx+4 Example 2: Divide 4xx3 6xx xx+1 YOU TRY a) xx 2 +8xx+12 xx+1 = b) 3xx3 5xx 2 32xx+7 xx 4 = 234

241 c) 6xx 3 8xx 2 +10xx+103 2xx+4 = Chapter 8 Divide a polynomial by a polynomial - rewriting the remainder as an expression (Duration 5:10) Example: Divide xx3 +8xx 2 17xx 15 xx+3 YOU TRY Divide the polynomials and write the remainder as an expression a) xx 2 5xx+7 xx 2 = b) xx3 4xx 2 6xx+4 xx 1 = 235

242 3. Polynomial division with missing terms Sometimes when dividing with polynomials, there may be a missing term in the dividend. We do not ignore the term, we just write in 0 as the coefficient. Polynomial division with missing terms (Duration 5:00) Chapter 8 Divide polynomials Missing terms The exponents must. If one is missing, we will add. Example 1: 3xx3 50xx+4 xx 4 Example 2: 2xx3 +4xx 2 +9 xx+3 YOU TRY a) 2xx 3 4xx+42 xx+3 = b) 3xx3 3xx 2 +4 xx 3 = 236

243 EXERCISE Evaluate the expression for the given value. Show your work. Chapter 8 1. aa 3 aa 2 + 6aa 21 when aa = 4 2. nn 2 3nn 11 when nn = 6 3. nn 3 7nn nn 20 when nn = 2 4. nn 3 9nn nn 21 when nn = nn 4 11nn 3 9nn 2 nn 5 when nn = 2 6. xx 4 5xx 3 xx + 13 when xx = 1 7. xx 2 + 9xx + 23 when xx = 3 8. xx 3 + xx 2 xx + 11 when xx = 6 9. xx 4 6xx 3 + xx 2 24 when xx = mm 4 + mm 3 + 2mm mm + 5 when mm = 3 Simplify. Write the answer in standard form. Show your work. 11. (5pp 5pp 4 ) (8pp 8pp 4 ) 12. (3nn 2 nn 3 ) (2nn 3 7nn 2 ) 13. (8nn + nn 4 ) (3nn 4nn 4 ) 14. (1 + 5pp 3 ) (1 8pp 3 ) 15. (5nn 4 + 6nn 3 ) + (8 3nn 3 5nn 4 ) 16. (3 + bb 4 ) + (7 + 2bb + bb 4 ) 17. (8xx 3 + 1) (5xx 4 6xx 3 + 2) 18. (2aa + 2aa 4 ) (3aa 2 6aa + 3) 19. (4pp 2 3 2pp) (3pp 2 6pp + 3) 20. (4bb 3 + 7bb 2 3) + (8 + 5bb 2 + bb 3 ) 21. (3 + 2nn 2 + 4nn 4 ) + (nn 3 7nn 2 4nn 4 ) 22. (nn 5nn 4 + 7) + (nn 2 7nn 4 nn) 23. (8rr 4 5rr 3 + 5rr 2 ) + (2rr 2 + 2rr 3 7rr 4 + 1) 24. (6xx 5xx 4 4xx 2 ) (2xx 7xx 2 4xx 4 8) (8 6xx 2 4xx 4 ) Multiply and simplify. Show your work 25. 6(pp 7) 26. 5mm 4 (4mm + 4) 27. (8bb + 3)(7bb 5) 28. (3vv 4)(5vv 2) 29. (5xx + yy)(6xx 4yy) 30. (7xx + 5yy)(8xx + 3yy) 31. (6nn 4)(2nn 2 2nn + 5) 32. (8nn 2 + 4nn + 6)(6nn 2 5nn + 6) 33. 3(3xx 4)(2xx + 1) 34. 7(xx 5)(xx 2) 35. (6xx + 3)(6xx 2 7xx + 4) 36. (5kk 2 + 3kk + 3)(3kk 2 + 3kk + 6) 37. (2aa 2 + 6aa + 3)(7aa 2 6aa + 1) 38. 3nn 2 (6nn + 7) 39. (7uu 2 + 2uu 3)(uu 2 + 4) 40. 3xx 2 (2xx + 3)(6xx + 9) Find each product by applying the special products formulas. Show your work 41. (xx + 8)(xx 8) 42. (1 + 3pp)(1 3pp) 43. (1 7nn)(1 + 7nn) 44. (5nn 8)(5nn + 8) 45. (4xx + 8)(4xx 8) 46. (4yy xx)(4yy + xx) 47. (4mm 2nn)(4mm + 2nn) 48. (6xx 2yy)(6xx + 2yy) 49. (aa + 5) 2 237

244 Chapter (xx 8) (pp + 7) (7 5nn) (5mm 3) (5xx + 7yy) (2xx + 2yy) (5 + 2rr) (2 + 5xx) (4vv 7)(4vv + 7) 59. (nn 5)(nn + 5) 60. (4kk + 2) (aa 4)(aa + 4) 62. (xx 3)(xx + 3) 63. (8mm + 5)(8mm 5) 64. (2rr + 3)(2rr 3) 65. (bb 7)(bb + 7) 66. (7aa + 7bb)(7aa 7bb) 67. (3yy 3xx)(3yy + 3xx) 68. (1 + 5nn) (vv + 4) (1 6nn) (7kk 7) (4xx 5) (3aa + 3bb) (4mm nn) (8xx + 5yy) (mm 7) (8nn + 7)(8nn 7) 78. (bb + 4)(bb 4) 79. (7xx + 7) 2 Divide: Show your work xx4 +xx 3 +2xx 2 4xx nn4 +nn 3 +40nn 2 5nn xx4 +24xx 3 +3xx 2 6xx 83. 5xx5 +18xx 3 +4xx + 9 9xx 84. 3kk4 +4kk kk nn4 +5nn 3 +2nn 2 nn 2 Divide and write your remainder as an expression. Show your work 86. vv2 2vv 89 vv xx2 2xx 71 xx nn2 +13nn+32 nn xx2 19xx+9 10xx aa2 4aa 38 aa pp2 56pp+19 9pp bb2 +87bb+35 3bb rr2 rr 1 4rr nn2 4 nn xx3 26xx 41 xx xx2 4xx+2 2xx aa3 +5aa 2 4aa 5 aa pp3 +5pp 2 +3pp 5 pp xx3 46xx+22 xx xx 3 +12xx xx vv 3 +4vv+19 4vv rr 3 rr 2 16rr+8 rr nn 3 +12nn 2 15nn 4 2nn+3 238

245 CHAPTER REVIEW KEY TERMS AND CONCEPTS Chapter 8 Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson. Product rule of exponents Quotient rule of exponents Power rule of a product Power rule of a quotient Power rule of a Power Zero power rule Negative exponent rule Reciprocal of negative rule Negative power of a quotient rule Scientific notation Standard notation (Decimal notation) Polynomial Monomial 239

246 Chapter 8 Binomial Trinomial Leading Term Leading Coefficient Degree of a Polynomial Constant Term 240

247 CHAPTER 9: FACTORING EXPRESSIONS AND SOLVING BY FACTORING Chapter Objectives By the end of this chapter, students should be able to Factor a greatest common factor Factor by grouping including rearranging terms Factor by applying special-product formulas Factor trinomials by using a general strategy Solve equations and applications by factoring Chapter 9 SECTION 9.1: GREATEST COMMON FACTOR AND GROUPING A. FINDING THE GREATEST COMMON FACTOR B. FACTORING THE GREATEST COMMON FACTOR C. A BINOMIAL AS THE GREATEST COMMON FACTOR D. FACTOR BY GROUPING E. FACTOR BY GROUPING BY REARRANGING TERMS EXERCISE SECTION 9.2: FACTORING TRINOMIALS OF THE FORM x 2 + bx + c A. FACTORING TRINOMIALS OF THE FORM x 2 + bx + c B. FACTORING TRINOMIALS OF THE FORM x 2 + bx + c WITH A GCF EXERCISE SECTION 9.3: FACTORING TRINOMIALS OF THE FORM ax 2 + bx + c A. FACTORING TRINOMIALS OF THE FORM ax 2 + bx + c BY GROUPING B. FACTORING TRINOMIALS OF THE FORM ax 2 + bx + c BY THE BOTTOMS UP METHOD C. FACTORING TRINOMIALS OF THE FORM ax 2 + bx + c BY THE TRIAL AND ERROR METHOD D. FACTORING TRINOMIALS OF THE FORM ax 2 + bx + c WITH A GCF IN THE COEFFICIENTS EXERCISE SECTION 9.4: SPECIAL PRODUCTS A. DIFFERENCE OF TWO SQUARES B. PERFECT SQUARE TRINOMIALS C. FACTORING SPECIAL PRODUCTS WITH A GCF IN THE COEFFICIENTS D. A SUM OR DIFFERENCE OF TWO CUBES EXERCISE SECTION 9.5: FACTORING, A GENERAL STRATEGY EXERCISE SECTION 9.6: SOLVE BY FACTORING A. ZERO PRODUCT RULE B. SOLVE EQUATIONS BY FACTORING C. SIMPLIFY THE EQUATION EXERCISE CHAPTER REVIEW

248 Chapter 9 SECTION 9.1: GREATEST COMMON FACTOR AND GROUPING A. FINDING THE GREATEST COMMON FACTOR In this lesson, we focus on factoring using the greatest common factor, GCF, of a polynomial. When we multiplied polynomials, we multiplied monomials by polynomials by distributing, such as 4xx 2 (2xx 2 3xx + 8) = 8xx 4 12xx xx We work out the same problem, but backwards. We will start with 8xx 2 12xx and obtain its factored form. First, we have to identify the GCF of a polynomial. We introduce the GCF of a polynomial by looking at an example in arithmetic. The method in which we obtained the GCF between numbers in arithmetic is the same method we use to obtain the GCF with polynomials. Definition The factored form of a number or expression is the expression written as a product of factors. The greatest common factor (GCF) of a polynomial is the largest polynomial that is a factor of all terms in the polynomial. Determine the GCF of Two Monomials (Duration 2:32) Find the GCF of 88rr 18 and 24rr rr 18 = 24rr 13 = 88 /\ 24 /\ GCF = YOU TRY Find the GCF a) 24xx 3 and 56xx 15 b) 12yy 5, 6yy 20 and 21yy 7 242

249 Determine the GCF of two monomials (Two variables) (Duration 3:45) Chapter 9 Find the GCF of 108xx 5 yy 3 and 96xx 7 yy xx 5 yy 3 = 96xx 7 yy 2 GCF= 108 /\ 96 /\ YOU TRY Find the GCF: a) 15mm 3 nn 6 and 45mm 4 nn 2 b) 24xx 4 yy 2 zz, 18xx 2 yy 4, and 12xx 3 yyzz 5 Find common factor with smaller coefficients (Duration 2:28) Greatest common factor: On variables we use Example: Find the greatest common factor. a) 15aa aa 2 25aa 5 b) 4aa 4 bb 7 12aa 2 bb aabb 9 YOU TRY Find the GCF: a) 4xx 2 20xx + 10 b) 6xx 4 15xx 3 + 9xx 2 243

250 B. FACTORING THE GREATEST COMMON FACTOR Factor using the product method (Duration 6:39) Chapter 9 Identify the greatest common factor. Then factor. a) 18xx 5 + 6xx xx 3 18 /\ 6 /\ 24 /\ GCF = b) 60aa 4 bb 3 15aa 3 bb aa 2 bb 5 60 /\ 15 /\ 48 /\ GCF = YOU TRY Factor: a) 25xx 4 15xx xx 3 b) 12xx 3 yy 2 zz 20xx 4 yy 3 zz 5 16xxyy 4 244

251 Factor using the division method (Duration 4:08) Chapter 9 a (b + c) = Put in the front, and divide. What is left goes in the. Example: Factor a) 9xx 4 12xx 3 + 6xx 2 b) 21aa 4 bb 5 14aa 3 bb 7 + 7aa 2 bb 2 YOU TRY Factor using the division method. a) 21xx xx 2 + 7xx b) 4xx 2 20xx + 16 CHECK YOUR SOLUTION: To check your answer, you can distribute your GCF back into the parenthesis. Using distributive property to verify the factored form (Duration 2:46) Example: a) 5xx 2 (6xx 2 2xx + 5) b) 3xx 4 (6xx 3 + 2xx 7) YOU TRY Check your answer from the YOU TRY in the section Factoring using the division method above. a) b) 245

252 Steps for factoring out the greatest common factor Step 1. Find the GCF of the expression. Step 2. Rewrite each term as a product of the GCF and the remaining factors. Step 3. Rewrite as a product of the GCF and the remaining factors in parenthesis. Step 4. Verify the factored form by multiplying. The product should be the original expression. Chapter 9 C. A BINOMIAL AS THE GREATEST COMMON FACTOR As part of a general strategy for factoring, we always look for a GCF. Sometimes the GCF is a monomial, like in the previous examples, or a binomial. Here we discuss factoring a polynomial where the GCF is a binomial. Binomial GCF (Duration 2:20) GCF can be a. Example: a) 5xx(2yy 7) + 6yy(2yy 7) b) 3xx(2xx + 1) 7(2xx + 1) YOU TRY Factor: a) 3aa(2aa + 5bb) 4bb(2aa + 5bb) b) (9xx 2 2)3yy (9xx 2 2)5xx D. FACTOR BY GROUPING When we have polynomials that have at least 4 terms. Sometimes, we can factor them by using a process known as factor by grouping. Steps for factoring by grouping To factor by grouping, we first notice the polynomial expression obtains four terms. Step 1. Group two sets of two terms, e.g., ax + ay + bx + by = (ax + ay) + (bx + by). Step 2. Factor the GCF from each group, e.g., a(x + y) + b(x + y) Step 3. Factor the GCF from the expression, e.g., (x + y)(a + b). Factoring by grouping (Duration 4:01) Grouping: GCF of the and Then factor out (if it matches!) 246

