In this discussion, you are assigned two rational expressions to work on. Remember to factor all polynomials completely. Read the following instructions in order and view the example to complete this discussion. • • • • • Explain in your own words what the meaning of domain is. Also, explain why a denominator cannot be zero. Find the domain for each of your two rational expressions. Write the domain of each rational expression in set notation (as demonstrated in the example). Do both of your rational expressions have excluded values in their domains? If yes, explain why they are to be excluded from the domains. If no, explain why no exclusions are necessary. Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing. Do not write definitions for the words; use them appropriately in sentences describing your math work. o Domain o Excluded value o Set o Factor o Real numbers Your initial post should be at least 250 words in length. Support your claims with examples from required material(s) and/or other scholarly resources, and properly cite any references. Problem 1) w2+11w+30 7w Problem 2) 2n-1 4n2-9 9/14/10 12:39 PM Page 381 Chapter dug84356_ch06a.qxd 6 Rational Expressions Advanced technical developments have made sports equipment faster, lighter, and more responsive to the human body. Behind the more flexible skis, lighter bats, and comfortable athletic shoes lies the science of biomechanics, which is the study of human movement and the factors that influence it. Designing and testing an athletic shoe go hand in hand. While a shoe is being designed, it is tested in a multitude of ways, including long-term wear, rear foot stability, and strength of materials. Testing basketball shoes usually includes an evaluation of the force applied to the ground by the foot during running, jumping, and landing. Many biome­ chanics laboratories have a 5 6.1 Reducing Rational Expressions 6.2 Multiplication and Division 6.3 Finding the Least Common Denominator 6.4 Addition and Subtraction 6.5 Complex Fractions 6.6 Solving Equations with Rational Expressions 6.7 Applications of Ratios and Proportions 6.8 Applications of Rational Expressions sure the force exerted when a player cuts from side to side, as well as the force against the bottom of the shoe. Force exerted in landing from a lay­ up shot can be as high as 14 times the weight of the body. Side-to-side force is usu­ ally about 1 to 2 body weights Force (thousands of pounds) special platform that can mea- 4 3 2 1 0 50 100 150 200 Weight (pounds) 250 300 in a cutting movement. In Exercises 53 and 54 of Section 6.7 you will see how designers of athletic shoes use proportions to find the amount of force on the foot and soles of shoes for activities such as running and jumping. dug84356_ch06a.qxd 382 9/14/10 12:39 PM Page 382 6-2 Chapter 6 Rational Expressions 6.1 In This Section U1V Rational Expressions and Functions 2 U V Reducing to Lowest Terms U3V Reducing with the Quotient Rule for Exponents U4V Dividing a - b by b - a U5V Factoring Out the Opposite of a Common Factor 6 U V Writing Rational Expressions E X A M P L E 1 Reducing Rational Expressions Rational expressions in algebra are similar to the rational numbers in arithmetic. In this section, you will learn the basic ideas of rational expressions. U1V Rational Expressions and Functions A rational number is the ratio of two integers with the denominator not equal to 0. For example, 0 3 -9 -7 , , , and 1 2 4 -6 are rational numbers. Of course, we usually write the last three of these rational num­ 3 bers in their simpler forms , -7, and 0. A rational expression is the ratio of two poly­ 2 nomials with the denominator not equal to 0. Because an integer is a monomial, a rational number is also a rational expression. As with rational numbers, if the denom­ inator is 1, it can be omitted. Some examples of rational expressions are x 2 - 1 3a 2 + 5a - 3 3 , , , and 9x. x+8 a-9 7 A rational expression involving a variable has no value unless we assign a value to the variable. If the value of a rational expression is used to determine the value of a second variable, then we have a rational function. For example, x2 – 1 3a2 + 5a – 3 and w = y= x+8 a-9 are rational functions. We can evaluate a rational expression with or without function notation as we did for polynomials in Chapter 5. Evaluating a rational expression a) Find the value of 4x - 1 for x = -3. x+2 U Calculator Close-Up V To evaluate the rational expression in Example 1(a) with a calculator, first use Y = to define the rational expres­ sion. Be sure to enclose both numera­ tor and denominator in parentheses. Then find y1(-3). 3x + 2 b) If R(x) = 2x - 1, find R(4). Solution a) To find the value of 4x - 1 for x = -3, replace x by -3 in the rational expression: x+2 4(-3) - 1 -13 = = 13 -3 + 2 -1 So the value of the rational expression is 13. The Calculator Close-Up shows how to evaluate the expression with a graphing calculator using a variable. With a scientific or graphing calculator you could also evaluate the expression by entering (4(-3) - 1)/ (-3 + 2). Be sure to enclose the numerator and denominator in parentheses. b) R(4) is the value of the rational expression when x = 4. To find R(4), replace x by 4 3x + 2 in R(x) = 2x - 1: 3(4) + 2 R(4) = 2(4) - 1 14 R(4) = = 2 7 So the value of the rational expression is 2 when x = 4, or R(4) = 2 (read “R of 4 is 2”). Now do Exercises 1–6 dug84356_ch06a.qxd 9/14/10 12:39 PM Page 383 6-3 6.1 Reducing Rational Expressions 383 5 An expression such as 0 is undefined because the definition of rational numbers does not allow zero in the denominator. When a variable occurs in a denominator, any real number can be used for the variable except numbers that make the expression undefined. E X A M P L E 2 Ruling out values for x Which numbers cannot be used in place of x in each rational expression? a) x2 - 1 x+8 b) x+2 2x + 1 c) x+5 x2 - 4 Solution a) The denominator is 0 if x + 8 = 0, or x = -8. So -8 cannot be used in place of x. (All real numbers except -8 can be used in place of x.) 1 1 b) The denominator is zero if 2x + 1 = 0, or x = -2. So we cannot use -2 in place 1 of x. (All real numbers except -2 can be used in place of x.) c) The denominator is zero if x 2 - 4 = 0. Solve this equation: x-2=0 x=2 x2 - 4 = 0 (x - 2)(x + 2) = 0 Factor. or x + 2 = 0 Zero factor property or x = -2 So 2 and -2 cannot be used in place of x. (All real numbers except 2 and -2 can be used in place of x.) Now do Exercises 7–14 In Example 2 we determined the real numbers that could not be used in place of the variable in a rational expression. The domain of any algebraic expression in one variable is the set of all real numbers