Essentials of Intermediate Algebra

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1 Essentials of Intermediate Algebra BY Tom K. Kim, Ph.D. Peninsula College, WA Randy Anderson, M.S. Peninsula College, WA 9/24/2012

2 Contents 1 Review 1 2 Rules of Exponents Multiplying Two Exponentials with the Same Base Practice Problems Raising an Exponential to Another Power (Power to a Power) A Product Raised to a Power A Quotient Raised to a Power Quotient of Two Exponentials (Dividing Two Exponentials) Negative Exponents and Zero as an Exponent Scientic Notation Converting to/from Scientic Notation Arithmetic with Scientic Notation Polynomials Monomials Multiplication and Division of Monomials Addition and Subtraction of Monomials Polynomials Evaluating Polynomials Addition and Subtraction of Polynomials Multiplying Polynomials Dividing Polynomials by Monomials Linear Equations and Inequalities in One Variable Inverses Additive Inverse Multiplicative Inverse Addition Property of Equality Multiplication Property of Equality Summary of Solving Linear Equations in One Variable General Equations Equations with Fractions Special Cases Identity Inconsistent Applications Percentage Mixture Problems Distance/Rate Problems Linear Inequalities in One Variable Graphing Inequality Solutions on the Number Line Addition/Multiplication Property of Inequalities Compound Inequalities i

3 CONTENTS ii 5 Equations of Lines Cartesian Coordinate System Graphing Lines by Points Slope of a Line Equations of Lines X and Y Intercepts Slope-Intercept form of the Line Point-Slope form of the Line Parallel and Perpendicular Lines Applications of Lines Linear Inequalities in Two Variables Systems of Linear Equations in Two Variables Graphing Method Substitution Method Elimination Method Applications Distance/Rate in Moving Medium Miscellaneous Problems Practice Problems Systems of Linear Inequalities in Two Variables Linear and Non-Linear Graphs Scaling Practice Problems Linear and Nonlinear Growth and Decay Getting Data Points from Equations Linear and Non-Linear Growth Linear and Non-Linear Decay Practice Problems Applications of Graphs Choosing Appropriate Axes Interpreting Graph Shapes to Physical Events Comparison of Two Graphs Practice Problems

4 Chapter 1 Review This is just a place holder until the review chapter has been written. 1

5 Chapter 2 Rules of Exponents Recall that 5 3 means multiply 5 three times. 5 3 = = is referred to as an exponential expression or an exponential, the 3 is referred to as the exponent or power, and the 5 is referred to as the base. Denition In general is referred to as an exponential expression or exponential, where the b is referred to as the base of the exponential, and n is referred to as the exponent or power. For a positive integer n, n tells you how many times to multiply b. b n b n = b b b }{{} n times 2.1 Multiplying Two Exponentials with the Same Base View the Video Tutorial for this Section here. Suppose we wish to multiply If we use the denition of exponentials = }{{} } {{ 7 7 } 2 3 = = 7 5 This example would seem to indicate that if you multiply two exponentials, you add the exponents. To get the general rule, notice that the base did not have to be 7. For example or in general, for any base b, = }{{} } {{ 9 9 } 2 3 = = 9 5 b 2 b 3 = }{{} b b b } {{ b } b 2 3 = b 2+3 = b 5 2

6 CHAPTER 2. RULES OF EXPONENTS 3 So far, in all our examples above, the two exponentials being multiplied had the same base (7, 9, and b.) What would happen if the bases of the two exponentials being multiplied are not the same? a 2 b 3 = }{{} a a b } {{ b } b 2 3 = a 2 b 3 Notice that we can do nothing. So, in order to multiply two exponentials, we have established that the bases must be the same. As for the exponents, notice that they do not have to be 2 and 3. b 4 b 5 = b } b {{ b } b b } b {{ b b } b 4 5 = b 4+5 = b 9 Notice that the exponents do not have to be the same. So, in general, we have Proposition Multiplying two exponentials b m b n = b m+n The two exponentials must be multiplied. The bases of the two exponentials must be the same. Example Multiply and simplify, if possible. 1) 2) ( 5) 4 ( 5) ) ( ) 1 6 x5 y (2 3 x 5 y 7) 4) ( 5 2 x 6 y 5) ( 4x 2) + ( 3 3 xy 9) (xy) 5) (x 3) 4 (x 3) 7 6) ( 2 3n ) ( 2 5n) Solutions: 1) ( 5) 4 ( 5) 3 = ( 5) 4+3 = ( 5) 7 = ) Note that Proposition is written with just two exponentials being multiplied. A natural question to ask is, what happens if you multiply three or more exponentials with the same base. The short answer is that you multiply two at a time. In other words = }{{} = = 7 11 In general, if you are multiplying three or more exponentials with the same base, then you simply add all the exponents.

7 CHAPTER 2. RULES OF EXPONENTS 4 3) ( 1 6 x5 y ) (2 3 x 5 y 7) 4) = 1 6 x5 y 2 3 x 5 y 7 Since everything inside the parentheses are multiplied, the parentheses are not necessary. = x 5 x 5 y y 7 = 8 6 x5+5 y 1+7 y is the same as y 1. = 4 3 x10 y 8 ( 5 2 x 6 y 5) ( 4x 2) + ( 3 3 xy 9) (xy) = x 6 x 2 y x x y 9 y = 25(4)x 6+2 y x 1+1 y 9+1 = 100x 8 y x 2 y 10 Note that we can do nothing with the addition, as Proposition only applies when multiplying exponentials. 5) One very common misunderstanding that students have when presented with a mathematical statement is that they interpret variables as being just a number. For example, when students see Proposition 2.1.1, they think that b must be one variable or number. There is nothing that says that b has to be a number or just one variable. All that Proposition requires is that the bases of the two exponentials that are being multiplied be the same. Note that for this problem, the two exponentials are being multiplied and the bases are the same (x 3). So we leave the base alone and add the exponents. (x 3) 4 (x 3) 7 = (x 3) 4+7 = (x 3) 11 6) Since the two exponentials 2 3n and 2 5n are multiplied together and the bases are the same, we can use Proposition to add the exponents 3n and 5n. Since these are like terms, 3n + 5n = 8n. ( 2 3n ) ( 2 5n) = 2 3n+5n = 2 8n In reading and using Proposition 2.1.1, it should be remembered that the proposition can be read from left to right, which is how we used it in Example 2.1.1, or it could also be read from right to left. In other words, b m+n = b m b n What this tells us is that if the exponent of an exponential can be written as a sum of two terms, then the exponential can be split into two exponentials with the same base. For example = Example Rewrite as a product of exponentials, if possible. 1) x 2+4 2) b 2x+1 Solutions: 1) x 2+4 = x 2 x 4 2) b 2x+1 = b 2x b 1 = b b 2x

8 CHAPTER 2. RULES OF EXPONENTS Practice Problems Multiply the exponentials and simplify, if possible. 1) (x 4) 2 (x 4) 5 2) 3 n 3 n 3

9 CHAPTER 2. RULES OF EXPONENTS Raising an Exponential to Another Power (Power to a Power) View the Video Tutorial for this Section here. Again, we begin with the denition of exponentials. ( 7 2 ) 3 = = = 7 3(2) = 7 6 We note that the 7, 2 and 3 could have been any other numbers. ( 9 5 ) 4 = = = 9 4(5) = 9 20 So, in general, when you raise an exponential to another power (what we call power to a power), you leave the base alone and multiply the two exponents. Proposition Raising an exponential to another power (b m ) n = b m n Example Perform the indicated operations and simplify, if possible. 1) ( ( 5) 3 ) 3) 2 ( x3 x 7) 5 2) ( x 5 ) 2n 4) ( y4 y 3n) 6 Solutions: 1) (( 5) 3) 2 = ( 5) 2 3 = ( 5) 6 = Note that when we are simplifying exponentials, the general rule is that we multiply out any numbers that we get. 2) ( x 5 ) 2n = x 5 2n = x 10n 3) ( x3 x 7) 5 = ( x 3+7) 5 = ( x 10) 5 = x 10 5 = x 50

10 CHAPTER 2. RULES OF EXPONENTS 7 4) ( y4 y 3n) 6 = ( y 4+3n) 6 = y 6(4+3n) = y 24+18n Notice that in the above example, we used Proposition by reading it from left to right, (b m ) n = b m n. Now, if we read the same Proposition from right to left, we get b m n = (b m ) n. This means that 2 3n = ( 2 3) n = 8 n or 2 3n = 2 n 3 = (2 n ) 3 Example Find the values of a and b so that the following are true. 1) 5 2n = a n 3) x n2 = (x a ) b 2) 5 2n = ( a b) 2 Solutions: 1) To rewrite 5 2n as a n means we want n to be the outer exponent. So a = n = ( 5 2) n = 25 n 2) In this case, we want the 2 to be the outer exponent. Since multiplication is commutative, 2n = n 2, we have So for (5 n ) 2 = ( a b) 2, a = 5 and b = n. 5 2n = 5 n 2 = (5 n ) 2 3) Note that by denition of exponentials, n 2 = n n. So and for (x n ) n = (x a ) b, a = n and b = n. x n2 = x n n = (x n ) n

11 CHAPTER 2. RULES OF EXPONENTS A Product Raised to a Power View the Video Tutorial for this Section here. As with the other rules, we start with an example and apply the denition of exponentials. (5 7) 3 = (5 7)(5 7)(5 7) = }{{} }{{} = Note that the 5 and 7 could have been any other numbers. ( 4 15) 3 = ( 4 15)( 4 15)( 4 15) = ( 4) ( 4) ( 4) = ( 4) Also, the exponent did not have to be 3. So in general, we have Proposition Product raised to a power (a b) n = a n b n The expression inside the parentheses must be a product. Example Perform the indicated operations and simplify, if possible. 1) ( 4x) 2 3) (3 ( x 5 y 4) 3 ) 2 2) ( 5x 2 y ) 3 4) ( y4 y 3n) 6 Solutions: 1) ( 4x) 2 = ( 4) 2 x 2 = 16x 2 2) ( 5x 2 y ) 3 = ( 5) 3 ( x 2) 3 y 3 = 125x 2 3 y 3 = 125x 6 y 3 3) Here we are faced with nested parentheses. There are a couple of dierent approaches that we can take. First, if we follow order of operations, then we would work from the inner parentheses out. (7 ( x 5 y 4) 3 ) 2 = ( 7 ( x 5) 3 ( y 4 ) 3 ) 2 note that the exponent 3 only applies to x 5 and y 4. = ( 7x 15 y 12) 2 = 7 2 ( x 15) 2 ( y 12 ) 2 = 49x 30 y 24

12 CHAPTER 2. RULES OF EXPONENTS 9 Another approach would be to consider ( x 5 y 4) as one quantity. For example, suppose that instead of ( x 5 y 4) 3 it was z 3 then ( 7z 3) 2 = 7 2 z 3 2. Using this approach, we get The actual problem. (7 ( x 5 y 4) 3 ) 2 ( = 7 2 ( x 5 y 4) ) 3 2 = 49 ( x 5 y 4) 3 2 = 49 ( x 5 y 4) 6 = 49 ( x 5) 6 ( y 4 ) 6 = 49x 30 y 24 The same problem only we think of ( x 5 y 4) as z. ( 7z 3 ) 2 = 7 2 ( z 3) 2 = 49z 3 2 = 49z 6 So these rst three lines are done exactly the same for both problems. Note that we would not actually do it this way with the z's. This side is written to show how we are viewing the problem in our mind. 4) Note that this problem was simplied in Example (4), using Proposition and Proposition There are often more than one way to simplify an exponential expression, as we will demonstrate with this problem. Here, we will use Proposition followed by Proposition and Proposition ( y4 y 3n) 6 = ( y 4) 6 ( y 3n ) 6 = y 24 y 18n = y 24+18n Notice that the nal answer is the same as in Example As with the other propositions, we note that Proposition can either be read from left to right, (ab) n = a n b n, which is how we used it in the above example, or from right to left, a n b n = (ab) n. Note that to use this property, there are three conditions. The exponentials must be multiplied. The exponents must be the same. The bases can be dierent. This is dierent from Proposition 2.1.1, which requires that you multiply two exponentials with the same base and dierent exponents. So = (2 3) 4 = 6 4 using Proposition while = = 4 5 using Proposition Example Find a and n, such that the following can be written as a n. 1) 9x 2 2) 125(x 7) 3 3) 25x6 Solutions: 1) Note that 9 = 3 2. So 9x 2 = 3 2 x 2 = (3x) 2 and for (3x) 2 = a n, a = 3x and n = 2. 2) 125(x 7) 3 = 5 3 (x 7) 3 = [5(x 7)] 3 So a = 5(x 7) and n = 3.

13 CHAPTER 2. RULES OF EXPONENTS 10 3) 25x 6 = 5 2 x 2 3 = 5 2 ( x 3) 2 = ( 5x 3) 2 So a = 5x 3 and n = 2.

14 CHAPTER 2. RULES OF EXPONENTS A Quotient Raised to a Power View the Video Tutorial for this Section here. As with the previous rules, we begin with an example where we apply the denition of exponentials. ( ) 3 5 = = = Note that 5 and 7 could have been any other number or expression and that goes for the exponent 3 as well. So in general, we have Proposition Quotient raised to a power ( a b ) n = a n b n Example Perform the indicated operations and simplify, if possible. 1) 2) ( ) ( 6x 5 y 8 ) 2 3) ( ) 5 x + 3 (x 1) 4 7z 3 Solutions: 1) ( ) 4 10 = ( 10) = ) Note that the a and b in Proposition should not be viewed as one number or variable, but rather as the entire numerator (top of the fraction) for a and the entire denominator (bottom of the fraction) for b. In this problem the entire numerator is 6x 5 y 8. So we think of that as a and the entire denominator 7z 3 as b. ( 6x 5 y 8 ) 2 ( 6x 5 y 8) 2 = 3) 7z 3 (7z 3 ) 2 = 62 ( x 5) 2 ( y 8 ) (z 3 ) 2 = 36x10 y 16 49z 6 ( ) 5 x + 3 (x 1) 4 = = = (x + 3) 5 ((x 1) 4 ) 5 (x + 3)5 (x 1) 4 5 (x + 3)5 (x 1) 20

15 CHAPTER 2. RULES OF EXPONENTS 12 At this point we have no way of simplifying (x + 3) 5 or (x 1) 20. A mistake that students often make here is that they will try to apply Proposition and raise each term, the x and 3, to the fth power. However, if you look carefully, Proposition only applies when the base of the exponential is made up of a product, not a sum. So if the numerator had been a product, (x 3) 5 = x 5 3 5, but since the numerator is a sum, (x + 3) 5, we can do nothing at this point. Note that we will learn in a later section how to multiply this expression out. As with the other propositions, Proposition can be read from right to left, which gives us a n ( a ) n b n = b Note that to use this property, the exponents must be the same, while the bases can be dierent. Example Find a, b, and n such that the following problems can be written as ( a b ) n. 1) 2) x 2 16 (x 1) ) 27x 3 y 6 Solutions: 1) x 2 16 = x2 4 2 = ( x 4 ) 2 So if we want ( x 4 ) 2 = ( a b ) n, then a = x, b = 4, and n = 2. 2) Note that for problem (1), we used the fact that 16 = 4 2 because we wanted to rewrite 16 as something squared. However, in this problem we want to rewrite 16 as something to the fourth power since the numerator is raised to the fourth power. So we use 16 = 2 4. (x 1) 4 Here, we have that a = x 1, b = 2, and n = = = (x 1)4 2 4 ( x 1 2 ) 4 3) 27x 3 y 6 = 33 x 3 y 2 3 = (3x)3 (y 2 ) 3 ( 3x = y 2 ) 3 So a = 3x, b = y 2 and n = 3.

16 CHAPTER 2. RULES OF EXPONENTS Quotient of Two Exponentials (Dividing Two Exponentials) View the Video Tutorial for this Section here. As with the other rules, we begin rst by looking at an explicit example to try and understand the general rule. Again, we fall back to the denition of exponentials = = = = 7 2 Note that three of the 7's in the numerator will cancel the three 7's in the denominator. In other words from the ve 7's in the numerator, three of the 7's are subtracted } 7 {{ } = 7 } {{ 7 7 } 3 = = 7 2 To generalize this, we note rst that the base of the exponent in both the numerator and denominator must be the same (7 in this example.) So, for the general case, we will use b. The exponents in the numerator (5 for this example) and denominator (3 for this example) do not have to be the same and so we use m and n respectively for the exponents. Proposition Quotient of two exponentials (Dividing two exponentials) b m b n = bm n The base of the exponentials in both the numerator and denominator must be the same. You always subtract the exponent in the denominator, n, from the exponent in the numerator, m. Example Perform the indicated operations and simplify, if possible. 1) 2) 2 9 ( 2) 5 ( 2 3) 6 ( 2 3) 4 3) 4) ( 5x 3 y 2) 4 (10xy 2 ) 2 ( A 3 (r + h) 2) 7 (A 4 (r + h)) 4 Solutions:

17 CHAPTER 2. RULES OF EXPONENTS 14 1) Note that ( 2) 5 = ( 1 2) 5 = ( 1) 5 (2) 5 = = 2 5. So 2 9 ( 2) 5 = = = 24 1 = 2 4 = 16 Notation: remember that but ( 2) 4 = ( 2) ( 2) ( 2) ( 2) = = = = 16 2) ( 2 3) 6 ( 2 3) 4 = = ( 2 3 ( 2 3 ) 6 4 ) 2 = 4 9 3) ( 5x 3 y 2) 4 ( ) (10xy 2 ) 2 = 54 x 3 4 ( ) y x 2 (y 2 ) 2 = = 5 4 x 3 4 y 2 4 (2 5) 2 x 2 y x 12 y x 2 y 4 = 54 2 x 12 2 y = 52 x 10 y = 25x10 y 4 4 4) As we did in Example , we treat (r + h) as one quantity ( A 3 (r + h) 2) 7 (A 4 (r + h)) 4 = ( A 3 ) 7 ( (r + h) 2 ) 7 (A 4 ) 4 (r + h) 4 = A21 (r + h) 14 A 16 (r + h) 4 = A (r + h) 14 4 = A 5 (r + h) 10 A natural question to ask when viewing Proposition is what happens if the exponent in the denominator, n, is larger than the exponent in the numerator, m. For example, what happens when we have We will discuss this in the next section. x 3 x 8

18 CHAPTER 2. RULES OF EXPONENTS 15 As with the other propositions, we now observe what happens if we read this one from right to left, instead of from left to right. b m n = bm b n Example Find a, b, and n such that the following can be written as an b 1) 3 5x 2 2) (x 2) x 1 Solutions: 1) 2) So a = 3, b = 9, and n = 5x. 3 5x 2 = 35x 3 2 = 35x 9 (x 2) x 1 = = (x 2)x (x 2) 1 (x 2)x x 2

19 CHAPTER 2. RULES OF EXPONENTS Negative Exponents and Zero as an Exponent View the Video Tutorial for this Section here. The denition of exponentials allows us to interpret 2 3 = = 8. However, if we have a negative exponent, for example 2 3, the question is how do we interpret this. Note that multiplying -2 three times means ( 2) ( 2) ( 2) = ( 2) 3. To interpret negative exponents, we rst decide that we want Proposition to be true for any exponents. In other words, = = 7 2 But from the denition of exponentials, we can simplify the above as = = = Since both are equal to , we must have 7 2 = In general we have Proposition Negative exponents. b n = 1 b n n does not necessarily mean a negative number. It just means the opposite of whatever n is. Note that 1 = bn b n So this rule says that any time you cross the fraction (division) bar, the sign of the exponent changes (from positive to negative or from negative to positive.) There are several ways to get the second part of the rule, 1 b n = b n. One way is to read Proposition from right to left 1 b n = b n Now if we replace all the n s with n, we get 1 b n = b ( n) = b n Example Simplify. Write your answer without any negative exponents. 1) 5x 2 3 3) 4) x 3 y 9 5x 2 y 4 ( 50x 3 yz 6 ) 5 2) ( 2 3 x 3 y 5) 2 ( 3x 2 y 4) 3 100x 4 y 10 z 12

20 CHAPTER 2. RULES OF EXPONENTS 17 Solutions: 1) Since there are no parentheses around 5x, the -2 power only applies to the x (order of operations says to do exponents before multiplication.) 5x 2 3 = 5 3x 2 2) ( 2 3 x 3 y 5) 2 ( 3x 2 y 4) 3 ( 2 3 x 3 y 5) 2 ( 3x 2 y 4) 3 = ( 2 3) 2 ( x 3 ) 2 ( y 5 ) 2 ( 3) 3 ( x 2) 3 ( y 4 ) 3 = 2 6 x 6 y 10 ( 3) 3 x 6 y 12 Since everything is multiplied, it is better to leave the negative exponents and simplify and then get rid of the negative exponents later. = 2 6 ( 3) 3 x 6+( 6) y 10+( 12) Proposition works for negative exponents as well. So x 6 x 6 = x 6+( 6) and so forth. = 2 6 ( 3) 3 x 12 y 2 = 2 6 ( 3) 3 x 12 y 2 Now that all the multiplication of exponentials are done, we get rid of the negative exponents by moving them across the fraction bar. = 64 27x 12 y 2 = 64 27x 12 y 2 3) When dividing exponentials, instead of using Proposition directly, we will use the idea that was used to get the proposition. For example, if we are simplifying y4 y, we recall that y 4 means y multiplied 9 4 times while y 9 means y is multiplied 9 times. After canceling out 4 of the y's, we are left with 5 y's. Since there were more y's in the denominator, the 5 y's are left in the denominator. So y4 y = 1 9 y. Note 5 that this method only works if the exponents are positive. x 3 y 9 5x 2 y 4 In the case of division, it is usually better to get rid of the negative exponents rst. = x3 x 2 y 4 5y 9 Get rid of the negative exponents by moving them up or down. = x5 5y 5 First note that a negative divided by a negative is positive. 4 y's in the numerator and 9 y's in the denominator, so 4 cancel out and we are left with 5 y's in the denominator.