253 Example: a) 15xxxx + 10yy 18xx 12 b) 6xx 2 + 3xxxx + 2xx + yy Chapter 9 YOU TRY Factor by grouping: a) 10aaaa + 15bb + 4aa + 6 b) 6xx 2 + 9xxxx 14xx 21yy c) 5xxxx 8xx 10yy + 16 d) 12aaaa 14aa 6bb + 7 E. FACTOR BY GROUPING BY REARRANGING TERMS Sometimes after completing Step 2, the binomials are not identical (by more than a negative sign). At this point, we must return to the original problem and rearrange the terms so that when we factor by grouping, we obtain identical binomials in Step 2. Factor by grouping rearranging terms (Duration 4:41) If binomials don t match: Example: a) 12aa 2 7bb + 3aaaa 28aa b) 6xxxx xx 15yy YOU TRY Factor: a) 4aa 2 21bb 3 + 6aaaa 14aabb 2 b) 8xxxx 12yy xx 247

254 EXERCISE Chapter 9 Factor the greatest common factor (GCF) if possible. If not, write No common factor. Check your answer by multiplying the factors. 1) 9bb + 8bb 2 2) 45xx ) 56 35pp 4) 7aaaa 35aa 2 bb 2 5) 3aa 2 bb + 6aa 3 bb 2 6) 5xx 2 5xx 3 15xx 4 7) 20xx 4 30xx ) 28mm mm ) 30bb 9 + 5aaaa 15aa 2 10) 48aa 2 bb 2 56aa 3 bb 50aa 5 bb 11) 20xx 8 yy 2 zz xx 5 yy 5 zz + 35xx 3 yy 3 zz 12) 3mmnn mm 13) 30qqqqqq 5qqqq + 5qq 14) 50xx 2 yy + 10xxxx xxzz 2 15) 1 + 2nn 2 16) 18mmmm 5 + 3mmmm 3 21mm 3 nn 2 + 3mmmm Factor completely. Check your answer using distributive property. 17) 40rr 3 8rr 2 25rr ) 3nn 3 2nn 2 9nn ) 15bb bb 2 35bb 49 20) 3xx xx 2 + 2xx ) 35xx 3 28xx 2 20xx ) 7xxxx 49xx + 5yy 35 23) 32xxxx + 40xx yy + 15xx 24) 16xxxx 56xx + 2yy 7 25) 2xxxx 8xx 2 + 7yy 3 28yy 2 xx 26) 40xxxx + 35xx 8yy 2 7yy 27) 35xx 3 10xx 2 56xx ) 16xxxx 3xx 6xx 2 + 8yy 29) 6xx 3 48xx 2 + 5xx 40 30) 14vv vv 2 7vv 5 31) 7nn nn 2 5nn 15 32) 28pp pp pp ) 15aaaa 2bb 2 6aa + 5bb 3 34) 42rr rr rr 35) 5mmmm mm 25nn 36) 8mm + 15nn mmmm 37) 4uuuu + 14uu vv + 42uu 38) 56xx yy + 8xxxx 7 39) 56aaaa aa 16bb 40) 24xxxx + 25yy 2 20xx 30yy 3 248

255 SECTION 9.2: FACTORING TRINOMIALS OF THE FORM x 2 + bx + c Chapter 9 A. FACTORING TRINOMIALS OF THE FORM x 2 + bx + c Factoring with three terms, or trinomials, is the most important technique, especially in further algebra. Since factoring is a product of factors, we first look at multiplying to develop the process of factoring trinomials. Steps for factoring trinomials of the form xx 22 + bbxx + cc Step 1. Find two numbers, pp and qq, that pp + qq = bb and pp qq = cc Step 2. Rewrite the expression so that the middle term is split into two terms, pp and qq. Step 3. Factor by grouping. Step 4. Verify the factored form by finding the product. Factoring a trinomial with leading coefficient of 1 (ac method) (Duration 10:33 ) Factoring trinomials with a leading coefficient of 1. xx 22 + bbbb + cc 1. Make two sets of parentheses and put the factors of xx 2 in the first position of each set of parentheses. (x )(x ) 2. The second positions are the factors of c that add to b. Example: Factor. Factors of c a) xx 2 + 8xx + 12 b) xx 2 4xx 32 Factors of c c) yy 2 13yy + 36 d) xx 2 + xx 12 e) xx xx + 75 f) 3xx 2 27xx + 42 g) 2yy 3 12yy 2 80yy 249

256 Factoring trinomials 2 (Duration 5:01 ) Chapter 9 Factor: a) xx xx + 32 b) xx xx 60 c) xx 2 9xx + 20 d) xx 2 5xx 24 YOU TRY a) xx 2 + 9xx + 18 b) xx 2 4xx + 3 c) xx 2 8xx 20 d) aa 2 9aaaa + 14bb 2 Factoring trinomials X box method (Duration 3:21) If there is a in front of xx 2, then the ac method gives us. a) xx 2 2xx 8 b) xx 2 + 7xxxx 8yy 2 250

257 YOU TRY Chapter 9 a) uu 2 8uuuu + 15vv 2 b) mm 2 + 2mmmm 8nn 2 PRIME POLYNOMIALS: If a trinomial (or polynomial) is not factorable, then we say we the trinomial is prime. For example: factor xx 2 + 2xx + 6. We identify b = 2 and c = 6 Factor of c Sum 2, = 5, not b 2, = 5, not b 1, = 7, not b 1, = 7, not b We can see from the table that there are not any factors of 6 whose sum is 2. In this case, we call this trinomial not factorable, or better yet, the trinomial is prime. B. FACTORING TRINOMIALS OF THE FORM x 2 + bx + c WITH A GCF Factoring the GCF is always the first step in factoring expressions. If all terms have a common factor, we first, factor the GCF and then factor as usual. Factoring trinomials with factoring GCF first (Duration 3:39) a) 7xx xx 70 b) 4xx 4 yy + 36xx 3 yy xx 2 yy 3 YOU TRY Factor: a) 3xx 2 24xx + 45 b) 4xx xx

258 EXERCISE Factor completely. If a trinomial is not factorable, write prime. Chapter 9 1) pp pp ) nn 2 9nn + 8 3) xx 2 9xx 10 4) bb bb ) xx 2 9xx 10 6) nn 2 15nn ) pp pp ) nn 2 8nn ) xx 2 11xxxx + 18yy 2 10) xx 2 + xxxx 12yy 2 11) xx 2 + 4xxxx 12yy 2 12) 5aa aa ) 6aa aa ) 6xx xxxx + 12yy 2 15) 6xx xxxx + 378yy 2 16) xx 2 xx 72 17) xx 2 + xx 30 18) xx xx ) mm xx 36 20) 6mm 2 36mmmm 162nn 2 21) bb 2 17bb ) xx 2 3xx 18 23) aa 2 6aa 27 24) pp 2 + 7pp 30 25) mm 2 15mmmm + 50nn 2 26) xx 2 + xx ) xx xxxx + 16yy 2 28) uu 2 9uuuu + 14vv 2 29) xx xxxx + 45yy 2 30) 5nn 2 45nn ) 5vv vv 25 32) mm 2 2mmmm + nn 2 252

259 SECTION 9.3: FACTORING TRINOMIALS OF THE FORM ax 2 + bx + c Chapter 9 A. FACTORING TRINOMIALS OF THE FORM ax 2 + bx + c BY GROUPING When the leading coefficient a is not 1, it takes a few more steps to factor the trinomial. There are many ways to factor this type of trinomials. You are going to learn 2 methods in this section. The first one is factor by grouping and the second one is the bottoms- up method. First, let s take a look at the grouping method. Steps for factoring trinomials of the form ax 2 + bx + c using the grouping method Step 1. Find two numbers, pp and qq, that pp + qq = bb and pp qq = aa cc Step 2. Rewrite the expression so that the middle term is split into two terms, p and q. Step 3. Factor by grouping. Step 4. Verify the factored form by finding the product. Factor trinomials when leading coefficient is not 1 - Grouping method (Duration 4:21) a) 4xx 2 4xx 15 b) 20xx xx + 3 YOU TRY Factor using the grouping method and verify your answer by multiplying the two binomials. Show your work. a) 3xx xx + 6 b) 8xx 2 2xx

260 B. FACTORING TRINOMIALS OF THE FORM ax 2 + bx + c BY THE BOTTOMS UP METHOD Chapter 9 Steps for factoring trinomials of the form ax 2 + bx + c using the bottoms-up method Step 1. Multiply aa cc, then write a new trinomial in the form of x 2 + bx + a c Step 2. Factor as you normally would with trinomials with the leading coefficient of 1. Step 3. Divide the constants in each binomial factor by the original value of a. Step 4. Simplify the fractions formed. Step 5. If the simplified fractions does not have the denominator of 1, move the denominator to the coefficient of the variable. Step 6. Verify the factored form by finding the product Factor trinomials when the leading coefficient is NOT 1 - Bottoms up method (Duration 4:20) a) 4xx 2 4xx 15 b) 20xx xx + 3 YOU TRY Factor the trinomials using the bottoms up method. Show your work. c) 3xx xx + 6 d) 8xx 2 2xx

261 Chapter 9 C. FACTORING TRINOMIALS OF THE FORM ax 2 + bx + c BY THE TRIAL AND ERROR METHOD Factoring by trial-and-error is just a guess and check process when you try to add up different products to get the middle term bx. This sometimes works out faster than other methods above and sometimes not. If you want a step-by-step process that always works, this method may not be the best method for you. Factor trinomials when the leading coefficient is NOT 1 Trial and error method (Duration 5:22) a) 4xx 2 4xx 15 (2xx )(2xx ) (4xx )(4xx ) b) 20xx xx + 3 (5xx )(4xx ) (10xx )(2xx ) (20xx )(xx ) (5xx )(4xx ) (10xx )(2xx ) (20xx )(xx ) YOU TRY Factor the trinomials below by using the trial-error method: a) 10xx 2 27xx

262 Chapter 9 D. FACTORING TRINOMIALS OF THE FORM ax 2 + bx + c WITH A GCF IN THE COEFFICIENTS As always, when factoring, we will first look for a GCF in the coefficients, factor the GCG, then factor the trinomial as usual. Factoring trinomials of the form ax 2 + bx + c with GCF in the coeffients (Duration 1:45) Stop at 7:00 Example: 18xx 2 15xx 12 YOU TRY a) 24xx 2 22xx + 4 b) 18xx 3 33xx xx 256

263 EXERCISE Factor completely by using grouping method. Show your work. Chapter 9 1) 7xx 2 48xx ) 7bb bb + 2 3) 5aa 2 13aa 28 4) 2xx 2 5xx + 2 5) 2bb 2 bb 3 6) 5kk kk + 6 7) 3xx 2 17xx ) 6xx 2 39xx 21 9) 6xx 2 29xx ) 4xx 2 + 9xx ) 4xx xxxx + 3yy 2 12) 3uu uuuu 10vv 2 Factor completely by using the bottoms up method. Show your work. 13) 4mm 2 9mm 9 14) 4xx xx ) 6pp 2 11pp 7 16) 4rr 2 + rr 3 17) 4rr 2 + 3rr 7 18) 3rr 2 4rr 4 19) 3xx xx 8 20) 2xx 2 5xx 3 21) 2yy yy ) 7aa 2 11aa ) 4xx xx ) 24aa 2 30aa ) 10xx xx 2 10xx 26) 2xx 3 yy + 12xx 2 yy + 18xxxx 27) 5tt tt ) 2xx 2 + 8xx ) 7xx 2 2xxxx 5yy 2 30) 24xx^2 52xxxx + 8yy^2 31) 4xx xxxx + 3yy 2 32) 3uu uuuu 10vv 2 257

264 SECTION 9.4: SPECIAL PRODUCTS In the previous chapter, we recognized two special products: difference of two squares and perfect square trinomials. In this section, we discuss these special products to factor expressions. Chapter 9 A. DIFFERENCE OF TWO SQUARES When we see a binomial where both the 1 st and 2 nd terms are perfect square and one subtracts another, you have the difference of two squares. You can apply the following formula to factor quickly. Difference of two squares aa 22 bb 22 = (aa + bb)(aa bb) Factor a Difference of Squares (Duration 4:19) Example: Factoring binomials a) xx 2 36 b) xx 2 16 c) 100 9xx 2 d) 2xx 2 18 Warning: The sum of squares aa 22 + bb 22 does NOT factor. It is always prime. Factoring a difference of squares with two variables (Duration 1:48) Factor: xx 2 49yy 2 YOU TRY Factor completely: a) xx 2 9 b) 25 mm 2 c) 9aa 2 25bb 2 d) 5xx 2 45yy 2 258

265 Chapter 9 B. PERFECT SQUARE TRINOMIALS In this section, we discuss two special types of trinomials that are called the perfect square trinomials. In order to have a perfect square trinomial, we need to have the 1 st and the 2 nd terms squared and the middle term is twice the 1 st and the 2 nd terms. This pattern allows us to be more efficient when we factor trinomials. Perfect square trinomials aa 2 + 2aaaa + bb 2 = (aa + bb) 2 aa 2 2aaaa + bb 2 = (aa bb) 2 Factor perfect square trinomials (Duration 5:53) Given a perfect square trinomial, factor the trinomial into the square of a binomial: Example: 1. 36xx xx xx 2 28xx + 4 YOU TRY Factor by using the perfect square formula: a) xx 2 6xx + 9 b) 4xx xxxx + 25yy 2 259

266 C. FACTORING SPECIAL PRODUCTS WITH A GCF IN THE COEFFICIENTS Sometimes, we have to factor a GCF out of a polynomial before we can apply the difference of two squares binomial or the perfect square trinomial formulas. Factor a difference of squares with a common factor (Duration 3:05) Chapter 9 a) Example: 4xx 2 36 YOU TRY a) Factor completely: 72xx 2 8 b) Factor the GCF and apply the perfect square formula: 3xx 2 18xx + 27 D. A SUM OR DIFFERENCE OF TWO CUBES Sum or difference of two cubes There are special formulas for a sum or difference of two cubes. Difference of two cubes: aa 3 bb 3 = (aa bb)(aa 2 + aaaa + bb 2 ) Sum of two cubes: aa 3 + bb 3 = (aa + bb)(aa 2 aaaa + bb 2 ) We can also use the acronym SOAP for the formulas for factoring a sum or difference of two cubes. Same: binomial has the same sign as the expression Opposite: middle term of the trinomial has the opposite sign than the expression Always Positive: last term of the trinomial is always positive SOAP is an easier way of remembering the signs in the formula because the formulas for the sum and difference of two cubes are the same except for the signs. Let s take a look: 260

