OPTIONAL: Watch the Flash version of the video for Section 6.1: Rational Expressions (19:09).

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1 UNIT V STUDY GUIDE Rational Expressions and Equations Course Learning Outcomes for Unit V Upon completion of this unit, students should be able to: 3. Perform mathematical operations on polynomials and rational expressions. 3.1 Find the restrictions on replacement values and the least common denominator. 3.2 Simplify rational expressions and complex rational expressions. 3.3 Perform mathematical operations including addition, subtraction, multiplication, and division on rational expressions. 3.4 Solve equations and word problems involving rational expressions. Reading Assignment Chapter 6: Rational Expressions and Equations Unit Lesson These lesson notes are to be used in conjunction with your textbook. They provide additional information and examples for each chapter covered. Links to optional videos have also been provided in case you wish for additional information or to see step-by-step examples of how certain types of problems can be solved. Chapter 6: Rational Expressions and Equations In this unit, we will be looking at rational expressions. We will learn how to simplify, add, subtract, multiply, and divide them. Lastly, we will use them in solving equations and word problems. Section 6.1: Rational Expressions OPTIONAL: Watch the Flash version of the video for Section 6.1: Rational Expressions (19:09). Restriction for replacement values: Recall that anytime we divide by 0, this is said to be undefined. When dealing with rational expressions, the same holds true. Our denominator cannot equal zero. To find what values make our rational expression undefined, we need to set the denominator equal to 0 and solve. For many of the rational expressions we will see in this section, we will need to factor. Refer to your textbook and the lesson video for some great examples in addition to the ones shown here. x 5 Example: 3x 2 x 14 First, we set the denominator equal to zero. Then, we factor and solve for x: 3x 2 x 14 = 0 To factor 3x 2 x 14, we will use the grouping method. To do this, we multiply the first and last coefficients, 3 and -14. The product is -42. We need to find two numbers whose product is -42 and whose sum is -1. Since the product is negative, one number will be negative and one will be positive. Since the sum is negative, the larger of the two numbers MAT 1302, Algebra I 1

2 will need to be negative. We will begin by listing all pairs of numbers whose product UNIT x is STUDY -42, letting GUIDE the larger number be negative: 1, -42 2, -21 3, -14 6, -7 6*(-7) = 42 and 6 + (-7) = -1 Next, we replace x with + 6x 7x. Then, we factor by grouping: (3x 2 + 6x) + ( 7x 14) = 0 We now factor out of each parentheses the GCF of that parentheses: Finally, we set each factor equal to 0 and solve: 3x(x + 2) 7(x + 2) = 0 (3x 7)(x + 2) = 0 3x 7 = 0 x + 2 = 0 3x = x = 0 2 3x = 7 3x 3 = 7 3 x = 2 x = 7 3 Therefore, the numbers for which the rational expression is undefined are 7/3 and -2. Sometimes, we have cases where the denominator cannot equal zero, and thus there are no restrictions for replacement values. For example, suppose we have the rational expression x + 1 x The denominator of x is prime. If we try to set this equal to 0, we will not arrive at a real number. Therefore, there are no restrictions to what x can be. No matter what we choose for the value of x, we will never have a denominator equal to 0. Simplifying rational expressions: Be sure to go through the lesson video and textbook for some great examples and instruction for simplifying rational expressions. Here, we will look at one that is a little more indepth than the examples in the textbook. Click here for an example of simplifying rational expressions. Section 6.2: Multiplication and Division OPTIONAL: Watch the Flash version of the video for Section 6.2: Multiplication and Division (21:06). Multiplying rational expressions: When multiplying rational expressions, we first factor everything that can be factored. Then, we simplify by crossing out what the diagonals have in common as well as what each individual rational expression has in common (top and bottom of each fraction). Example: 2x 2 xy 9x2 6xy 8y 2 6x 2 +7xy+2y 2 3xy 4y 2 First, we factor each numerator and denominator. We will provide multiple examples of factoring polynomials by showing how to factor each polynomial in our example: Factor 2x 2 xy by factoring out the GCF, x: 2x 2 xy = x(2x y) MAT 1302, Algebra I 2