that can be used in place of the variable. For rational expressions, the domain must exclude any real numbers that cause the denom­ inator to be zero. E X A M P L E 3 Domain Find the domain of each expression. a) x2 - 9 x+3 b) x x2 - x - 6 c) x-5 4 Solution a) The denominator is 0 if x + 3 = 0, or x = -3. So -3 can’t be used for x. The domain is the set of all real numbers except -3, which is written in set notation as {x 1 x e -3}. b) The denominator is 0 if x2 - x - 6 = 0: x2 - x - 6 = 0 (x - 3)(x + 2) = 0 x-3=0 or x + 2 = 0 x=3 or x = -2 So -2 and 3 can’t be used in place of x. The domain is the set of all real numbers except -2 and 3, which is written as {x 1 x e -2 and x e 3}. dug84356_ch06a.qxd 384 9/14/10 12:39 PM Page 384 6-4 Chapter 6 Rational Expressions c) Since the denominator is 4, the denominator can’t be 0 no matter what number is used for x. The domain is the set of all real numbers, R. Now do Exercises 15–22 Note that if a rational expression is used to define a function, then the domain of the rational expression is also called the domain of the function. For example, the 2 domain of the function y = x – 9 is the set of all real numbers except -3 or {x I x ¥ -3}. x +3 When dealing with rational expressions in this book, we will generally assume that the variables represent numbers for which the denominator is not zero. U2V Reducing to Lowest Terms Rational expressions are a generalization of rational numbers. The operations that we perform on rational numbers can be performed on rational expressions in exactly the same manner. Each rational number can be written in infinitely many equivalent forms. For example, 3 6 9 12 15 = = = = = . 5 10 15 20 25 U Helpful Hint V How would you fill in the blank in 3 — 5 = 10 ? Most students learn to divide 5 into 10 to get 2, and then multiply 3 by 2 to get 6. In algebra, it is better to multiply the numerator and denomi­ 3 nator of 5 by 2, as shown here. Each equivalent form of 3 is obtained from 3 by multiplying both numerator and 5 5 denominator by the same nonzero number. This is equivalent to multiplying the frac­ tion by 1, which does not change its value. For example, 3 3 3 2 6 = 1= = 5 5 5 2 10 and 3 3 3 9 = = . 5 5 3 15 If we start with 6 and convert it into 3, we say that we are reducing 6 to lowest terms. 10 5 10 We reduce by dividing the numerator and denominator by the common factor 2: /2 3 3 6 = = 10 2/ 5 5 A rational number is expressed in lowest terms when the numerator and the denomi­ nator have no common factors other than 1. CAUTION We can reduce fractions only by dividing the numerator and the denom­ inator by a common factor. Although it is true that 2+4 6 = , 10 2 + 8 we cannot eliminate the 2’s, because they are not factors. Removing them from the sums in the numerator and denominator would not result in 3. 5 Reducing Fractions If a e 0 and c e 0, then ab b = . ac c dug84356_ch06a.qxd 9/14/10 12:39 PM Page 385 6-5 6.1 Reducing Rational Expressions 385 To reduce rational expressions to lowest terms, we use exactly the same procedure as with fractions: Reducing Rational Expressions 1. Factor the numerator and denominator completely. 2. Divide the numerator and denominator by the greatest common factor. Dividing the numerator and denominator by the GCF is often referred to as dividing out or canceling the GCF. E X A M P L E 4 Reducing Reduce to lowest terms. a) 30 42 b) x2 - 9 6x + 18 c) 3x2 + 9x + 6 2x2 - 8 Solution a) 30 2/ /3 5 = Factor. 42 2/ 3/ 7 = 5 7 Divide out the GCF: 2 3 or 6. b) Since 9 = 9 1 = 1 , it is tempting to apply that fact here. However, 9 is not a common 18 9 2 2 2 factor of the numerator and denominator of x - 9 , as it is in 9 . You must factor 6x + 18 18 the numerator and denominator completely before reducing. (x - 3)(x + 3) x2 - 9 = 6(x + 3) 6x + 18 Factor. x-3 Divide out the GCF: x + 3. 6 This reduction is valid for all real numbers except -3, because that is the domain of the original expression. If x = -3, then x + 3 = 0 and we would be dividing out 0 from the numerator and denominator, which is prohibited in the rule for reducing fractions. = c) 3x2 + 9x + 6 3(x + 2)(x + 1) = 2(x + 2)(x - 2) 2x2 - 8 Factor completely. 3x + 3 Divide out the GCF: x + 2. 2(x - 2) This reduction is valid for all real numbers except -2 and 2, because that is the domain of the original expression. = Now do Exercises 23–46 CAUTION In reducing, you can divide out or cancel common factors only. You +3 cannot cancel x from xx + 2 , because it is not a factor of either x + 3 or x + 2. But x is a common factor in 32xx , and 32xx = 23 . Note that there are four ways to write the answer to Example 3(c) depending on whether the numerator and denominator are factored. Since 3x + 3 3(x + 1) 3(x + 1) 3x + 3 = = = , 2(x - 2) 2(x - 2) 2x - 4 2x - 4 dug84356_ch06a.qxd 386 9/14/10 12:39 PM Page 386 6-6 Chapter 6 Rational Expressions any of these four rational expressions is correct. We usually give such answers with the denominator factored and the numerator not factored. With the denominator factored you can easily spot the values for x that will cause an undefined expression. U3V Reducing with the Quotient Rule for Exponents To reduce rational expressions involving exponential expressions, we use the quotient rule for exponents from Chapter 4. We restate it here for reference. Quotient Rule for Exponents If a e 0, and m and n are any integers, then am = am-n. an E X A M P L E 5 Using the quotient rule in reducing Reduce to lowest terms. a) 3a15 6a7 b) 6x4y2 4xy5 Solution a) 3a15 3/a15 Factor. 7 = 6a 3/ 2 a7 a15-7 Quotient rule = for exponents 2 = a8 2 b) 6x4y2 2/ 3x 4y2 = 4xy5 2/ 2xy 5 3x4-1y2-5 = 2 = Factor. Quotient rule for exponents 3x3y-3 3x3 = 3 2 2y Now do Exercises 47–58 The essential part of reducing is getting a complete factorization for the numerator and denominator. To get a complete factorization, you must use the techniques for fac­ toring from Chapter 5. If there are large integers in the numerator and denominator, you can use the technique shown in Section 5.1 to get a prime factorization of each integer. E X A M P L E 6 Reducing expressions involving large integers Reduce 420 to lowest terms. 