267 Factor a Sum or Difference of Cubes (Duration 4:58) Chapter 9 Special product Cubes Sum of cubes: Difference of cubes: Example: Factor. a) mm b) 8aa 3 27yy 3 YOU TRY a) mm 3 27 b) 125pp 3 + 8rr 3 Sometimes, you have to factor out the common factor in the coefficients before you can apply the Sum or Difference of Cubes formulas. Factor a Sum or Difference of Cubes when coefficients have common factor (Duration 3:34) Factor completely: c) 4xx 3 32 d) 2xx

268 Factor a Sum or Difference of Cubes Caution (Duration 5:22) Chapter 9 Example: Factor completely 64xx yy /\ 216 /\ YOU TRY a) 5xx 3 40 b) 128aa 4 bb aabb 5 262

269 EXERCISE Name the special product and factor completely. Show your work. Chapter 9 Name Name 1) rr ) vv ) pp ) 9kk 2 4 5) 33xx ) 16xx ) 1111aa bb 22 8) aa 2 2aa + 1 9) xx ) xx 2 6xx ) kk ) kk 2 4kk ) 2222pp ) 8xx 2 24xxxx + 18yy 2 15) 88 mm 33 16) xx ) uu 33 18) 125aa ) xx yy 33 20) xx ) 3333mm ) xx ) 44aa bb 22 24) 9aa ) xx ) 125xx yy 3 27) 44vv ) 25aa aaaa + 9bb 2 29) 55nn ) xx 2 + 8xxxx + 16yy 2 31) 44mm nn 22 32) 20xx xxxx + 5yy 2 33) nn ) xx ) xx ) 64xx ) xx ) 18mm 2 24mmmm + 8nn 2 39) xx 44 yy 44 40) zz

270 Chapter 9 SECTION 9.5: FACTORING, A GENERAL STRATEGY A General Strategy To Factoring Step 1. Factor out the greatest common factor, if possible. Step 2. Determine the number of terms in the polynomial. Step 3. a) Two Terms Difference of two squares: aa 2 bb 2 = (aa + bb)(aa bb) Difference of two cubes: aa 3 bb 3 = (aa bb)(aa 2 + aaaa + bb 2 ) Sum of two cubes: aa 3 + bb 3 = (aa + bb)(aa 2 aaaa + bb 2 ) b) Three Terms Perfect square trinomial: aa 2 + 2aaaa + bb 2 = (aa + bb) 2 or aa 2 2aaaa + bb 2 = (aa bb) 2 Old fashion way: xx 2 + bbbb + cc = (xx + pp)(xx + qq): using the ac method aaaa 2 + bbbb + cc: Factor by grouping or by the bottoms up method. c) Four Terms Factor by grouping, rearranging terms, if needed. Step 4. Check your work by multiplying out the product of factors. General factoring strategy (Duration 5:00) Always do first! 2 terms: 3 terms: 4 terms: Example: a) 25xx 2 16 b) xx 2 xx 20 c) xxxx + 2yy + 5xx

271 Chapter 9 EXERCISE Apply the general factoring strategy to factor the following polynomials completely. Show your work. 1) 24aaaa 18aah + 60yyyy 45yyh 2) 2xx 3 128yy 3 3) 54uu ) xx 2 4xxxx + 3yy 2 5) mm 2 4nn 2 6) xx 3 7) nn 3 + 7nn nn 8) 5xx 2 + 2xx 9) mmmm 12xx + 3mm 4xxxx 10) 27mm 2 48nn 2 11) 16xx xxxx + 36yy 2 12) 2xx 3 + 5xx 2 yy + 3yy 2 xx 13) 5xx 2 22xx 15 14) xx 3 27yy 3 15) 3mm 3 6mm 2 nn 24nn 2 mm 16) 3aaaa + 15aadd 2 + xx 2 cc + 5xx 2 dd 2 17) 16aa 2 9bb 2 18) 32xx 2 18yy 2 19) vv 2 vv 20) 9nn 3 3nn 2 265

272 Chapter 9 SECTION 9.6: SOLVE BY FACTORING When solving linear equations, such as 2x 5 = 21, we can solve by isolating the variable on one side and a number on the other side. However, in this chapter, we have an x 2 term, so if it looks different, then it is different. Hence, we need a new method for solving trinomial equations. One method is using the zero product rule. There are other methods for solving trinomial equations, but that is for a future chapter. Definition A polynomial equation is any equation that contains a polynomial expression. A trinomial equation is written in the form aaaa 22 + bbbb + cc = 00 where a, b, c are coefficients, and a 0. If the trinomial equations have the highest power is 2, they are also called as quadratic equations. A. ZERO PRODUCT RULE Zero product rule If aa; bb are non-zero factors, then aa bb = 00 implies aa = 00 or bb = 00 or both aa = bb = 00. Solve equations by using the Zero product rule (Duration 4:04) ) Zero product rule: To solve we set each equal to zero Example: a) (5xx 1)(2xx + 5) = 0 b) 2xx(xx 6)(2xx + 3) = 0 YOU TRY Solve for x: a) xx(xx + 7) = 0 b) (2x 3)(5x + 1) = 0 266

273 B. SOLVE EQUATIONS BY FACTORING Chapter 9 Steps for solving trinomial equations Step 1. Write the given equation in the form aaaa 2 + bbbb + cc = 0. Step 2. Factor the left side of the equation into a product of factors. Step 3. Use the zero product rule to set each factor equal to zero and then solve for the unknown. Step 4. Verify the solution(s). Solve quadratic equations by factoring the GCF (Duration 2:38) Example: Solve a) xx 2 + 4xx = 0 b) 14xx 2 35xx = 0 YOU TRY Solve the equations by factoring: a) nn 2 9nn = 0 b) 7nn 2 28nn = 0 Factor and solve quadratic equations when a = 1 (Duration 5:24 ) Example: Solve a) xx 2 + 6xx + 8 = 0 b) xx xx 50 = 0 c) xx 2 8xx + 15 = 0 d) xx 2 5xx 24 = 0 267

274 YOU TRY Chapter 9 Solve the equations by factoring a) cc 2 + 5cc + 6 = 0 b) yy 2 9yy + 14 = 0 Factor and solve quadratic equation with a negative leading coefficient (Duration 5:10) Example: Solve by factoring. a) xx 2 + 7xx + 18 = 0 b) xx 2 12xx 36 = 0 YOU TRY Solve by factoring: a) xx 2 + xx + 6 = 0 b) yy 2 3yy + 18 = 0 268

275 Chapter 9 Factor & solve quadratic equations with common factor in the coefficients (Duration 4:27) Example: Solve by factoring. a) 3xx xx + 18 = 0 b) 8xx 2 72xx = 0 YOU TRY Solve by factoring: c) 3xx 2 24xx + 45 = 0 d) 4xx xx = 0 Factor and solve a quadratic equation when the leading coefficient is NOT 1 (Duration 6:17) Example: Solve by factoring. a) 4xx xx 21 = 0 b) 3xx 2 23xx + 30 = 0 269

276 YOU TRY Chapter 9 Solve by factoring a) 3xx xx + 6 = 0 b) 8xx 2 2xx 15 = 0 Factor and solve a quadratic equation using the difference of 2 squares (Duration 3:58) Example: Solve by factoring a) xx 2 49 = 0 b) 4xx 2 = 81 Factor and solve a quadratic equation using the difference of 2 squares with GCF (Duration 5:03) Example: Solve by factoring c) 48xx 2 75 = 0 d) 2xx 2 =

277 YOU TRY Chapter 9 Factor binomials and solve the equation a) xx 2 9 = 0 b) 8xx 2 = 50 Factor and solve a quadratic equation using the perfect square trinomials (Duration 4:52 ) Example: Solve by factoring. a) xx 2 + 4xx + 4 = 0 b) xx 2 10xx + 25 = 0 c) 4xx 2 12xx + 9 = 0 YOU TRY Solve by factoring: a) xx 2 2xx + 1 = 0 b) 9xx 2 + 6xx + 1 = 0 271

278 Chapter 9 C. SIMPLIFY THE EQUATION Sometimes the equation isn t so straightforward. We may have to do some preliminary work so that the equation takes the form of a trinomial equation and then we can use the zero product rule. Solve by factoring Simplify first (Duration 4:57) Example: a) 2xx(xx + 4) = 3xx 3 b) (2xx 3)(3xx + 1) = 8xx 1 YOU TRY Simplify the following equations and solve by factoring. a) (xx 7)(xx + 3) = 9 3xx 2 + 4xx 5 = 7xx 2 + 4xx

279 EXERCISE Solve each equation by factoring. Show your work. Chapter 9 1) (kk 7)(kk + 2) = 0 2) (xx 1)(xx + 4) = 0 3) 6xx = 0 4) 2nn nn 28 = 0 5) 7xx xx + 15 = 0 6) 5nn 2 9nn 2 = 0 7) xx 2 4xx 8 = 8 8) xx 2 5xx 1 = 5 9) 49pp pp 163 = 5 10) 7xx xx 20 = 8 11) 7rr = 49rr 12) xx 2 6xx = 16 13) 3vv 2 + 7vv = 40 14) 35xx xx = 45 15) 4kk kk 23 = 6kk 7 16) 9xx xx = 7xx + 8xx ) 2mm mm + 40 = 2mm 18) 40pp pp 168 = pp + 5pp 2 19) (aa + 4)(aa 3) = 0 20) (2xx + 5)(xx 7) = 0 21) pp 2 + 4pp 32 = 0 22) mm 2 mm 30 = 0 23) 40rr 2 285rr 280 = 0 24) 2bb 2 3bb 2 = 0 25) vv 2 8vv 3 = 3 26) aa 2 6aa + 6 = 2 27) 7kk kk + 13 = 5 28) 7nn 2 28nn = 0 29) 6bb 2 = 5 + 7bb 30) 9nn nn = 36 31) aa 2 + 7aa 9 = 3 + 6aa 32) xx xx + 30 = 6 33) 5nn nn + 40 = 2 34) 24xx xx 80 = 3xx 273

280 CHAPTER REVIEW KEY TERMS AND CONCEPTS Chapter 9 Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson. Factored form Greatest common factor (GCF) 274

281 Chapter 10 CHAPTER 10: RATIONAL EXPRESSIONS Chapter Objectives By the end of this chapter, students should be able to: Evaluate rational expressions Obtain the excluded values of the expression Reduce rational expressions Multiply rational expressions with and without factoring Divide rational expressions with and without factoring Find least common denominators Add and subtract rational expressions with and without common denominators Contents CHAPTER 10: RATIONAL EXPRESSIONS SECTION 10.1: REDUCE RATIONAL EXPRESSIONS A. EVALUATE RATIONAL EXPRESSIONS B. FIND EXCLUDED VALUES OF RATIONAL EXPRESSIONS C. REDUCE RATIONAL EXPRESSIONS WITH MONOMIALS D. REDUCE RATIONAL EXPRESSIONS WITH POLYNOMIALS EXERCISE SECTION 10.2: MULTIPLY AND DIVIDE RATIONAL EXPRESSIONS A. MULTIPLY AND DIVIDE RATIONAL EXPRESSIONS WITH MONOMIALS B. MULTIPLY AND DIVIDE RATIONAL EXPRESSIONS WITH POLYNOMIALS C. MULTIPLY AND DIVIDE RATIONAL EXPRESSIONS IN GENERAL EXERCISE SECTION 10.3 OBTAIN THE LOWEST COMMON DENOMINATOR A. OBTAIN THE LCM IN ARITHMETIC REVIEW B. OBTAIN THE LCM WITH MONOMIALS C. OBTAIN THE LCM WITH POLYNOMIALS D. REWRITE FRACTIONS WITH THE LOWEST COMMON EXERCISE SECTION 10.4: ADD AND SUBTRACT RATIONAL EXPRESSIONS A. ADD OR SUBTRACT RATIONAL EXPRESSIONS WITH A COMMON DENOMINATOR B. ADD AND SUBTRACT RATIONAL EXPRESSIONS WITH UNLIKE DENOMINATORS EXERCISE CHAPTER REVIEW

282 SECTION 10.1: REDUCE RATIONAL EXPRESSIONS A. EVALUATE RATIONAL EXPRESSIONS Definition Chapter 10 A rational expression is a ratio of two polynomials, i.e., a fraction where the numerator and denominator are polynomials. Evaluate rational expressions (Duration 4:18 ) Rational Expression: Quotient of two a) xx2 2xx 8 xx 4 when xx = 4 b) xx 2 xx 6 xx 2 when xx = 2 +xx 12 YOU TRY Evaluate: a) xx 2 4 xx 2 +6xx+8 wwheeee xx = 6 b) 3xx xx 2 when xx = 2 +12xx 2 276

283 Chapter 10 B. FIND EXCLUDED VALUES OF RATIONAL EXPRESSIONS Rational expressions are special types of fractions, but still hold the same arithmetic properties. One property of fractions we recall is that the fraction is undefined when the denominator is zero. Determine the excluded value(s) of a rational expression Note: A rational expression is undefined when the denominator is zero. Step 1. Set the denominator of the rational expression equal to zero. Step 2. Solve the equation for the given variable. Step 3. The values found in the previous step are the values excluded from the expression. Find excluded value(s) of a rational expression (Duration 2:24 ) a) xx 2 1 3xx 2 +5xx YOU TRY Find the excluded value(s) of the expression. a) 3zz zz+5 b) xx 2 1 3xx 2 +5xx 277

284 Chapter 10 C. REDUCE RATIONAL EXPRESSIONS WITH MONOMIALS Rational expressions are reduced, just as in arithmetic, even without knowing the value of the variable. When we reduce, we divide out common factors as we discussed with polynomial division with monomials. Now, we use factoring techniques and exponent properties to reduce rational expressions. Reducing rational expressions If PP, QQ, KK are non-zero polynomials and PPPP is a rational expression, then QQQQ PP. KK QQ. KK = PP QQ We call a rational expression irreducible if there are no more common factors among the numerator and denominator. Reduce monomials (Duration 2:44) Quotient rule of exponents: aamm = aann a) 16aa 5 12xx 9 b) 15aa3 bb 2 25aabb 5 It is important to note that we were only able to use the quotient rule when. YOU TRY Simplify: a) 2xx 2 4xx 3 b) 15xx4 yy 2 25xx 2 yy 6 278