21 CHAPTER 2. RULES OF EXPONENTS 18 4) ( 50x 3 yz 6 100x 4 y 10 z 12 ) 5 Note that the expression inside the parentheses can be simplied. In cases like this, it is usually better to simplify rst = 1 2. ( ) get rid of all the negative exponents y y 10 5 = simplied the z's. 6 z's on top and 12 z's on 2x 4 x 3 z 6 the bottom, so canceling out 6 z's leaves 6 z's in the bottom. Since everything inside the parentheses are ( ) y 11 5 multiplied and divided, using Proposition 2.4.1, = 2x 7 z , and implies that we simply multiply every exponent inside the parentheses by 5. = y x 35 z 30 = 25 x 35 z 30 y 55 = 32x35 z 30 y 55 Now we need to determine what it means to take something to the zero power. As with negative exponents, we want to be sure that we dene b 0 in such a way that it does not contradict any of the other propositions. So we look at the following example = = 1 Since anything divided by itself (other than zero) is 1. However, if we use Proposition 2.5.1, we have = 52 2 = 5 0 Since both 5 0 and 1 are equal to the same thing, then they must equal each other So, we have Proposition Zero exponent. for any b 0. b 0 = = 1 Example Simplify. Your answer should not have any negative exponents.

22 CHAPTER 2. RULES OF EXPONENTS 19 1) 2) 3xy 0 (x 2 2y 3 ) 0 3) ( 5x 2 y 3) Solutions: 1) Since there are no parentheses, the zero power only applies to y. 3xy 0 = 3x(1) = 3x 2) Here, since we have parentheses around x 2 2y 3, the whole expression is raised to the zero power. ( x 2 2y 3) 0 = 1 3) ( 5x 2 y 3) = 2 3 = 2 3 = 8

23 CHAPTER 2. RULES OF EXPONENTS Scientic Notation One commonly used application of exponentials is scientic notation. Scientic notation is a way to write really large or small numbers in a more ecient way. By more ecient, we mean that not only is it a quicker way to write the number, but it also allows us to carry out calculations with the number easier. Before computers, these reasons justied the use of scientic notation. With the advent of computers, scientic notation became a crucial concept to store numbers on computers. We will discuss this further after our discussion of how to convert to and from scientic notation Converting to/from Scientic Notation To understand scientic notation, we note the following properties when working with the number 10 and powers of 10. For positive powers of 10: If you raise 10 to some positive integer power n, 10 n, then that is the same as the number 1 followed by n zeros = = 1 1 {}}{ = = 1 2 {}}{ = = 1 3 {}}{ 000 {}}{ 10 4 = = Multiplying any number by 10 n just moves the decimal point of the number by n places to the right, where n is a positive integer. For negative powers of 10: = = = = If we let n be a positive integer then 10 n multiplied out will be the number 1.0 with the decimal moved n places to the left = 1 10 =. 1 = = =. 0 1 = = = = Multiplying any number by 10 n just moves the decimal point of the number n places to the left, where n is a positive integer. Example Compute the product = = = = ) ) ) Solutions: 4

24 CHAPTER 2. RULES OF EXPONENTS 21 1) Multiplying by 10 7 means move the decimal 7 places to the right (add 7 zeros.) Recall that can also be written as = = ) 10 4 means move the decimal 4 places to the left = = ) 10 2 means move the decimal 2 places to the left. We dene scientic notation as follows: = = 10.3 Denition Scientic Notation A nonzero number is in scientic notation if it is written as a 10 n where a is a real number where there is always exactly one digit to the right of the decimal, in other words 1 a < 10 if a > 0 (is positive) 10 < a 1, if a < 0 (is negative) n is an integer. means multiplication (the old multiplication symbol from elementary school.) Example Write the following in scientic notation. 1) 3 2) ) Solutions: 1) According to denition 2.7.1, since 3 is between 1 and 10, it can represent a. Since we do not want to change 3, we multiply it by 10 0 which is just the number 1. So the number 3 in scientic notation is ) To make the number be a number between -10 and -1, we need to move the decimal 6 places to the left to get a. a = = Now, to go from back to the original number , we would have to move the decimal 6 places to the right of So = ) To get the number to be between 1 and 10, we need to move the decimal 4 places to the right. a = = 9.01 To go from 9.01 back to the original number , we would need to move the decimal in 9.01, 4 places to the left, so =

25 CHAPTER 2. RULES OF EXPONENTS 22 Scientic notation and Calculator Displays Depending on your calculator, numbers in scientic notation will be displayed in dierent ways. The only way to know what your calculator is doing is for you to test it out yourself or read the manual. Here are a couple of common ways that calculators display scientic notation. Note that this is not a comprehensive list. The actual number to display Display E 4 Display Arithmetic with Scientic Notation The arithmetic of interest are multiplication, division, and powers (exponentials.) As mentioned at the beginning of the chapter, scientic notation allows us to work with really large or small numbers more eciently. Suppose you wish to multiply 574, 000, 000, 000, 000, , 150, 000, 000, 000, 000. Note that depending on your calculator, you may not even be able to enter these numbers into it. If you think that these numbers are unrealistic in real life, note that if you removed three of the zeros, the numbers would be in the trillions, which is the range where our (United States) decit is currently at. Also, the distance from the earth to the nearest star is about 24,200,000,000,000 miles. 1 In the following examples we will see how to use scientic notation to carry out these types of calculations. Example Compute the following and write your answer in scientic notation. 1) (574,000,000,000,000,000) (320,150,000,000,000,000) 2) (23, 000, 000, 000) ( ) 3) ( ) ( ) Solutions: 1) 574, 000, 000, 000, 000, 000 = , 150, 000, 000, 000, 000 = Convert both numbers to scientic notation. ( ) ( ) = ( )( ) = = Since everything is multiplication, we multiply the decimal numbers together rst and the powers of 10 together. We use the exponential rules to multiply the powers of {}}{ = Note that is no longer in scientic notation. So we rewrite in scientic notation as , leaving the power alone. 1 This is the distance from the earth to Proxima Centauri, according to Wolfram Alpha.

26 CHAPTER 2. RULES OF EXPONENTS 23 = = Use the exponential rules to multiply 10 1 and ) (23, 000, 000, 000) ( ) = ( ) ( ) = (2.3 5)( ) = ( 16) = = = ) ( ) ( ) We rst rewrite the division as a fraction = Since the numerator and denominator both consist of only multiplication, we can divide the decimal numbers and the powers of 10 separately {}}{ = = Since is not between 1 and 10, we rewrite it as and then simplify. Comment on scientic notation and computers As mentioned at the beginning of the chapter, because it can represent a really large or small number more eciently, scientic notation found an important role in the development of computers. Although computer memory size has increased signicantly over the years, going from kilobytes (thousands of bytes) to megabytes (millions of bytes) to gigabytes (billions of bytes), it is still nite. Because of the limited memory, a computer will assign a xed size to store numbers. Note that it is beyond the scope of this book and the author's knowledge to go into the details of computer architecture, so we limit ourselves to the following simplied view. Suppose a computer assigns only enough memory to store eight digits for each number. If the computer stored the numbers directly, then it would not be able to store any numbers in the range of a hundred million or more. For example the number 100,000,000 could not be used because it has nine digits. However, if the computer used scientic notation, such that the rst six digits represented the decimal number a, and the last two digits represented n, the exponent of 10 (see denition 2.7.1), then 100, 000, 000 = could be stored by the computer since a = 1 and n = 8 means we really only need to store two digits. So by using scientic notation, a computer can store much bigger numbers in a xed amount of space compared to the regular representation of numbers.

27 Chapter 3 Polynomials 3.1 Monomials View the Video Tutorial for this Section here. Denition A monomial is dened as a product of numbers and variables, where the variables are raised to positive integer powers. The product of all the numbers is called the coecient. If the monomial has just one variable, then the exponent of the variable is called the degree of the monomial. ax n a is the coecient n is the degree (must be a non-negative integer.) Note that a monomial can have more than one variable and there is a denition of degree for a multivariable monomial, but we will not use it. The reason for not introducing it is that it is of no practical use for us and its denition can cause confusion with our power rules. Example Determine which of the following are monomials and identify the coecient. If the monomial has just one variable, also indicate its degree. 1) 2) Solutions: 2 3 x ) 4) 2x 3 y ( 3x 2 ) ( 5y 7) 1) 2 3 is a monomial. Since there is no variable, we use our zero power rule, Proposition to rewrite 2 3 = So the coecient is 2 3 and the degree is zero. = 2 3 x0 2) This is also an monomial since we can rewrite the division by a number to a multiplication by a number as follows. x 5 10 = 1 10 x5 24

28 CHAPTER 3. POLYNOMIALS 25 So the coecient is 1 10 and the degree is 5. 3) 2x 3 y is not a monomial because we are dividing by y which is the same as multiplying by y 1 (using the negative power rule, Proposition 2.6.1) By denition the variables in a monomial must have nonnegative powers. This problem is really the quotient of two monomials. 4) ( 3x 2) ( 5y 7) = 15x 2 y 7, so this is a monomial. The coecient is 15. However, since it has two variables, we have not dened the degree of such a monomial. Now that we have dened what a monomial is, we can talk about operations using monomials Multiplication and Division of Monomials To multiply and divide monomials simply involves using the rules of exponents from Chapter 2. Example Perform the indicated operation and simplify. 1) ( 5x 3 y 8) ( 10xy 5) 2) 3x 7 y 4 12x 3 y Solutions: 1) ( 5x 3 y 8) ( 10xy 5) = 50x 4 y 13 2) 3x 7 y 4 12x 3 y = x4 y Addition and Subtraction of Monomials Recall that the distributive property states that (b + c)a = ba + ca (b c)a = ba ca If we read these from left to right, we get what we usually refer to as the distributive property. However, we could also read them from right to left, i.e. ba + ca = (b + c)a ba ca = (b c)a This states that if two terms that are added or subtracted share a common factor (a, in this case), then that factor can be taken out of both terms. We refer to this as factoring out a common factor. So if we have 3x + 5x = (3 + 5)x x is common to both terms and so can be factored out. = 8x 3x and 5x are called like terms because when you factor out the common variable, you are just left with numbers that can be added. Now, if we have 3x 2 + 5x = (3x + 5)x since only one x is common to both terms. Notice that what is left in the parentheses cannot be combined because they are not both just numbers. So 3x 3 and 5x are not like terms and therefore cannot be combined.

29 CHAPTER 3. POLYNOMIALS 26 So 3x 2 and 5x 2 are like terms. 3x 2 + 5x 2 = (3 + 5)x 2 x 2 is common to both terms and so factored out. = 8x 2 Denition To be able to add or subtract two terms, you must be able to factor out all the variables. This means that the variables must be exactly the same and to the same power for both terms. We refer to these as like terms. Example Perform the indicated operation and simplify. 1) 3x 4 4xy + 7x 2 + xy 2) x 3 y 5xy 3 2x 3 y + 4x 3 y 5xy 3 Solutions: 1) 3x 4 4xy + 7x 2 + xy 3x 4 4xy + 7x 2 + xy We change all the subtractions to adding by the opposite. This way, when we start moving the terms around, we do not loose tract of the negatives. = 3x 4 + ( 4)xy + 7x 2 + xy = 3x 4 + 7x 2 + ( 4 + 1)xy The only like terms are the ones with xy in them. So we can combine the 4xy and xy. = 3x 4 + 7x 2 + ( 3)xy = 3x 4 + 7x 2 3xy 2) x 3 y 5xy 3 2xy + 4x 3 y 5xy 3 x 3 y 5xy 3 2xy + 4x 3 y 5xy 3 = x 3 y + ( 5)xy 3 + ( 2)x 3 y + 4x 3 y + ( 5)xy 3 There are three terms with x 3 y, so we combine those three just like we would for two terms. = [1 + ( 2) + 4]x 3 y + [( 5) + ( 5)]xy 3 = 3x 3 y + ( 10)xy 3 = 3x 3 y 10xy 3

30 CHAPTER 3. POLYNOMIALS Polynomials View the Video Tutorial for this Section here. Denition A polynomial is a sum of one or more monomials. The general form for an n-th degree polynomial is a n x n + a n 1 x n a 1 x + a 0 where a n,..., a 0 represent real numbers with a n 0 and n is a non-negative integer (0, 1, 2,...) Leading term: is the highest degree monomial. (a n x n ) in the above notation. Leading coecient: is the coecient of the leading term. a n in the above notation. Degree: of a polynomial is the degree of the leading term. n for the above. So given the polynomial 3x 4 7x 3 + x 9 if we were to write it in the general form given in Denition 3.2.1, we rst rewrite it as 3x 4 7x 3 + x 9 = 3x 4 + ( 7)x 3 + 0x 2 + x + ( 9) Since there is no x 2 term, we use 0x 2 which is 0. So n = 4, a 4 = 3, a 3 = 7, a 2 = 0, a 1 = 1, a 0 = 9. The leading term is 3x 4, the leading coecient is 3, and the degree of the polynomial is 4. Denition Given an n-th degree polynomial a n x n + a n 1 x n a 1 x + a 0 the i-th degree term is the monomial of i-th degree, a i x i note that the number i cannot be bigger than n, i.e. i n. Given this denition, for the above 4-th degree polynomial 3x 4 7x 3 + x 9 7x 3 0x 2 x is the third degree term. This polynomial does not have a second degree term. If we need to explicitly write it, then we would say that the second degree term is 0x 2. is the rst degree term. 9 is the zero degree term or also called the constant term. Recall that we call it the zero degree because 9 = 9x 0.

31 CHAPTER 3. POLYNOMIALS Evaluating Polynomials When evaluating polynomials, replacing the variables with numbers or other expressions, we need to keep in mind that Order of Operations states that exponents come before multiplication. This means that ( 5) 2 = ( 5)( 5) = = = = 25 So one must be careful of where the parentheses are when dealing with negatives and even powers. Note that it is not as crucial for odd powers, because raising a negative number to an odd power still keeps it negative, e.g. ( 5) 3 = ( 5)( 5)( 5) = = = 125 Example Evaluating polynomials. 1) Given y = 3x 2 6x + 5, nd y, if x = 1 3 x = 3 x = t 2) Given y = x 3 3x 2 4x 1, nd y, if x = 1 x = ξ x = 2t Solutions: 1) Given y = 3x 2 6x + 5, We rst note that we can rewrite the polynomial as y = 3x 2 + ( 6)x + 5 for x = 1 3, ( ) 2 ( ) 1 1 y = 3 + ( 6) ( ) 1 = 3 + ( 2) = ) = ( 3 3 = = 8 3 for x = 3, y = 3( 3) 2 + ( 6)( 3) + 5 = 3(9) = = 4 for x = t, y = 3( t) 2 + ( 6)( t) + 5 = 3t 2 + 6t + 5 2) Given y = x 3 3x 2 4x 1, we rst note that this is the same as y = x 3 + ( 3)x 2 + 4x + ( 1).