616 Solution Use the method of Section 5.1 to get a prime factorization of 420 and 616: � � 2 420 2 616 � � 2 210 2 308 � � 3 105 2 154 � � 5 35 7 77 7 11 dug84356_ch06a.qxd 9/14/10 12:39 PM Page 387 6-7 6.1 387 Reducing Rational Expressions The complete factorization for 420 is 22 3 5 7, and the complete factorization for 616 is 23 7 11. To reduce the fraction, we divide out the common factors: 420 22 3 5 7 = 3 2 7 11 616 3 5 = 2 11 15 = 22 Now do Exercises 59–66 U4V Dividing a - b by b - a In Section 4.5 you learned that a - b = -(b - a) = -1(b - a). So if a - b is divided by b - a, the quotient is -1: a - b = -1(b - a) b-a b-a = -1 We will use this fact in Example 7. E X A M P L E 7 Expressions with a - b and b - a Reduce to lowest terms. a) 5x - 5y 4y - 4x b) m2 - n2 n-m Solution a) Factor out 5 from the numerator and 4 from the denominator and use x - y = -1: y-x 5x - 5y 5(x - y) 5 5 = = (-1) = 4y - 4x 4(y - x) 4 4 Another way is to factor out -5 from the numerator and 4 from the denominator and then use y - x =1: y-x 5x - 5y -5(y - x) -5 5 = = (1) = 4(y - x) 4 4y - 4x 4 -1 b) m2 - n2 (m - n)(m + n) = Factor. n-m n-m m-n = -1(m + n) = -m - n n-m = -1 Now do Exercises 67–74 CAUTION We can reduce a - b to -1, but we cannot reduce a - b. There is no factor b-a a+b that is common to the numerator and denominator of a-b a+b or a+b . a-b U5V Factoring Out the Opposite of a Common Factor If we can factor out a common factor, we can also factor out the opposite of that com­ mon factor. For example, from -3x - 6y we can factor out the common factor 3 or the common factor -3: -3x - 6y = 3(-x - 2y) or -3x - 6y = -3(x + 2y) To reduce an expression, it is sometimes necessary to factor out the opposite of a common factor. dug84356_ch06a.qxd 388 9/14/10 12:39 PM Page 388 6-8 Chapter 6 Rational Expressions E X A M P L E 8 Factoring out the opposite of a common factor Reduce -3w - 3w2 w2 - 1 to lowest terms. Solution We can factor 3w or -3w from the numerator. If we factor out -3w, we get a common factor in the numerator and denominator: -3w - 3w 2 -3w(1 + w) = Factor. w2 - 1 (w - 1)(w + 1) -3w = Since 1 + w = w + 1, we divide out w + 1. w-1 3w = Multiply numerator and denominator by -1. 1-w The last step is not absolutely necessary, but we usually perform it to express the answer with one less negative sign. Now do Exercises 75–84 The main points to remember for reducing rational expressions are summarized in the following reducing strategy. Strategy for Reducing Rational Expressions 1. Factor the numerator and denominator completely. Factor out a common fac­ tor with a negative sign if necessary. 2. Divide out all common factors. Use the quotient rule if the common factors are powers. U6V Writing Rational Expressions Rational expressions occur in applications involving rates. For uniform motion, rate is distance divided by time, R = DT . For example, if you drive 500 miles in 10 hours, your 00 rate is 510 or 50 mph. If you drive 500 miles in x hours, your rate is 500 x mph. In work prob­ W lems, rate is work divided by time, R = . For example, if you lay 400 tiles in 4 hours, T 400 your rate is 400 4 or 100 tiles/hour. If you lay 400 tiles in x hours, your rate is x tiles/hour. E X A M P L E 9 Writing rational expressions Answer each question with a rational expression. a) If a trucker drives 500 miles in x + 1 hours, then what is his average speed? b) If a wholesaler buys 100 pounds of shrimp for x dollars, then what is the price per pound? c) If a painter completes an entire house in 2x hours, then at what rate is she painting? Solution 500 a) Because R = D, he is averaging x + 1 mph. T b) At x dollars for 100 pounds, the wholesaler is paying or x dollars/pound. 100 c) By completing 1 house in 2x hours, her rate is 1 2x x 100 dollars per pound house/hour. Now do Exercises 107–112 dug84356_ch06a.qxd 9/17/10 8:06 PM Page 389 6-9 6.1 389 ▼ Fill in the blank. 1. A rational number is a ratio of two with the denominator not 0. 2. A rational expression is a ratio of two with the denominator not 0. 3. A rational expression is reduced to lowest terms by the numerator and denominator by the GCF. 4. The rule is used in reducing a ratio of monomials. 5. The expressions a - b and b - a are . 6. If a rational expression is used to determine y from x, then y is a function of x. True or false? 7. A complete factorization of 3003 is 2 · 3 · 7 · 11 · 13. 8. A complete factorization of 120 is 23 · 3 · 5. x+1 9. We can’t replace x by -1 or 3 in --. x-3 2x 10. For any real number x, -- = x. 2 a2 + b2 11. Reducing -- to lowest terms yields a + b. a+b Exercises U Study Tips V • If you must miss class, let your instructor know. Be sure to get notes from a reliable classmate. • Take good notes in class for yourself and your classmates. You never know when a classmate will ask to see your notes. U1V Rational Expressions and Functions Evaluate each rational expression. See Example 1. 3x - 3 x+5 1. Evaluate -- for x = -2. 3x + 1 - for x = 5. 2. Evaluate 4x - 4 2x + 9 , find R(3). 3. If R(x) = -x -20x - 2 -, find R(-1). 4. If R(x) = x-8 x-5 -, find R(2), R(-4), R(-3.02), and 5. If R(x) = x+3 R(-2.96). Note how a small difference in x (-3.02 to -2.96) can make a big difference in R(x). x - 2x - 3 2 -, find R(3), R(5), R(2.05), 6. If R(x) = x-2 and R(1.999). Which numbers cannot be used in place of the variable in each rational expression? See Example 2. x 7. -x+1 7a 9. -3a - 5 2x + 3 11. -x2 - 16 p-1 13. -2 3x 8. -x-7 84 10. -3 - 2a 2y + 1 12. 2 y -y-6 m + 31 14. -5 Find the domain of each rational expression. See Example 3. x2 + x 15. -x-2 x+4 16. -x-5 6.1 Warm-Ups Reducing Rational Expressions dug84356_ch06a.qxd 390 17. 9/14/10 12:39 PM Page 390 6-10 Chapter 6 Rational Expressions x x + 5x + 6 U3V Reducing with the Quotient Rule 2 for Exponents x2 + 2 18. 2 x - x - 12 19. 20. Reduce each expression to lowest terms. Assume that all variables represent nonzero real numbers, and use only positive exponents in your answers. See Example 5. x2 - 4 2 x2 - 3x 9 21. x-5 x 22. x2 - 3 x+9 47. x10 x7 48. y8 y5 49. z3 z8 50. w9 w12 51. 4x7 -2x5 52. -6y3 3y9 53. -12m9n18 8m6n16 54. -9u9v19 6u9v14 55. 6b10c4 -8b10c7 56. 9x20y -6x25y3 57. 30a3bc 18a7b17 58. 15m10n3 24m12np U2V Reducing to Lowest Terms Reduce each rational expression to lowest terms. Assume that the variables represent only numbers for which the denominators are nonzero. See Example 4. 23. 6 27 24. 14 21 25. 42 90 26. 42 54 27. 36a 90 28. 56y 40 78 30w 30. 6x + 2 6 32. 2x + 4y 6y + 3x 34. 35. 3b - 9 6b - 15 36. 3m + 9w 3m - 6w 37. w2 - 49 w+7 38. a2 - b2 a-b 39. a -1 a2 + 2a + 1 40. x -y x2 + 2xy + y2 67. 3a - 2b 2b - 3a 68. 5m - 6n 6n - 5m 41. 2x2 + 4x + 2 4x2 - 4 42. 2x2 + 10x + 12 3x2 - 27 69. h2 - t 2 t-h 70. r 2 - s2 s-r 43. 3x2 + 18x + 27 21x + 63 44. x 3 - 3x 2 - 4x x 2 - 4x 71. 2g - 6h 9h2 - g2 72. 5a - 10b 4b2 - a2 45. 2a3 + 16 4a + 8 46. w3 - 27 w2 - 3w 73. x2 - x - 6 9 - x2 74. 1 - a2 a +a- 2 29. 31. 33. 2 Reduce each expression to lowest terms. Assume that all variables represent nonzero real numbers, and use only positive exponents in your answers. See Example 6. 59. 60. 68 44y 210 264 616 660 61. 62. 2w + 2 2w 231 168 936 624 63. 64. 5x - 10a 10x - 20a 630x5 300x9 96y2 108y5 65. 