285 D. REDUCE RATIONAL EXPRESSIONS WITH POLYNOMIALS However, if there is a sum or difference in either the numerator or denominator, we first factor the numerator and denominator to obtain a product of factors, and then reduce. Reduce polynomials (Duration 5:00) Chapter 10 To reduce polynomials, we common. This means we must first. a) 2xx 2 +5xx 3 2xx 2 5xx+2 b) 9xx2 30xx+25 9xx 2 25 Note: you can use the bottoms-up method to factor the binomials. YOU TRY Simplify: a) b) c) xx xx xx+1111 Warning: You cannot reduce terms, only factors. This means we cannot reduce anything with a + or between the parts. In examples above, we are not allowed to divide out the xx s because they are terms (separated by + oooo ) not factors (separated by multiplication). 279

286 EXERCISE Evaluate the expression for the given value. 1) 4vv wwheeee vv = 6 2) xx 3 xx 2 4xx+3 wwheeee xx = 4 3) bb+2 bb 2 wwheeee bb = 0 +4bb+4 Chapter 10 4) bb 3 3bb+9 wwheeee bb = 2 5) aa+2 aa 2 +3aa+2 wwheeee aa = 1 6) nn2 nn 6 nn 3 wwheeee nn = 4 Find the excluded value(s). 7) 3kk2 +30kk kk+10 10) rr2 +3rr+2 5rr+10 xx+10 13) 8xx 2 +80xx 8) 15nn 2 10nn+25 11) bb2 +12bb+32 bb 2 +4bb 32 14) 10xx+16 6xx+20 9) 10mm2 +8mm 10mm 27pp 12) 18pp 2 36pp 15) 6nn2 21nn 6nn 2 +3nn Simplify each expression. 16) 21xx2 18xx 19) 18mm xx 2 22) 28xx 2 +28xx 25) 12xx2 42xx 30xx 2 42xx nn 9 28) 9nn 81 17) 24aa 40aa ) 4pp+2 23) nn2 +4nn 12 nn 2 7nn+10 26) 6aa 10 10aa+4 29) 28mm ) 32xx3 8xx 4 xx+1 21) xx 2 +8xx+7 9vv+54 24) vv 2 4vv 60 27) 2nn2 +19nn 10 9nn+90 30) 49rr+56 56rr 31) bb2 +14bb+48 bb 2 +15bb+56 9pp+18 34) pp 2 +4pp+4 32) 30xx 90 50xx+40 35) 3xx2 29xx+40 5xx 2 30xx 80 33) kk2 12kk+32 kk ) 8mm+16 20mm 12 37) 2xx2 10xx+8 3xx 2 7xx+4 38) 7nn2 32nn+16 4nn 16 39) nn2 +2nn+1 6nn+6 40) 4kk3 2kk 2 2kk 9kk 3 18kk 2 + 9kk 280

287 Chapter 10 SECTION 10.2: MULTIPLY AND DIVIDE RATIONAL EXPRESSIONS We use the same method for multiplying and dividing fractions to multiply and divide rational expressions. A. MULTIPLY AND DIVIDE RATIONAL EXPRESSIONS WITH MONOMIALS Recall. When we multiply two fractions, we divide out the common factors, e.g., = = We multiply rational expressions using the same method. Multiply and divide monomials (Duration 4:49) With monomials, we can use. aa mm aa nn = aa mm aann = a) 6xx 2 yy 5 5xx 3 10xx4 3xx 2 yy 7 b) 4aa5 bb 9aa 4 6aabb4 12bb 2 YOU TRY a) Multiply: 25xx2 24yy4 8yy 8 55xx 7 b) Divide: aa4 bb 2 aa bb

288 B. MULTIPLY AND DIVIDE RATIONAL EXPRESSIONS WITH POLYNOMIALS When we multiply or divide polynomials in rational expressions, we first factor using factoring techniques, then reduce out the common factors. Chapter 10 Warning: We are not allowed to reduce terms, only factors. Multiply and divide rational expressions with polynomials (Duration 5:00 ) To divide out factors, we must first! a) xx 2 +3xx+2 4xx 12 xx2 5xx+6 xx 2 4 b) 3xx2 +5xx 2 6xx2 +xx 1 xx 2 +3xx+2 xx 2 3xx 4 YOU TRY xx 2 9 a) Multiply: xx2 8xx+16 xx 2 + xx 20 3xx+9 b) Divide: xx2 xx 12 5xx2 +15xx xx 2 2xx 8 xx 2 +xx 2 282

289 Chapter 10 C. MULTIPLY AND DIVIDE RATIONAL EXPRESSIONS IN GENERAL We can combine multiplying and dividing rational expressions in one expression, but, remember; we reciprocate the fraction that directly precedes the division sign and then change the division to multiplication. Lastly, we can reduce the common factors. Warning: We are not allowed to reduce terms, only factors. Multiply and divide rational expressions together (Duration 4:53) To divide:. Be sure to before. a) xx 2 +3xx 10 2xx2 xx 3 8xx+20 xx 2 +6xx+5 2xx 2 +xx 6 6xx+15 YOU TRY Simplify: a) aa 2 +7aa+10 aa+1 aa 1 aa 2 + 6aa+5 aa 2 +4aa+4 aa+2 283

290 EXERCISE Simplify each expression. Watch for special products to help with factoring more quickly. 1) 8xx ) 9nn 2nn 7 5nn Chapter 10 3) 5xx ) 7(mm 6) mm 6 5mm(7mm 5) 7mm 5 5) 7rr rr 6 7rr(rr+10) (rr 6) 2 25nn+25 6) nn+30 7) xx xx+21 35xx+21 8) xx2 6xx 7 xx+5 xx+5 xx 7 9) 8kk 1 24kk 2 40kk 15kk 25 10) (nn 8) 6 10nn 80 11) 4mm+36 mm+9 mm 5 5mm 2 12) 3xx 6 12xx 24 (xx + 3) bb+2 13) 40bb 2 24bb (5bb 3) 14) nn 7 6nn nn nn 2 13nn+42 15) 27aa+36 9aa+63 6aa ) xx2 12xx+32 xx 2 6xx 16 7xx2 +14xx 7xx 2 +21xx 17) (10mm mm) 18mm3 36mm 2 20mm 2 40mm 18) 10bb 2 30bb+20 30bb+20 2bb 2 +10bb 19) 10pp xx (xx+4) 20) xx 3 (xx 3)(xx 6) 6xx (xx 6) 21) vv vv 2 11vv+10 22) pp 8 1 pp 2 12pp+32 pp 10 23) 2rr rr+6 2rr 7rr+42 24) vv2 +10vv+9 3vv+4 vv 9 3vv+4 25) kk 7 7kk2 28kk kk 2 kk 12 8kk 2 56kk 26) nn 7 9nn+54 nn 2 2nn 35 10nn+50 27) nn2 +2nn+1 nn nn2 16 5nn+4 28) xx2 1 xx 2 4 xx2 +xx 2 2xx 4 xx 2 xx 2 3xx 6 29) aa aa 6bb aa2 4bb 2 aa 2 +3aaaa+2bb 2 3aa+9 aa+2bb 30) xx2 +3xx 10 2xx2 xx 3 8xx+20 xx 2 +6xx+5 2xx 2 +xx 6 6xx

291 Chapter 10 SECTION 10.3 OBTAIN THE LOWEST COMMON DENOMINATOR As with fractions in arithmetic, the least common denominator or LCD is the lowest common multiple (LCM) of the denominators. Since rational expressions are fractions with polynomials, we use the LCD to add and subtract rational expression with different denominators. In this section, we obtain LCDs of rational expressions. First, let s take a look at the method in finding the LCM in arithmetic. A. OBTAIN THE LCM IN ARITHMETIC REVIEW To find the LCM using the prime factorization: 1) Find the prime factorization for each number by using the factor tree 2) Write each number in the exponential form 3) Collect all prime factors that show up in all numbers with the highest exponent 4) Multiply all the prime factors that collected in step 3 to find the LCM Determining the Least Common Multiple Using Prime Factorization (Duration 4:41) Determine the least common multiple (LCM): a) 16 and 18 b) 72 and 54 YOU TRY Find LCM. a) Find LCM of 3, 6, and 15 using the prime factorization method. b) Find LCM of 25, 315 and 150 using the prime factorization method. 285

292 B. OBTAIN THE LCM WITH MONOMIALS Find the LCM with monomials (Duration 2:55) Chapter 10 To find the LCM/LCD of monomials: Use factors with exponents. Find the LCM of the monomials below. a) 5xx 3 yy 2 and 4xx 2 yy 5 b) 7aabb 2 cc and 3aa 3 bb YOU TRY Find LCM: a) 4xx 2 yy 5 and 6xx 4 yy 3 zz 6 b) 12aa 2 bb 5 and 18aaaaaa C. OBTAIN THE LCM WITH POLYNOMIALS We use the same method, but now we factor using factoring techniques to obtain the LCM between polynomials. Recall, all factors are contained in the LCM. Find the LCM of polynomials (Duration 4:45) To find the LCM/LCD of polynomials: Use factors with exponents. This means we must first. Find the LCM of the following polynomials. a) xx 2 + 3xx 18 and xx 2 + 4xx 21 b) xx 2 10xx + 25 and xx 2 xx

293 YOU TRY Chapter 10 Find the LCM of the following polynomials. a) xx 2 + 2xx 3 aaaaaa xx 2 xx 12 b) xx 2 10xx + 25 aaaaaa xx 2 14xx + 45 D. REWRITE FRACTIONS WITH THE LOWEST COMMON Identify LCD and build up to matching denominators (Duration 4:59 ) Example: 5aa a) 4bb 3 cc and 3cc 6aa 2 bb b) 5xx and xx 2 xx 2 5xx 6 xx 2 +4xx+3 YOU TRY Find the LCD between the two fractions. Rewrite each fraction with the LCD. a) 5aa aaaaaa 3cc 4bb 3 cc 6aa 2 bb b) 5xx xx 2 5xx 6 aaaaaa xx 2 xx 2 +4xx+3 287

294 EXERCISE Find the equivalent numerator. Chapter 10 1) 3 8 =? 48 3) aa =? 5 5aa 5) (xx 4) (xx+2) =? xx 2 +5xx+6 2) aa xx =? xxxx 4) 6) 2 =? xx+4 xx =? 3aa 2 bb 2 cc 9aa 5 bb 2 cc 4 Find the lowest common multiple. 7) 2aa 3, 6aa 4 bb 2 aaaaaa 4aa 3 bb 5 8) xx 2 3xx, xx 3 aaaaaa xx 9) xx + 2 aaaaaa xx 4 10) xx 2 25 aaaaaa xx ) xx 2 + 3xx + 2 aaaaaa xx 2 + 5xx ) 5xx 2 yy aaaaaa 25xx 3 yy 5 zz 13) 4xx 8, xx 2 aaaaaa 4 14) xx, xx 7 aaaaaa xx ) xx 2 9 aaaaaa xx 2 6xx ) xx 2 7xx + 10, xx 2 2xx 15, aaaaaa xx 2 + xx 6 Find the LCD and rewrite each fraction with the LCD. 17) 3aa 5bb 2 and 2 10aa 3 bb 18) xx+2 xx 3 and xx 3 xx+2 xx 19) xx 2 16 and 3xx xx 2 8xx+16 20) xx+1 xx 2 36 and 2xx+3 xx 2 +12xx+36 4xx 21) xx 2 xx 6 and xx+2 xx 3 22) 3xx xx 4 and 2 xx ), 2 xx 2 6xx xx 3 and xx 6 24) 5xx+1 xx 2 3xx 10 and 4 xx 5 25) 3xx+1 xx 2 xx 12 and 2xx xx 2 +4xx+3 3xx 26) xx 2 6xx+8 xx 2 xx 2 +xx 20 and 5 xx 2 +3xx

295 Chapter 10 SECTION 10.4: ADD AND SUBTRACT RATIONAL EXPRESSIONS Adding and subtracting rational expressions are identical to adding and subtracting with fractions. Recall, when adding with a common denominator, we add across numerators and keep the same denominator. This is the same method we use with rational expressions. Note, methods never change, only problems. Helpful tips when adding and subtracting rational expressions: For adding and subtracting with rational expressions, here are some helpful tips: Identify the denominators: are they the same or different? Combine the rational expressions into one expression. Once combined into one expression, then reduce the fraction, if possible. A fraction is reducible only if there is a GCF in the numerator. A. ADD OR SUBTRACT RATIONAL EXPRESSIONS WITH A COMMON DENOMINATOR Recall. We can use the same properties for adding or subtracting fractions with common denominators also for adding and subtracting rational expressions with common denominators: aa cc ± bb aa ± bb = cc cc Add/ subtract rational expressions with common denominator (Duration 5:00) Add/subtract rational expressions Add the and keep the When subtracting, we will first the negative. Don t forget to Example: a) xx 2 +4xx xx 2 2xx 15 + xx+6 xx 2 2xx 15 b) xx 2 +2xx 6xx+5 2xx 2 9xx 5 2xx 2 9xx 5 289

296 YOU TRY Chapter 10 Evaluate. a) Add: xx 4 xx 2 2xx 8 + xx+8 xx 2 2xx 8 b) Subtract: 6xx 12 15xx 6 3xx 6 3xx 6 B. ADD AND SUBTRACT RATIONAL EXPRESSIONS WITH UNLIKE DENOMINATORS Recall. We can use the same properties for adding and subtracting integer fractions with unlike denominators for adding and subtracting rational expressions with unlike denominators: aa bb ± cc aaaa ± bbbb = dd bbbb Add rational expressions with different denominators (Duration 4:56) Add/subtract rational expressions with different denominators To add or subtract, we the denominators by by the missing This means we have to to find the LCD. Example: a) 2xx xx xx 2 +xx 6 290