32 CHAPTER 3. POLYNOMIALS 29 for x = 1, y = ( 1) 3 + ( 3)( 1) 2 + ( 4)( 1) + ( 1) = 1 + ( 3) 1 + (4) + ( 1) = 1 for x = ξ, y = ( ξ) 3 + ( 3)( ξ) 2 + ( 4)( ξ) + ( 1) = ξ 3 + ( 3)ξ 2 + 4ξ + ( 1) = ξ 3 3ξ 2 + 4ξ 1 for x = 2t, y = (2t) 3 + ( 3)(2t) 2 + ( 4)(2t) + ( 1) = 8t 3 + ( 3)4t 2 + ( 8)t + ( 1) = 8t 3 12t 2 8t 1

33 CHAPTER 3. POLYNOMIALS Addition and Subtraction of Polynomials View the Video Tutorial for this Section here. Since a polynomial is just a sum of monomials, the concept behind adding and subtracting polynomials is the same as that used for monomials. We can only add or subtract like terms, Denition Example Perform the indicated operation and simplify. 1) ( 3x 4 + 5x 3 x + 2 ) ( ) x x3 + x ) (x 5 8x 3 + 3x2 4 5 ) 3 x (x 3 + x x2 7 ) 3 Solutions: 1) ( 3x 4 + 5x 3 x + 2 ) + 2) ( ) x 4 3 9x3 + x = 3x 4 + 5x 3 x x4 3 9x3 + x Since the polynomials are added, we can ignore the parentheses = 3x 4 + 5x 3 + ( 1)x x4 + ( 9)x 3 + x We change all the subtractions to adding by the opposite. Also, dividing by 3 is the same as multiplying by 1 x4 3, so 3 = 1 3 x4. = [ 3 + 3] 1 x 4 + [5 + ( 9)]x 3 + x 2 + ( 1)x + [2 + 4] = 8 3 x4 4x 3 + x 2 x + 6 ( ) ( ) x 5 8x 3 + 3x x x 3 + x x2 7 3 ( ) = x 5 8x 3 + 3x x + ( 1) x 3 + x x2 7 3 Since the two polynomials are subtracted, we can ignore the parentheses around the rst polynomial but not the one following the subtraction sign. So we convert the subtraction to adding by the opposite. = x 5 + ( 8)x 3 + 3x2 4 + ( ) 5 3 x + ( 1)x 3 + x x Distributing the 1 inside the second polynomial changes all the signs. = x 5 +( 8)x x2 + ( 3) 5 x+( 1)x 3 + ( 2) 1 x x = [ 1 + ( 2)] 1 x 5 + [ 8 + ( 1)] x 3 + [ ] 1 x 2 + ( ) 5 3 x Add the like terms. = 1 2 x5 9x x2 5 3 x Example Let Evaluate: P = x 3 3x 2 + x and Q = x3 2 + x2 5x ) 2P + Q 2) P 2Q 3) 4P 2Q 4) 4Q 6(P + 5) Solutions: 1) 2P + Q

34 CHAPTER 3. POLYNOMIALS 31 Q P {}}{{}}{ = 2 x 3 3x 2 + x x x2 5x Since P = x 3 3x 2 + x This means that where ever we see P, we can replace it with x 3 3x 2 + x Likewise for Q. = 2 [x 3 + ( 3)x 2 + x ] x3 2 + x2 + ( 5)x = 2 x 3 +2 ( 3)x 2 +2 x x3 +x 2 + ( 5) 2 x+3 Since we are adding, the square bracket around Q can be removed. The square brackets around P cannot be removed because of the 2 that is multiplied to P. To remove these brackets, we must distribute the 2 into P. = 2x 3 + ( 6)x 2 + x x3 + x 2 + ( 5) 2 x + 3 = [ ] [ ( 1 x 3 +[ 6+1]x )] x+[2+3] 2 = 5 2 x3 5x 2 2x + 5 2) P 2Q Q P {}}{{}}{ x 3 3x 2 + x x x2 5x = [x 3 + ( 3)x 2 + x ] [ ( x ( 2) 2 + x2 + 5x ) ] We rst convert all the subtractions to adding by the opposite. = x 3 +( 3)x 2 + x ( x3 +1+( 2) 4 2 +( 2) x2 +( 2) 5x ) +( 2) 3 2 Distribute -2 into Q. = x 3 + ( 3)x x ( x 3) + ( 2)x 2 + 5x + ( 6)

35 CHAPTER 3. POLYNOMIALS 32 [ ] 1 = [1 + ( 1)]x 3 + [ 3 + ( 2)]x x + [1 + ( 6)] = 5x x 5 3) 4P 2Q = 4 [x 3 3x 2 + x ] [ x = 4 [x 3 + ( 3)x 2 + x ] ( 2) 2 + x2 5x ] [ x x2 + ( 5 ) ] x = 4 x ( 3)x x ( 2)x3 2 + ( 2) x2 + 2 = [4 + ( 1)]x 3 + [ 12 + ( 2)]x 2 + [1 + 5]x + [4 + ( 6)] = 3x 3 14x 2 + 6x 2 ( 5 ) x + ( 2)3 2 4) 4Q 6(P + 5) [ x 3 = x2 5x = x3 + 4 x ] 6 ([x 3 3x 2 + x ] ( 5 2 ) x ( 6) ) + 5 (x 3 + ( 3)x ) x = 2x 3 + 4x 2 + ( 10)x ( 6) x 3 + ( 6)( 3)x 2 + ( 6) ( = 2x 3 + 4x 2 + ( 10)x ( 6)x x = [2 + ( 6)]x 3 + [4 + 18]x 2 + [ 10 + ( 3 4 ) x + ( 36) ) ]x + [12 + ( 36)] ( ) 1 x + ( 6)(6) 4 = 4x x x 24

36 CHAPTER 3. POLYNOMIALS Multiplying Polynomials View the Video Tutorial for this Section here. The key property used to multiply any two polynomials is the Distributive property. Proposition Distributive Property. a(b + c) = ab + ac and a(b c) = ab ac It is important to remember that any time you read a mathematical statement such at the one above, that a, b, and c do not necessarily have to represent just one number or variable. They can represent an expression. Note that in the above proposition, we would say that the a is distributed into b + c. Example Multiply and then simplify, if possible. 1) 2xy 2 ( x 2 + 3x 3 y 5) 2) (x + 3)(x 7) 3) (x 5) 2 4) (x 2)(x 2 3x + 5) Solutions: 1) 2xy 2 ( x 2 + 3x 3 y 5) In this case we apply the distributive property, Property 3.2.1, with a = 2xy 2, b = x 2, and c = 3x 3 y 5. = 2xy 2 x 2 + 2xy 2 3x 3 y 5 We distribute (multiply) 2xy 2 into x 2 + 3x 3 y 5 = 2x 3 y 2 + 6x 4 y 7 2) Note that the parentheses means that we can consider the expressions inside each parentheses as a single quantity. This means that we can use the distributive property by distributing (x + 3) into x 7 or distribute (x 7) into x + 3. (x + 3)(x 7) = (x + 3)(x + ( 7)) Here we will distribute (x + 3) into x + ( 7) using Property with a = (x + 3), b = x and c = 7. = (x + 3)x + (x + 3)( 7) }{{}}{{} 1st term 2nd term = x x + 3 x + x ( 7) + 3( 7) Now we distribute x into x + 3 in the rst term and 7 into x + 3 in the second term. = x 2 + 3x + ( 7)x + ( 21) = x 2 + ( 4)x + ( 21) = x 2 4x 21 3) The most common mistake with this problem is that students will try to apply Proposition and bring the power inside the parentheses. But Proposition states that you can only bring the power inside the parentheses, if what is inside the parentheses are multiplied together, not added or subtracted. The only thing we can do at this point is to apply the denition of exponents and then use the Distributive property to multiply everything out. (x 5) 2 = (x 5)(x 5) = (x + ( 5))(x + ( 5)) = (x + ( 5))x + (x + ( 5))( 5) = x 2 + ( 5)x + ( 5)x + 25 = x 2 10x + 25

37 CHAPTER 3. POLYNOMIALS 34 4) Here we will distribute (x 2) into x 2 3x + 5. (x 2)(x 2 3x + 5) = (x + ( 2))(x 2 + ( 3)x + 5) = (x + ( 2))x 2 + (x + ( 2))( 3)x + (x + ( 2))5 = x 3 + ( 2)x 2 + ( 3)x 2 + 6x + 5x + ( 10) = x 3 5x x 10

38 CHAPTER 3. POLYNOMIALS Dividing Polynomials by Monomials View the Video Tutorial for this Section here. In order to divide a polynomial by a monomial, we need to recall the rule for adding fractions, which says that you can only add fractions if they have the same denominator, in which case the numerators are added together and placed over the common denominator. If we read this rule from right to left, it states that a b + c b = a + c b Denition Rule for adding fractions read backwards. Similarly, for subtraction we have a + c b a c b = a b + c b = a b c b In other words, we can break down a fraction that has a sum in the numerator to the sum of two fractions with the same denominator. The signicance of this is as follows. Simplify If you applied your arithmetic correctly, you should have gotten 5, since you add the 4 and the 6 which gives you 10 and then divide it by 2. Now, if you think about the order of operations, you might wonder what happened here, since division is suppose to come before addition. Although the fraction sign means division, there is a little more to it than that. It is understood that whatever is in the numerator and denominator of a fraction are actually inside parentheses. Normally, we do not write them in because it is understood that they are implied to be there by the fraction sign. In other words really means (4+6) 2. This means that we must add the 4 and 6 rst before we can divide by 2. However, if we use Denition 3.2.3, we can rewrite the fraction as follows = = = 5 in which case we do the division rst, before the addition. Note that both 4 and 6 must be divided by 2. So Denition gives us a way to do the division before addition, when dealing with fractions. Example Divide and simplify, if possible. 1) 2) 3x 4 12 x 2 5x x 3) 6(x 3) (x 3) 9 3(x 3) Solutions: 1) 3x 4 12 = 3x = x 4 1 3

39 CHAPTER 3. POLYNOMIALS 36 2) x 2 5x x = x2 10x 5x 10x x = x x 3) 6(x 3) (x 3) 9 3(x 3) = 6(x 3)2 3(x 3) = 2(x 3) x 3 = 2x x 3 = 2x 2 3 x (x 3) 3(x 3) 9 3(x 3)

40 Chapter 4 Linear Equations and Inequalities in One Variable View the Video Tutorial for an overview of Linear Equations. With the introduction of variables in algebra, we can now talk about equations which could not be done with just arithmetic. An equation consists of two expressions, at least one of which contains a variable, set equal to each other. To solve an equation is to nd all values that make the equation true. For example to solve x + 3 = 5 means nd all numbers, x, such that if you add 3 to it, you get 5. As we will nd, in the following sections, the only number that satises this condition is 2. So we would say that the solution is x = 2. It should be noted that our description of an equation is not the most general denition, but does describe the type of equations that we will work with. There are many types and classications of equations which depend on the number of variables in the equation and what type of operations are involved. The creation of equations and the methods of solving them is the primary goal of mathematics. The ability to create and solve equations allows us to answer a vast range of real life problems. Problems such as how much a person should be saving each month if they wish to retire in 20 years, what will the weather be like tomorrow, to how long will radio-active waste be dangerous, and what should be the holding pattern and descent of a passenger jet plane to minimize passenger discomfort. All the mathematical concepts that we learn are to help us solve various types of equations. In this chapter, we begin the exploration of equations by studying the most fundamental type of equation, the linear equation in one variable. The concepts and techniques that you will learn in this chapter will be used over and over again throughout the rest of your mathematical studies. 4.1 Inverses To understand how we solve equations, we must rst understand the idea of inverses. In mathematics, an inverse is something that will undo some operation. So, in talking about inverses, we need to also talk about what operation we are undoing Additive Inverse To understand what an additive inverse does, we begin with an example. Start with the number 3. If we add 5 to it, we get 8. Now, if we add 5 to 8, we get back to the number that we started with, 3. Start with the Add 5 Now add 5 We get the number we number started with = ( 5) 3 So adding the 5 undoes the adding of 5. This means that 5 is the additive inverse of 5. In general, we have that 37

41 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 38 Denition Additive Inverse The additive inverse of a number is the opposite of the number. In other words, the additive inverse of a is a. Example Find the additive inverse of the following: 1) 2 3 2) 3 3) 4x Solutions: 1) The opposite of 2 3 is ) Note that in the denition of additive inverse, Denition 4.1.1, when we say that a is the additive inverse of a, a does not mean a negative number. a just means the opposite of whatever a is. In this case, since a = 3, the opposite of 3 is 3, i.e. a = ( 3) = 3. So the additive inverse of 3 is 3. 3) The opposite of 4x is 4x Multiplicative Inverse Just as the additive inverse undoes addition, a multiplicative inverse would undo multiplication. We use the following example to illustrate what the multiplicative inverse does. Start with the number Multiply by 5 Now multiply by = ( 1 5 So multiplying by 1 5 undoes the multiplication by 5, which means that In general, we have ) We get the number we started with 3 is the multiplicative inverse of Denition Multiplicative Inverse The multiplicative inverse of a number is its reciprocal. In other words, the multiplicative inverse of a is 1 a Another way to say this is that the multiplicative inverse of a b is b a. Example Find the multiplicative inverse. 1) 2 3 2) 3 3) 0 Solutions: 1) The multiplicative inverse of 2 3 is its reciprocal ) The multiplicative inverse of 3 is its reciprocal 1 3. Note that we need to pay attention to what type of inverse we want. The additive inverse of 3 is 3 but the multiplicative inverse is ) The reciprocal of 0 is 1 0, which is undened. This means that 0 has no multiplicative inverse. This makes sense because if you multiply any number by 0 you get 0. For example 2 0 = 0 and 3 0 = 0. Now if 0 had a multiplicative inverse, we should be able to undo the multiplication by 0 and get back to the number before we multiplied by 0. But then the question would be what number to we go back to? 2 or 3? There is no way to know and so we cannot undo the multiplication by 0, hence 0 has no multiplicative inverse.

42 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE Addition Property of Equality View the Video Tutorial for this section and the next section here. To solve equations requires two concepts. The idea of undoing an operation, which we discussed with the inverses (Additive and Multiplicative) and the idea of balance. The idea of balance is as follows. Suppose we have x = 2 This means that x is the same as 2. One way to visualize this is as follows. We label one box of unknown weight as x and the other that weights 2 pounds with 2. x = 2 would mean that if we place the box labeled x on one side of the scale and the box labeled 2 on the other, then the scale would be balanced. Now if we added 3 pounds to the left side (x), the scale would no longer be balanced. But, if we added the 3 pounds to both sides of the scale, it would remain balanced. So mathematically this means that if we start with x = 2 and add 3 to both sides of the equation, it would still be equal. x + 3 = or x + 3 = 5 Now there are two implications to the above example. The rst is that if we add the same thing to both sides of the equation, then they are still equal to each other. The second is a little more subtle. The solution(s)

43 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 40 to an equation, is the number(s) that make the equation true. So the solution to the equation x = 2 would be 2, since if we replace the x with 2, we get x = 2 (2) = 2 which is true. Notice that 2 is also the solution to the new equation we got since x + 3 = 5 (2) + 3 = 5 5 = 5 So the second implication is that if you add the same thing to both sides of an equation, then the old equation and the new equation will have the same solutions. If we generalize this, we get Proposition Addition Property of Equality. If then a = b a + c = b + c The addition property has two implications. If you have an equation and you add the same thing to both sides of the equation, then they are still equal. The old equation and the new equation have the same solutions. Now, if you have had any experience solving equations or read the beginning of this chapter, you may be scratching your head because my example above seems to be going in reverse order. I started with x = 2 and got x + 3 = 5. You would be justied in your quandary. Normally, when we solve an equation, the goal is to get the variable by itself on one side of the equation and everything else on the other side. Keep in mind that the above example is of the addition property, Proposition 4.2.1, not about solving an equation. We now turn our attention to solving equations using the addition property of equality and the inverses. To solve the equation left side {}}{ x + 3 = right side {}}{ 5 as we stated before, we must try to get the variable, in this case x, by itself on one side of the equation. So we need to get rid of the 3. Since the 3 is connected to x by addition, to undo the addition by 3, we need to add the additive inverse 3. But according to the addition property of equality, Proposition 4.2.1, if we add 3 to the left side of the equation, then we must also add it to the right side to keep the equation balanced. (x + 3) + ( 3) = 5 + ( 3) Add -3 to both sides of the equation. There are parentheses around x + 3 because we are adding the -3 to the entire side, which consists of both the x and the 3. x ( 3) = 2 However, recall that for addition, the parentheses do nothing and can be ignored (note that this will not be true when we talk about multiplication in the next section.) x + 0 = 2 x = 2 Note that we can check to see if this is the correct answer for the equation by plugging in 2 for any x we see in the original equation and see if the two sides actually come out the same. So, the check is as follows:

44 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 41 x + 3 = 5 Example Solve. (2) + 3 = 5 Plug in 2 for the x. 5 = 5 Since this is true, x = 2 is the solution to the equation. 1) x 7 = 9 3) x + a b = c, solve for x. 2) x = 2 4) x + 3 = 7 + 2x Solutions: 1) x 7 = 9 x + ( 7) = 9 x + ( 7) +7 x = 16 = 9 +7 We rewrite the subtraction as adding the opposite. Since -7 is connected to x by addition, we add 7 to undo it. To balance the equation, we add the 7 to both sides of the equation. Notation: We write what we want to add to both sides of the equation below the equation itself. 2) Since 2 5 is added to x, we undo the addition by adding 2 3 to both sides of the equation. 2 + x = x = ( 2 3 ) Notation: expressions written below an equation are understood to be added to the equation. Here, we are adding 2 3 to both sides of the equation. To add fractions, they must ( have the same denominator. So 2 = 2 3 ) 1 3 = 6 3 x = 4 3 3) We treat the variables a, b, and c just as we would numbers. x + a b = c x + a + ( b) = c x + a a + ( b) +b = c a+b Convert subtraction to addition by the opposite. Add the additive inverse of a and b, which are a and b, to both sides of the equation. x = c a + b Since there are no like terms, there is nothing to combine here. 4) The goal of solving an equation is to get the variable we want to solve for by itself on one side of the equation and everything else on the other side. In this problem, notice that x is on both sides of the equation. So we need to gather all the x's on one side of the equation.

45 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 42 x + 3 = 7 + 2x x x To get rid of the x on the left side, since it is added to 3 we add its opposite, x to both sides of the equation 3 7 = x 4 = x

46 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE Multiplication Property of Equality View the Video Tutorial for this section and the revious section here. As stated in the previous section, to solve equations requires two concepts. The idea of undoing an operation (inverses) and the idea of balancing an equation. In the last section, we discussed how adding the same quantity to both sides of an equation keeps the equation balanced. Here we will discuss how an equation is balanced when multiplication is involved. As before, we begin with x = 2 which we can visualize as two boxes that balance a scale. If we triple the left side, i.e. multiply x by 3, note that the scale would be o balance. To balance the scale, we would have to multiply both sides by 3. Which gives us 3x = 3(2) 3x = 6 As with the addition property of equality, there are two implications to the multiplication property. First, that multiplying both sides of the equation by the same number or expression keeps the equation balanced and that the new and old equations both have the same answers. With the multiplication property, there are a couple of subtle points that we must be careful of that did not exist with the addition property. We will discuss these after rst stating the multiplication property.

47 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 44 Proposition Multiplication Property of Equality. If then a = b c(a) = c(b) for any c 0. The multiplication property has two implications. If you have an equation and you multiply the same nonzero expression to both sides of the equation, they are still equal. The old equation and the new equation have the same solutions. As we stated before, there are a couple of subtle but important points to keep in mind when using the multiplication property of equality. The rst is that it does not apply if you multiply both sides of the equation by the number zero. The reason for this is that if you multiply any number by 0, you get 0. So a false equation can be turned into a true statement. For example, 2 = 5 We know that 2 does not equal -5. So this equation is never true. 0(2) = 0( 5) Now multiply both sides by zero. 0 = 0 This equation is always true. So we went from an equation, 2 = 5, which is never true to an equation, 0 = 0, which is always true. These equations cannot be equivalent to each other and so we see that multiplying both sides of an equation by zero can lead to misleading answers. As to the second issue, it can be better discussed through an example. We start with the equation x + 3 = 5 Suppose now that we wish to double the right side, i.e. multiply the 5 by 2. The question is, what do we do to the left side. Note that if we just double the x that will not be enough to balance out the doubling of 5 since it take both x and 2 to balance out 5. In other words, 2x To balance the equation, if we double the right side, then we must double everything on the left side of the equation as well. In other words we must double the 3 as well as the x.

48 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 45 In algebraic form, we mean that x + 3 = 5 Given this equation, if we double the right side, then we must double the entire left side. 2(x + 3) = 2(5) Notice that unlike the addition property, the parentheses cannot be ignored when applying the multiplication property of equality. 2x + 6 = 10 It should be noted here that the above example is to explain the nuances of the multiplication property of equality and not an example of how to solve equations. The above is just a demonstration of how you multiply both sides of the equation by a number and still maintain the equality. To solve equations using the multiplication property, we need to choose carefully what we will multiply both sides of the equation by. The choice of what to multiply by is determined by what multiplication we wish to undue, which involves the use of multiplicative inverses. Example Solve. 1) 2) Solutions: 3x = 5 x 5 = 4 3) 4) 2 5 x = 6 2x 5 = 6 5) 6) x = 3 (a + 3)x = 5, solve for x 1) Remember that to solve an equation, we must get the variable, in this case x, by itself on one side of the equation. 3x = 5 Since the 3 is multiplied to x, to get rid of it, we need to multiply by the multiplicative inverse of 3, which is ( 3x) = 1 3 ( 5) x = 5 3 If we multiply 1 3 on one side, we must also multiply it on the other side of the equation to keep it balanced. 2) x 5 can be viewed as 1 5 x and so we multiply by 5 to undo the multiplication by 1 5. So 3) The multiplicative inverse of 2 5 is 5 2, so x ( 5 x ) 5 5 = 4 = 5(4) x = x = 6 5 ( 25 ) 2 x = 5 2 (6) x = 15

49 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 46 4) Note that 2x 5 is the same as 2 5x, so this equation is really the same equation as the previous example and can of course be solved the same way. However, an alternate way to solve the equation would be to view the x as being multiplied by 2 and divided by 5 and so undo those two operations instead of viewing it as one number 2 5 that is multiplied to x. 2x 5 = 6 ( ) 2x 5 = 5(6) 5 Since dividing by 5 really means multiply by 1 5, we undo it by multiplying by 5. Multiplicative inverse of -2 is ( 2x) = 1 2 (30) x = ) Recall that x can be viewed as 1 x. Note that the reciprocal of -1 is 1 = 1. So x = 3 1( x) = 1(3) x = 3 6) Since we are solving for x, we need to get rid of the expression (a + 3) which is multiplied to x. To undo this multiplication we need to multiply by the multiplicative inverse of a + 3, which is 1 a+3. So (a + 3)x = 5 1 a + 3 (a + 3)x = 1 a + 3 (5) x = 5 a + 3 Note here that we do not want to distribute the x into (a + 3). Although there is nothing algebraically wrong with distributing in the x, it would not help us to isolate the x. In fact, if we distribute the x, then we end up with two terms that have x in them. Since we are trying to get x by itself on one side of the equation, this would be counter productive.