924a23 448a19 66. 270b75 165b12 2 U4V Dividing a - b by b - a Reduce each expression to lowest terms. See Example 7. 2 2 dug84356_ch06a.qxd 9/14/10 12:39 PM Page 391 6-11 6.1 U5V Factoring Out the Opposite of a Common Factor 103. Reducing Rational Expressions y3 - 2y2 - 4y + 8 y2 - 4y + 4 104. 391 mx + 3x + my + 3y m2 - 3m - 18 Reduce each expression to lowest terms. See Example 8. 75. -x - 6 x+6 77. -2y - 6y 3 + 9y 76. -5x - 20 3x + 12 78. y - 16 -8 - 2y 2 79. -3x - 6 3x - 6 80. 8 - 4x -8x - 16 81. -12a - 6 2a2 + 7a + 3 82. -2b2 - 6b - 4 b2 - 1 83. a -b 2b2 - 2ab 84. x -1 x - x2 3 3 3 Reduce each expression to lowest terms. See the Strategy for Reducing Rational Expressions box on page 388. 85. 2x12 4x8 86. 4x2 2x9 87. 2x + 4 4x 88. 2x + 4x2 4x 89. a-4 4-a 90. 2b - 4 2b + 4 91. 2c - 4 4 - c2 92. -2t - 4 4 - t2 93. x2 + 4x + 4 x2 - 4 94. 3x - 6 x - 4x + 4 -2x - 4 95. 2 x + 5x + 6 2x + 2w - ax - aw x3 - xw2 106. x2 + ax - 4x - 4a x2 - 16 -2x - 8 96. 2 x + 2x - 8 2q8 + q7 2q6 + q5 99. u - 6u - 16 u2 - 16u + 64 100. v + 3v - 18 v2 + 12v + 36 a3 - 8 2a - 4 102. 4w2 - 12w + 36 2w3 + 54 98. 2 U6V Writing Rational Expressions Answer each question with a rational expression. Be sure to include the units. See Example 9. 107. If Sergio drove 300 miles at x + 10 miles per hour, then how many hours did he drive? 108. If Carrie walked 40 miles in x hours, then how fast did she walk? 109. If x + 4 pounds of peaches cost $4.50, then what is the cost per pound? 110. If nine pounds of pears cost x dollars, then what is the price per pound? 111. If Ayesha can clean the entire swimming pool in x hours, then how much of the pool does she clean per hour? 2 97. 101. 105. 2 8s12 12s - 16s5 112. If Ramon can mow the entire lawn in x  3 hours, then how much of the lawn does he mow per hour? 6 2 Applications Solve each problem. 113. Annual reports. The Crest Meat Company found that the cost per report for printing x annual reports at Peppy Printing is given by the formula 150 + 0.60x , x where C(x) is in dollars. C(x) = dug84356_ch06a.qxd 392 9/14/10 12:39 PM Page 392 6-12 Chapter 6 Rational Expressions given by the formula, C(p) = 500,000 . 100 - p a) Use the accompanying graph to estimate the cost for removing 90% and 95% of the toxic chemicals. b) Use the formula to find C(99.5) and C(99.9). c) What happens to the cost as the percentage of pollutants removed approaches 100%? 0.80 0.70 0.60 0.50 Annual cost (hundred thousand dollars) Cost per report (dollars) a) Use the accompanying graph to estimate the cost per report for printing 1000 reports. b) Use the formula to find C(1000), C(5000), and C(10,000). c) What happens to the cost per report as the number of reports gets very large? 0.40 1 2 3 4 5 Number of reports (thousands) Figure for Exercise 113 5 4 3 2 1 0 90 91 92 93 94 95 96 97 98 99 Percentage of chemicals removed 114. Toxic pollutants. The annual cost in dollars for removing p% of the toxic chemicals from a town’s water supply is 6.2 In This Section U1V Multiplication of Rational Numbers 2 U V Multiplication of Rational Expressions U3V Division of Rational Numbers 4 U V Division of Rational Expressions 5 U V Applications Figure for Exercise 114 Multiplication and Division In Section 6.1, you learned to reduce rational expressions in the same way that we reduce rational numbers. In this section, we will multiply and divide rational expressions using the same procedures that we use for rational numbers. U1V Multiplication of Rational Numbers Two rational numbers are multiplied by multiplying their numerators and multiplying their denominators. Multiplication of Rational Numbers If b e 0 and d e 0, then a c ac = . b d bd dug84356_ch06a.qxd 9/14/10 12:39 PM Page 393 6-13 E X A M P L E 6.2 1 Multiplication and Division 393 Multiplying rational numbers 6 Find the product 7 14 . 15 Solution U Helpful Hint V The product is found by multiplying the numerators and multiplying the denominators: Did you know that the line separating the numerator and denominator in a fraction is called the vinculum? 6 14 84 = 7 15 105 = 21 4 Factor the numerator and denominator 21 5 4 Divide out the GCF 21. 5 The reducing that we did after multiplying is easier to do before multiplying. First factor all terms, reduce, and then multiply: = 6 14 2 3/ 2 7/ = 7 15 /7 /3 5 = 4 5 Now do Exercises 1–8 U2V Multiplication of Rational Expressions Rational expressions are multiplied just like rational numbers: factor, reduce, and then multiply. A rational number cannot have zero in its denominator and neither can a rational expression. Since a rational expression can have variables in its denominator, the results obtained in Examples 2 and 3 are valid only for values of the variable(s) that would not cause a denominator to be 0. E X A M P L E 2 Multiplying rational expressions Find the indicated products. a) 9x 10y 5y 3xy b) -8xy4 15z 3z3 2x5y3 Solution a) 9x 10y 3 3//x 2 5//y = Factor. 5y 3xy 5//y 3//xy = b) 6 y -8xy4 15z -2 2 2/xy4 /3 5z Factor. 3 5 3= 3z 2x y 3/z3 2/x5y3 = -20xy4z z3x5y3 Reduce. = -20y z2x4 Quotient rule Now do Exercises 9–18 dug84356_ch06a.qxd 394 9/14/10 12:39 PM Page 394 6-14 Chapter 6 Rational Expressions E X A M P L E 3 Multiplying rational expressions Find the indicated products. a) 2x - 2y 2x 4 x2 - y2 b) x 2 + 7x + 12 x 2x + 6 x2 - 16 c) a+b 8a2 2 6a a + 2ab + b2 Solution a) 2x - 2y 2/(x - y) 2x = 4 x2 - y2 2/ 2/ = b) c) 2/ x (x - y)(x + y) x x+y Factor. Reduce. x2 + 7x + 12 x (x + 3) (x + 4) = 2 2(x + 3) 2x + 6 x - 16 x = 2(x - 4) a+b 8a2 a + b 2 4a2 2 2= 6a a + 2ab + b 2 3a (a + b)2 4a = 3(a + b) x (x - 4)(x + 4) Factor. Reduce. Factor. Reduce. Now do Exercises 19–26 U3V Division of Rational Numbers By the definition of division, a quotient is found by multiplying the dividend by the re­ c d ciprocal of the divisor. If the divisor is a rational number , its reciprocal is simply . d c Division of Rational Numbers If b e 0, c e 0, and d e 0, then a c a d � = . b d b c E X A M P L E 4 Dividing rational numbers Find each quotient. a) 5 � 1 2 b) 6 3 � 7 14 b) 6 3 6 14 2 3/ 2 /7 � = = =4 7 14 7 3 /7 3/ Solution a) 5 � 1 = 5 2 = 10 2 Now do Exercises 27–34 U4V Division of Rational Expressions We divide rational expressions in the same way we divide rational numbers: Invert the divisor and multiply. dug84356_ch06a.qxd 9/14/10 12:39 PM Page 395 6-15 E X A M P L E 6.2 5 Multiplication and Division 395 Dividing rational expressions Find each quotient. 5 5 � a) 3x 6x b) x7 � (2x 2) 2 c) 4 - x2 x-2 � x2 + x x 2 - 1 Solution U Helpful Hint V A doctor told a nurse to give a patient half of the usual 500-mg dose of a drug. The nurse stated in court, “dividing in half means dividing by 1/2, which means multiply by 2.” The nurse was in court because the patient got 1000 mg instead of 250 mg and died (true story). Dividing a quantity in half and divid­ ing by one-half are not the same. a) 5 5 5 6x � = 3x 6x 3x 5 = Invert the divisor and multiply. 