3 UNIT x STUDY GUIDE Factor 6x 2 + 7xy + 2y 2 by using the grouping method. We multiply the first and last coefficients, 6*2 = 12. We need two numbers whose product is 12 and whose sum is 7. The numbers we need are 4 and 3 since 4*3 = 12 and 4+3=7. Rewrite 7xy as 4xy + 3xy: 6x 2 + 7xy + 2y 2 = 6x 2 + 4xy + 3xy + 2y 2 Factor by grouping method: (6x 2 + 4xy) + (3xy + 2y 2 ) = 2x(3x + 2y) + y(3x + 2y) = (2x + y)(3x + 2y) Factor 9x 2 6xy 8y 2 by using the grouping method. We multiply the first and last coefficients, 9*(-8) = -72. We need two numbers whose product is -72 and whose sum is -6. The numbers we need are -12 and 6 since -12*6 = -72 and = -6. Rewrite -6xy as -12xy + 6xy: 9x 2 6xy 8y 2 = 9x 2 12xy + 6xy 8y 2 Factor by using the grouping method: (9x 2 12xy) + (6xy 8y 2 ) = 3x(3x 4y) + 2y(3x 8y) = (3x + 2y)(3x 8y) Factor 3xy 4y 2 by factoring out the GCF, y: 3xy 4y 2 = y(3x 4y). Now, our problem becomes x(2x y) (3x 4y)(3x + 2y) (2x + y)(3x + 2y) y(3x 4y) Our next step is to cross out what the diagonals have in common. This means that we cross out what any numerator and denominator have in common: x(2x y) (3x 4y)(3x + 2y) (2x + y)(3x + 2y) y(3x 4y) = x(2x y) (2x + y) 1 y Then we multiply straight across: = x(2x y) y(2x + y) Dividing rational expressions: The only difference between multiplying and dividing rational expressions is the extra step of KEEP CHANGE FLIP. Keep the first fraction the same. Change from division to multiplication. Flip the second fraction. Example: ab2 +2b 2 +a+2 ab a 3b+3 ab+a+2b+2 ab 2 3b 2 a+3 First, KEEP CHANGE FLIP: ab 2 + 2b 2 + a + 2 ab a 3b + 3 ab2 3b 2 a + 3 ab + a + 2b + 2 We factor each numerator and denominator. Notice that all of these polynomials have four terms. This means that each polynomial will have to be factored by grouping: Factor ab 2 + 2b 2 + a + 2 by grouping: MAT 1302, Algebra I 3

4 (ab 2 + 2b 2 ) + (a + 2) b 2 (a + 2) + 1(a + 2) (b 2 + 1)(a + 2) UNIT x STUDY GUIDE Factor ab a 3b + 3 by grouping: Factor ab 2 3b 2 a + 3 by grouping: Notice that we can factor b 2 1 further: Factor ab + a + 2b + 2 by grouping: (ab a) + ( 3b + 3) a(b 1) 3(b 1) (a 3)(b 1) b 2 (a 3) 1(a 3) (b 2 1)(a 3) (b 1)(b + 1)(a 3) (ab + a) + (2b + 2) a(b + 1) + 2(b + 1) (a + 2)(b + 1) Now, our division problem becomes (b 2 + 1)(a + 2) (b 1)(b + 1)(a 3) (a 3)(b 1) (a + 2)(b + 1) Now, we cross out what the diagonals have in common, which means to cross out what any numerator and denominator have in common: (b 2 + 1)(a + 2) (b 1)(b + 1)(a 3) (a 3)(b 1) (a + 2)(b + 1) = b Section 6.3: Addition, Subtraction, and Least Common Denominators OPTIONAL: Watch the Flash version of the video for Section 6.3: Addition, Subtraction, and Least Common Denominators (29:06). When we are adding and subtracting rational expressions with like denominators, we simply add or subtract the numerators, keeping the denominator the same. Lastly, we always check to see if our answer can be simplified. We do this by factoring the numerator and denominator, if possible, and crossing out what the numerator and denominator have in common. Example: t2 8t 15 t 3 t 3 First, we distribute the negative sign to 8t 15: Next, we factor t 2 8t + 15 into (t 5)(t 3): t 2 8t 15 = t2 (8t 15) t 2 (8t 15) = t2 8t + 15 MAT 1302, Algebra I 4

5 yt 2 8t + 15 = (t 5)() UNIT x STUDY GUIDE Our final step is to cross out what the numerator and denominator have in common, which is t 3: (t 5)() = t 5 Finding the least common denominator Factor each denominator completely. The LCD will be the product of all unique factors, each raised to its highest exponent (if applicable). Click here for an example of finding the least common denominator. Section 6.4: Addition and Subtraction with Unlike Denominators OPTIONAL: Watch the Flash version of the video for Section 6.4: Addition and Subtraction with Unlike Denominators (15:47). Now, we will apply all principles that we learned in the previous section as we add and subtract rational expressions with unlike denominators. Example: 5x+3y 2x 2 y 3x+4y xy 2 First, let us find the LCD, which will be the product of all unique factors. The unique factors in our denominators are 2, x and y. Now, each one raised to its highest exponent will mean that the x and y will both be squared, since that is the highest exponent used. So, the LCD is 2x 2 y 2. Now, one way we can think of the next step is Multiply the numerator and denominator of each fraction by what it is missing from the LCD in the denominator. In our first fraction, we are missing that extra y, so we multiply the numerator and denominator by y. The second fraction is missing the 2 and another x, so we multiply the numerator and denominator by 2x: y(5x + 3y) y(2x 2 y) 2x(3x + 4y) 2x(xy 2 ) 5xy + 3y 2 2x 2 y 2 6x2 + 8xy 2x 2 y 2 5xy + 3y 2 (6x 2 + 8xy) 2x 2 y 2 Our next step is to distribute the negative sign in the numerator to the parentheses: 5xy + 3y 2 6x 2 8xy 2x 2 y 2 Our final steps are to combine like terms in the numerator and write them in descending order: 5xy + 3y 2 6x 2 8xy 2x 2 y 2 6x 2 3xy + 3y 2 2x 2 y 2 Although we can factor out a negative 3 from the numerator, we will not be able to simplify any further. We can stop here. MAT 1302, Algebra I 5