50 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE Summary of Solving Linear Equations in One Variable General Equations View the Video Tutorial for this section here. In general, the equations that we encounter will have a combination of addition, subtraction, multiplication and division within the equation, which will require us to use a combination of the addition and multiplication properties of equality as well as the various inverses. The point of confusion for students often comes down to what do you do rst, because there are often more than one way to solve an equation. Recall that to solve an equation, we use the additive and multiplicative inverses to undo the various operations to get the variable by itself. Usually, when you want to undo a sequence of events, you work in the reverse order. For example, after you shower, the sequence that you follow to dress would be to put on your undergarments rst, then your outer garments. To take a shower, you would have to undo what you did to get dressed. To do this, you would start by removing your outer garments rst and then your undergarments. So to undress, you reverse the sequence you followed to get dressed. Likewise, when we solve equations, we use the order of operations in reverse. To solve the equation 2x + 3 = 11 note that on the left side of the equation, if x had been a number and you were just evaluating 2x + 3, order of operations would say that you rst multiply by 2 then add the 3. To solve for x means we must undo the multiplying by 2 and the addition of 3. To do this, we work in the reverse of the order of operations and rst undo the addition by 3. So we add 3 to both sides of the equation. Now, we undo the multiplication. Example Solve. 2x = x = (2x) = 1 2 (8) x = 4 1) 4 6x = 8 2) x 3[x (2 x)] = 2 3) 1 + 4(x 3) = 3(3 2x) 2x 4) a(x 2) = bx 3, solve for x Solutions: 1) Note that 4 6x can be rewritten as 4 + ( 6x). If we were evaluating this expression, according to order of operations, -6 is rst multiplied to x and then the 4 is added last. So to undo these operations and get x by itself, we work in the reverse order and rst undo the addition of 4 (by adding -4) and then undo the multiplication by -6 (by multiplying by 1 6.) 4 6x = = 8 4 6x = ( 6x) = 1 6 ( 4) x = 4 6 x = 2 3 2) Here we need to get rid of all the brackets and parentheses (by using the distributive property) so that we can combine all the terms that have x in them.

51 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 48 x 3[x (2 x)] = 2 x + ( 3)[x + ( 1)(2 + ( x))] = 2 x + ( 3)[x + ( 2) + x)] = 2 x + ( 3)[2x + ( 2)] = 2 x + ( 6x) + 6 = 2 5x = 2 6 5x = ( 5x) = 1 5 ( 8) Convert all the subtractions to adding by the opposites. With multiple nested parentheses and brackets, we work from the inside out. Now that we only have one term with x in it, we can solve for x. We rst undo the addition by 6, following the reverse of the order of operations, by adding -6 to both sides of the equation. Now we undo the multiplication by -5 by multiplying by the reciprocal 1 5. x = 8 5 3) In order to get x by itself on one side of the equation, we need to combine all the x's together. To do this, we need to get rid of the parentheses as they prevent us from combining all the x's (x 3) = 3(3 2x) 2x To get rid of the parentheses, we distribute the 4 and 3 into x 3 and 3 2x respectively x + ( 12) = 9 + ( 6)x + ( 2)x 4x + ( 11) +8x +11 = ( 8)x +8x 12x = (12x) = 1 12 (20) x = 5 3 We need to get all the x's on one side of the equation. Since the 8x is added to 9, we undo the addition by adding 8x to both sides. This will remove the x term on the right side of the equation. Likewise, we add 11 to get rid of the -11 on the left side of the equation. 4) Since we are solving for x, we treat the other variables, a and b, like numbers, which means that they must also be on the other side of the equation from x. a(x 2) = bx 3 a(x + ( 2)) = bx + ( 3) ax bx + ( 2)a +2a = bx bx + ( 3) +2a ax bx = 3 + 2a (a b)x = 3 + 2a 1 1 a b (a b)x = a b ( 3 + 2a) In order to combine all the x's, we need to get rid of the parentheses by distributing in the a since x is inside the parentheses. We get rid of the bx, on the right side, by adding bx to both sides and we get rid of ( 2)a, on the left side, by adding 2a to both sides. Now, although we cannot technically add ax and bx, recall that combining like terms was really the distributive property used in the reverse direction (see Sec ), which we call factoring. Here we can factor out the x. We need the parentheses around 3 + 2a since 1 a b needs to be multiplied to the entire side, i.e. to both terms.

52 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 49 x = 3+2a a b Equations with Fractions View the Video Tutorial for this section here. Given an equation that contains fractions, the multiplication property of equality, Proposition 4.3.1, gives us a technique for getting rid of the fractions. We will describe the technique through the following example. x = 3 2 First we nd the least common denominator (LCD) of all the fractions on both sides of the equal sign, which in this case is 6 ( x ) ( ) 3 = Now we multiply both sides of the equation by the LCD. ( x ) ( ) = Remember that we must multiply the entire side by 6. That means that every term in the equation is multiplied by 6. 2x + ( 5) +5 = 9 +5 Now that all the fractions are gone, we solve as we have done before. 2x = (2x) = 1 2 (14) x = 7

53 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 50 Summary of Technique: When solving equations that contain fractions, we can use the multiplication property of equality to remove all the fractions as follows: Find the least common denominator (LCD) of all the fractions on both sides of the equation. Multiply both sides of the equation by the LCD. Note that this technique only works for equations, because the multiplication property of equality only applied to equations. Equation We can apply the technique as shown above. Expression We cannot get rid of the fractions. Here, you make each fraction have the same LCD and then carry out the subtraction. x = 3 2 x ( x ) x 5 6 Example Solve. 1) x = x 4 2) 5x 12 x 1 = 2x Solutions: 1) x = x 4 Since this is an equation, we can get rid of the fractions by multiplying through by the LCD, which in this case is 4. ( ) x 2 ( x ) 4 5 = Remember that we must multiply the entire side of the equation by 4. ( ) x 2 ( x ) 4 + 4( 5) = (x 2) + ( 20) = x Since there are x's on both sides of the equation, we need to distribute the 2 inside the parentheses so that we can collect all the x's together on one side of the equation. 2x + ( 4) + ( 20) = x 2x + ( 24) = x 2x 2x Since the 2x is added to -24, to undo the addition, we add 2x to both sides. Note that this places all the x's on the right side of the equation.

54 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE = x 1( 24) = 1( x) Recall that the multiplicative inverse of -1 is = x 2) 5x 12 x 1 = 2x Here the LCD is 12. But before we multiply by it, keep in mind that the fraction bar implies parentheses around the numerator and denominator. So when we multiply by 12, we must multiply the entire numerator by 12. ( 5x x 1 ) ( ) 2x + 3 = ( ) ( ) 5x x ( 12) = 4(2x + 3) x + ( 6)(x 1) = 4(2x + 3) 5x + ( 6x) + 6 = 8x + 12 When we reduce the right side of the equation 12 3 to 4, keep in mind that the 4 must still be multiplied to the whole numerator 2x + 3. Likewise, the subtraction on the left side of the equation applies to the entire numerator, so it can be viewed as +( 1) (x 1) 2 which when multiplied by 12 gives us the -12. x + 6 +x 12 = 8x x = 9x 1 9 ( 6) = 1 9 (9x) 2 3 = x Special Cases Identity In the above examples, when we solved the equations, we ended up with one answer. Now we will discuss a couple of special situations that can arise while solving equations. We start by solving the following equation. 2x 15 = 3(x 5) x 2x + ( 15) = 3x + ( 15) + ( x) 2x + ( 15) 2x = 2x 2x 15 = 15 Note here that all the terms with x in them have disappeared. The question is then what is the solution to the original equation. This is where we have to remember that the addition and multiplication property of equality stated that when you add or multiply the same number to both sides of the equation, not only are they still equal, but the solutions to the new equation and the old are the same. In other words, the values of x that make 15 = 15 are also the values of x that make 2x 15 = 3(x 5) x. Since the equation 15 = 15 is always true and does not depend on what value x is, this means that the original equation will also always be true, no matter what x is. In other words, no matter what x value you choose, if you work both the left and right side out separately, the numbers will always equal each other. In such cases, we say that the solution to the equation 2x 15 = 3(x 5) x is all real numbers. We call equations that are always true, an Identity.

55 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 52 Denition An equation that is always true is called an Identity. An equation is an Identity when you try to solve it for the variable, but the variable ends up being eliminated from the equation and you end up with the same number on both sides of the equation, like 2 = 2, 0 = 0 or 5 = 5. For such equations, we say that the solution is all real numbers Inconsistent The second special case occurs when the variable disappears from the equation, but what remains do not equal each other. For example 2x 5 = 3(x 5) x 2x + ( 5) = 3x + ( 15) + ( x) 2x + ( 5) 2x = 2x 2x 5 = 15 Since -5 never equals -15, this means that there is no x value that makes the equation 5 = 15, which in turn means that there is no x value that will make the equation 2x 5 = 3(x 5) x true. In other words, no matter what value you pick for x, if you work out the left and right side out separately, the two sides will never give you the same number. In such cases, we say that there is no solution. We call equations that are never true, inconsistent. Denition An equation that is never true is called Inconsistent. An equation is inconsistent when you try to solve it and the variable ends up being eliminated from the equation and you end up with dierent numbers on either side of the equation, like 0 = 1, 3 = 3, or 2 = 5. For such equations, we say that it has no solutions.

56 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE Applications Percentage Recall that percentages are a way to talk about a part or piece of something. For example, 50% means half of something and 25% means a quarter. Now, suppose someone came up to you and said if you gave him your car, he would give you half. Would you do it? A natural question to ask would be half of what. Would that not make a dierence in how you would answer the proposition? Suppose the person that came up to you was Bill Gates and he was saying that he would give you half his wealth for your car. What if it was Bill Gates saying that he would give you half his lunch? Notice that your answer will probably depend on what he meant by half. If you are talking about a part or piece then you need to be thinking about or identifying the whole from which you take the piece. So when we talk about percentages, the rst question that should pop into your head is, percentage of what? In other words, what is the whole from which we are taking a piece or part. The other thing to keep in mind with percentages is that the part or piece can be referred to either as a percentage or the actual amount. For example, if you slice an apple pie into ten even slices, the we can talk about 20% of the pie (percentage of the whole that we want) or two slices (the actual amount of the whole pie that we want.) Both are referring to the same amount of the pie. The formula for calculating percentages is Denition The formula for calculating percentages is part is percentage of whole part = percentage (%) whole Remember that a percentage is not a number and so must be converted to its decimal form before using it in the above formula. If we know any two of the three quantities that make up the above formula (part, percentage or whole), we can solve for the third quantity. Example Write out the appropriate formula before solving the problem. 1) 12.5% of what is 7? 2) If 62.5% of the students in the class are female and there are 12 males, how many females are in the class? 3) Suppose the total bill at a restaurant came to $92.88 including an 8% sales tax. What the price of the meal before tax? Solutions: 1) Here 7 is the part and.15 is the percentage and our unknown is the whole. So if we let x be the whole, then.125x = (.125x) = (7) x = 56 2) Our problem here is that we are given the percentage of female students but the part given is the number of male students. The 62.5% refers to the percentage of all students that are female. In other words, the whole when talking about 62.5% is the total number of students which we do not know. So in terms of females, all we know is the percentage. However, since the class can only be made up of females and males, if 62.5% are female then or 37.5% are males. Since we are told there are 12 male students, that represents the part when the percentage is 37.5%. Since we have the part and the percentage, we can nd the whole. Note that whether we are talking about the 62.5% females or the 37.5% males, the whole that those percentages refer to is the total number of students, both male and female. So if we let x be the total number of students, then we have 12 =.375x = x 32 = x

57 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 54 So there are 32 students in total. Since the problem asked for the number of females, we take the total number of students and subtract the total number of males. So there are 20 female students in the class = 20 3) In solving word problems, the most common mistake that students make is to try and write an equation right away. The rst step should be to understand what the problem is talking about and to write out in words the relationship between the various information given. For example, in this problem, we are talking about the cost of a meal. Most of us understand that when you buy most items, you not only pay the cost of the item, but you must also pay a sales tax on the item. So the total bill at the restaurant is made up of the price of the meal plus the sales tax for that meal, total cost = cost of the meal + sales tax (8% of the meal cost) We are given that the total cost of the meal is $ If we let x be the cost of the meal, then we have So the cost of the meal before taxes is $ Mixture Problems View the Video Tutorial for this section here = x +.08x = 1.08x = x 86 = x To talk about mixture problems, we need to rst discuss what concentrations of various material (e.g. sugar, salt or acid) in a medium (usually liquid) means. Suppose you go into a convenience store and buy a 20 oz fruit drink which claims that it is made with 5% real fruit juice (i.e. the drink has 5% concentration of fruit juice.) If you drink the whole thing, then 5% of what you drank is fruit juice or another way to say it is that you drank.05(20) = 1 ounce of actual fruit juice. The other 19 ounces was water, sugar, and various other chemicals. For your health, what is in the other 19 ounces would be important, but for our discussion of solving mixture problems, what is important to remember is that we can talk about the fruit juice in the fruit drink two dierent ways. Either as 5% of the whole drink or 1 ounce, the actual amount of the whole drink that is fruit juice. The mixture problems that we will deal with always involves mixing two solutions of dierent concentrations to get third, dierent concentration. The diagram used to organize the given information of each problem will be the same. The dierence will be in the actual values used. So to be able to solve these types of problems, one must rst understand how to build the diagram. Example Be sure to write a sentence at the end to answer each question. 1) How many liters of a 15% acid solution should we mix to 8 liters of a 40% acid solution to obtain a 30% acid solution? Round the answer to four decimal places. 2) How many gallons each of iced tea that contains 35% sugar should be mixed with iced tea that has 5% sugar to make 6 gallons of iced tea with 10% sugar? Solutions: 1) To solve mixture problems, we will always use the following diagram. This row is just labeling each bottle. bottle 1 bottle 2 bottle 3 15% acid 40% acid 30% acid solution solution solution Total amount of liquid in each bottle. x + 8 = x + 8 Actual amount of acid in each bottle..15x +.4(8) =.3(x + 8) If we let x be the amount of the 15% acid solution mixed, then 15% of x or.15x is the actual amount of acid in this bottle. So for each bottle, the actual amount of acid is found by multiplying

58 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 55 the corresponding percentage to the total amount of liquid in each bottle. Since we get the 30% acid solution by mixing the other two, the total amount of liquid in the 30% bottle comes from the other two bottles, i.e. the total amount of liquid in the 30% bottle is the sum of the other two bottles. Likewise, the acid in the 30% bottle can only come from the other two. This means that.15x +.4(8) =.3(x + 8).15x =.3x.15x x =.15x.8.15 = x x You need to mix liters of the 15% acid solution to the 40% solution to get a 30% acid solution. 2) In this problem, we do not know how much of the two dierent iced teas we need to mix together. However, we do know the total when mixed together must be 6 gallons, i.e. the sum of the two must be 6. So if we let x be the amount of the iced tea with 35% sugar, then 6 x must be the amount of the iced tea with 5% sugar that we mix in since x + (6 x) = 6 (Note that we could have let x be the amount of the iced tea with 5% sugar, in which case 6 x would have been the amount of the iced tea with 35% sugar. You should try solving the problem this way to verify that you still get the same nal answers. This is why it is important that you write a sentence at the end to state your answer as the variable you solve for may or may not be what was asked for, depending on how you had set up the problem.) bottle 1 bottle 2 bottle 3 This row is just labeling each bottle. 35% 5% 10% sugar sugar sugar Total amount of iced tea in each bottle. x + 6 x = 6 Actual amount of sugar in each bottle of iced tea..35x +.05(6 x) =.1(6) As with the previous example and all the mixture problems that we encounter, we get our equation from the actual amount of whatever concentration the percentages are based on (in this case sugar.) Since the sugar in the nal mixture can only come from the two bottles that we mix together, we have that.35x +.05(6 x) =.1(6).35x x =.6.3x =.6.3.3x =.3 x =.3.3 x = 1 So we need to mix 1 gallon of iced tea with 35% sugar to 6 1 = 5 gallons of iced tea with 5% sugar Distance/Rate Problems In order to solve what we call Distance/Rate problems, it is important to rst understand the relationship between distance, rate, and time. Suppose a car is traveling at a constant speed (rate) 70 miles per hour (mph). This means that every hour the car has gone another 70 miles in distance. So

59 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 56 number of hours traveling distance traveled (in miles) = 2(70) 2(70) {}}{ = 3(70). t. t(70) So, in this case, the distance traveled, d, in time t hours is d = 70t. In general we say that Denition The formula relating distance, rate and time is where d = r t r is the constant rate (speed) at which the object travels. t is the travel time, (the length of time the object has traveled at speed r.) d is the distance traveled in time t and at speed r. Note that the units for speed, time, and distance must be the same, e.g. if the speed is in miles per hour, then time must be in hours and distance must be in miles whereas if the speed was in feet per second, then time must be in seconds and the distance must be in feet. Just as with the mixture problems, the table we use to organize the information from any distance/rate problem will be the same. So understanding what the table says and how to build it, is the key step to understanding how to solve these word problems. Example Be sure to write a sentence at the end to answer the question. 1) Two hikers are on the opposite ends of a 15 mile trail that starts at the base and ends at the top of a mountain. At 6:00 a.m. both hikers start their hike, with the hiker at the base of the mountain walking up the trail at 2 miles per hour, while the hiker coming down from the top is walking down at 3 m.p.h. At what time will they meet? 2) A passenger train leaves the station at noon and travels east at a rate of 50 m.p.h. One hour later, an express leaves the same station chasing after the passenger train at 65 m.p.h. How long will it take the express to catch up to the passenger train? Solutions: 1) For distance/rate problems, we build a table with one column being the speed, another column being the time traveled, and a third column representing distance traveled. The distance column is always lled by multiplying the rate (speed) and time. rate * time = distance rate time distance in miles per hour in hours in miles hiker at the base 2 m.p.h. t 2t hiker at the top 3 m.p.h. t 3t Since the hikers started and stopped at the same time, the time that they have each hiked is the same. As we do not know that time, we let it be t. Recall that we said previously that the distance in the above table is always lled by multiplying the rate and time. So 2t is the distance traveled by the hiker going to the base to the top of the mountain, while 3t is the distance traveled by the hiker coming down the mountain. The equation that we build to solve distance/rate problems will always come from some relationship between the distances. In this case, since the total distance between the two hikers is 15 miles, when they meet on the trail, the sum of the distances, traveled by each hiker, should result in 15 miles. See gure below.