5/ 2 3//x Factor. /3/x /5 =2 b) x7 x7 1 � (2x 2) = 2 2 2x 2 = c) Divide out the common factors. x5 4 Invert and multiply. Quotient rule 4 - x2 x2 - 1 4 - x2 x-2 � 2 = 2 2 x +x x-2 x +x x -1 Invert and multiply. -1 = (2 - x) (2 + x) (x + 1)(x - 1) Factor. x(x + 1) x-2 = -1(2 + x)(x - 1) x 2-x = -1 x-2 = -1(x2 + x - 2) x Simplify. = -x 2 - x + 2 x Now do Exercises 35–48 We sometimes write division of rational expressions using the fraction bar. For example, we can write a+b 3 a + b 1 as —. � 1 3 6 6 No matter how division is expressed, we invert the divisor and multiply. E X A M P L E 6 Division expressed with a fraction bar Find each quotient. a+b 3 a) — 1 6 x2 - 1 2 b) — x-1 3 a2 + 5 3 c) — 2 dug84356_ch06a.qxd 396 9/20/10 11:45 AM Page 396 6-16 Chapter 6 Rational Expressions Solution a+b 3 a) — 1 6 a+b 3 1 6 Rewrite as division. a+b 6 3 1 Invert and multiply. a + b 2 3p Factor. p3 1 (a + b)2 Reduce. 2a + 2b x2 - 1 2 b) — x-1 3 x2 - 1 2 x-1 3 x2 - 1 3 2 x-1 Rewrite as division. Invert and multiply. (x - 1)(x + 1) 3 Factor. x-1 2 3x + 3 Reduce. 2 U Helpful Hint V In Section 6.5 you will see another technique for finding the quotients in Example 6. a2 + 5 3 c) — 2 a2 + 5 3 2 a2 + 5 1 3 2 Rewrite as division. a2 + 5 6 Now do Exercises 49–56 U5V Applications We saw in Section 6.1 that rational expressions can be used to represent rates. Note that there are several ways to write rates. For example, miles per hour is written mph, mi/hr, or mi. The last way is best when doing operations with rates because it helps us hr reconcile our answers. Notice how hours “cancels” when we multiply miles per hour and hours in Example 7, giving an answer in miles, as it should be. E X A M P L E 7 Using rational expressions with uniform motion Shasta drove 200 miles on I-10 in x hours before lunch. a) Write a rational expression for her average speed before lunch. b) She drives for 3 hours after lunch at the same average speed. Write a rational expression for her distance after lunch. Solution a) Because R D , T her rate before lunch is 200 miles x hours or 200 x mph. dug84356_ch06a.qxd 9/14/10 12:39 PM Page 397 6-17 6.2 Multiplication and Division b) Because D = R T, her distance after lunch is the product of 3 hours (her time): D= 200 x 397 mph (her rate) and 200 mi 600 3 hr = mi x hr x Now do Exercises 77–78 The amount of work completed is the product of rate and time, W = R T. So if a machine washes cars at the rate of 12 per hour and it works for 3 hours, the amount of W work completed is 36 cars washed. Note that the rate is given by R = T . E X A M P L E 8 Using rational expressions with work It takes x minutes to fill a bathtub. a) Write a rational expression for the rate at which the tub is filling. b) Write a rational expression for the portion of the tub that is filled in 10 minutes. Solution W a) The work completed in this situation is 1 tub being filled. Because R = T , the rate 1 tub 1 at which the tub is filling is x min or x tub/min. b) Because W = R T, the work completed in 10 minutes or the portion of the tub that 1 is filled in 10 minutes is the product of x tub/min (the rate) and 10 minutes (the time): W= 1 tub 10 10 min = tub x min x Now do Exercises 79–80 Warm-Ups ▼ Fill in the blank. 1. 2. expressions are multiplied by multiplying their numerators and multiplying their denominators. can be done before multiplying rational expressions. 3. To rational expressions, invert the divisor and multiply. True or false? 4. One-half of one-fourth is one-sixth. 2 5 10 5. = 3 7 21 6. The product of x-7 6 and is -2. 3 7-x 7. Dividing by 2 is equivalent to multiplying by 8. For any real number a, 9. 2 1 4 � = 3 2 3 a a �3= . 3 9 1 . 2 6.2 dug84356_ch06a.qxd 9/17/10 8:07 PM Page 398 Exercises U Study Tips V • Personal issues can have a tremendous effect on your progress in any course. If you need help, get it. • Most schools have counseling centers that can help you to overcome personal issues that are affecting your studies. 24. 12 4x + 10 8 35 3. 15 24 25. 16a + 8 2a2 + a - 1 4a2 - 1 5a2 + 5 25 56 6. 48 35 26. 6x - 18 4x 2 + 4x + 1 2x - 5x - 3 6x + 3 U1V Multiplication of Rational Numbers Perform the indicated operation. See Example 1. 2 1. 3 5 6 3 2. 4 3 8 4. 4 21 7. 24 2 5 12 51 5. 17 10 7 20 8. 3 35 10 Perform the indicated operation. See Example 4. 27. Perform the indicated operation. See Example 2. 2x 3 5x2 11. 6 5 4x 3y 7 21 2y 9x 12. 10 5 x2 10. 3 x 2 U3V Division of Rational Numbers U2V Multiplication of Rational Expressions 9. (4x2 + 20x + 25) 1 4 1 2 30. 32 33. 40 3 28. 1 4 1 6 1 2 29. 12 5 15 7 14 22 34. 9 9 31. 12 32. 13. 5a 3ab 12b 55a 14. 3m 35p 7p 6mp U4V Division of Rational Expressions 15. -2x6 21a2 7a5 6x 16. 5z3w -6y5 -9y3 20z9 35. x2 4 17. 15t3y5 24t5w3y2 20w7 18. 22x2y3z 37. 5x 2 3 39. 8m3 n4 41. y-6 2 43. x 2 + 4x + 4 8 (x + 2)3 16 Perform the indicated operation. See Example 5. 6x5 33y3z4 Perform the indicated operation. See Example 3. x 2 36. 3 2a2 6 2a 10x 21 38. 4u2 3v 14u 15v6 (12mn2) 40. 2p4 3q3 (4pq5) 6-y 6 42. 4-a 5 19. 2x + 2y 7 20. 3 a +a 21. 3a + 3b 10a 15 a2 - b2 44. a2 + 2a + 1 3 a2 - 1 a 22. b3 + b 10 2 5 b +b 45. t 2 + 3t - 10 t 2 - 25 (4t - 8) 46. w2 - 7w + 12 w2 - 4w 2 15 6x + 6y 2a + 2 6 23. (x2 - 6x + 9) 3 x-3 3 4 (w2 - 9) a2 - 16 3 2 5 15 2 dug84356_ch06a.qxd 9/14/10 12:39 PM Page 399 6-19 6.2 47. (2x2 - 3x - 5) � 48. (6y2 - y - 2) � 2x - 5 x-1 2y + 1 3y - 2 67. 2mn4 3m5n7 � 2 4 6mn2 mn 68. rt2 rt2 2� 32 rt rt 69. 3x2 + 16x + 5 x2 2 x 9x - 1 70. x 2 + 6x + 5 x4 x 3x + 3 71. a2 - 2a + 4 a2 - 4 72. w2 - 1 w-1 (w - 1)2 w2 + 2w + 1 73. 2x2 + 19x - 10 4x2 - 1 � 2 2 x - 100 2x - 19x - 10 74. x3 - 1 9x 2 + 9x + 9 � x2 - x x2 + 1 Perform the indicated operation. See Example 6. x - 2y 5 49. — 1 10 3m + 6n 50. —8— 3 4 399 (a + 2)3 2a + 4 x2 - 4 12 51. — x-2 6 6a2 + 6 5 52. — 6a + 6 5 x2 + 9 3 53. — 5 1 a-3 54. — 4 75. 9 + 6m + m2 9 - 6m + m2 76. 3x + 3w + bx + bw x2 - w2 x 2 - y2 55. — x-y 9 x 2 + 6x + 8 56. —— x+2 x+1 U5V Applications m2 - 9 m + mk + 3m + 3k 2 6 - 2b 9 - b2 Solve each problem. Answers could be rational expressions. Be sure to give your answers with appropriate units. See Examples 7 and 8. 77. Marathon run. Florence ran 26.2 miles in x hours in the Boston Marathon. a) Write a rational expression for her average speed. Miscellaneous Perform the indicated operation. 57. x-1 9 3 1-x 58. 2x - 2y 1 3 y-x 59. 3a + 3b 1 a 3 60. a-b 2b - 2a b 61. a 1 2 Multiplication and Division 62. 2 5 2g 3h 1 h 63. 6y � (2x) 3 64. 