297 Subtract rational expressions with different denominators (Duration 5:00) Chapter 10 Example: b) 2xx+7 xx 2 2xx 3 3xx 2 xx 2 +6xx+5 Warning: We are not allowed to reduce terms, only factors. YOU TRY 7aa a) Add + 4bb 4 3aa 2 bb 6aaaa4 b) Subtract 7bb 5aa 4aa 2 6 c) Add + 3aa 8aa+4 8 d) Subtract xx+1 xx+1 xx 4 xx 2 7xx

298 EXERCISE Add or subtract the rational expressions. Simplify completely. 2 1) + 4 aa+3 aa+3 2) tt2 +4tt tt 1 3) 5 6rr 5 8rr 4) + 2tt 7 tt tt 2 6tt 2 Chapter 10 5) aa+2 aa ) xx 1 2xx+3 4xx xx 7) 5xx+3yy 2xx 2 yy 3xx+4yy xxxx 2 8) 2zz zz 1 3zz zz+1 9) 8 3 xx 2 4 xx+2 10) tt 5 tt 3 4tt 12 11) 2 4 5xx 2 +5xx 3xx+3 12) tt yy yy tt yy+tt 13) xx 2 xx 2 +5xx+6 xx 2 +3xx+2 14) 2xx 4 xx 2 1 xx 2 +2xx 3 15) 4 aa2 aa 2 aa aa 17) xxxx 2 xx 2 yy 16) xx2 6xx 8 xx 2 xx 2 18) 2aa 1 3aa 2 + 5aa+1 9aa 19) 2cc dd cc 2 dd cc+dd cccc 2 20) 2 xx xx+1 21) 23) 25) 27) 29) xx 5 4xx 3aa 4aa aa 6aa 30 2xx + 5 xx 2 9 xx 2 +xx 6 xx 1 xx 2 +3xx+2 + xx+5 xx 2 +5xx+4 2rr rr 2 ss 2 rr+ss rr ss 22) 24) 26) 4xx xx xx xx+5 2xx 3 xx 2 1 xx 2 +5xx+4 4xx 3 xx 2 2xx 3 xx 2 5xx+6 28) 3xx+2 3xx ) xx 4 xx 2 xx+2 + 4xx+5 xx 2 4xx+3 xx 2 +4xx 5 292

299 CHAPTER REVIEW KEY TERMS AND CONCEPTS Chapter 10 Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson. Rational expression Undefined rational expression 293

300 Chapter

301 Chapter 11 CHAPTER 11: RATIONAL EQUATIONS AND APPLICATIONS Chapter Objectives By the end of this chapter, students should be able to: Identify extraneous values Apply methods of solving rational equations to solve rational equations Solve applications with rational equations including revenue, distance, and work-rate problems Contents CHAPTER 11: RATIONAL EQUATIONS AND APPLICATIONS SECTION 11.1: RATIONAL EQUATIONS A. EXCLUDED VALUES REVIEW B. SOLVE RATIONAL EQUATIONS BY CLEARING DENOMINATORS WITH THE LCD C. FACTORING DENOMINATORS D. SOLVING RATIONAL EQUATIONS WITH EXTRANEOUS SOLUTIONS EXERCISE SECTION 11.2: WORK-RATE PROBLEMS A. ONE UNKNOWN TIME B. TWO UNKNOWN TIMES EXERCISE SECTION 11.3: UNIFORM MOTION PROBLEMS A. UNIFORM MOTION PROBLEMS B. UNIFORM MOTION PROBLEMS WITH STREAMS AND WINDS EXERCISE SECTION 11.4: REVENUE PROBLEMS EXERCISE CHAPTER REVIEW

302 SECTION 11.1: RATIONAL EQUATIONS When solving rational equations, we can solve by using the same strategy we used to solve linear equations with fractions: clearing denominators. However, we first need to revisit excluded values. A. EXCLUDED VALUES REVIEW Chapter 11 Note A rational expression is undefined where the denominator is zero. Find excluded values of a rational expression (Duration 4:20) Find values where a rational expression is undefined. a) 17 4xx 32 b) xx+19 (xx 10)(xx 4) c) 9xx+10 xx 2 9xx

303 Chapter 11 Definition Recall, the excluded values are values which make the expression undefined. Hence, when solving a rational equation, the solution(s) is any value(s) except the excluded values. If we obtain a solution that is an excluded value, we call this an extraneous solution. B. SOLVE RATIONAL EQUATIONS BY CLEARING DENOMINATORS WITH THE LCD Steps for solving rational equations with the same denominator Step 1. Determine the excluded values of the equation. Step 2. Clear denominators by multiplying each term by the lowest common denominator. Step 3. Solve the equation. Step 4. Verify that the solutions obtained are not an excluded value. Solve rational equations by clearing the denominators-part 1 (Duration 9:15) Example: Solve the rational equations. a) xx = 3xx b) 4 xx = 3 4 6: 8: LCD: LCD= Excluded values: Excluded values: YOU TRY a) Solve for xx: 2 3 xx 5 6 = 3 4 b) Solve for xx: 9 xx =

304 Solve rational equations by clearing the denominators - part 2 (Duration 4:32) Chapter 11 Example: Solve the rational equations. c) 2 5xx 3 = 4 xx xx d) 2xx 16 xx = 4 xx YOU TRY a) Solve for xx: 5xx+5 xx+2 +3xx = xx 2 xx+2 b) Solve for xx : xx xx xx+1 = 5 (xx+1)(xx+2) 298

305 C. FACTORING DENOMINATORS Often we will need to factor denominators before finding the LCD. Chapter 11 Solve rational equations with factoring the denominators first- Part 1 (Duration 4:27) Example: Solve the rational equations. a) xx + xx = xx+3 xx 2 xx 2 4 xx+2 Excluded values: Solve rational equations with factoring the denominators first Part 2 (Duration 4:45) Example: Solve the rational equations. b) xx 3 xx+6 + xx 2 xx 3 = xx 2 xx 2 +3xx 18 Excluded values: 299

306 YOU TRY a) Solve for tt: tt 1 = 11 tt 1 tt 2 tt 2 3tt+2 b) Solve for xx: 2xx xx+1 + 3xx xx+1 = xx2 xx 2 +2xx+1 Chapter 11 D. SOLVING RATIONAL EQUATIONS WITH EXTRANEOUS SOLUTIONS Solve a rational equation with no solution (Duration 5:07 ) Solve the rational equations. xx a) 6xx 36 9 = 1 xx 6 Excluded values: 300

307 Chapter 11 Solve a rational equation with extraneous solutions (Duration 5:00 ) Rational equations extraneous Because we are working with fractions, the cannot be. Solve the rational equations. xx b) 2 = 3xx+56 xx 8 xx 4 xx 2 12xx+32 c) xx + 2 = xx 2 xx 4 4xx 12 xx 2 6xx+8 YOU TRY a) Solve for nn: nn nn+5 2 nn 9 = 11nn+15 nn 2 4nn

308 EXERCISE Solve. Be sure to verify all solutions. Chapter 11 1) 3xx xx = 0 2) xx + 20 xx 4 = 5xx xx 4 2 3) xx + 6 = 2xx xx 3 xx 3 4) 2xx = 4xx+5 3 3xx 4 6xx 1 3xx 4 5) 3mm 7 = 3 2mm 5 3mm+1 2 6) 4 xx 1 xx = 12 3 xx 7) = 3 3 xx 2 4 xx 8) xx 8 xx = 1 4 9) xx + 1 = xx+1 10) xx 4 xx 1 = 12 xx ) 4xx 4 = 1 2xx 6 5xx ) xx 1 xx 3 + xx+2 xx+3 = ) xx 2 xx+3 1 xx 2 = 1 xx 2 +xx 6 14) 3xx 5 5xx 5 + 5xx 1 7xx 7 xx 4 1 xx = 2 15) xx 5 xx 9 + xx+3 xx 3 = 4xx2 xx 2 12xx+27 16) 2xx xx+1 3 xx+5 = 8xx2 xx 2 +6xx+5 17) 2xx xx xx 4 = 3xx xx 2 2xx 8 18) xx+2 3xx 1 1 xx = 3xx 3 3xx 2 xx 19) 6xx = 2xx 2 2xx xx 2 1 3xx xx ) xx 3 xx+6 + xx 2 xx 3 = xx 2 xx 2 +3xx

309 SECTION 11.2: WORK-RATE PROBLEMS Chapter 11 Work-rate equation If the first person does a job in time A, a second person does a job in time B, and together they can do a job in time T (total). We can use the work-rate equation: 11 AA + 11 BB jjjjjj pppppp tttttttt AA jjoooo pppppp tttttttt BB jjjjjj pppppp tttttttt TT(tttttttt) = 11 TT A. ONE UNKNOWN TIME Work- rate problem (Duration 4:45) Adam does a job in 4 hours. Each hour he does of the job. Betty does a job in 12 hours. Each hour she does of the job. Together, each hour they do of the job. This means it takes them, working together, hours to do the entire job. Work Equation: Use! Example 1: Catherine can paint a house in 15 hours. Dan can paint it in 30 hours. How long will it take them working together? Catherine: hours Dan: hours Team: hours Example 2: Even can clean a room in 3 hours. If his sister Faith helps, it takes them 2 2 hours. How long 5 will it take Faith working alone? Even: hours Faith: hours Team: hours 303

310 YOU TRY Chapter 11 a) If worker A can do a piece of work alone in 6 days and worker B can do it alone in 4 days, how long will it take the two working together to complete the job? Time Job per day Worker A Worker B Together a) Adam can assemble a furniture set in 5 hours. If his sister Maria helps, they can finish it in 3 hours. How long will it take Maria to do the job alone? Time Job per hour Adam Maria Together Fill and drain problem (Duration 0:54) Example: One inlet pipe can fill an empty pool in 8 hours, and a drain can empty the pool in 12 hours. How long will it take the pipe to fill the pool if the drain is left open? Inlet pipe Drain Together Time Rate Fill-drain equation : 304

311 YOU TRY Chapter 11 a) A sink can be filled by a pipe in 5 minutes, but it takes 7 minutes to drain a full sink. If both the pipe and the drain are opened, how long will it take to fill the sink? Time Fill per minute Fill the sink Drain the sink Together B. TWO UNKNOWN TIMES Solve work- rate problems with 2 unknowns. (Duration 7:57 ) Example: If Alfonso does a job in 30 hours less than Zoe, and they can do the job together in 8 hours, long will it take each to do the job alone? Time (hours) Work rate Alfonso Zoe Together 305

312 YOU TRY Chapter 11 a) Mike takes twice as long as Rachel to complete a project. Together, they can complete a project in 10 hours. How long will it take each of them to complete a project alone? Time Project/ hour Mike Rachel Together b) Brittney can build a large shed in 10 days less than Cosmo. If they built it together, it would take 12 days. How long would it take each of them working alone? Time Built per day Cosmo Brittney Together 306

313 Chapter 11 EXERCISE 1) A tank can be filled by one pipe in 20 minutes and by another in 30 minutes. How long will it take both pipes together to fill the tank? 2) Tim can finish painting his barn in 10 hours. It takes his wife JoAnn only 8 hours to do the same job. If they work together, how long will it take them to complete the job? 3) Adan can do a piece of work in 3 days, Bernie in 4 days, and Cynthia in 6 days each working alone. How long will it take them to do it working together? 4) A carpenter and his assistant can do a piece of work in 3 days. If the carpenter himself could do the work alone in 5 days, how long would the assistant take to do the work alone? 5) A sink can be filled from the faucet in 5 minutes. It takes only 3 minutes to empty the sink when the drain is open. If the sink is full and both the faucet and the drain are open, how long will it take to empty the sink? 6) Of two inlet pipes, the smaller pipe takes 5 hours longer than the larger pipe to fill a pool. When both pipes are open, the pool is filled in 6 hours. If only the larger pipe is open, how many hours are required to fill the pool? 7) It takes John 16 minutes longer than Sally to mow the lawn. If they work together they can mow the lawn in 15 minutes. How long will it take each to mow the lawn if they work alone? 8) Bill s father can paint a room in 3 hours less than Bill can paint it. Working together they can complete the job in 2 hours. How much time would each require working alone? 9) Two workers, a trainer and a trainee, working together can do a job in 3 hours. The trainer is 3 times faster than the trainee to complete the same job. How long will it take the trainee to finish the same job? 10) The faucet alone can fill the sink in 6 minutes, while it takes 8 minutes to empty it with the drain. How long will it take to fill the sink? 11) It takes Roberto 8 hours longer than Paula to repair a transmission. If it takes them 3 hours to do the job if they work together, how long will it take each of them working alone? 12) A water tank is being filled by two inlet pipes. Pipe A can fill the tank in 4 hours, while both pipes together can fill the tank in 2 hours. How long does it take to fill the tank using only Pipe B? 13) Cheng takes 10 hours longer to pave a driveway than Sammy to do a job. Working together they can do the job in 12 hours. How long does it take each working alone? 307

314 Chapter 11 SECTION 11.3: UNIFORM MOTION PROBLEMS We can recall uniform motion problems in the word problems chapter. We used the formula rr tt = dd and organized the given information in a table. Now, we use the equation as: tt = dd rr We apply the same method in this section only the equations will be rational equations. A. UNIFORM MOTION PROBLEMS Uniform motion applications part 1: set up the equation (Duration 4:09) Uniform motion applications part 2: solve the equation (Duration 5:26) Caitlyn went on a 56 mile trip to a soccer game. On the way back, due to road construction she had to drive 28 miles per hour slower. This made the trip take 1 hour longer. How fast did she drive to the soccer game? Recall: To r t d From 308

315 YOU TRY Chapter 11 a) Greg went to a conference in a city 120 miles away. On the way back, due to road construction he had to drive 10 mph slower which resulted in the return trip taking 2 hours longer. How fast did he drive on the way to the conference? 309