60 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 57 So, we have that 2t + 3t = 15 5t = 15 t = 15 5 t = 3 This means that it takes 3 hours for the two hikers to meet. Since they started at 6:00 a.m., they meet at 6+3 or 9:00 a.m. 2) Since the express leaves one hour after the passenger train and they both stop when they meet, if we let t be the total time that the passenger train has been traveling when the two meet, then the total time the express train has been traveling must be one hour less, so t 1. The table that we build for this problem is as follows: rate * time = distance rate time distance in miles per hour in hours in miles passenger train 50 m.p.h. t 50t express 65 m.p.h. t 1 65(t 1) Since both trains leave the same station and stop when they meet, the distance traveled by each train is the same. So 50t = 65(t 1) 50t = 65t 65 65t 65t 15t = 65 t = t = 13 3 Note that t = 13 3 hours is the time the passenger train travels. To nd the total time the express travels, we need to calculate t 1 = = = 10 3 So the express travels for 10 3 or hours.

61 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE Linear Inequalities in One Variable The techniques used to solve linear inequalities are similar to those we used to solve equations. There are, however, some subtleties in dealing with inequalities. One is that the solution will usually involve an innite number of numbers and so we need a way to represent those answers. The other has to do with multiplying by negative numbers Graphing Inequality Solutions on the Number Line We begin by looking at how to represent the solutions. Given two numbers, in comparing them, one of the following three statements must be true. The two numbers are equal (=) in value, the rst number is greater than (>) the second number, or the rst number is less than (<) the second number. This is called the trichotomy of numbers. When we write x = 2 it is understood that the solution is 2 as that is the only number that you can replace x with and still have a true statement. x = 2 We say that 2 is the solution because if you replace x with any other number, the statement is not true. (2) = 2 Replace x with 2 and the statement is true. (3) = 2 Replace x with 3 and the statement is not true. ( 2) = 2 Replace x with -2 and the statement is not true. However, if we have x > 2, note that there are many numbers, in fact an innite quantity of numbers, that will make the statement x > 2 true. We call all these numbers the solutions to the inequality x > 2. When students are asked to think of all numbers greater than 2, they often think the answers must be whole numbers, 3, 4, 5,... However, the decimal number 2.01 is also greater than 2. Note that if you collect all the real numbers between 2 and 3, there will be more there than all the whole numbers, 1, 2, 3,... x > 2 This means all numbers x such that they are greater than 2. There are an innite quantity of numbers that will satisfy this statement. (2.01) > 2 Replace x with 2.01 and the statement is true. (2.001) > 2 Replace x with and the statement is true (3) > 2 Replace x with 3 and the statement is true (2) > 2 Replace x with 2 and the statement is not true, since 2 is not greater than 2. This also shows us that although x > 2 has an innite number of solutions, not every number will be a solution. Since it is not possible to list all the numbers that satisfy the statement x > 2, another way to express this statement is with a graph on the number line. The hollow circle around the 2 indicates that we start at 2 but do not include it, but we do include all the numbers that are close to 2 but greater than 2, such as 2.01, 2.001, etc. and the line going to the right indicates that we are talking about numbers greater than 2. Note that nowhere on the number line do we write x. That is because it is understood that the whole graph is talking about x. The general rule for graphing the solutions on the number line is as follows.

62 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 59 Denition Inequality Signs (<,, >, ) When graphing the solution to inequalities on the number line, remember that on the number line, anything to the left is smaller and anything to the right is bigger. In the following, we assume that the variable is x and the number is a. x < a means all numbers less than a, so we use a hollow circle around a and shade to the left of a. x a means all numbers less than or equal to a, so we use a lled circle around a and shade to the left of a. x > a means all numbers greater than a, so we use a hollow circle around a and shade to the right of a. x a means all numbers greater than or equal to a, so we use a lled circle around a and shade to the right of a. Example Graph the solution on the number line. 1) x 3 2) x < 5 3 3) 3 x Solutions: 1) Since we can include -3, we use a lled circle. 2) Note that there is no need to but tick marks at 0, 1, 2 and then place 5 3 to write is 5 3 and use a hollow circle. between 1 and 2. All we need 3) It is usually easier to read the inequality with the variable on the left side. So we ip the entire expression around. Think of 3 x as being written on a transparent piece of paper and you ip the paper to view the back side. You would see x 3. This is the same as 3 x. The way to check that you ipped it correctly, is that whatever the inequality was pointing to, it should still point to it after the ip. In this case, the pointy part of the inequality sign is going towards x Addition/Multiplication Property of Inequalities The addition property of inequality works just like the addition property of equality. Proposition Addition Property of Inequality If you add the same quantity to both sides of an inequality, the inequality is still true. if a > b then a + c > b + c if a < b then a + c < b + c Note that the solutions to the new and old inequalities must also be the same.

63 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 60 Just like with equations, this makes sense if you think of a scale. Only this time, the scale is not balanced. One side is heavier than the other. If you add or take away the same amount of weight from both sides of the scale, the same side will still be heavier than the other. The multiplication property of inequality is a little more subtle than its equality counterpart. The dierence occurs when dealing with multiply both sides by a negative number. For example, 2 < 5 is a true statement. If we multiply both sides by 3, we get 3(2) < 3(5) 6 < 15 which is also true. Now, if we had multiplied by -3 instead of 3, we get 3(2) < 3(5) 6 < 15 which is not true. To make the statement true, we must change the sign of the inequality to So, in general we have 6 > 15 Proposition Multiplication Property of Inequality When multiplying both sides of an inequality by a nonzero number, if the number you are multiplying is positive, then the inequality sign does not change. For c > 0 if a < b if a > b then c a < c b then c a > c b if the number you are multiplying is negative, then the inequality sign must change (go from < to >or from > to <). For c < 0 if a < b if a > b then c a > c b then c a < c b When solving inequalities, the general strategy is the same as for equations. We must get the variable that we want to solve for by itself on one side of the inequality sign. Example Solve. Graph your solution on the number line. 1) 3x + 5 < x 3 2) 1 4x 5 3) x 2 6 > 2 3 x 1 2 Solutions: 1) As with equations, we must get all the terms with variables on one side of the inequality. left side {}}{ 3x + 5 < 3x x right side {}}{ x 3 < x x + ( 3) 5 2x < (2x) < 1 2 ( 8) x < 4 To get all the x's to the left side, we add x to both sides. To also add -5 to both sides so that only the x's remain on the left side. Note that since we added (the x and -5), the inequality sign does not change.

64 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 61 So the graph of the solution on the number line is 2) 1 4x ( 4)x 5 1 4x ( 4x) 1 4 (4) x 1 Since we added the -1 to both sides, the inequality sign does not change. Since we multiplied both sides by a negative number,, the inequality sign has changed direction. 1 4 So the graph of the solution on the number line is 3) As with the equations, we can use the multiplication property of inequalities to get rid of the fractions by multiplying both sides by the least common denominator (LCD). ( ) x ( 2 > 6 3 x 1 ) 2 Multiply both sides by the LCD, which is 6 in this case. Since we are multiplying by a positive number, the inequality sign does not change directions. ( ) ( ) 2 1 x 2 > 6 3 x + ( 6) 2 Remember that we must distribute the 6 in on the right side because 2 3 x and 1 2 are two separate terms, where as the x 2 6 is one term (fraction) and so we can cancel the 6's out right away. x 2 > 2(2x) + ( 3) x + ( 2) 4x +2 > 4x 4x + ( 3) +2 3x > ( 3x) < 1 3 ( 1) Since we multiply both sides by a negative number,, the inequality sign has changed direction. 1 3 x < 1 3 So the graph of the solution on the number line is

65 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE Compound Inequalities To talk about compound inequalities, we must rst talk about the conjunctions and and or. These words have mathematical meaning when we are talking about a collection of things, such as numbers. For example, x > 3 and x < 7 we are talking about all numbers x that are greater than 3 and at the same time less than 7. In other words, we are talking about all the numbers between 3 and 7. We refer to the word and as intersection. Why we call it intersection can been seen by graphing the above two inequalities on the number line. The line on top represents all numbers less than 7 while the line on bottom represents all numbers greater than 3. So all the numbers that are both greater than 3 and less than 7 are where the two lines overlap (intersect) each other, the shaded region on the graph. So the nal graph that represents x > 3 and x < 7 is This can also be written with the double inequality sign The word or is referred to as a union. 3 < x < 7 x < 3 or x > 5 means all number that are either less than -3 or greater than 5. This includes all the numbers described by both inequalities, hence the union of both answers. The graph would simple be the graph of both inequalities. So we can summarize the above as follows. Denition Intersection (and)/union (or) and (Intersection) means that both conditions connected by the word and must be satised. it is referred to as an intersection and is denoted by the symbol. the combined notation a < x < b means x > a and x < b. Note that if a is not smaller than b, the above makes no sense. or (Union) means that either conditions connected by the word or can be satised. it is referred to as a union, and is denoted by the symbol. unlike the intersection (and), there is no combined notation for unions. Example Solve and graph your solutions on the number line.

66 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 63 1) x > 3 x < 5 2) x < 2 x > 5 2 3) 2x 5 3x + 2 2(x + 6) + 1 < 3 x 4) 2 < 3 2x 7 5) 3x < 5x (x 3) Solutions: 1) Remember that union,, means or. So if we graph both inequalities we see that the two lines cover the entire number line. This means that the answer is all real numbers. Graphically, we shade the entire number line for our answer. 2) As with the previous problem, since the variable x is already by itself on one side of the inequality, we simply graph both of them. Remember that 5 2 = > 2, which means that 5 2 is to the right of 2 on the number line. Since mean intersection and we can see that the two lines to not intersect anywhere, this means that there is no solution to the inequalities. Note that this makes sense, since 5 2 is bigger than 2, there is no number that is bigger than 5 2 and at the same time smaller than 2. So there is no graph for this one. We simply say that there is no solution or we can use the symbol, which is called the empty set (i.e. the collection of all the solutions to this problem is empty.) 3) Here we need to solve each inequality for x rst. 2x 5 3x + 2 2(x + 6) + 1 < 3 x 3x + 5 3x + 5 2x < 3 x x 7 2x + 13 < 3 + ( x) +x x 1( x) 1(7) x 7 3x < (3x) < 1 3 ( 10) 10 x 7 x < 3 Now we graph the two inequalities. Note that 10 3 = number line. > 7. So 10 3 is to the right of -7 on the The answer is then all the numbers between -7 and , including -7 but not including 3.

67 CHAPTER 4. LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE 64 4) Recall that the double inequality sign means intersection (and). The goal here is to get the x by itself in the middle. left side middle right side {}}{{}}{{}}{ 2 < 3 2x 7 Note that there are three sides to this problem. The addition and multiplication properties still apply, only we apply them to all three sides instead of two. To undo the 3 we add -3 to all three sides. 2 < 3 2x < 2x ( 5) > 1 2 ( 2x) 1 2 (4) To undo the multiplication by -2, we multiply by 1 2 to all three sides. Since we are multiplying by a negative number, the equality signs must change direction. 5 2 > x 2 2 x < 5 2 We ip the entire inequality around. So the answers are all numbers greater than or equal to -2 and less than ) We solve the two inequalities for x. 3x 5x (x 3) 5x 5x 2x ( 3)(x + ( 3)) ( 3x) ( 2x) 1 2 (3) ( 3x) x x 1 3 (5) 1 3 ( 3x) x x x 5 3 Since 3 2 > 5 3, 3 2 is to the right of 5 3 on the number line. So the graph of the solution on the number line is

68 Chapter 5 Equations of Lines 5.1 Cartesian Coordinate System View the Video Tutorial for this section here. Suppose you work for a window manufacturer wants to make windows of the following shape. You are in charge of programming the computer controlled saw that will cut the glass to shape. The question is, how do you describe the shape to the computer in order to have it cut the glass to the correct shape and dimensions? Note that these types of machines with robotic arms exist and are used in manufacturing. So we are not talking about some futuristic AI (articial intelligence) machine. The famous French mathematician and philosopher René Descartes supposedly had a similar problem. The story goes, that while looking up at a ceiling, he observed a y crawling around. In trying to describe the y's position, he came up with what we now refer to as the Cartesian coordinate system. The idea is that we draw two lines, one going horizontally (left and right), called the x-axis, and the other going vertically (up and down), called the y-axis, and describe the location of a point by its horizontal and vertical distance from the point where the two lines intersect, called the origin. So we refer to the point by these two distances, which together are called the coordinates of the point. The notation we use to describe a point is called an ordered pair, (x,y). Here are some examples of points on the Cartesian coordinate plane. We note the following about the above graph. The tick marks (the griding) must be evenly spaced in each of the x and y directions. They do not have to be the same in both directions. For example the tick marks along the x-axis might each represent 65

69 CHAPTER 5. EQUATIONS OF LINES 66 5 feet, while the tick marks along the y-axis might each represent 1 2 foot. The point A has coordinates (2, 3). The rst number, 2, is the horizontal distance from the origin and the second number, 3, is the vertical distance from the origin. So, to get to point A from the origin, we go 2 units to the right and up 3. Positive means go up or to the right and negative means go down or to the left. So point C with coordinates ( 3, 4) means from the origin go left 3 and down 4 to get to C. To gure out the coordinates of point B, if we start from B and go down vertically to the x-axis, the number on the x-axis represents the x-coordinate. For B, we see that it is -5. From B, if we go horizontally until we hit the y-axis, we get the y-coordinate, which in this case is 2. So, we know the coordinates of B are ( 5, 2). We refer to the coordinates as an ordered pair, because the order in which the numbers appear makes a dierence. Notice that the point B has coordinates ( 5, 2), but if we switch the ordering, we get (2, 5) which are the coordinates of point D. So, going back to our original problem of the window, if we set the origin be at the center bottom point of the window, then label each point where the window makes a new angle and setup the tick marks as shown below, we can nd the coordinates of those points. A = (0, 0), B = ( 5, 2), C = ( 4, 6), D = ( 2, 9), E = (2, 9), F = (4, 6), G = (5, 2). To cut the glass, these coordinates would be given to the computer and the computer told to cut straight across from point A to B to C to... It should be noted here that all of our discussion above was in 2 dimensions and everything that we will do in this class will always be in two dimensions. Remember that in 2 dimensions every point is identied by two coordinate values. If we wish to talk about points in 3 dimensions, then we would need another axis and hence another coordinate, 4 dimensions would require 4 coordinate values, etc...

70 CHAPTER 5. EQUATIONS OF LINES Graphing Lines by Points View the Video Tutorial for this section here. Recall that a solution to an equation is the value that if you substitute in for the variable, it makes the equation true, e.g. the equation x + 3 = 5 has x = 2 as the solution because if you replace x with 2, you get (2) + 3 = 5 which is true. Now suppose we have an equation with two variables, such as x + y = 5 In this case, the solution to the equation must involve two numbers. One value for x and one for y. So there are multiple combinations of numbers that will work, in fact there are an innite number of answers. For example, if x = 5 and y = 0 or x = 1 and y = 4 or x = 0.5 and y = 4.5 or x = 1 and y = 6, etc... If we write the solutions as ordered pairs, (5, 0), (1, 4), (0.5, 4.5), ( 1, 6), and plotted all the possible solutions on the Cartesian coordinate plane, all the points together would form a line. Again, keep in mind that when we draw the line, we are saying that all the points along the line are solutions to the equation. Students often think that this means just the points with integer coordinates, such as (0, 5) and (1, 4) in the above example. However, you should understand that there are more points, on the line, between (0, 5) and (1, 4) that there are points with just integer coordinate values all along the line. In general, we have the following denition of the equation of lines. Denition Equation of Lines in Two Variables (General Form) In two dimensions, an equation of the form Ax + By + C = 0 where A, B, and C are real numbers and A and B are not both zeros, the solutions of such an equation, (x, y), will graph as a line. Conversely, the equation of any line can be written in this form. Example Verify that the given equation represents a line by putting in the general form, Ax + By + C = 0, and identifying A, B, and C. Then graph it by nding two points on the line. 1) 2x 3y = 12 2) y = 2 3x 1 3) 2x = 8 Solutions: 1) 2x 3y = x + ( 3)y + ( 12) = 0 So A = 2, B = 3, C = 12 and from Denition 5.2.1, we know that 2x 3y = 12 must graph as a line. From geometry, we know that two points dene a line. Therefore, if we can nd two points or solutions of the equation 2x 3y = 12, then we can graph the line by connecting the two points. Note that we can use whatever points we want, as long as they are solutions of the equation. So the

71 CHAPTER 5. EQUATIONS OF LINES 68 question is, how do we nd two points? You could just try random guessing for the x and y values, but that would not be very ecient. Instead, we note that since the points must be solutions of the given equation, if we were to let x be some value, say x = 0, then plugging this into the equation would allow us to solve for a y value. 2(0) 3y = 12 3y = 12 y = 4 This tells us that the point (0,-4) is a solution of the given equation. Now to nd the second point, we could use x = 1, but note that if you do, the y values will be a fraction ( ) y = Although there is nothing wrong with this, the fraction will be a little harder to graph. So, instead of using x = 1, what if we let y = 0? Remember that our only condition is that the x and y values satisfy the given equation. There is nothing that says that we have to set the x value rst. So, if y = 0, then 2x 3(0) = 12 2x = 12 x = 6 We now have two points, (0,-4) and (6,0). So if we graph them, the line that goes through them will be the line representing the equation 2x 3y = 12. When graphing, be sure to label the axes and the tick marks at the minimum. 2) The equation y = 2 3x 1, will also graph as a line. We know this because y = x+1 3 x x 2 3 x + y + 1 = 0 So A = 2 3, B = 1, C = 1, which according to Denition graphs as a line. Since the equation is already solved for y, this makes it easier for us to pick the x values and then solve for the y values. Again, the choice of x values to use is up to us. So we want to try to use numbers that will make the calculations easier. Also, this form of the equation makes it easier to use what we call a T chart to list the solutions. x y or 2 3 x 1 (x,y) (0) 1 = 1 (0,-1) (3) 1 = 2 1 = 1 (3,1) So the line is

72 CHAPTER 5. EQUATIONS OF LINES 69 3) Since this equation has no y in it, we can view the y term as being 0y, as 0y is really just zero and adding zero does not change anything. 2x + 0y 8 = 8 8 2x + 0y + ( 8) = 0 So A = 2, B = 0, C = 8. We note here that this in not the most helpful way to look at the equation. To better understand how this particular equation graphs, it is more helpful to solve this equation for x, 1 2 (2x) = 1 2 (8) x = 4 Now this is where it is important for us to remember that we are working in 2-dimensions. That means every point must have both an x and a y value. Since this equation only species that the x-coordinate value must always be 4 and says nothing about the y values, we can use whatever y value we want, as long as we use 4 for the x value. For example the points (4,-2) and (4,2) are both solutions to the equation. If we plot all the points that have an x-coordinate of 4, we end up with the following vertical line.