8x � (18x) 9 65. a3b4 a5b7 -2ab2 ab 66. -2a2 20a 3a2 15a3 b) She runs at the same average speed for 12 hour in the Cripple Creek Fun Run. Write a rational expression for her distance at Cripple Creek. 78. Driving marathon. Felix drove 800 miles in x hours on Monday. a) Write a rational expression for his average speed. b) On Tuesday he drove for 6 hours at the same average speed. Write a rational expression for his distance on Tuesday. dug84356_ch06a.qxd 400 9/14/10 12:39 PM Page 400 6-20 Chapter 6 Rational Expressions 82. Area of a triangle. If the base of a triangle is 8x + 16 yards and its height is 1 yards, then what is the area of the 79. Filling the tank. Chantal filled her empty gas tank in x minutes. a) Write a rational expression for the rate at which she filled her tank. x+2 triangle? b) Write a rational expression for the portion of the tank that is filled in 2 minutes. 1 — x+2 yd 8x + 16 yd 80. Magazine sales. Henry sold 120 magazine subscriptions in x days. a) Write a rational expression for the rate at which he sold the subscriptions. Figure for Exercise 82 Getting More Involved 83. Discussion Evaluate each expression. b) Suppose that he continues to sell at the same rate for 5 more days. Write a rational expression for the number of magazines sold in those 5 days. c) One-half of 81. Area of a rectangle. If the length of a rectangular flag is x meters and its width is 5 meters, then what is the area of x the rectangle? b) One-third of 4 4x 3 d) One-half of 3x 2 84. Exploration Let R = 5 — x 1 a) One-half of 4 6x2 + 23x + 20 2x + 5 . and H = 24x2 + 29x - 4 8x - 1 a) Find R when x = 2 and x = 3. Find H when x = 2 and x = 3. b) How are these values of R and H related and why? m xm Figure for Exercise 81 6.3 In This Section U1V Building Up the Denominator U2V Finding the Least Common Denominator 3 U V Converting to the LCD Finding the Least Common Denominator Every rational expression can be written in infinitely many equivalent forms. Because we can add or subtract only fractions with identical denominators, we must be able to change the denominator of a fraction. You have already learned how to change the denominator of a fraction by reducing. In this section, you will learn the opposite of reducing, which is called building up the denominator. U1V Building Up the Denominator To convert the fraction 2 3 into an equivalent fraction with a denominator of 21, we factor 21 as 21 = 3 7. Because 2 3 already has a 3 in the denominator, multiply dug84356_ch06a.qxd 9/14/10 12:39 PM Page 401 6-21 6.3 Finding the Least Common Denominator 401 the numerator and denominator of 2 by the missing factor 7 to get a denominator 3 of 21: 2 2 7 14 = = 3 3 7 21 For rational expressions the process is the same. To convert the rational expression 5 x+3 into an equivalent rational expression with a denominator of x2 - x - 12, first factor x2 - x - 12: x2 - x - 12 = (x + 3)(x - 4) From the factorization we can see that the denominator x + 3 needs only a factor of x - 4 to have the required denominator. So multiply the numerator and denominator by the missing factor x - 4: 5 5(x - 4) 5x - 20 = = x + 3 (x + 3)(x - 4) x2 - x - 12 E X A M P L E 1 Building up the denominator Build each rational expression into an equivalent rational expression with the indicated denominator. a) 3 = ? 12 b) 3 ? = w wx c) 2 ? 3= 3y 12y8 Solution a) Because 3 = 3, we get a denominator of 12 by multiplying the numerator and 1 denominator by 12: 3 3 12 36 3= = = 1 1 12 12 b) Multiply the numerator and denominator by x: 3 3 x 3x = = w w x wx c) Note that 12y8 = 3y3 4y5. So to build 3y3 up to 12y8 multiply by 4y5: 2 2 4y5 8y5 3= 3 5= 3y 3y 4y 12y8 Now do Exercises 1–20 In Example 2 we must factor the original denominator before building up the denominator. E X A M P L E 2 Building up the denominator Build each rational expression into an equivalent rational expression with the indicated denominator. a) 7 ? = 3x - 3y 6y - 6x b) x-2 ? = x + 2 x2 + 8x + 12 402 9/14/10 12:39 PM Page 402 6-22 Chapter 6 Rational Expressions U Helpful Hint V Notice that reducing and building up are exactly the opposite of each other. In reducing you remove a factor that is common to the numera­ tor and denominator, and in building up you put a common factor into the numerator and denominator. Solution a) Because 3x - 3y = 3(x - y), we factor -6 out of 6y - 6x. This will give a factor of x - y in each denominator: 3x - 3y = 3(x - y) 6y - 6x = -6(x - y) = -2 3(x - y) To get the required denominator, we multiply the numerator and denominator by -2 only: 7 7(-2) = 3x - 3y (3x - 3y)(-2) -14 = 6y - 6x b) Because x2 + 8x + 12 = (x + 2)(x + 6), we multiply the numerator and denominator by x + 6, the missing factor: x - 2 (x - 2)(x + 6) = x + 2 (x + 2)(x + 6) = x2 + 4x - 12 x2 + 8x + 12 Now do Exercises 21–32 CAUTION When building up a denominator, both the numerator and the denomina­ tor must be multiplied by the appropriate expression. U2V Finding the Least Common Denominator We can use the idea of building up the denominator to convert two fractions with different denominators into fractions with identical denominators. For example, 5 6 1 4 and can both be converted into fractions with a denominator of 12, since 12 = 2 6 and 12 = 3 4: 5 5 2 10 = = 6 6 2 12 1 1 3 3 = = 4 4 3 12 The smallest number that is a multiple of all of the denominators is called the least common denominator (LCD). The LCD for the denominators 6 and 4 is 12. To find the LCD in a systematic way, we look at a complete factorization of each denominator. Consider the denominators 24 and 30: 24 = 2 2 2 3 = 23 3 30 = 2 3 5 Any multiple of 24 must have three 2’s in its factorization, and any multiple of 30 must have one 2 as a factor. So a number with three 2’s in its factorization will have enough to be a multiple of both 24 and 30. The LCD must also have one 3 and one 5 in its factorization. We use each factor the maximum number of times it appears in either factorization. So the LCD is 23 3 5: 24 2 3 � dug84356_ch06a.qxd � 3 5 = 2 2 2 3 5 = 120 30 dug84356_ch06a.qxd 9/14/10 12:39 PM Page 403 6-23 6.3 Finding the Least Common Denominator 403 If we omitted any one of the factors in 2 2 2 3 5, we would not have a multiple of both 24 and 30. That is what makes 120 the least common denominator. To find the LCD for two polynomials, we use the same strategy. Strategy for Finding the LCD for Polynomials 1. Factor each denominator completely. Use exponent notation for repeated factors. 2. Write the product of all of the different factors that appear in the denominators. 