316 Chapter 11 B. UNIFORM MOTION PROBLEMS WITH STREAMS AND WINDS Another type of uniform motion problem is where a boat is traveling in a river with the current or against the current (or an airplane is flying with the wind or against the wind). If a boat is traveling downstream, the current will push it or increase the rate by the speed of the current. If a boat is traveling upstream, the current will pull against it or decrease the rate by the speed of the current. Uniform motion Streams & winds Part 1: set up the equation (Duration 4:28) Uniform motion Streams & winds Part 2: solve the equation (Duration 5:55) Alicia rows a boat downstream for 182 miles. The return trip upstream took 12 hours longer. If the current flows at 3 mph, how fast does Alicia row in still water? Recall: Uniform motion: Rational equation: Downstream (faster) Upstream (slower) r t d 310

317 Chapter 11 YOU TRY a) A man rows downstream for 30 miles then turns around and returns to his original location, the total trip took 8 hours. If the current flows at 2 miles per hour, how fast would the man row in still water? 311

318 Chapter 11 EXERCISE 1) An athlete plans to row upstream a distance of 6 kilometers and then return to his starting point in a total time of 4 hours. If the rate of the current is 2 km/hr., how fast should he row? 2) A pilot flying at a constant rate against a headwind of 30 km/hr flew for 720 kilometers, then reversed direction and returned to his starting point. He completed the round trip in 10 hours. What was the speed of the plane? 3) Tim, an open water swimmer, is training for the Olympics. To do so, he swims in a stream that is 3m/h. Tim finds that he can swim 4 miles against the current in the same amount of time that he can swim 10 miles with the current. How fast can Tim swim with no current? 4) A plane flies against the wind 288 miles from LAX to San Jose and then returns home with the same wind. The wind speed is 60m/h. If the total flying time was 2 hours, what is the speed of the plane? 5) A salmon is swimming in a river that is flowing downstream at a speed of 2 miles per hour. The salmon can swim 12 miles upstream in the same amount of time it would take to swim 24 miles downstream. What is the speed of the salmon in still water? 312

319 Chapter 11 SECTION 11.4: REVENUE PROBLEMS Revenue problems are the problems where a person buys a certain number of items for a certain price per item. If we multiply the number of items by the price per item, we will get the total value. We can recall revenue problems in the word problems chapter. We used the formula AAAAAA: Amount Value = Total or Number of items Price = Revenue n p = R Rational equation applications Revenue problems (Duration 9:16) Revenue table: Number of items n Price p Revenue R To solve: Divide by Example 1: A group of college students bought a couch for $80. However, five of them failed to pay their share so the others had to each pay $8 more. How many students were in the original group? n p R Example 2: A merchant bought several pieces of silk for $70. He sold all but two of them at a profit of $4 per piece. His total profit was $18. How many pieces did he originally purchase? n p R 313

320 YOU TRY Chapter 11 a) A man buys several fish for $56. After three fish die, he decides to sell the rest at a profit of $5 per fish. His total profit was $4. How many fish did he buy to begin with? b) A group of students bought a couch for their dorm at cost $96. However, 2 students failed to pay their share, so each student had to pay $4 more. How many students were in the original group? 314

321 Chapter 11 EXERCISE 1) A merchant bought some pieces of silk for $900. He marked up $15 more per each piece and made $75 profit. Find the number of pieces purchased. 2) A group of students planned to chip in to raise a total of $100 for one of their friends birthday party but 5 persons couldn t come so they didn t pay. This increased the share of the others by $1 each. Find the amount that each person paid after. 3) A merchant bought a number of barrels of apples for $120. He kept 2 barrels and sold the remainder at a profit of $2/barrel making a total profit of $34. How many barrels did he originally buy? 4) A dealer bought a number of sheep for $440. After 5 had died, he sold the remainder at a profit of $2 each, making a profit of $60 for the sheep. How many sheep did he originally purchase? 5) Orlando bought a number of hats at equal cost for $500. He sold all but 2 for $540 at a profit of $5 for each item. How many hats did he buy? 6) A fashion store bought a lot of suits for $750. The store sold all of them for $1,000 making a profit of $10 on each suit sold. How many suits did the store buy? 7) A group of schoolboys tried to raise for $4,500 for a local animal shelter. Five boys transferred to different schools before they had a chance to raise money, hence each remaining boy was compelled to raise $45 more. How many boys were in the original group and how much had each agreed to raise? 8) The total expenses of a camping party were $72. If there had been 3 fewer persons in the party, it would have cost each person $2 more than it did. How many people were in the party and how much did it cost each one? 315

322 CHAPTER REVIEW KEY TERMS AND CONCEPTS Chapter 11 Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson. Extraneous solution Work-rate equation 316

323 Chapter 12 CHAPTER 12: RADICALS Chapter Objectives By the end of this chapter, students should be able to: Simplify radical expressions Rationalize denominators (monomial and binomial) of radical expressions Add, subtract, and multiply radical expressions with and without variables Solve equations containing radicals Contents CHAPTER 12: RADICALS SECTION 12.1 INTRODUCTION TO RADICALS A. INTRODUCTION TO PERFECT SQUARES AND PRINCIPAL SQUARE ROOT B. INTRODUCTION TO RADICALS C. SIMPLIFY RADICALS WITH PERFECT PRINCIPAL nnnnnn ROOT D. SIMPLIFY RADICALS WITH PERFECT PRINCIPAL nnnnnn ROOT USING EXPONENT RULE E. SIMPLIFY RADICALS WITH NO PERFECT ROOT F. SIMPLIFY RADICALS WITH COEFFICIENTS G. SIMPLIFY RADICALS WITH VARIABLES WITH NO PERFECT RADICANTS EXERCISE SECTION 12.2: ADD AND SUBTRACT RADICALS A. ADD AND SUBTRACT LIKE RADICALS B. SIMPLIFY, THEN ADD AND SUBTRACT LIKE RADICALS EXERCISE SECTION 12.3: MULTIPLY AND DIVIDE RADICALS A. MULTIPLY RADICALS WITH MONOMIALS B. DISTRIBUTE WITH RADICALS C. MULTIPLY RADICALS USING FOIL D. MULTIPLY RADICALS WITH SPECIAL-PRODUCT FORMULAS E. SIMPLIFY QUOTIENTS WITH RADICALS EXERCISE SECTION 12.4: RATIONALIZE DENOMINATORS A. RATIONALIZING DENOMINATORS WITH SQUARE ROOTS B. RATIONALIZING DENOMINATORS WITH HIGHER ROOTS C. RATIONALIZE DENOMINATORS USING THE CONJUGATE EXERCISE SECTION 12.5: RADICAL EQUATIONS

324 Chapter 12 A. RADICAL EQUATIONS WITH SQUARE ROOTS B. RADICAL EQUATIONS WITH TWO SQUARE ROOTS C. RADICAL EQUATIONS WITH HIGHER ROOTS EXERCISE CHAPTER REVIEW

325 SECTION 12.1 INTRODUCTION TO RADICALS A. INTRODUCTION TO PERFECT SQUARES AND PRINCIPAL SQUARE ROOT Chapter 12 Introduction to square roots (Duration 7:03 ) Some numbers are called. It is important that we can recognize when working with square roots. 1 2 = 1 1 = 6 2 = 6 6 = 2 2 = 2 2 = 7 2 = 7 7 = 3 2 = 3 3 = 8 2 = 8 8 = 4 2 = 4 4 = 9 2 = 9 9 = 5 2 = 5 5 = 10 2 = = To determine the square root of a number, we have a special symbol. 9 The square root of a number is the number times itself that equals the given number. 9 = 36 = 49 = 81 = You can think of the square root as the opposite or inverse of squaring. Actually, numbers have two square roots. One is positive and one is negative. 5 5 = 25 and 5 5 = 25 To avoid confusion 25 = 5 and 25 = 5 What about these square roots?

326 YOU TRY Chapter 12 a) Find the perfect square of: 11 2 = 12 2 = 13 2 = 14 2 = 15 2 = 16 2 = 17 2 = 18 2 = 19 2 = 20 2 = b) Find the square root of: 441 = 484 = 529 = 576 = 625 = 676 = 729 = 784 = 841 = 900 = Principal n th square roots vs. general square roots (Duration 5:23 ) Note: In this class, we will only consider the principal nn tttt roots when we discuss radicals. B. INTRODUCTION TO RADICALS Radicals are a common concept in algebra. In fact, we think of radicals as reversing the operation of an exponent. Hence, instead of the square of a number, we square root a number; instead of the cube of a number, we cube root a number to reverse the square to find the base. Square roots are the most common type of radical used in algebra. Definition If aa is a positive real number, then the principal square root of a number aa is defined as aa = bb if and only if aa = bb 22 The is the radical symbol, and aa is called the radicand If given something like aa, then 3 is called the root or index; hence, aa third root of aa. In general, nn aa = bb if and only if aa = bb nn is called the cube root or If nn is even, then aa and bb must be greater than or equal to zero. If nn is odd, then aa and bb must be any real number. Here are some examples of principal square roots: 1 = = 11 4 = = 25 9 = 3 81 is not a real number The final example 81 is not a real number. Since square root has the index is 2, which is even, the radicand must be greater than or equal to zero and since 81 < 0, then there is no real number in which we can square and will result in 81,i.e.,? 2 = 81. So, for now, when we obtain a radicand that is negative and the root is even, we say that this number is not a real number. There is a type of number where we can evaluate these numbers, but just not a real one. 320

327 Introduction to square roots, cube roots, and N th roots (Duration 9:09) Chapter 12 The principal nn tttt root of aa is the nn tttt root that has the same sign as aa, and it is denoted by the radical symbol. nn aa We read this as the,, or. The positive integer of the radical. If nn = 2, the index. The number. 4 = 4 16 = Square roots (n = 2) 1 = 4 = 9 = 16 = 25 = 4 = 4 16 = 1 = 4 = 9 = 16 = 25 = Cube roots (n = 3) = 1 = 8 = 8 = 3 27 = = 64 = 64 = = = Example: Simplify 1) 36 = 2) 81 = 3) 4 9 = 3 4) 64 = 5 5) 32 3 = 6) 8 = 321

328 Inverse properties of nn tttt Powers and nn tttt Roots If aa has a principal nn tttt root, then. Chapter 12 If nn is odd, then. If nn is even, then. We need the for any nn tttt root with an exponent for which the index is to assure the nn tttt root is. Example: Simplify 1) xx 2 3 2) xx 9 4 3) xx 8 4 4) yy 12 C. SIMPLIFY RADICALS WITH PERFECT PRINCIPAL nn tttt ROOT Simplify perfect nn tttt roots (Duration 4:04 ) Example: a) 81 b) c) 16 d)

329 Simplify perfect nn tttt roots negative radicands (Duration 4:32 ) Chapter 12 Example: Simplify each of the following. 4 a) 16 5 b) 32 6 c) 64 = = = YOU TRY Simplify. Show your work. a) 36 b) c) d) 1 D. SIMPLIFY RADICALS WITH PERFECT PRINCIPAL nn tttt ROOT USING EXPONENT RULE There is a more efficient way to find the nn tth root by using the exponent rule but first let s learn a different method of prime factorization to factor a large number to help us break down a large number into primes. This alternative method to a factor tree is called the stacked division method. Prime factorization stacked division method (Duration 3:45) a) 1,350 b)

330 Simplify perfect root radicals using the exponent rule (Duration 5:00 ) Chapter 12 nn Roots: mm where nn is the Roots of an expression with exponents: the by the. Example: Simplify. a) 46,656 = 5 b) 1,889,568 = Simplify perfect root radicals with variables (Duration 5:43 ) Example: Simplify. 3 a) zz 9 b) mm 6 5 c) nn 10 YOU TRY Simplify the following radicals using the exponent rule. Show your work. 6 a) 64 3 b) 729 c) xx 2 yy 4 zz 10 7 d) xx 21 yy

331 Chapter 12 E. SIMPLIFY RADICALS WITH NO PERFECT ROOT Not all radicands are perfect squares, where when we take the square root, we obtain a positive integer. For example, if we input 8 in a calculator, the calculator would display and even this number is a rounded approximation of the square root. To be as accurate as possible, we will leave all answers in exact form, i.e., answers contain integers and radicals no decimals. When we say to simplify an expression with radicals, the simplified expression should have a radical, unless the radical reduces to an integer a radicand with no factors containing perfect squares no decimals Following these guidelines ensures the expression is in its simplest form. Product rule for radicals If aa, bb are any two positive real numbers, then aaaa = aa bb In general, if aa, bb are any two positive real numbers, then Where nn is a positive integer and nn 22. nn aaaa nn = aa nn bb Simplify square roots with not perfect square radicants (Duration 7:03) Recall: The square root of a square For a non-negative real number, aa: aa 22 = aa For example: 25 = 5 5 = 5 2 = 5 The product rule for square roots Given that aa and bb are non-negative real numbers,. 45 =. Example: 8 = 48 = 150 = 1,350 = 325

332 Simplify radicals with not perfect radicants using exponent rule (Duration 4:22) Chapter 12 To take roots we the by the index aa 2 bb = nn aa nn bb = When we divide if there is a remainder, the remainder. Example: a) 72 3 b) 750 YOU TRY Simplify. Show your work. a) 75 3 b) 200 F. SIMPLIFY RADICALS WITH COEFFICIENTS Simplify radicals with coefficients (Duration 3:52) If there is a coefficient on the radical: by what. Example: a) b) 3 96 YOU TRY Simplify. a) 5 63 b)

333 G. SIMPLIFY RADICALS WITH VARIABLES WITH NO PERFECT RADICANTS Simplify radicals with variables (Duration 4:22) Chapter 12 Variable in radicals: the by the Remainders: Example: 4 a) aa 13 bb 23 cc 10 dd 3 b) 125xx 4 yyzz 5 YOU TRY Simplify. Assume all variables are positive. a) xx 6 yy 5 b) 5 18xx 4 yy 6 zz 10 c) 20xx 5 yy 9 zz 6 327