73 CHAPTER 5. EQUATIONS OF LINES Slope of a Line View the Video Tutorial for this section here. One of the characteristic features of lines is the concept called the slope of a line. Most of us already have an intuitive idea of what a slope is. For example, if you go skying at a mountain resort, there is the Bunny Slope (gentle incline) for beginners and the Widow Maker (steep incline) for the more advance skier and/or crazy person. In order for the mathematical denition of slope to match our intuitive idea of slope, we dene it as follows. Denition Slope of a line. Let A and B be two points on the line. Then the slope of a line, denoted with the letter m, is dened as the following ratio. m = vertical distance from point A to B (or from B to A) horizontal distance from point A to B (or from B to A) some other ways this is dened is m = rise change in y = run change in x positive value means go up or to the right and a negative value means go down or to the left. Slope Formula: given two points (x 1, y 1 ) and (x 2, y 2 ), the formula to calculate slope is m = y 2 y 1 x 2 x 1 There are two intuitive concepts about slopes that we want to be sure that our denition above captures. 1) Slope of the steeper line should be greater than the slope of a less steep line. 2) The slope of a line should be constant and so it should not matter what two points you use to nd the slope and it should also not matter in which direction you go between the two chosen points. We use the following graph of two lines to explore these two concepts. Two points on the widow maker are (2, 12) and (4, 24). To calculate the slope we can just count o how much we need to up and to the right to go from (2, 12) to (4, 24). go up 12 m = go right 2 m = 12 2 m = 6

74 CHAPTER 5. EQUATIONS OF LINES 71 {}}{{}}{{}}{{}}{ or we can also use the formula for the slope with ( 2, 12 ) and ( 4, 24 ) x 1 y 1 m = y 2 y 1 x 2 x = 4 2 = 12 2 = 6 In either case, we see that the slope for the widow maker is 6. For the bunny slope, we know that it goes through the points (0, 0) and (8, 4). So, from the origin to (8, 4), we go up 4 and to the right 8. x 2 y 2 m = 4 8 = 1 2 Notice that the bunny slope has a slope of 1 2 which is smaller than the slope of the Widow Maker, m = 6. This coincides with our notion of steepness of the two lines. Now, to see that the order in which we use the points does not matter when calculating the slope, note that we went from the point (2, 12) to (4, 24) when calculating the slope of the Widow Maker. If we go in the reverse direction, from the point (4, 24) to (2, 12), we get x 1 y 1 go down 12 m = go left 2 = 12 2 = 6 {}}{{}}{{}}{{}}{ or using the formula with ( 4, 24 ) and ( 2, 12 ) x 2 y 2 m = y 2 y 1 x 2 x = 2 4 = 12 2 = 6 So the slope is still 6. See the blue lines in the graph below. To see that the choice of points does not matter, we calculate the slope going from the point (2, 12) to the

75 CHAPTER 5. EQUATIONS OF LINES 72 new point (6, 36), see purple lines in the gure above. m = go up 24 go right 4 = 24 4 = 6 So, we see that it does not matter in what direction or what two points you use to calculate the slope of the line, as long as both points are on the line. Example Graph the line, given a point on the line and its slope. You must graph the line with in the provided grid. 1) m = 3 4, ( 3, 2) 2) m = 6 5, (5, 4) Solutions: 1) One way to view slope is as the directions for going from one point on the line to another where the numerator tells you how far to go up or down and the denominator tells you how far to go right or left. The slope here is m = 3 4 = 3 4 = 3 4 If we view it as 3 4, then this means we go down by 3 and to the right 4, where as if we the slope as 3 3 4, we go up by 3 and to the left by 4. So, using m = 4, and starting from the point ( 3, 2), we have the following. 2) The problem here is that when we start from the point (5, 4), we do not have enough space to go up by 6 and across the right by 5. Therefore, we will view the slope as the ratio of -6 to -5 instead of 6 to 5. m = 6 5 = 6 5

76 CHAPTER 5. EQUATIONS OF LINES 73 Which means that if we go down by 6 then we must go left by 5. Example Find the slope of the line given the following information. 1) The line goes through the points (5, 3) and ( 5, 7). 2) The equation of the line is 2x 3y = 12. 3) The line goes through the points ( 4, 9) and ( 6, 9). 4) The equation of the line is x = 3 Solutions: 1) Given two points, we can use the slope formula in Denition Note that it does not matter which we consider the rst and the second point. So we will let (5, 3) be the rst point and ( 5, 7) be the second point. So the slope is m = 2 5. ( x1 5, y1 3 ) ( x2 5, y2 7 ) m = 7 ( 3) 5 5 = = 4 10 = 2 5 2) Given the equation, one way to nd the slope is to nd two points on the line. Note that we will later learn another way to nd the slope, given the equation. This is the same line (equation) in Example 5.2.1, from which we know that the points (0, 4) and(6, 0) are on the line. Using these two points, we have m = = 4 6 = 2 3 where we let (6,0) be the rst point and (0,-4) be the second point. So the slope is m = ) Here we are again given two points. We let (-4,-9) be the rst point and (-6,-9) be the second point. 9 ( 9) m = 6 ( 4) = = 0 2 = 0

77 CHAPTER 5. EQUATIONS OF LINES 74 So the slope is m = 0. Note that if you use the two points to graph the line, it is a horizontal line. This makes sense since the slope measures the incline of a line and a horizontal line is at, i.e. it has no incline. 4) Given the equation of the line, x = 3, we need to nd two points on the line. Since this equation has no restriction on y, we can use whatever value we want for the y-coordinate. So (-3,0) and (-3,2) are two solutions to the equation and hence points on the line. Using the slope formula m = = ( 3) = 2 0 Recall that a fraction with zero in the denominator is undened. This means that the slope of the line is undened. Note that x = 3 graphs as a vertical line. In general, a vertical line has an undened slope. Some may think that since slope measures steepness and a vertical line is innitely steep, the slope of the vertical line must be innite. The problem with this reasoning is that slope not only indicates steepness, but also direction. If you take a horizontal line, which has slope zero and rotate it counter-clockwise, the slope of the line becomes a greater and greater positive number (see gure below.) Now, if you take the same horizontal line and rotate it clock-wise, the lines become more and more vertical, but the slopes get more negative (see gure below.) So, rotating counter-clockwise, it looks like the slope of a vertical line should be positive innity, but if you rotate clockwise, it looks like the slope of a vertical line should be negative innity. Since the slope cannot be both positive and negative at the same time, we say that the slope of a vertical line is undened. Comment on negative slope and steepness From Example 5.3.1, we can see that a positive slope means that as you look at the line from left to right, the line is increasing (going up.) Negative slope means, as you look from left to right, the line is decreasing (going down.) Also, if you have a sharp eye, you may have noticed that for negative sloped lines, the steeper you decrease, the more negative the slope (the value of the slope is decreasing.) This seems to go against our intuitive idea that steeper means greater slope. Remember that -7 is smaller than -1. The problem is that when we think of steepness, we are only thinking of the incline and not the direction in which the line inclines. However, slope takes into account both the incline and direction, positive means an increasing line and a negative slope means a decreasing line. To ignore the direction and just talk about the steepness, we look at the absolute value of the slope, m. For example, a line with slope m = 7 versus a line with slope m = 1. Since 7 = 7 which is bigger than 1 = 1, would tell us that the line with a slope of 7 is steeper than the line with a slope of 1.

78 CHAPTER 5. EQUATIONS OF LINES 75

79 CHAPTER 5. EQUATIONS OF LINES Equations of Lines In sec. 5.2, we discussed the relationship between an equation and the graph that represents the equation. Denition told us that any line can be written in the form Ax + By = C. In this section, we will discuss two other ways of writing the equation of a line that are more useful when you are trying to nd the equation of a line X and Y Intercepts View the Video Tutorial for this section here. Before talking about the other ways to write the equation of a line, we mention here two special types of points called x and y intercepts. In general, the graph of an equation can have more than one x and y intercepts. But for lines, there will be at most one x-intercept and one y-intercept. The x-intercept is where the graph intersects the x-axis. Recall that for any point on the x-axis, the y-coordinate is 0. Therefore, to nd the x-intercept on a line, we would set y = 0 and solve for x. Likewise, the y-intercept is where the graph intersects the y-axis. So in order to nd the y-intercept, we set x = 0 and solve for y. Denition X and Y Intercepts Given any equation and its graph (does not have to be a line), in terms of the graph The x-intercept(s) is a point(s) where the graph crosses the x-axis. The y-intercept(s) is a point(s) where the graph crosses the y-axis. in terms of the equation Find the x-intercept(s) by setting y = 0 and solving for x. Find the y-intercept(s) by setting x = 0 and solving for y. Note that for lines, there will only be one x-intercept and one y-intercept. Example Find the intercepts of the following lines. 1) x 3y = 6 2) y = 2x 3) y = 2 Solutions: 1) To nd the x-intercept, we set y = 0. x 3(0) = 6 x = 6 So the x-intercept is (6, 0). Note that since the intercepts are points, you must give the ordered pair as your answer. To nd the y-intercept, we set x = 0. (0) 3y = 6 3y = 6 y = 6 3 y = 2 So the y-intercept is (0, 2). Note that you were not asked to graph this. The graph is shown to help visualize what we have done.

80 CHAPTER 5. EQUATIONS OF LINES 77 2) x-intercept: (0) = 2x 0 2 = x 0 = x So we get (0, 0) as the x-intercept. Note that this must also be the y-intercept as the x-coordinate is also 0. The origin, (0, 0), is the only point on the Cartesian Coordinate system that can be both the x and y intercept. 3) y = 2 means that the y-value must always be -2 but that the x-value can be anything. So, since the y-value can never be 0, there can be no x-intercept. However, to nd the y-intercept, we set x = 0 but the y value is still -2. So the point (0, 2) is the y-intercept. Note that the line y = 2 will graph as a horizontal line.

81 CHAPTER 5. EQUATIONS OF LINES Slope-Intercept form of the Line View the Video Tutorial for this section here. Finding the equations of lines is a key topic in math because of its many uses in real life applications. To nd the equation of a line will always require two things. 1) The slope of the line. 2) A point on the line. Given these two things, we can nd the equation of the line in the form called the slope-intercept form of the line. Denition Slope-Intercept form of the Line y = mx + b is called the slope-intercept form of the line, and any line (other than the vertical line) can be written in this form. This is called the slope-intercept form because the slope of the line and the y-intercept can be read directly from the equation. m is the slope of the line (0, b) is the y-intercept of the line. Example Show that in the slope intercept form of the line, y = mx + b, m is the slope of the line. Solution: To calculate the slope using the slope formula, Denition 5.3.1, we need two points. One of the points we will use will be the y-intercept. Recall that to nd the y-intercept, you set x = 0 and solve for y. y = m(0) + b y = b So, we can see that (0, b) is the y-intercept. To see that the number m in front of x must represent slope, we note that since y = mx + b this means that we can replace y with mx + b. So any point (x, y) y {}}{ on this line can be written as (x, mx + b). Since we also know that (0, b) is a point on the line, we have our two points, (x, mx + b) and (0, b), and can use the slope formula, Denition 5.3.1, to calculate the slope. (mx + b) b slope = x 0 = mx x = m We have just shown that the line y = mx + b has m as its slope. Example Find the equation of the line given the following information. 1) The line goes through the point (2, 1) with a slope of ) 2) Goes through the points ( 1, 2) and (9, 3). Solutions:

82 CHAPTER 5. EQUATIONS OF LINES 79 1) Using the slope-intercept form of the line, y = mx + b, we have found the line if we can nd the values for m and b. From Denition 5.4.2, we know that m is the slope of the line. So here m = 2 3. That means our equation so far is y = 2 3 x + b Now to nd b, we use that fact that (2, 1) is a point on the line. This means that (2, 1) is a solution of the above equation, i.e. if we replace x with 2 and y with 1, the above equation must be true. (1) = 2 3 (2) + b = b 4 3 = b = b = b So the equation of the line is y = 2 3 x 1 3 2) As stated above, to nd the equation of a line, we always need the slope of the line and a point on the line. Here, we are given two points instead of the slope and a point. However, given two points we can calculate the slope using the slope formula, given in Denition So given the two points ( 1, 2) and (9, 3), we have m = ( 1) = 5 10 = 1 2 and our line so far is y = 1 2 x + b To nd b, we can use either of the two given points. We will use the point ( 1, 2) just because the numbers look to be easier to deal with. Again, as in the previous problem, since ( 1, 2) is a solution of the equation, we can replace the x and y with -1 and 1. (2) = 1 2 ( 1) + b So the line is = b 1 2 = b = b = b y = 1 2 x ) Since we are just given the graph of the line, we must make some reasonable assumptions. Namely that we can read o the coordinates of the following two points, A and B, from the graph as being integer coordinates.

83 CHAPTER 5. EQUATIONS OF LINES 80 So A = ( 2, 0) and B = (1, 1). Using the slope formula, we get as the slope m = ( 2) = 1 3 Note that we can also just read the slope o the graph. However, we need to be careful when reading the slope from the graph since in this case the vertical tick marks (grids) represent 1 2 units while the horizontal tick marks represent 1 unit. So, although to go from point A to B, you go up 2 grid lines, that only represents 1 unit, while going 3 grids to the right represent 3 units. So the slope is again m = 1 3. The line that we want is then y = 1 3 x + b and using point A = ( 2, 0), we have (0) = 1 3 ( 2) + b The equation of the line is then = = b + b y = 1 3 x Point-Slope form of the Line View the Video Tutorial for this section here.

84 y 1 +1 = 2 3 x CHAPTER 5. EQUATIONS OF LINES 81 As state before, to nd the equation of a line, you need the slope and a point on the line. In the previous subsection, we found the equation of the line using the slope-intercept form, Denition Here, we will discuss another way to write the equation of a line, called the Point-Slope form or formula. We begin by showing how the Point-Slope form comes from the slope formula, Denition Recall that the slope formula, m = y2 y1 x 2 x 1, lets you nd the slope of the line using any two points on the line. So given the slope, m, and a point (x 1, y 1 ) on the line, if we let (x, y) be any other point on the line, then, using the slope formula, we have m = y y 1 x x 1 (x x 1 )m = y y 1 x x 1 (x x 1 ) Note that (x, y) is not just any arbitrary point, but an arbitrary point on the line. So, since both (x, y) and (x 1, y 1 ) are points on the same line, the dierence in the y's over the dierence in the x's must be a constant, which we call the slope of the line. Get rid of the denominator on the right side by multiplying both sides by (x x 1 ) m(x x 1 ) = y y 1 y y 1 = m(x x 1 ) So for any point (x, y) on the line, the above equation must be true. Another way to view it is that any solution (x, y) to the equation is a point on the line. Hence, the above is the equation of the line with slope m and going through the point (x 1, y 1 ). Denition Point-Slope form of the Line Given the slope m and a point (x 1, y 1 ) on the line, the equation of the line is y y 1 = m(x x 1 ) Note that like the slope-intercept form of the line, the point-slope form can represent any line except the vertical line. Example Find the equation of the line given the following information. 1) The line goes through the point (2, 1) with a slope of ) Goes through the points ( 1, 2) and (9, 3). Solutions: 1) We have already solved this problem using the slope-intercept form of the line. Here we will use the point-slope form of the line. We are given m = 2 3, so y (1) = 2 (x (2)) 3 y 1 = 2 (x 2) 3 Note that we can put this in slope-intercept form by solving for y. y 1 = 2 (x 2) 3 y = 2 3 x 1 3

85 CHAPTER 5. EQUATIONS OF LINES 82 This is the same as in Example ) Since we are given two points, we can nd the slope using the slope formula. m = ( 1) = 5 10 = 1 2 Using the point ( 1, 2) (note it does not matte which point is used, as they are both on the same line), y 2 = 1 (x ( 1)) 2 y 2 = 1 (x + 1) 2 Summary of the Equation of Lines We have discussed three dierent ways of writing the equation of a line. Each has its usefulness in dierent situations. General Form: Ax + By + C = 0 Any line can be represented in general form, including the vertical line. Useful form when using the elimination method to solve systems of equations (discussed in the next chapter.) Form is not unique, meaning dierent equations can represent the same line. For example x + y + 1 = 0 and 2x + 2y + 2 = 0 both represent the same line. Slope-Intercept Form: y = mx + b Every line except the vertical line can be represented in this form. Can identify the slope and y-intercept immediately, which lets us graph the line easily. This form is unique, meaning if two equations in this form dier in either m or b, then they represent two dierent lines. Point_Slope Form: y y 1 = m(x x 1 ) Every line except the vertical line can be represented in this form. Quickest way to write the equation of a line. This form is not unique, meaning two dierent equations in this form can represent the same line. (All you have to do is use a dierent point on the line for (x 1, y 1 ).)

86 CHAPTER 5. EQUATIONS OF LINES Parallel and Perpendicular Lines View the Video Tutorial for this section here. Geometrically, we say that two lines are parallel if they never intersect each other. Algebraically, we say that two lines are parallel if they have the same slope. This makes sense since the slope indicates incline and direction and you would expect two parallel lines to be going in the same direction with the same incline, see graph below. Perpendicular lines are lines that intersect each other at 90 degrees. Algebraically, we say that two lines are perpendicular if the product of the slopes equals -1 or another way of saying this is that the slopes are negative reciprocals of each other. For example, if the slope of one line is 2 3, then the slope of the perpendicular line is 3 2. So to summarize, we have Proposition Slope of parallel and perpendicular lines. Parallel lines have the same slope. Perpendicular lines The product of their slopes equals -1. (This is useful when you know the slopes and want to check to see if the lines are perpendicular.) The slopes are negative reciprocals of each other, i.e. if the slope of one of the lines is b a, then the slope of the other line is a b. (This is useful when you want to nd the slope of the perpendicular line.) Horizontal and vertical lines are perpendicular to each other. Before going on to some examples, we will rst show that the above two properties of perpendicular lines are equivalent. To show this, we start with the property that the product of the slopes to two perpendicular lines equals -1. If we let the slope of the two perpendicular lines be m 1 and m 2, then m 1 m 2 = 1 1 m 1 (m 1 m 2 ) = 1 m 1 ( 1) Solve for m 2. m 2 = 1 m 1 This says that the slope of one line is the negative reciprocal of the other.