6 Example: x 2 9 x 2 x 6 UNIT x STUDY GUIDE First, we will need to factor both denominators. The polynomial x 2 9 is the difference of squares. This will factor into (x 3)(x + 3). The polynomial x 2 x 6 factors into (x 3)(x + 2): 3 (x 3)(x + 3) + 2 (x 3)(x + 2) Our unique factors are (x 3), (x + 3), and (x + 2). Since each factor appears at most once in each denominator, then each factor will only appear once in our LCD. The LCD is (x 3)(x + 3)(x + 2). Our next step is to multiply the numerator and denominator of each fraction by what is missing in the denominator from the LCD. For this example, we multiply the first fraction s numerator and denominator by (x + 2), and multiply the second fraction s numerator and denominator by (x + 3): 3(x + 2) (x 3)(x + 3)(x + 2) + 2(x + 3) (x 3)(x + 2)(x + 3) Distribute 3 and 2, respectively: 3x + 6 (x 3)(x + 3)(x + 2) + 2x + 6 (x 3)(x + 2)(x + 3) Add the numerators: 3x (2x + 6) (x 3)(x + 3)(x + 2) 5x + 12 (x 3)(x + 3)(x + 2) Since the numerator cannot be factored and it has nothing in common with the denominator, our rational expression is completely simplified. Click here for another example of adding/subtracting rational expressions with unlike denominators. Section 6.5: Complex Rational Expressions OPTIONAL: Watch the Flash version of the video for Section 6.5: Complex Rational Expressions (29:43). When dealing with complex rational expressions, simplify the numerator and denominator separately as much as possible. Then, divide the numerator by the denominator. You can also think of the KEEP CHANGE FLIP phrase: Keep the numerator the same. Change to multiplication. Flip the denominator. Click here to see an example of simplifying complex rational expressions. Section 6.6: Rational Equations OPTIONAL: Watch the Flash version of the video for Section 6.6: Rational Equations (17:01). The textbook gives one method for solving rational expressions. This is another way we can look at these problems. When dealing with equations involving the equality of two rational expressions (two single fractions) such as the problem below, we can use the process of cross multiplication to solve (the same idea as solving proportions). Example: n+2 = n+1 n 3 n 2 First, we establish what restrictions we have on our variables by setting each denominator equal to zero and solving. The variable n cannot be 3 or 2 since those two numbers cause one of our denominators to be equal to 0. MAT 1302, Algebra I 6

7 First, we cross multiply: UNIT x STUDY GUIDE (n 3)(n + 1) = (n + 2)(n 2) n 2 2n 3 = n 2 4 Next, we subtract n 2 from both sides: Now, we solve for n: n 2 2n 3 n 2 = n 2 4 n 2 2n 3 = 4 2n = n = 1 2n 2 = 1 2 n = 1 2 When we are dealing with a rational equation involving multiple fractions, as in the example below, we can rewrite all fractions using the LCD, then just work with the numerator. The idea is that if the denominators are all the same, the numerators will also be equal to each other. Here is an example using this method: Example: 5 3a 2a+2 = 3 a a 2 +4a+3 a+3 a+1 First, let s establish what restrictions we have in the denominators. We do this by setting each denominator equal to 0 and solving for a. In our first fraction, we factor a 2 + 4a + 3 as (a + 3)(a + 1). Therefore, a 3, 1. The denominators of the other two fractions will have the same restrictions since they include these factors. The LCD will be the product of all unique factors. In this problem, we have only two unique factors, (a + 3) and (a + 1). Therefore, the LCD is (a + 3)(a + 1). Multiply 2a+2 a+3 by (a + 1) since this is what is missing from the denominator, and multiply 3 a by (a + 3), since this is what is missing from the denominator: a+1 5 3a (a + 1)(2a + 2) (a + 3)(3 a) = (a + 3)(a + 1) (a + 1)(a + 3) (a + 3)(a + 1) Now that all of our denominators are the same, we can just look at the numerators to solve for a: 5 3a (a + 1)(2a + 2) = (a + 3)(3 a) Then, we multiply (a + 1)(2a + 2) and (a + 3)(3 a): 5 3a (2a 2 + 4a + 2) = 9 a 2 Next, we distribute the negative sign to the parentheses: 5 3a 2a 2 4a 2 = 9 a 2 Then combine like terms on the left hand side: 3 7a 2a 2 = 9 a 2 MAT 1302, Algebra I 7