3. On each factor, use the highest power that appears on that factor in any of the denominators. E X A M P L E 3 Finding the LCD If the given expressions were used as denominators of rational expressions, then what would be the LCD for each group of denominators? c) a2 + 5a + 6, a2 + 4a + 4 b) x3yz2, x5y2z, xyz5 a) 20, 50 Solution a) First factor each number completely: 50 = 2 52 20 = 22 5 The highest power of 2 is 2, and the highest power of 5 is 2. So the LCD of 20 and 50 is 22 52, or 100. b) The expressions x 3yz 2, x 5y 2z, and xyz 5 are already factored. For the LCD, use the highest power of each variable. So the LCD is x5y2z 5. c) First factor each polynomial. a2 + 5a + 6 = (a + 2)(a + 3) a2 + 4a + 4 = (a + 2)2 The highest power of (a + 3) is 1, and the highest power of (a + 2) is 2. So the LCD is (a + 3)(a + 2)2. Now do Exercises 33–46 U3V Converting to the LCD When adding or subtracting rational expressions, we must convert the expressions into expressions with identical denominators. To keep the computations as simple as possible, we use the least common denominator. E X A M P L E 4 Converting to the LCD Find the LCD for the rational expressions, and convert each expression into an equivalent rational expression with the LCD as the denominator. a) 4 2 , 9xy 15xz b) 5 1 3 , , 6x2 8x3y 4y2 Solution a) Factor each denominator completely: 9xy = 32xy 15xz = 3 5xz dug84356_ch06a.qxd 404 9/14/10 12:39 PM Page 404 6-24 Chapter 6 Rational Expressions U Helpful Hint V What is the difference between LCD, GCF, CBS, and NBC? The LCD for the denominators 4 and 6 is 12. The least common denominator is greater than or equal to both numbers.The GCF for 4 and 6 is 2. The greatest common fac­ tor is less than or equal to both num­ bers. CBS and NBC are TV networks. The LCD is 32 5xyz. Now convert each expression into an expression with this denominator. We must multiply the numerator and denominator of the first rational expression by 5z and the second by 3y: ⎫ ⎪ ⎬ Same denominator 2 2 3y 6y ⎪ = = 15xz 15xz 3y 45xyz ⎭ 4 4 5z 20z = = 9xy 9xy 5z 45xyz b) Factor each denominator completely: 6x 2 = 2 3x 2 8x3y = 23x3y 4y2 = 22y 2 The LCD is 23 3 x3y2 or 24x3y2. Now convert each expression into an expression with this denominator: 5 4xy2 20xy2 5 2= 2 2= 24x3y2 6x 6x 4xy 1 1 3y 3y = 3 = 3 8x y 8x y 3y 24x3y2 3 6x3 18x3 3 = = 4y2 4y2 6x3 24x3y2 Now do Exercises 47–58 E X A M P L E 5 Converting to the LCD Find the LCD for the rational expressions 5x x2 - 4 and 3 x2 + x - 6 and convert each into an equivalent rational expression with that denominator. Solution First factor the denominators: x2 - 4 = (x - 2)(x + 2) x2 + x - 6 = (x - 2)(x + 3) The LCD is (x - 2)(x + 2)(x + 3). Now we multiply the numerator and denominator of the first rational expression by (x + 3) and those of the second rational expression by (x + 2). Because each denominator already has one factor of (x - 2), there is no reason to multiply by (x - 2). We multiply each denominator by the factors in the LCD that are missing from that denominator: 5x 5x2 + 15x 5x(x + 3) = = x -4 (x - 2)(x + 2)(x + 3) (x - 2)(x + 2)(x + 3) 2 3 3x + 6 3(x + 2) = = x2 + x - 6 (x - 2)(x + 3)(x + 2) (x - 2)(x + 2)(x + 3) ⎫ ⎪ Same ⎬ denominator ⎪ ⎭ Now do Exercises 59–70 dug84356_ch06a.qxd 9/17/10 8:08 PM Page 405 6-25 6.3 405 ▼ Fill in the blank. 1. To the denominator of a fraction, we multiply the numerator and denominator by the same nonzero real number. 2. The is the smallest number that is a multiple of all denominators. 3. The LCD is the product of every factor that appears in the factorizations, raised to the power that appears on the factor. 2 2+5 5. -- = -3 3+5 6. The LCD for the denominators 25 · 3 and 24 · 32 is 25 · 32. 1 1 7. The LCD for -- and -- is 60. 10 6 1 1 8. The LCD for -- and -- is x2 - 4. x-2 x+2 1 1 - and -- is a2 - 1. 9. The LCD for a2 - 1 a-1 True or false? 2 2·5 4. -- = -3 3·5 Exercises U Study Tips V • Try changing subjects or tasks every hour when you study. The brain does not easily assimilate the same material hour after hour. • You will learn more from working on a subject one hour per day than seven hours on Saturday. U1V Building Up the Denominator Build each rational expression into an equivalent rational expression with the indicated denominator. See Example 1. 1 ? 1. - = 3 27 2 ? 2. - = 5 35 3 ? 3. -- = -4 16 3 ? 4. -- = -7 28 ? 5. 1 = -7- ? 6. 1 = -3x- ? 7. 2 = -6 ? 8. 5 = -12 5 ? 9. -- = -x ax x ? 10. -- = -3 3x ? 11. 7 = -2x ? 12. 6 = -4y 5 ? 13. -- = -b 3bt 7 ? 14. -- = -2ay 2ayz -9z ? 15. -- = -2aw 8awz -7yt ? 16. -- = -3x 18xyt ? 2 17. -- = -3 3a 15a 7b ? 18. --5 = --8 12c 36c 4 ? 19. --2 = -5xy 10x 2y 5 ? 5y2 20. -= -8x3z 24x5z3 6.3 Warm-Ups Finding the Least Common Denominator dug84356_ch06a.qxd 406 9/14/10 12:39 PM Page 406 6-26 Chapter 6 Rational Expressions Build each rational expression into an equivalent rational expression with the indicated denominator. See Example 2. 41. x2 - 16, x 2 + 8x + 16 5 ? 21. = x+3 2x + 6 43. x, x + 2, x - 2 44. y, y - 5, y + 2 45. x 2 - 4x, x 2 - 16, 2x 4 ? 22. = a-5 3a - 15 23. 42. x2 - 9, x 2 + 6x + 9 46. y, y 2 - 3y, 3y 5 ? = 2x + 2 -8x - 8 U3V Converting to the LCD 24. 3 ? = m - n 2n - 2m Find the LCD for the given rational expressions, and convert each rational expression into an equivalent rational expression with the LCD as the denominator. See Example 4. 25. 8a ? = 20b2 - 20b3 5b2 - 5b 47. 1 3 , 6 8 48. 5 3 , 12 20 26. 5x ? = -6x - 9 18x2 + 27x 49. 1 5 , 2x 6x 50. 3 1 , 5x 10x 27. 3 ? = 2 x+2 x -4 51. 2 1 , 3a 2b 52. y x , 4x 6y 28. a ? = 2 a+3 a -9 53. 5 3 , 84a 63b 29. 3x ? = 2 x + 2x + 1 x+1 54. -7x ? 30. = 2 2x - 3 4x - 12x + 9 31. 32. 55. 4b 6 , 75a 105ab 1 3 , 3x 2 2x 5 ? y-6 = 2 y-4 y + y - 20 56. ? z-6 = 2 z - 2z - 15 z+3 57. x y 1 , , 2 5 3 9y z 12x 6x y 58. 5 1 3b , 3, 6 12a b 14a 2ab3 U2V Finding the Least Common Denominator If the given expressions were used as denominators of rational expressions, then what would be the LCD for each group of denominators? See Example 3. See the Strategy for Finding the LCD for Polynomials box on page 403. 33. 12, 16 34. 28, 42 35. 12, 18, 20 36. 24, 40, 48 2 2 37. 6a , 15a 4 8a b , 5 6a2c 59. 2x 5x , x-3 x+2 60. 2a 3a , a-5 a+2 3 2 39. 2a b, 3ab , 4a b 40. 4m3nw, 6mn5w8, 9m6nw 3 9 Find the LCD for the given rational expressions, and convert each rational expression into an equivalent rational expression with the LCD as the denominator. See Example 5. 38. 18x , 20xy 6 3 dug84356_ch06a.qxd 9/14/10 12:39 PM Page 407 6-27 6.4 61. 4 5 , a-6 6-a 62. 4 5x , x - y 2y - 2x 63. x 5x , x 2 - 9 x 2 - 6x + 9 5x -4 , 2 64. 2 x - 1 x - 2x + 1 65. w +2 -2w , w2 - 2w - 15 w2 - 4w - 5 66. z-1 z+1 , z2 + 6z + 8 z2 + 5z + 6 67. Addition and Subtraction 69. 