87 CHAPTER 5. EQUATIONS OF LINES 84 To make the above look exactly like Proposition 5.5.1, use b a for m 1 and solve for m 2. Example Find the equation of the line described below and write it in slope-intercept form, if possible. 1) The line goes through the point (-2,3) and is parallel to the line 3x 4y = 5 2) The line perpendicular to the line 5x + 3y = 9 and intersecting that line at the y-intercept. 3) The line going through the point (-3,-4) that is perpendicular to the line y = 5. Solutions: 1) Since the line we want is parallel to the line 3x 4y = 5, that means that they have the same slope. To nd the slope, we rewrite the given equation in slope-intercept form, i.e. we solve for y. 3x 4y = 5 3x 3x 4y = 3x ( 4y) = 1 ( 3x + 5) 4 y = 3 4 x 5 4 So the slope of the line we want is m = 3 4. Now that we have the slope and a point (-2,3) we can nd the line by using either the point-slope form (Denition 5.4.3) or the slope-intercept form (Denition ) We will use the slope-intercept form since the instructions are to put the line in this form at the end. y = 3 4 x + b Since we found the slope to be = 3 4 ( 2) + b To nd b, we plug in the point (-2,3) and solve for b. 3 = b = b So the line is y = 3 4 x ) The rst thing we want to do is rewrite the given equation in slope-intercept form so that we can read o the slope. 5x + 3y = 9 5x 5x 3y = 5x + 9 y = 1 ( 5x + 9) 3 y = 5 3 x + 3 So the slope of this line is 5 3 and its y-intercept is (0, 3). Since the line we want is perpendicular to this line, that means that the slope of the line that we want must be the negative reciprocal, m = ( 5) 3 = 3 5. So the line we want is y = 3 5 x + b To nd b, we note that this line is to have the same y-intercept as the given line, which is (0, 3). In other words, b is still 3. So the line that we want is y = 3 5 x + 3 3) The line y = 5 is a horizontal line which means that the any vertical line will be perpendicular to it. Since we get a vertical line by xing the x value and we want the vertical line to go through the point (-3,-4), the line that we want is x = 3

88 CHAPTER 5. EQUATIONS OF LINES 85 Example Determine if the following lines are perpendicular. Solution: We assume that the points A, B, C, and D are on the two lines and all fall on the grid lines and hence have the following integer coordinates. Using points A and B, the slope of Line 1 is Using point C and D, the slope of Line 2 is Since the product of the slopes is not -1 the lines are not perpendicular. A = ( 4, 3), B = (1, 4), C = ( 3, 4), D = (1, 1) m 1 m 2 = 7 5 m 1 = 4 ( 3) 1 ( 4) = 7 5 m 2 = ( 3) = 5 4 ( 5 ) =

89 CHAPTER 5. EQUATIONS OF LINES Applications of Lines In studying lines, we have seen that the equation of lines involves a relationship between two variables. To understand this relationship, we use the slope formula as follows. Let (x 1, y 1 ), (x 2, y 2 ), (x 3, y 3 )... be points on the same line where the x-coordinates are all equally spaced (where we call the equal spacing h), i.e. Then using the slope formula, we have the following: m = y 2 y 1 x 2 x 1 x 2 x 1 = x 3 x 2 = = h Using the slope formula with points (x 1, y 1 ), (x 2, y 2 ). m(x 2 x 1 ) = y 2 y 1 Clear the fraction by multiplying both sides by x 2 x 1. m h = y 2 y 1 Since we picked our x-values so that x 2 x 1 = h and so we can replace x 2 x 1 with h. m = y 3 y 2 x 3 x 2 Now we calculate the slope using points (x 2, y 2 ), (x 3, y 3 ). Note that it will still lead to the same slope value m. m(x 3 x 2 ) = y 3 y 2 Clear the fractions as before by multiplying both sides by x 3 x 2. m h = y 3 y 2 y 3 y 2 = y 2 y 1 Since they both equal m h. What we have just shown is that if the dierence in the x-values are constant then the dierence in the y-values is also constant. This means that if the relationship between two quantities is linear, then as one increases by a constant amount the other also increases (or decreases) by a constant amount. So being told that the relationship between two quantities is linear implies two things. As one quantity increases by a constant amount, the other also increases (or decreases) by a constant amount. The formula connecting the two quantities is the equation of a line. Example If a car is traveling at a constant speed of 50 mph, then the relationship between the distance traveled and the time spent traveling is linear. Find the equation that relates the distance to time and calculate how long it took to travel 413 miles. Solution: We let the x-coordinate be time,t, and the y-coordinate be the distance, d. Since the relationship between the distance and time is linear, this means that if we were to plot (t, d), the points would graph as a line. To write the equation of a line, we always need two things, the slope of the line and a point on the line. Now at time 0, since the car has not yet moved, the distance is 0, and so we have our rst point, (0,0). After one hour, t = 1, the car will have traveled 50 miles and so we get our second point (1,50). Using these two points, we can calculate the slope. m = = 50 and since (0, 0) is the y-intercept, writing the equation in slope-intercept form, we have y = mx + b d = 50t + 0 d = 50t

90 CHAPTER 5. EQUATIONS OF LINES 87 So the equation that relates distance to time when traveling at a constant speed of 50 mph, is d = 50t. To calculate how long it took to travel 413 miles, we simply replace d with 413 and solve for t. So it took 8.26 hours to travel 413 miles at 50 mph. 413 = 50t = t 8.26 = t Example A 1000 gallon septic tank is being drained by a pump at a constant rate. It takes 45 minutes to remove 423 gallons. How long will it take to drain the tank completely? Round your answer to two decimal places. Solution: If we let A, the amount of liquid in the septic tank, be the y-coordinate and t, the time the pump is on, be the x-coordinate, then since the liquid is being drained at a constant rate, the relationship between A and t is linear. In other words, the points (t, A) would graph as a line. We are given two points on this line. At time 0, the tank is full, so (0,1000) is one point on the line. After 45 minutes, 423 gallons are removed, which means that = 577 gallons remain in the tank. So the second point is (45,577). Given the two points, we can nd the equation of the line that relates A and t. First we nd the slope using the two points m = 45 0 = The point (0,1000) is the y-intercept, so if we write the equation of the line in slope-intercept form, we get y = mx + b A = t To calculate the time it takes to empty the tank, we set A = 0 and solve for t = t = t ( 1000) = t ( t So it takes approximately minutes to drain the tank completely. Example The relationship between temperature measured in degrees Fahrenheit and degrees Celsius is linear. Also, water freezes at 32 Fahrenheit or 0 Celsius and boils at 212 Fahrenheit or 100 Celsius at sea level. Find the formula for converting degrees Celsius to degrees Fahrenheit. Solution: Since we want to convert Celsius into Fahrenheit, we will let C, degrees in Celsius, be the x-coordinate and F, degrees in Fahrenheit, be the y-coordinate. Then we know that the points (C, F ), will graph as a line since we are told that the relationship between them is linear. The information about when water freezes and boils gives us two points on this line, (0, 32) and (100, 212). So the slope of the line is m = = = 9 5 Since (0, 32) is the y-intercept, the equation in slope-intercept form is y = mx + b F = 9 5 C + 32 )

91 CHAPTER 5. EQUATIONS OF LINES Linear Inequalities in Two Variables View the Video Tutorial for this section here. Linear inequalities in two variables is of a similar form to the equations of lines with the dierence being that the equal sign is replaced by an inequality sign. So in general, a linear inequality in two variables can be written in the form < Ax + By + C 0 > As with the linear inequalities in one variable, the answers to these inequalities cannot be written down explicitly and so we graph the solutions. There are two techniques to graph the solutions to these inequalities. One involves an understanding of what the equation and inequalities are saying on a graph and the other technique is a more straight forward plugging in values and calculating method. Both are valid techniques and both have their usefulness. The student is encouraged to try to understand both techniques, but the method you need to learn will depend on your instructor and/or you. Example Graph the solutions of the following inequalities. 1) 2x + 3y 1 > 0 2) 2x 3y + 1 > 0 3) 4x 2y + 6 4) x + 3 > 0 Solutions: 1) Method 1: First solve the inequality for y. 2x + 3y 1 2x +1 > 0 2x (3y) > 1 ( 2x + 1) 3 y > 2 3 x Now we graph the line y = 2 3 x and note that for any point, (x, y), on the line, y and 2 3 x will always be equal to each other. For example, point A = ( 1, 1) is on the line and so if we use these x and y values, we get y 2 3 x ( 1) so y, which in this case is 1 is equal to 2 3 x + 1 3, for x = 1. Now if we draw a vertical line, say at x = 1, any point on this line will have the same x value. This means that for points A, B and C in the above graph, 2 3 x + 1 3, will be the same value for all three points. In the case of point A, the y-coordinate of A will be equal to this value. However, for

92 CHAPTER 5. EQUATIONS OF LINES 89 point B, the y coordinate will be greater than that of A and hence greater than 2 3 x + 1 3, while for point C, the y coordinate will be less than that of A and hence less than 2 3 x For B=(-1,2) For C=(-1,-2) y 2 3 x y 2 3 x ( 1) ( 1) > 1 or y > 2 3 x < 1 or y < 2 3 x Since we want y > 2 3 x We want to include all the points that are above the line, such as point B. So we shade the region above the line. Note that we use a dashed line to indicate that we do not include any of the points on the line y = 2 3 x So the graph of the solution looks like the following. Method 2: The idea of method 2 is that once we have graphed the line, the line separates the plane into two pieces and the solution to the inequality must be one of the pieces. To gure out which one, we pick a test point that we know for certain is on one side of the line and plug it into the inequality. If the inequality is true, then we shade the side of the line that contains our test point. If the inequality is false, then we shade the side that does not contain the test point. So if we use as our test point, (-1,2), then we have 2x + 3y 1 > 0 2( 1) + 3(2) 1 > > 0 3 > 0 which is true and so we shade the side of the graph that contains this test point. This will give us the same graph we got using method 1. Some notes on the 2 methods: For method 2 it is not necessary to solve the inequality for y rst. All that is required is that you graph the related line in some way (nd two points, slope and a point, etc...) Although method 1 requires more calculations and is more conceptually involved, it is this conceptual complexity that makes this method more useful in helping you to understand another interpretation of what the equation of a line means. Also, the concept of what is above and below a graph is an important one, and knowing how to identify them using a vertical line will be of use later on in this class as well as other math classes. 2) We will focus on the use of method 1 and leave method 2 for students to try out. So the rst thing we

93 CHAPTER 5. EQUATIONS OF LINES 90 need to do is to solve the inequality for y. 2x 3y+1 2x 1 > 0 2x ( 3y) > 1 ( 2x 1) 3 y < 2 3 x Note that inequality sign changed direction since we multiplied both sides by a negative number, 1 3. Now we graph the line y = 2 3 x+ 1 3 and draw a vertical line at x = 2. Since point A is on the line, that means its x and y values will make the two sides of the inequality equal to each other. Since we want y to be less than 2 3 x+ 1 3, that means we want to be below the line and in the region where point B is located. 3) First, we must solve for y. 4x 6 2y ( 4x 6) 1 2 ( 2y) 2x 3 y y 2x 3 Now we graph the line y = 2x 3 and we graph it as a solid line since the inequality says that y can equal as well as be greater than 2x 3. To determine the greater than part, we draw an arbitrary vertical line, in this case we use x = 1. Along the vertical line, x = 1, point A is above the line y = 2x 3 which means that y > 2x 3 when x and y are replaced with the x and y coordinates of point A. So we shade the side that contains point A.

94 CHAPTER 5. EQUATIONS OF LINES 91 Note that point B is below the line y = 2x 3 and so if you use the coordinates of point B, you will get y < 2x 3. 4) Since this inequality only contains one variable, we will rst solve for it. x > 0 3 x > 3 When dealing with equations that have only one variable, we must always keep in mind what dimension we are working in. In one dimension the solution of x > 3 would graph on the number line as we did in a previous chapter. However, in this chapter we are solving in two dimensions and so there is an implied y coordinate value. We know that in two dimensions x = 3 graphs as a vertical line. We use a dashed line since x cannot equal -3. Here, we cannot use the vertical line method we used in the other problems. However, if we step back for a moment and think about what this problem wants, it is fairly straightforward to see what the answer must be. We want to shade the region where the x coordinate values are greater than -3. That means we want to shade all the points on the right side of the vertical line x = 3.

95 Chapter 6 Systems of Linear Equations in Two Variables Up to this point, when solving equations, we have always solved one equation for an answer. However, in the previous chapter, we saw that the equation of a line Ax + By + C = 0 which contains two variables, x and y, does not have one answer. In fact, it has an innite number of answers and the plot of all these answers forms a line in the 2-dimensional Cartesian Coordinate System. There is a property in mathematics that states that if you want to solve an equation that has multiple variables in it for one specic answer, then in order to even have a chance to solve for the one answer, the number of equations involving the variables must be the same as the number of variables. So in the case of linear equations in two variables, if we wish to get just one answer, then we must have two dierent equations. These two equations are what we call a system of equations. Denition System of equations In general, a system of equations means a collection of equations and the solution to the system of equations is the set of values that solves all the equations in the system at the same time. In this chapter we discuss some techniques for solving systems of linear equations in two variables. 6.1 Graphing Method View the Video Tutorial for this section here. Although the graphing method is the least practical in getting the exact solution, it is the best method for illustrating the conceptual meaning of the solution to a system of equations. Proposition Graphing Method for solving a System of Linear Equations in Two Variables. Graph the solutions of each equation. Since they are linear, they will graph as lines. The point where the two lines intersect is the point whose x and y coordinates will satisfy both equations and hence will be the solution to the system of equations. Example Solve the system of equations by using the graphing method. 92

96 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 93 1) { x 2y = 8 3x 4y = 6 2) { 2x 3y = 12 4x + 6y = 12 3) { 15x + 18y = 6 5x 6y = 2 Solutions: 1) We know that both equations will graph as lines. We have two methods for graphing a line and the method chosen will depend on the equation. In this case, for the line x 2y = 8 It is easy to nd the x and y intercepts and so graph the line using those two points. x-intercept y-intercept x 2(0) = 8 x = 8 x = 8 (8,0) (0) 2y = ( 2y) = 1 2 ( 8) y = 4 (0,4) For the second equation, if we solve for the x and y intercepts, we get x-intercept 3x 4(0) = 6 3x = 6 x = 2 (-2,0) y-intercept 3(0) 4y = 6 4y = 6 y = 6 4 = 3 2 (0, 3 2 ) Here we run into one of the weaknesses of the graphing technique. Since we are reading the solution, the intersection point, from the graph, we need to be sure that the graph is accurate. Although we could label the tick marks in fractional values, it could become unwieldy if the numbers get large. Notice that to plot the point (8,0) would require 16 tick marks along the x-axis if going by 1 2. So, in general, we try to avoid fractional coordinates. So we can try to nd a second point with integer coecients by trial and error or since we have one nice point, (-2,0), we can use the slope to nd another point. To nd the slope, we will use the slope-intercept form. So we rst solve for y. 3x 4y = 6 4y = 3x 6 y = 3 4 x y = 3 4 x The slope of the line is m = 3 4 and since we know (-2,0) is on the line, we can start from there and go up 3 and to the right 4, to get another point, (2,3), see graph below.

97 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 94 Now, if we graph the two line together, we get the following graph. From the graph, it would appear that the point (2,3) is the solution to the system. We can check this by plugging in 2 for x and 3 for y into the two equations to see if it satises both equations. x 2y = 8 (2) 2(3) = 8 3x 4y = 6 3(2) 4(3) = 6 So (2,3) is a solution to both equations and hence the solution to the system. 2) In this problem, graphing the lines by nding two points on each line leads to two lines that look parallel. The problem is that when you are looking at a graph, you are only seeing the graph for a xed range of x and y values and so it is possible that lines intersect at x and y values that are not shown on the graph. So we need an analytic way to ensure that the lines are indeed parallel. Recall that two lines are parallel if they have the same slopes. So we rst put both equations in slope-intercept form 2x 3y = 12 2x 2x 3y = 2x + 12 y = 1 ( 2x + 12) 3 y = 2 3 x 4 4x + 6y = 12 +4x +4x 6y = 4x + 12 y = 1 (4x + 12) 6 y = 2 3 x + 2

98 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 95 So both lines have the same slope m = 2 3, which means that the two lines are parallel. Since parallel lines never intersect, there is no solution to the system of equations. 3) In this problem, we again can graph the lines by nding two points on each line, but it will be more instructive to put both equations in slope-intercept form. 15x + 18y = 6 +15x +15x 18y = 15x 6 y = 1 (15x 6) 18 y = x 6 18 y = 5 6 x 1 3 5x 6y = 2 5x 5x 6y = 5x + 2 y = 1 ( 5x + 2) 6 y = 5 6 x 2 6 y = 5 6 x 1 3 Notice that, in slope intercept form, both equations are the same, which means that the two lines are actually on top of each other (or are the same line). In other words, the two lines intersect each other at every point along the line. The solution is every point on the line. Note that there are an innite number of solutions but that not every point is a solution. Only the points that satisfy the equation y = 5 6 x 1 3 are solutions to the system. This means that for any given x value, the y value is 5 6 x 1 3. For example, if x = 0 then y = 5 6 (0) 1 3 = 1 3. So ( 0, 5 6 (0) ( 3) 1 or 0, 1 3) is a solution. In general, we write the solution as the ordered pair ( x, 5 6 x 3) 1.

99 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES Substitution Method View the Video Tutorial for this section here. As mentioned before, one of the problems with the graphing method is that it is dependent on the accuracy of the graph. So if the solution to a system contained fractional values, the graphing method may not be able to get the exact answer. The substitution method is an analytical method that lets us nd the exact solution to a system of equations. The substitution method is based on the idea that if we say that A = B, then wherever we see A, we can replace it with B and likewise, wherever we see B we can replace it with A. We use the following example to describe the method. Example Solve { x 2y = 8 (Equation 1) 3x 4y = 6 (Equation 2) Solution: x 2y +2y = 8 +2y x = 2y 8 x = (2y 8) x = 2y + 8 Step 1: We choose one of the equations and solve for one of the variables. Here we choose the Equation 1 and solve for x as that will be the easiest to do. Step 2: 3x 4y = 6 3( 2y + 8) 4y = 6 6y y = 6 Now, in the equation that was not used above, in this case Equation 2, we substitute in for x the expression that we found, 2y + 8. By making this substitution we end up with an equation that contains only one variable which can then be solved our normal way. Note that you must use the equation that was not solved for in the rst step, in this case we must use Equation 2. If you substitute into the equation that has already been solved, in this case Equation 1, the variables will always drop out and you will end up with a statement that is always true, such as 0 = 0, which will tell us nothing. 6y y = 6 10y = y = 30 y = 1 10 ( 30) y = 3 continuation of Step 2: Solve for y. x = 2y + 8 x = 2(3) + 8 x = 2 Step 3: Substitute the 3 for y into the equation that we solved originally for x in the rst step.

100 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 97 The solution is therefore (2, 3). Example Solve the following using the substitution method. 1) { 39x + 4y = 5 y = 21x + 5 2) { 2x + 5y = x = 1 y 3) { y + 2 = 3 4 x 8 = 3x 4y Solutions: 1) Since the second equation in this system has already been solved for y, we will substitute 21x + 5 for y into the rst equation and solve for x. Substitute 1 3 in for x in the second equation. 39x + 4y = 5 39x + 4( 21x + 5) = 5 39x 84x + 20 = 5 45x = x = 15 x = 1 45 ( 15) x = = 1 3 y = ( ) = = 2 So the solution is the point ( 1 3, 2). 2) For step 1, we solve the second equation for y. Again, it does not matter which equation you solve or for what variable. 2 5 x 2 5 x+y = x y +y y = x Step 2, we substitute this in for y in the rst equation. 2x + 5y = 20 2x + 5(1 2 x) 5 = 20 2x + 5 2x = 20 5 = 20 Since 5 20, this tells us that this equation has no solution which means that the overall system of equations has no solution. In other words, the two equations represent parallel lines and so do not intersect anywhere. Note that if you put both equations in slope-intercept form, { y = 2 5 x + 4 y = 2 5 x + 1 they have the same slope but dierent y-intercepts, which means that the two lines are parallel. So again, we say that the system has no solution.

101 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 98 3) For this system, solving the rst equation for y would be easiest. Now we substitute this into the second equation. y + 2 = 3 4 x y = 3 4 x 2 8 = 3x 4y ( ) 3 8 = 3x 4 4 x 2 8 = 3x 3x = 8 Since this equation is always true, what this tells us is that the two equations represent the same line. 1 You can prove that they are the same line by putting both equations in slope-intercept form and seeing that you get the same equation in that form. So any point on the line, y = 3 4x 2, will be a solution to the system. As an ordered pair, this means that for any given x value, the y-coordinate value is 3 4x 2, we write the solutions as y {}}{ 3 (x, 4 x 2) 1 This is why it is important that you make sure that you substitute into the correct equation. If you make the substitution into the wrong equation, you will always get a solution similar to this one (all variables drop out and you get a true statement) and think the two equations represent the same line. For example, in the rst example problem, if you substituted the expression for y into the second equation, you would end up with all the variables dropping out and think the two equations represented the same line, when we know they do not.