334 EXERCISE Simplify. Show all your work. Assume all variables are positive. Chapter 12 1) 245 2) 36 3) 12 4) ) ) ) 192nn 8) 196vv 2 9) 252xx 2 10) 100kk 4 11) 7 64xx 4 12) 5 36mm 13) 4 175pp 4 14) 8 112pp 2 15) 2 128nn 16) 45xx 2 yy 2 17) 16xx 3 yy 3 18) 320xx 4 yy 4 19) 32xxxx 2 zz 3 20) 5 245xx 2 yy 3 21) 2 180uu 3 vv 22) 72aa 3 bb 4 23) 2 80hjj 4 kk 24) 6 50aa 4 bbbb 2 25) 8 98mmmm 26) 512aa 4 bb 2 27) 100mm 4 nn 3 28) 8 180xx 4 yy 2 zz 4 29) 2 72xx 2 yy 2 30) 5 36xx 3 yy 4 Simplify. Show all your work. Assume all variables are positive ) ) ) ) ) 224nn ) 2 48vv ) 32xx 4 yy ) 2 375uu 2 vv ) 6 648xx 5 yy 7 zz ) ) 224pp ) 7 320nn ) 256xx 4 yy ) 3 192aaaa 2 50) 9 9xx 2 yy 5 zz ) 648aa ) 3 896rr 3 42) 135xx 5 yy ) 7 81xx 3 yy ) 6 54mm 8 nn 3 pp 7 328

335 Chapter 12 SECTION 12.2: ADD AND SUBTRACT RADICALS Adding and subtracting radicals are very similar to adding and subtracting with variables. In order to combine terms, they need to be like terms. With radicals, we have something similar called like radicals. Let s look at an example with like terms and like radicals. 2xx + 5xx (2 + 5)xx 7xx (2 + 5) 3 Notice that when we combined the terms with 3, it was similar to combining terms with xx. When adding and subtracting with radicals, we can combine like radicals just as like terms. Definition If two radicals have the same radicand and the same root, then they are called like radicals. If this is so, then aa xx ± bb xx = (aa ± bb) xx, Where aa, bb are real numbers and xx is some positive real number. In general, for any root nn, nn aa xx nn ± bb xx nn = (aa ± bb) xx, Where aa, bb are real numbers and xx is some positive real number. Note: When simplifying radicals with addition and subtraction, we will simplify the expression first, and then reduce out any factors from the radicand following the guidelines in the previous section. A. ADD AND SUBTRACT LIKE RADICALS Add and subtract like radicals (Duration 3:11) 7 3 Simplify: 2xx 5yy + 3xx + 2yy Simplify: When adding and subtracting radicals, we can. Example: 3 3 a) b) YOU TRY Simplify 5 5 a) b)

336 B. SIMPLIFY, THEN ADD AND SUBTRACT LIKE RADICALS Add or subtract radicals requiring simplifying first (Duration 3:46) Chapter 12 Guidelines for adding and subtracting radicals Example: Simplify 2 50xx xx 5 50 /\ 18 /\ Add or subtract radicals requiring simplifying first (continue) (Duration 5:12) Example: 3 a) b) xx xx 2 yy yy xx 5 yy 2 YOU TRY Simplify. 3 a) b)

337 EXERCISE Simplify. In this section, we assume all variables to be positive. Chapter 12 1) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) 3 18xx 5 8xx xx xx 5 22) 2 2xxxx 2xxxx + 3 8xxxx + 3 8xxxx 23) 2 6xx 2 54xx xx 2 yy 3xx 2 yy 24) 2xx 20yy 2 + 7yy 20xx 2 3xxxx 25) 3 24tt 3 54tt 2 96tt tt 331

338 SECTION 12.3: MULTIPLY AND DIVIDE RADICALS Chapter 12 Recall the product rule for radicals in the previous section: Product rule for radicals If aa, bb are any two positive real numbers, then aaaa = aa bb In general, if aa, bb are any two positive real numbers, then Where nn is a positive integer and nn 22. nn aaaa nn = aa nn bb As long as the roots of each radical in the product are the same, we can apply the product rule and then simplify as usual. At first, we will bring the radicals together under one radical, then simplify the radical by applying the product rule again. A. MULTIPLY RADICALS WITH MONOMIALS Multiply monomial radical expressions (Duration 10:32 ) To multiply two radicals with the same index. Multiply the together and multiply the together. Then simplify. nn Product rule (with coefficients): p uu nn qq vv = Example 1: 2 3 = 3 3 Example 2: = Multiply: 3 3 a) 15 6 b) c) d) e)

339 Chapter 12 YOU TRY Simplify: a) b) Note: In this section, we assume all variables to be positive. Multiply monomial radicals with variables (Duration 4:58 ) Example: Multiply. a) 18xx 3 30xx 2 3 b) 16xx xx 2 YOU TRY Simplify. 5 a) 8xx 2 5 4xx 3 b) 60xx 4 6xx 7 333

340 Chapter 12 B. DISTRIBUTE WITH RADICALS When there is a term in front of the parenthesis, we distribute that term to each term inside the parenthesis. This method is applied to radicals. Multiply square roots using Distributive property (Duration 2:25 ) Example: Multiplying radical expressions with variables using Distributive property (Duration 6:57 ) Example: a) xx 2 xx 3 b) 4 yy 5 xxyy 3 yy c) zz zz zz zz 8 YOU TRY Simplify. a) 7 6 ( ) b) xx 3 + 8xx 60xx 334

341 C. MULTIPLY RADICALS USING FOIL Multiply binomials with radicals (Duration 4:10) Chapter 12 Recall: (aa + bb)(cc + aa) = Always be sure your final answer is. Example: 3 a) b) Multiply binomials with radicals with variables (Duration 5:29) Example: a) 2 xx xx 4 b) 3xx xx xx 1 YOU TRY Simplify. a) ( 5 2 3)( ) b) 3 vv vv

342 D. MULTIPLY RADICALS WITH SPECIAL-PRODUCT FORMULAS Multiply radicals using the perfect square formula (Duration 3:44) Chapter 12 Recall the Perfect Square formula: (aa + bb) 2 = Always be sure your final answer is Example: a) b) Conjugates Recall the Difference of for two squares formula: (aa bb)(aa + bb) = aa 22 bb 22 Notice in the 2 factors (aa bb) and (aa + bb) have the same first and second term but there is a sign change in the middle. When we have 2 binomials like that, we say they are conjugates of each other. Example: Binomials Its conjugate xx + 5 xx The product of two conjugates is the Difference of two squares. This result is very helpful when multiplying radical expressions and rationalizing radicals in the later section of this chapter. Multiply radicals using the difference of squares formula (Duration 1:27) The Difference of Squares formula: (aa bb)(aa + bb) = = = = = 336

343 YOU TRY Chapter 12 a) Simplify: ( ) 2 b) Simplify: (8 5)(8 + 5) E. SIMPLIFY QUOTIENTS WITH RADICALS Quotient rule for radicals If aa, bb are any two positive real numbers, where bb 00, then aa bb = aa bb If aa, bb are any two positive real numbers, where bb 00, then Where nn is a positive integer and nn 22. nn aa bb nn = aa nn bb Divide radicals (Duration 3:44) Note: A rational expression is not considered simplified if there is a fraction under the radical or if there is a radical in the denominator. Example: a) b)

344 Divide radicals with variables (Duration 4:34 ) Chapter 12 Examples: a) 100xx, assume xx is positive 5xx b) 3 64xx 2 yy 5 3, assume yy is not 0 4yy 2 Divide expressions with radicals (Duration 4:20 ) Simplify expressions with radicals: Always the first Before with fractions, be sure to first! Examples: a) b) YOU TRY Simplify. a) b) 44yy 6 aa 4 9yy 2 aa 8 c)

345 EXERCISE Simplify. Assume all variables are positive. Chapter 12 1) ) ) 5 10rr 2 5rr 3 4) 12mm 15mm 3 5) 3 4aa aa 3 3 6) 4xx 3 3 2xx 4 7) 6( 2 + 2) 8) 5 10(5nn + 2) 9) 5 15( ) 10) 5 15( ) 11) 10( 5 + 2) 12) 15( 5 3 3vv) 13) ( )( 3 + 2) 14) ( 2 + 3)( ) 15) ( 5 4 3)( 3 4 3) 16) ( 5 5)(2 5 1) 17) ( 2aa + 2 3aa)(3 2aa + 5aa) 18) (5 2 1)( 2mm + 5) 19) ) ) ) 12pp2 3pp 5 20) ) ) ) ) 3 28)

346 29) xx 2 31) 4xxxx 3xxxx 30) 5 4 5rr4 4 8rr 2 32) ( ) 2 Chapter 12 33) (xx xx 5) 2 34) ( 3 7) 2 35) ( ) 2 36) ( 2 5)( 2 + 5) 37) ( xx yy)( xx + yy) 38) (4 2 3)( ) 39) (xx yy 3)(xx + yy 3) 40) (9 xx + yy)(9 xx yy) 340

347 SECTION 12.4: RATIONALIZE DENOMINATORS A. RATIONALIZING DENOMINATORS WITH SQUARE ROOTS Chapter 12 Rationalizing the denominator with square roots To rationalize the denominator with a square root, multiply the numerator and denominator by the exact radical in the denominator, e.g., 11 xx xx xx Rationalize monomials (Duration 3:42) Example: Simplify by rationalizing the denominator. a) b) Rationalize monomials with variables (Duration 4:58) Rationalize denominators: No in the To clear radicals: by the extra needed factors in denominator (multiply by the same on top!) It may be helpful to first (both and ). Example: 7aaaa a) 6aacc 2 b) 5xxyy3 15xxxxxx YOU TRY Simplify. a) 6 5 b) c)

348 Chapter 12 B. RATIONALIZING DENOMINATORS WITH HIGHER ROOTS Radicals with higher roots in the denominators are a bit more challenging. Notice, rationalizing the denominator with square roots works out nicely because we are only trying to obtain a radicand that is a perfect square in the denominator. When we rationalize higher roots, we need to pay attention to the index to make sure that we multiply enough factors to clear them out of the radical. Rationalize higher roots (Duration 4:20) Rationalize Monomial higher root Use the To clear radicals by extra needed factors in denominator (multiply by the same on top!) Hint: numbers! Example: a) 7 5 bb 2 3 b) 7 9aa 2 bb YOU TRY Simplify. a) b)

349 Chapter 12 C. RATIONALIZE DENOMINATORS USING THE CONJUGATE There are times where the given denominator is not just one term. Often, in the denominator, we have a difference or sum of two terms in which one or both terms are square roots. In order to rationalize these denominators, we use the idea from a difference of two squares: Rationalize denominators using the conjugate (aa + bb)(aa bb) = aa 2 bb 2 We rationalize denominators of the type aa ± bb by multiplying the numerator and denominator by their conjugates, e.g., 1 aa bb aa + bb aa bb = aa bb (aa) 2 ( bb) 2 The conjugate for aa + bb is aa bb aa bb is aa + bb The case is similar for when there is something like aa ± bb in the denominator. Rationalize denominators using the conjugate (Duration 4:56) Rationalize Binomials What doesn t work: Recall: Multiply by the Example: 6 a) 5 3 b)

350 Rationalize denominators using the conjugate (Duration 2:59) Chapter 12 Example: Rationalize the denominator. 2 a) YOU TRY Simplify. a) b) c)

351 EXERCISE Simplify. Assume all variables are positive. Chapter 12 1) ) ) ) ) ) ) ) ) xx 2 10) 2xx 3 11) xx vv 4 2vv 3 12) 4 1 5xx 13) ) ) ) ) ) ) ) ) 5 + 5xx ) 2+ 5rr 3 23) ) ) ) ) aa bb 28) bb aa 7 29) aa+ bb 30) aa bb aa+ bb 345

352 Chapter 12 SECTION 12.5: RADICAL EQUATIONS Here we look at equations with radicals. As you might expect, to clear a radical we can raise both sides to an exponent. Recall, the roots of radicals can be thought of reversing an exponent. Hence, to reverse a radical, we will use exponents. Solving radical equations If xx 00 and aa 00, then xx = aa if and only if xx = aa 22 If xx 00 and aa is a real number, then nn xx = aa if and only if xx = aa nn We assume in this chapter that all variables are greater than or equal to zero. We can apply the following method to solve equations with radicals. Steps for solving radical equations Step 1. Isolate the radical. Step 2. Raise both sides of the equation to the power of the root (index). Step 3. Solve the equation as usual. Step 4. Verify the solution(s). (Recall, we will omit any extraneous solutions.) A. RADICAL EQUATIONS WITH SQUARE ROOTS Solve equations with one radical (Duration 6:47) Solving equations having one radical 1. the radical on of the equation. 2. of the equation to the of the. 3. the resulting equation. 4.. Some solutions might. The solutions that are called solutions. xx 2 = 3 xx 3 = 346

353 Example: Solve. a) xx 7 = 11 b) 3xx = 0 Chapter 12 3 c) 2 5xx 1 8 = 0 d) xx + 6 = xx YOU TRY Solve for xx. a) 7xx + 2 = 4 b) xx + 3 = 5 c) xx + 4xx + 1 = 5 d) xx + 6 = xx

354 B. RADICAL EQUATIONS WITH TWO SQUARE ROOTS Chapter 12 Solve equations with two radicals (Duration 5:11) Solving equations having two radicals 1. Put on of the. 2. to the of the. 3. If one radical, the remaining radical and raise to the of the index again. (If the radicals have been eliminated skip this step.) 4. the resulting equation. 5. Check for. Example: Solve. a) 2xx + 3 xx 8 = 0 b) 3 + xx 6 = xx

355 Solve equations with two radicals part 2 (Duration 4:33 ) Chapter 12 Example: Solve the equation. 1 8xx 16xx 12 = 1 Solve equations with two radicals part 3 check solutions (Duration 3:27) Check solutions 349

356 YOU TRY Chapter 12 Solve for xx and check solutions a) 2xx + 1 xx = 1 b) 2xx + 6 xx + 4 = 1 Check solutions Check solutions 350

357 C. RADICAL EQUATIONS WITH HIGHER ROOTS Chapter 12 Solve equations with radicals odd roots (Duration 2:42) The opposite of taking a root is to do an. 3 xx = 4 then xx = Example: 3 a) 2xx 5 5 = 6 b) 4xx 7 = 2 YOU TRY Solve for nn. 3 a) nn 1 4 = 4 b) xx 2 6xx = 2 351