8 UNIT x STUDY GUIDE Move everything to one side of the equal sign by subtracting 9 from both sides and adding a 2 to both sides: 3 7a 2a a 2 = 9 a a 2 6 7a a 2 = 0 Write in descending order: Factor out -1: Factor a 2 + 7a + 6 into (a + 6)(a + 1): a 2 7a 6 = 0 (a 2 + 7a + 6) = 0 (a + 6)(a + 1) = 0 Setting each factor equal to zero, we find solutions of -6 and -1. Since we already established in our first step that a 1, this means our solution is a = 6. We can plug -6 back into our original equation to check our solution. Click here for another example of solving equations involving rational expressions. Section 6.7: Applications Using Rational Equations and Proportions OPTIONAL: Watch the Flash version of the video for Section 6.7: Applications Using Rational Equations and Proportions (26:38). In this section, we will use the principles we learned in this chapter to solve word problems. Be sure to review all of the examples already given through the lesson video and textbook. Here are some additional problems to help you. Example: A community water tank can be filled in 18 hours by using the town office well, alone, and in 22 hours by using the high school well, alone. How long will it take to fill the tank if both wells are working? We will be using the work principle found at the bottom of page 415 of your textbook, where a will represent the time needed for the town office well to fill the tank (18) and b will represent the time needed for the high school well to fill the tank (22): t 18 + t 22 = 1 In order to add t + t, we need a common denominator. The LCD of 18 and 22 is t (11)18 + 9t (9)22 = 1 11t t 198 = 1 11t + 9t t 198 = 1 = 1 Multiply both sides by 198/20 and simplify: t 198 = t = = Click here to see an example of word problems involving motion. MAT 1302, Algebra I 8

9 Learning Activities (Non-Graded) Problem Solution Video Reruns UNIT x STUDY GUIDE The following videos are linked in the unit lesson. They are being included here for easy accessibility in case you want to review them and the step-by-step solutions they demonstrate: MAT1302 Section 6.1 Simplifying a Rational Expression MAT1302 Section 6.3 Finding the LCM MAT1302 Section 6.4 Subtracting Rational Expressions MAT1302 Section 6.5 Simplifying Complex Expressions MAT1302 Section 6.6 Solving Rational Expressions MAT1302 Section 6.7 Word Problem Involving Motion Practice What You Have Learned After reading Chapter 6, improve your mastery skills by working the odd problems in the exercise sets from each section: Section 6.1, pages Section 6.2, pages Section 6.3, pages Section 6.4, pages Section 6.5, pages Section 6.6, pages Section 6.7, pages After completing the odd problems, find out how well you did by checking your answers in the back of the book. These activities are non-graded, which means you do not have to submit them. If you experience difficulty in mastering any of the concepts, contact your instructor for additional information and guidance. Review What You Have Learned Before attempting the homework and the unit assessments, study the chapter summaries, review the concepts taught in the chapters, and work the odd-numbered problems in the review exercises: Chapter 6 Mid-Chapter Review, pages Chapter 6 Review Exercises, pages Chapter 6 Test (practice), page 433 If you would like to test the skills you learned in Chapters 1 through 6, you can try the Cumulative review for those chapters on page 434. You can check your answers to all of the above in the back of the book. NOTE: A Study Summary is provided at the end of each chapter to help you review the concepts taught in the chapters. If you experience difficulty in mastering any of the concepts, contact your instructor for additional information and guidance. Other Resources and Activities If you need additional guidance or information, access the Multimedia Library in MyMathLab. This library provides Animations that show how to solve selected types of problems, as well as Video Lectures, PowerPoints, and Interactive Figures for information about a variety of topics from this unit. The Multimedia Textbook is an electronic version of your textbook. To access the Multimedia Library, access MyMathLab, then click on Multimedia Library in the toolbar on the left side of the screen. When the dialog box comes up, select the chapter, section, and media types that you MAT 1302, Algebra I 9

10 would like to access from the dropdown menus, then click Find Now. This will UNIT bring x up STUDY a listing GUIDE of media assets that you can click on to view. Non-Graded Learning Activities are provided to aid students in their course of study. You do not have to submit them. If you have questions contact your instructor for further guidance and information MAT 1302, Algebra I 10