102 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES Elimination Method View the Video Tutorial for this section here. The elimination method is based on the concept that if you have two pairs of quantities that are equivalent to each other, A = B C = D and you add the left side together and the right side together, you still end up with the left and right side equaling each other, A + C = B + D The way we use this idea in solving a system of equations is to rewrite the equations such that when we add the two equations, all the terms containing one of the variables cancel each other out. We describe the method with the following example. Example Solve { x 2y = 8 (Equation 1) 3x 4y = 6 (Equation 2) Solution: { 3( x 2y) = 3( 8) 3x 4y = 6 { 3x 6y = 24 3x 4y = 6 Step 1: Rewrite one of the equations so that the terms containing the variable we wish to eliminate are exact opposites. Here we will get rid of the terms with the variable x. To do this, we multiply both sides of Equation 1 by 3. Since we multiplied both sides by 3, the new and old equation are equivalent. 3x 6y = x 4y = 6 6y + ( 4)y = 24 + ( 6) 10y = 30 y = 3 Step 2: Add the two equations together. If we did step 1 correctly, one of the variables should drop out in the resulting equation. Solve for the remaining variable. Here, the terms with x in them cancel each other out and so we solve for y. x 2y = 8 x 2(3) = 8 x 6 = 8 x = 2 x = 2 Step 3: Substitute the value found in step 2 into either of the equations and solve for the other variable. In this case, we substitute 3 in for y into equation 1. The solution is the point (2, 3). Before doing more examples, we need to discuss the idea of the least common multiple.

103 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 100 Denition Least Common Multiple The least common multiple of two whole numbers, a and b, is the smallest number that both a and b divide into. Let L be the least common multiple of a and b. Then L a L b = m = n where m and n are whole numbers. Note that if we solve the above for L, we get L = a m L = b n which means that the least common multiple of a and b is the smallest number that you can get by multiplying a and b by some other whole numbers. Example Find the least common multiple of the following pairs of numbers. 1) 4 and 2 2) 3 and 5 3) 6 and 8 Solutions: 1) The following are the steps to nd the least common factor. 4 = = 2 Step 1: Prime factor both numbers. Write both numbers as a product of prime numbers. Least common factor is 2 2 or 4. Step 2: The least common multiple must contain all the factors of both numbers. If they share a common factor, use the one that has more. Since both 4 and 2 share the common factor 2, we use the two 2's from 4. Note that 4 is the smallest number that you can get by multiplying some number to 4 and 2. In this case, 4 1 = 4 and 2 2 = 4. 2) For 3 and 5, since they are both prime numbers, there is nothing to do for step 1. For step 2, we just multiply both 3 and 5. So, the least common multiple is 3 5 = 15. 3) 6 = = 2 3 Step 1: Prime factor both numbers. Write both numbers as a product of prime numbers.

104 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 101 Least common factor is = 24 Step 2: The least common multiple must contain all the factors of both numbers. If they share a common factor, use the one that has more. Since both 6 and 8 share the common factor 2, we use the three 2's from 8. The reason for the above discussion of the least common multiple is that in example 6.3.1, we saw that in order to get rid of the x terms, they had to be exact opposites of each other and the only way to change the terms is to multiply the entire equation by some number. So the goal is to change both terms to the least common multiple with opposite signs. In example 6.3.1, equation 1 had x and equation 2 had 3x. Ignoring the negative sign for a moment, we are dealing with the numbers 1 and 3. The least common multiple is 3 and so if we multiply x by 3, the x terms will be exactly opposite to each other. Example Solve using the elimination method. 1) { 2x 3y = 1 5x 2y = 1 2) { 18x 6y = 15 9x + 3y = 4 3) { 3x + 6y = 18 y = 1 2 x + 3 Solutions: 1) We follow the steps of example { 2(2x 3y) = 2(1) 3(5x 2y) = 3(1) { 4x + 6y = 2 15x 6y = 3 Step 1: We will eliminate the y terms. The rst equation has a 3y while the second equation has 2y. Ignoring the signs for a moment, the least common multiple is 6. So if we multiply the rst equation by -2 and the second by 3, we will end up with the y terms being exact opposites. 4x + 6y = x 6y = 3 11x = 1 Step 2: Add the two equations and solve for x. x = ( 1 11 ) 3y = 1 3y = (1) y = 9 11 y = 1 ( ) 9 = Step 3: Substitute the x value into the rst equation and solve for y. 2) So the solution is the point ( 1 11, 3 11).

105 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 102 { 18x 6y = 15 2( 9x + 3y) = 2( 4) { 18x 6y = 15 18x + 6y = 8 Step 1: Here we will eliminate the y terms. Least common multiple is 6, so we multiply the second equation by 2. 18x 6y = x + 6y = 8 0 = 23 Step 2: Add the two equations. No solutions. Since 0 is never equal to -23, that means that the system of equations has no solutions. In other words the lines do not intersect. They are parallel lines. 3) { 3x + 6y = 18 y = 1 2 x + 3 { 3x + 6y = x + y = 3 First, we will rearrange the second equation so that the x and y terms line up in both equations. { 3x + 6y = 18 6 ( 1 2 x + y) = 6(3) { 3x + 6y = 18 3x 6y = 18 We will get rid of the y terms. Since the least common multiple of the y terms is 6, we multiply the second equation by -6. 3x + 6y = x 6y = 18 0 = 0 Add the two equations together. Since all the variables drop out and we end up with a statement that is always true, this means that the two equations represent the same line. (x, 12 x + 3 ) So the solutions are any points on the line y = 1 2 x + 3. As an ordered pair, this means that for any value x, the y value will be 1 2 x + 3.

106 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES Applications Distance/Rate in Moving Medium Recall that in a previous chapter we used the fact that if an object travels at a constant speed (rate), r, for some time, t, then the distance traveled, d, is the product of the speed and time, d = rt What we want to study now is what happens if the object is moving at a constant speed in a medium that is also moving. For example a boat moving in a owing river or an airplane ying in air that is also moving. In such cases, we have to understand the relationship between the speed of the object and the speed of the medium. To understand this relationship, we give the following illustration. Example Suppose there are three people, a newlywed couple on a train and the bride's ex standing at the station. The train is moving at 20 feet/sec. away from the station (and the ex) while the bride on the train walks towards the front of the train at 3 feet/sec. and the groom stays at the end of the train, see gure below. After 5 seconds, we have the following: To the groom, the bride would appear to be walking 3 feet/sec. away from him. To the ex, the groom would appear to be moving at 20 feet/sec. away from him, since the train is moving at 20 feet/sec. However, since the bride is walking on the train as the train moves, the bride would appear to be moving away from the ex faster than the groom, who is standing still on the train. To the ex, it would appear that the bride is moving away from him at the speed of the train plus her walking speed, = 23 feet/sec. We can verify this as follows. Since d = rt, if we solve for the rate, r, we get r = d t. Now after 5 sec. (t = 5), the distance between the ex and the groom is 100 ft., while the distance from the ex to the bride would be 115 ft. The speed of the groom, as viewed by the ex, is r = = 20 ft/sec. which is the speed of the train. This makes sense, since the groom is not moving on the train. The speed of the bride, as viewed by the ex, is r = = 23 ft/sec. After 10 seconds, we have a similar situation. The bride will be 30 feet away from the groom and so would appear to the groom as walking at a speed of r = = 3 ft/sec. To the ex, the bride would be 230 feet away and so would appear to be moving at a speed of r = = 23 ft/sec. or the speed of the train plus her walking speed. What this illustrates is that the measure of speed is relative to where you are when observing a moving object.

107 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 104 Proposition Relative Speed of Objects in a Moving Medium Let r s be the speed of the object in medium that is not moving (e.g. the speed of a boat in standing water.) Let r m be the speed of the moving medium (e.g. the speed of the river current.) Then to the observer who is standing still (not in the moving medium), if the object and medium are moving in the same direction(e.g. a boat going down river with the current), the speed of the object to the observer would be the sum r s + r m Whereas, if the object is moving in the opposite direction of the moving medium (e.g. a boat going up river against the current), the speed of the object to the observer would be the dierence r s r m if r s > r m r m r s if r s < r m We use which ever dierence gives us a positive value, since speed must be a positive number. Example A boat takes 7 hours to go up river 42 miles against the river current, while it takes 3 hours to cover the same distance going down river with the current. Find the speed of the boat in standing water and the speed of the river current. Solution: Let b be the speed of the boat in standing water and c the speed of the river current. Then according to Proposition above, we have that the overall speed going up river is b c and going down river is b + c, to someone watching the boat from the bank of the river. Note that we assume that b > c, since if the current is faster than the boat, then the boat could not go up river against the current. Now we build our distance, rate, and time table. rate time = distance up river b c 7 7(b c) down river b + c 3 3(b + c) The dierence between this problem and the distance/rate problems we did when talking about linear equations in one variable is that for those problems, we looked at the relationship between the two distances to create one equation, while here we have two unknowns, b and c, and so we need two equations instead of one. We know that the distance traveled up river in 7 hours is 42 miles and like wise the 42 miles was covered going down river in 3 hours. { 7(b c) = 42 3(b + c) = 42

108 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 105 An easy way to solve this system of equations is to rst get rid of the 7 and 3 in the two equations by dividing them out and then use the elimination method. { 1 7 (7)(b c) = 1 7 (42) 1 3 (3)(b + c) = 1 3 (42) { b c = 6 b + c = 14 b c = 6 + b + c = 14 2b = 20 b = 10 b + c = 14 (10) + c = 14 c = 4 So the speed of the boat is 10 mph while the speed of the river current is 4 mph Miscellaneous Problems The key to working out word problems is to rst restrain yourself from trying to get equations right away. Here is a general guideline for how to approach word problems. General Approach to Solving Word Problems: Keep in mind that you will have to read the problem at least two times. Step 1: On the rst read, ignore all the numbers and just focus on what the problem is talking about, e.g. is it talking about a car going from point A to B and back, two planes going in opposite directions, two dierent savings accounts, etc... Step 2: On the second read, write down on paper all the information given in the problem about what you identied in step 1. Just write the information down in words as done in the problem. The information should be organized so that you can, at a glance, determine what information belongs to what. A table to organize the information is often helpful. Step 3: Determine what the variable(s) should represent. Then convert the information found in step 2, into mathematical expressions using the variable(s) we just identied. Step 4: Now we step back and look at the information written out and try to nd the relationship between the quantities identied in step 1. This is the step at which we want to create equations. Step 5: Once the equation(s) are made, we solve them. Example How much 10% saline solution should be mixed with a 40% saline solution to get 50 ounces of a 28% saline solution? Find the amount of each saline solution used. Solution: Note that we have done problems like this before when discussing mixture problems. Here, we show a dierent approach to solving the problem using the idea of systems of equations. Step 1: The problem involves mixing two dierent types of liquid to get a third type. So we are talking about a 10%, 40%, and 28% saline solutions.

109 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 106 Step 2 and 3: Here we mix steps 2 and 3 a little because mixture problems are fairly straightforward, in that the table created for them always follows the same format. Since we do not know how much of either saline solution is mixed in, we let x = total amount of the 10% saline solution used y = total amount of the 40% saline solution used and make our table, as we did when we talked about mixture problems. total amount of solution in each bottle. actual amount of salt in each bottle. 10% 40% 28% x + y = x + 0.4y = 0.28(50) Step 4: So we end up with the system of equations, { x + y = x + 0.4y = 0.28(50) Step 5: Here we use the substitution method. x + y = 50 y = 50 x Solve the rst equation for y. 0.1x + 0.4(50 x) = 14 Substitute what we found for y above into the second equation. 0.1x x = x + 20 = x = 6 x = = 20 Solve for x. y = 50 (20) = 30 Substitute the x value into the equation above where we solved for y. So we need to use 20 ounces of the 10% saline solution and 30 ounces of the 40% saline solution. Example A total of $25,000 is to be distributed into two savings accounts which give a simple interest of 2% and 5%, respectively. How much should be invested into each account so that the total interest earned at the end of the year is $1300? Solution: Step 1: The problem involves two savings accounts, a 2% interest account and a 5% interest account. Step 2 & 3: Let x = amount invested at 2% y = amount invested at 5%

110 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 107 This problem is very similar to the previous example. The only dierence is that instead of talking about the total liquid and the amount of salt in each bottle, we instead deal with total dollar amounts invested and the interest earned in each account. We can make a table similar to that used for the previous mixture problem. 2% 5% $ amount in each account. x + y = 25,000 interest earned in each account 0.02x y = 1300 Step 4: So the system of equations that we need to solve is { x + y = x y = 1300 Step 5: We use the substitution method to solve the system where we solve the rst equation for y and substitute that into the second equation to solve for x. y = x 0.02x (25000 x) = x x = x = 50 x = = Before substituting this into the rst equation to nd y, we note that the above is saying that we need to invest -$ into the 2% account. The negative does not make any sense in the context of this problem and the way we set up the variables. This means that although there is a mathematical answer, it does not make any physical sense. So it is not possible to earn $1300 in interest under the given conditions (note that this makes sense, since even if you invested everything into the 5% account, you only get 0.05(25000) = 1250.) Example Randy needs to buy a new printer to print out iers for a new math class. An inkjet printer will cost $69.95 for the printer and 10 cents per page to print, while a laser printer will cost $ for the printer and cost 5 cents per page to print. Determine at what number of pages printed, the total cost will be the same in using either printer. Solution: Step 1: We are talking about two printers, an inkjet and laser. Step 2: inkjet laster printer cost $69.95 $ cost per page to print $0.1 $0.05 We want the total cost of using either printer to be the same. Step 3: Let x be the number of pages printed. Then the total cost to use each printer is the sum of the cost of the printer and the cost to print x pages. We let C I be the total cost of using the inkjet printer and C L the total cost of using the laser printer. Total cost = Cost to print x pages + Cost of printer C I = 0.1x C L = 0.05x Since we want to know when the total cost is the same, we also have the equation C I = C L So here, we have three equations and three unknowns.

111 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 108 Step 4: We use the substitution method, starting with the equation C I = C L 0.1x = 0.05x x x x = 50 x = = 1000 So if 1000 pages are printed, then the cost will be the same for either the inkjet or laser printer. Another way to approach this problem would have been to let the variable y represent the total cost, whether we are talking about the inkjet or the laser, just as x represents the number of pages printed in either printer. In this way, the printer we are talking about is simply identied by the equation rather than by the variables. y = 0.1x the inkjet printer y = 0.05x the laser printer So we get two lines and we seek the point at which the two lines intersect each other. If we use the substitution method (set the y's equal to each other), we end up with the same equation as above.

112 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES Practice Problems Be sure to write your answer as a sentence at the end. 1) The ACME Anvil Company sells two types of anvils. The base model is sold for $30 while the deluxe model is sold for $50. In conducting a market study, ACME found that their best customer, Mr. Wile E. Coyote, purchased a total of 350 anvils in the previous year which generated a total revenue of $12,700 for the company. Determine how many of the deluxe model Mr. Coyote purchased.

113 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES Systems of Linear Inequalities in Two Variables In the previous chapter we discussed how to solve a linear inequality in two variables. To solve a system of inequalities, we add just one more step at the end. We summarize the steps as follows: Steps to Solve a System of Linear Inequalities in Two Variables Graph the solution of each inequality on the same graph. Recall that this involves graphing the associated line which divides the cartesian plane in two. Then you shade in the half-plane that solves the inequality. The points at which the lines intersect are a key feature and so should be found analytically using the techniques we discussed in the previous sections of this chapter. Once you have shaded the solution to each inequality, since the solution of the system are the points that solve all the inequalities, the region where all the shadings from the rst step overlap will be the solution to the system. We illustrate the method with the following example. Example Solve { 2x + y < 4 3x + y 1 Solution: Solve each inequality. 2x + y < 4 3x + y 1 y < 2x 4 y 3x + 1 Graphing them together, we get

114 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 111 The solution is where the two shaded regions overlap each other. An important feature of the graphs is where the two lines intersect, as that forms a vertex (corner) of the region that represents the solution. We nd the intersection by converting the original system of inequalities into a system of equations. { 2x + y = 4 3x + y = 1 We will solve this system using the elimination method by multiplying the rst equation by -1. Substituting this into the rst equation, we get 2x y = 4 + 3x + y = 1 5x = 5 x = 1 2( 1) + y = 4 y = 2 So the point at which the two lines intersect is ( 1, 2). Note that this tells us that the region containing the solution is to the left of x = 1. The graph of the nal answer is Note that the points on the line y = 2x 4 are not included in the solution and so the point where the two lines intersect is also not included in the solution, which is why we use a hallow circle to represent that point. Example Graph the regions described by the following systems of linear inequalities. Be sure to nd and label the vertices (the corner points of the region.) 1) { 5x 2y > 20 y < 5x 15 2) x 4 y > 2 4x + 7y 26 Solutions: 1) The rst step is to graph the regions dened by each inequality. We do this by rst graphing the lines associated with each inequality. In the rst equation, we note that the x and y intercepts are easy to nd and they are integer values. x-intercept y-intercept 5x 2(0) = 20 5(0) 2y = 20 5x = 20 2y = 20 x = 4 y = 10 (4, 0) (0, 10)

115 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 112 So we can graph the line using the intercepts. Since the y-intercept is (0, 10), we know that the point (0,0) is above the line and so will use that as our test point. Plugging this into the rst inequality, we get 5(0) 2(0) > 20 0 > 20 which is a false statement. That means that the solutions lie on the other side (below) the line 5x 2y = 20, (see the gure below.) As for the second inequality, the associated line is y = 5x 15 which means it has a slope of m = 5 and y-intercept (0, 15). Using this we can graph the line and since the inequality says that y must be less than 5x 15, this means we want the region below the line y = 5x 15 (see the gure above.) Therefore, the solution is the overlapping region. We nd the point of intersection of the two lines (the vertex of the region) by solving the system of equations { 5x 2y = 20 y = 5x 15 Using the substitution method to substitute 5x 15 in for y in the rst equation, we get 5x 2(5x 15) = 20 5x 10x + 30 = 20 5x + 30 = 20 5x = 10 x = 2 To nd y, we substitute the 2 for x in the second equation, y = 5(2) 15 y = 5 So the lines intersect at (2, 5). This point is not included in the region and so we use a hallow circle. The graph of the nal solution is

116 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 113 2) The method for solving the system of inequalities is the same whether there are two, three, or more inequalities. The rst step is to graph the solution of each inequality. The line associated with x 4 is x = 4, which is a vertical line. We use a solid line and shade to the right of the vertical line, since x is greater than or equal to -4. The line associated with y > 2 is y = 2, which is horizontal line. We use a dashed line and shade everything above the line since y must be greater than 2. For the third inequality, we will rst solve it for y, to put it in slope-intercept form. 4x + 7y 26 7y 4x + 26 y 4 7 x So the associated line is y = 4 7 x Since the intercepts have fractional values, we seek an easier point to graph. If we use x = 3, y = 4 26 (3) = 7 = 14 7 = 2 Which means the point (3, 2) is on the line and using the slope m = 4 7, we can nd another point by going up 4 and to the left by 7, which gives us the point ( 4, 6) (see the gure below.) The inequality y 4 7 x denes the region below the line y = 4 7 x

117 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 114 Note that the three lines create a triangle and the region inside the triangle is where all the solutions overlap each other. So we need to nd the vertices (the corners of the triangle, point A, B, and C on the graph below.) Point A is where the lines x = 4 and y = 2 intersect, which is ( 4, 2). Point B is where the lines y = 2 and y = 4 7 x intersect. Using substitution with the variable y, we get 2 = 4 7 x + 26 ( 7 7(2) = x + 26 ) 7 14 = 4x = 4x 3 = x So point B is (3, 2). Point C is where the lines x = 4 and y = 4 7 x intersect. Using substitution with the variable x, we get y = 4 7 ( 4) = = 6 So point C is (3, 6). Note that looking at the graph above, it looked as though we could read the coordinates of A, B, and C from the graph. However, we need to check analytically (using the equations) to be sure of the exact coordinates as the graph may not be accurate. Note that points A and B are not included because points on the line y = 2 are not in the solution and so we use hallow circles to represent them. Point C is included because points on the line x = 4 and y = 4 7 x are in the solution and so we use a lled circle to represent C. So the graph of the region is

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