358 EXERCISE Solve. Be sure to verify all solutions. Chapter 12 1) 2xx = 0 2) 6xx 5 xx = 0 3) 3 + xx = 6xx ) 3 3xx 1 = 2xx 5) 4xx + 5 xx + 4 = 2 6) 2xx + 4 xx + 3 = 1 7) 2xx + 6 xx + 4 = 1 8) 6 2xx 2xx + 3 = 3 9) 5xx = 0 10) xx + 1 = xx ) xx 1 = 7 xx 12) 2xx + 2 = 3 + 2xx 1 13) 3xx + 4 xx + 2 = 2 14) 7xx + 2 3xx + 6 = 6 15) 4xx 3 = 3xx ) xx + 2 xx = ) xx = ) 5xx = ) 3 xx 3 = 12 20) 7xx + 15 = 1 352

359 CHAPTER REVIEW KEY TERMS AND CONCEPTS Chapter 12 Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson. Radicals Radicand Like-radicals Product rule for radicals Rationalize denominator process Conjugates To rationalize the denominator with square roots 353

360 Chapter

361 Chapter 13 CHAPTER 13: QUADRATIC EQUATIONS AND APPLICATIONS Chapter Objectives By the end of this chapter, students should be able to: Apply the Square Root Property to solve quadratic equations Solve quadratic equations by completing the square and using the Quadratic Formula Solve applications by applying the quadratic formula or completing the square Contents CHAPTER 13: QUADRATIC EQUATIONS AND APPLICATIONS SECTION 13.1: THE SQUARE ROOT PROPERTY A. SOLVE BASIC QUADRATIC EQUATIONS USING SQUAREROOT PROPERTY B. ISOLATE THE SQUARED TERM C. USE THE PERFECT SQUARE FORMULA EXERCISE SECTION 13.2: COMPLETING THE SQUARE A. COMPLETE THE SQUARE B. SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, a = C. SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, A EXERCISE SECTION 13.3: QUADRATIC FORMULA A. DETERMINANT OF A QUADRATIC EQUATION B. APPLY THE QUADRATIC FORMULA C. MAKE ONE SIDE OF AN EQUATION EQUAL TO ZERO EXERCISE SECTION 13.4: APPLICATIONS WITH QUADRATIC EQUATIONS A. PYTHAGOREAN THEOREM B. PROJECTILE MOTION C. COST AND REVENUE EXERCISE CHAPTER REVIEW

362 Chapter 13 We might recognize a quadratic equation from the factoring chapter as a trinomial equation. Although, it may seem that they are the same, but they aren t the same. Trinomial equations are equations with any three terms. These terms can be any three terms where the degree of each can vary. On the other hand, quadratic equations are equations with specific degree on each term. Definition A quadratic equation is a polynomial equation of the form aaaa 22 + bbbb + cc = 00 Where aaaa 22 is called the leading term, bbbb is call the linear term, and cc is called the constant coefficient (or constant term). Additionally, aa 00. SECTION 13.1: THE SQUARE ROOT PROPERTY A. SOLVE BASIC QUADRATIC EQUATIONS USING SQUAREROOT PROPERTY Square root property Let xx 00 and aa 00. Then xx 22 = aa if and only if xx = ± aa In other words, xx 22 = aa if and only if xx = aa or xx = aa Solve basic quadratic equations using square root property (Duration 2:53) Example: a) 8xx 2 = 648 b) xx 2 = 75 YOU TRY Solve. a) xx 2 = 81 b) xx 2 =

363 Solve equations with even exponents (Duration 4:26) Chapter 13 Consider: 5 2 = and ( 5) 2 = When we clear an even exponent, we have. Example: Solve. a) (5xx 1) 2 4 = 49 b) (3xx + 2) 4 = 81 YOU TRY Solve. a) (xx + 4) 2 = 25 b) (6xx 9) 2 =

364 Chapter 13 B. ISOLATE THE SQUARED TERM Let s look at examples where the leading term, or squared term, is not isolated. Recall, the squared term must be isolated to apply the square root property. Solve equations using square root property Isolating the squared term 1 st (Duration 5:00) Before we can clear an exponent, it must first be. Example: a) 4 2(2xx + 1) 2 = 46 b) 5(3xx 2) = 46 YOU TRY Solve. a) 5(3x 6) = 27 b) 5(r + 4) = 626 Note: When we have the other side of the equation of a squared term negative, the equation does not have a real solution. For example, the equation xx 2 = 1 does not have a real solution. There is a complex solution for this equation but we will not discuss it in this class. Example: Solve 2nn = 4 2nn 2 = 4 5 2nn 2 = 1 nn 2 = 1 2 This equation does not have a real solution. 358

365 Chapter 13 C. USE THE PERFECT SQUARE FORMULA In order for us to be able to apply the square root property to solve a quadratic equation, we cannot have the xx term in the middle because if we apply the square root property to the xx term, we will make the equation more complicated to solve. However sometimes, we have special cases that we can apply the perfect square formula to get rid of the xx term in the middle and then apply the square root property to solve the equations. Recall: Perfect square formula aa bb 22 = (aa + bb) 22 or aa bb 22 = (aa bb) 22 Solve equations using square root property Perfect Square formula (Duration 4:09) Example: a) xx 2 + 8xx + 16 = 4 b) 9xx 2 12xx + 4 = 25 YOU TRY Solve. a) xx 2 6xx + 9 = 81 b) 9xx xx + 25 = 4 359

366 EXERCISE Solve by applying the square root property. Chapter 13 1) (xx 3) 2 = 16 2) (xx 2) 2 = 49 3) (xx 7) 2 = 4 4) (ss 5) 2 = 16 5) (pp + 5) 2 = 81 6) (ss + 3) 2 = 4 7) (tt + 9) 2 = 37 8) (aa + 5) 2 = 57 9) (nn 9) 2 = 63 10) (rr + 1) 2 = ) (9rr + 1) 2 = 9 12) (7mm 8) 2 = 36 13) (3ss 6) 2 = 25 14) 5(kk 7) 2 6 = ) 5(gg 5) = ) 2nn = 5 17) (2ss + 1) 2 = 0 18) (zz 4) 2 = 25 19) 3nn 2 + 2nn = 2nn ) 8nn 2 29 = nn 2 21) 2(rr + 9) 2 19 = 37 22) 3(nn 3) = ) 7(2xx + 6) 2 5 = ) 6(4xx 4) 2 5 = 145 Apply the perfect square formula and solve the equations by using the square root property. 25) xx xx + 36 = 49 26) xx 2 + 6xx + 9 = 2 27) 16xx 2 40xx + 25 = 16 28) xx 2 + 4xx + 4 = 1 29) xx 2 14xx + 49 = 9 30) 25xx xx + 1 =

367 SECTION 13.2: COMPLETING THE SQUARE Chapter 13 When solving quadratic equations previously (then known as trinomial equations), we factored to solve. However, recall, not all equations are factorable. Consider the equation xx 2 2xx 7 = 0. This equation is not factorable, but there are two solutions to this equation: and 1 2. Looking at the form of these solutions, we obtained these types of solutions in the previous section while using the square root property. If we can obtain a perfect square, then we can apply the square root property and solve as usual. This method we use to obtain a perfect square is called completing the square. Recall. Special product formulas for perfect square trinomials: (aa + bb) 22 = aa bb 22 oooo (aa bb) 22 = aa bb 22 We use these formulas to help us solve by completing the square. A. COMPLETE THE SQUARE We first begin with completing the square and rewriting the trinomial in factored form using the perfect square trinomial formulas. Complete the square (Duration 5:00) Complete the square. Find cc. aa bb 22 is easily factored to To make xx 22 + bbbb + cc a perfect square, cc = Example: a) xx xx + cc b) xx 2 7xx + cc c) xx 2 3 xx + cc d) 7 xx2 + 6 xx + cc 5 361

368 Chapter 13 Note To complete the square of any trinomial, we always square half of the linear term s coefficient, i.e., bb oooo bb 22 We usually use the second expression when the middle term s coefficient is a fraction. YOU TRY Complete the square by finding cc: a) xx 2 + 8xx + cc b) xx 2 7xx + cc c) xx xx + cc 3 B. SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, a = 1 Steps to solving quadratic equations by completing the square Given a quadratic equation aaaa 22 + bbbb + cc = 00, we can use the following method to solve for xx. Step 1. Rewrite the quadratic equation so that the coefficient of the leading term is one, and the original constant coefficient is on the opposite side of the equal sign from the leading and linear terms. aaaa 22 + bbbb = cc + Step 2. If aa 11, divide both sides of the equation by aa Step 3. Complete the square, i.e., bb oooo bb 22 and add the result to both sides of the quadratic equation. Step 4. Rewrite the perfect square trinomial in factored form. Step 5. Solve using the square root property. Step 6. Verify the solution(s). Solve quadratic equation by completing the square (Duration 8:40) Solve the quadratic equation using the square root principle. (xx 5) 2 =

369 Example: a) xx 2 + 6xx 9 = 0 b) xx 2 5xx + 10 = 0 Chapter 13 c) 3xx 2 + 2xx 9 = 0 YOU TRY Solve. a) xx xx + 24 = 0 b) nn 2 8nn + 4 = 0 C. SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, A 1 Solve quadratic equation by completing the square a 1 (Duration 4:59) To complete the square: aaaa 2 + bbbb + cc = 0 1. Separate and 2. Divide by (everything!) 3. Find and to 363

370 Example a) 3xx 2 15xx + 18 = 0 b) 8xx + 32 = 4xx 2 Chapter 13 YOU TRY Solve. a) 3xx 2 36xx + 60 = 0 b) 2kk 2 + kk 2 = 0 364

371 EXERCISE Complete the square. Chapter 13 1) xx 2 30xx + 2) mm 2 36mm + 3) xx 2 15xx + 4) yy 2 yy + 5) aa 2 24aa + 6) xx 2 34xx + 7) rr rr + 8) pp 2 17pp + Solve each equation by completing the square. If the solution is not a real solution, then state not a real solution. 9) 6rr rr 24 = 6 10) 6nn 2 12nn 14 = 4 11) aa 2 56 = 10aa 12) xx 2 + 8xx + 15 = 8 13) 8nn nn = 64 14) nn 2 + 4nn = 12 15) xx 2 16xx + 55 = 0 16) nn 2 = nn 17) 4bb 2 15bb + 56 = 3bb 2 18) bb 2 + 7bb 33 = 0 19) xx xx 57 = 4 20) nn 2 8nn 12 = 0 21) nn 2 16nn + 67 = 4 22) xx 2 = 10xx 29 23) 5kk 2 10kk + 48 = 0 24) 7nn 2 nn + 7 = 7nn + 6nn 2 25) 2xx 2 + 4xx + 38 = 6 26) 8bb bb 37 = 5 27) 5xx 2 + 5xx = 31 5xx 28) vv 2 + 5vv + 28 = 0 29) kk 2 7kk + 50 = 3 30) 5xx 2 + 8xx 40 = 8 31) 8rr rr = 55 32) 2xx 2 + 3xx 5 = 4xx 2 33) 8aa aa 1 = 0 34) pp 2 16pp 52 = 0 365

372 SECTION 13.3: QUADRATIC FORMULA The quadratic formula is derived from the method of completing the square. If we took a general quadratic equation aaaa 2 + bbbb + cc = 0 and solved for xx by completing the square, we would obtain the quadratic formula. Let s try this. Deriving the Quadratic Formula (Duration 4:04) Chapter 13 Quadratic formula Given the quadratic equation aaaa 2 + bbbb + cc = 0. Then xx = bb ± bb2 4aaaa 2aa is called the quadratic formula. The quadratic formula is a formula for solving quadratic equations in terms of the coefficients. A. DETERMINANT OF A QUADRATIC EQUATION To make the quadratic formula easier to manage, we should calculate the discriminant first. Then substitute it into the quadratic formula to find xx. Discriminant Given the quadratic equation aaaa 2 + bbbb + cc = 0, then its discriminant, denoted as the upper-case Greek letter delta, Δ, is defined as ΔΔ = bb ΔΔ = bb Quadratic formula: xx = bb ± ΔΔ 2222 Case 1: If Δ is positive, the equation has 2 solutions. Case 2: If Δ is zero, the equation has 1 solution. Case 3: If Δ is negative, the equation has no real solution. (The equation has two complex solutions but we will not discuss complex numbers in this class). 366

373 Chapter 13 Determinant (Duration 4:58) aaaa 2 + bbbb + cc = 0 xx = If bb 2 4aaaa If bb 2 4aaaa If bb 2 4aaaa Example: Determine the number of solutions to the quadratic equation. xx xx + 49 = 0 Determinant examples (Duration 4:58) Example: Describe the type of solutions to the quadratic equation. a) xx 2 3xx 28 = 0 aa = bb = cc = a) xx 2 4xx + 12 = 0 aa = bb = cc = b) 2xx 2 + xx + 5 = 0 aa = bb = cc = c) 2xx 2 + 8xx + 8 = 0 aa = bb = cc = 367

374 YOU TRY Chapter 13 Find the determinant and determine how many solutions each of the following equation has. a) xx 2 + 3xx 4 = 0 b) 2xx 2 + 4xx + 1 = 0 c) xx 2 4xx + 4 = 0 d) 8aa 2 + 5aa + 1 = 0 B. APPLY THE QUADRATIC FORMULA Solve equations uing quadratic formula two real rational solutions (Duration 3:14 ) Example: Solve using the quadratic formula. 6xx 2 xx 15 = 0 368

375 Solve equations uing quadratic formula two real irrational solutions (Duration 4:16 ) Chapter 13 Example: Solve using the quadratic formula. 2xx 2 4xx 3 = 0 Solve equations uing quadratic formula one solution (Duration 2:50 ) Example: Solve using the quadratic formula. 4xx 2 12xx + 9 = 0 YOU TRY Solve. a) xx 2 + 3xx + 2 = 0 b) 3xx 2 + 2xx 7 = 0 369

376 C. MAKE ONE SIDE OF AN EQUATION EQUAL TO ZERO Chapter 13 Set one side of an equation equal to 0 before solving (Duration 4:46) Before using the quadratic formula, the equation must equal. Example: a) 2xx 2 = 15 7xx b) 3xx 2 5xx + 2 = 7 YOU TRY Solve. a) 25xx 2 = 30xx + 11 b) 3xx 2 + 4xx + 8 = 2xx 2 + 6xx 5 370