2 3 4 , , 2q 2 - 5q - 3 2q 2 + 9q + 4 q 2 + q - 12 70. p -3 2 , , 2p2 + 7p - 15 2p2 - 11p + 12 p2 + p - 20 407 Getting More Involved 71. Discussion Why do we learn how to convert two rational expressions into equivalent rational expressions with the same denominator? x 3 -5 , , 6x - 12 x 2 - 4 2x + 4 72. Discussion 68. Which expression is the LCD for 3 2b -5 , , 4b - 9 2b + 3 2b2 - 3b 2 3x - 1 22 3 x2(x + 2) 6.4 In This Section U1V Addition and Subtraction of Rational Numbers 2 U V Addition and Subtraction of Rational Expressions U3V Applications and 2x + 7 ? 2 32 x(x + 2)2 a) 2 3 x(x + 2) b) 36x(x + 2) c) 36x2(x + 2)2 d) 23 33x3(x + 2)2 Addition and Subtraction In Section 6.3, you learned how to find the LCD and build up the denominators of rational expressions. In this section, we will use that knowledge to add and subtract rational expressions with different denominators. U1V Addition and Subtraction of Rational Numbers We can add or subtract rational numbers (or fractions) only with identical denominators according to the following definition. dug84356_ch06a.qxd 408 9/14/10 12:39 PM Page 408 6-28 Chapter 6 Rational Expressions Addition and Subtraction of Rational Numbers If b e 0, then a c a+c a c a-c + = and - = . b b b b b b E X A M P L E 1 Adding or subtracting fractions with the same denominator Perform the indicated operations. Reduce answers to lowest terms. a) 1 7 + 12 12 b) 1 3 4 4 b) 1 3 -2 1 - = =4 4 4 2 Solution a) 1 7 8 4/ 2 2 + = = = 12 12 12 4/ 3 3 Now do Exercises 1–8 If the rational numbers have different denominators, we must convert them to equivalent rational numbers that have identical denominators and then add or subtract. Of course, it is most efficient to use the least common denominator (LCD), as in Example 2. E X A M P L E 2 Adding or subtracting fractions with different denominators Find each sum or difference. a) U Helpful Hint V Note how all of the operations with rational expressions are performed according to the rules for fractions. So keep thinking of how you perform operations with fractions, and you will improve your skills with fractions and with rational expressions. 3 7 + 20 12 b) 1 4 6 15 Solution a) Because 20 = 22 5 and 12 = 22 3, the LCD is 22 3 5, or 60. Convert each fraction to an equivalent fraction with a denominator of 60: 3 7 3 3 7 5 + = + 20 12 20 3 12 5 9 35 = + 60 60 44 = 60 4 11 = 4 15 11 = 15 Build up the denominators. Simplify numerators and denominators. Add the fractions. Factor. Reduce. b) Because 6 = 2 3 and 15 = 3 5, the LCD is 2 3 5 or 30: 1 4 1 4 = 6 15 2 3 3 5 = Factor the denominators. 1 5 4 2 Build up the denominators. 2 3 5 3 5 2 dug84356_ch06a.qxd 9/14/10 12:39 PM Page 409 6-29 6.4 Addition and Subtraction = 5 8 30 30 Simplify the numerators and denominators. = -3 30 Subtract. = -1 3 10 3 Factor. 1 10 Reduce. =- 409 Now do Exercises 9–18 U2V Addition and Subtraction of Rational Expressions Rational expressions are added or subtracted just like rational numbers. We can add or subtract only when we have identical denominators. All answers should be reduced to lowest terms. Remember to factor first when reducing, and then divide out any com­ mon factors. E X A M P L E 3 Rational expressions with the same denominator Perform the indicated operations and reduce answers to lowest terms. a) 2 4 + 3y 3y b) 2x 4 + x+2 x+2 c) 2x + 1 x2 + 2x (x - 1)(x + 3) (x - 1)(x + 3) Solution a) 2 4 6 + = Add the fractions. 3y 3y 3y = b) 2 y Reduce. 2x 4 2x + 4 + = x+2 x+2 x+2 = Add the fractions. 2(x + 2) Factor the numerator. x+2 =2 Reduce. x 2 + 2x - (2x + 1) x + 2x 2x + 1 = (x - 1)(x + 3) (x - 1)(x + 3) (x - 1)(x + 3) 2 c) Subtract the fractions. = x2 + 2x - 2x - 1 (x - 1)(x + 3) Remove parentheses. = x2 - 1 (x - 1)(x + 3) Combine like terms. = (x - 1)(x + 1) (x - 1)(x + 3) Factor. = x+1 x+3 Reduce. Now do Exercises 19–30 dug84356_ch06a.qxd 410 9/14/10 12:39 PM Page 410 6-30 Chapter 6 Rational Expressions CAUTION When subtracting a numerator containing more than one term, be sure to enclose it in parentheses, as in Example 3(c). Because that numerator is a binomial, the sign of each of its terms must be changed for the subtraction. In Example 4, the rational expressions have different denominators. E X A M P L E 4 Rational expressions with different denominators Perform the indicated operations. U Helpful Hint V You can remind yourself of the difference between addition and multiplication of fractions with a simple example: If you and your spouse each own 1/7 of Microsoft, then together you own 2/7 of Microsoft. If you own 1/7 of Microsoft, and give 1/7 of your stock to your child, then your child owns 1/49 of Microsoft. a) 5 2 + 2x 3 c) a+1 a-2 6 8 b) 4 2 + x 3y xy 3 Solution a) The LCD for 2x and 3 is 6x: 5 2 5 3 2 2x + = + 2x 3 2x 3 3 2x Build up both denominators to 6x. = 15 4x + 6x 6x Simplify numerators and denominators. = 15 + 4x 6x Add the rational expressions. b) The LCD is x 3y 3. 4 2 4 y2 2 x2 + 3= 3 2 + xy3 x 2 3 x y xy xy y Build up both denominators to the LCD. = 4y2 2x2 3 3+ 3 3 xy xy Simplify numerators and denominators. = 4y 2 + 2x 2 x3y3 Add the rational expressions. c) Because 6 = 2 3 and 8 = 23, the LCD is 23 3, or 24: a + 1 a - 2 (a + 1)4 (a - 2)3 = 6 8 6 4 8 3 Build up both denominators to the LCD 24. = 4a + 4 3a - 6 24 24 Simplify numerators and denominators. = 4a + 4 - (3a - 6) 24 Subtract the rational expressions. = 4a + 4 - 3a + 6 24 Remove the parentheses. = a + 10 24 Combine like terms. Now do Exercises 31–46 dug84356_ch06a.qxd 9/14/10 12:39 PM Page 411 6-31 E X A M P L E 6.4 5 Addition and Subtraction 411 Rational expressions with different denominators Perform the indicated operations: a) U Helpful Hint V 1 2 + 2 x - 9 x + 3x b) 2 4 2 5-a a-5 Solution 1 2 1 2 The LCD is x(x - 3)(x + 3). + = + x2 - 9 x2 + 3x (x - 3)(x + 3) x(x + 3) Needs x � a) � Once the denominators are factored as in Example 5(a), you can simply look at each denominator and ask, “What factor does the other denomi­ nator(s) have that is missing from this one?” Then use the missing factor to build up the denominator. Repeat until all denominators are identical, and you will have the LCD. Needs x - 3 = 1 x 2(x - 3) + (x - 3)(x + 3)x x(x + 3)(x - 3) = x 2x - 6 + x(x - 3)(x + 3) x(x - 3)(x + 3) = 3x - 6 x(x - 3)(x + 3) We usually leave the denominator in factored form. b) Because -1(5 - a) = a - 5, we can get identical denominators by multiplying only the first expression by -1 in the numerator and denominator: 4 2 4(-1) 2 = 5 - a a - 5 (5 - a)(-1) a - 5 = -4 2 a-5 a-5 = -6 -4 - 2 = -6 a-5 6 =a-5 Now do Exercises 47–64 In Example 6, we combine three rational expressions by addition and subtraction. E X A M P L E 6 Rational expressions with different denominators Perform the indicated operations. x+1 2x + 1 1 + x2 + 2x 6x + 12 6 Solution The LCD for x(x + 2), 6(x + 2), and 6 is 6x(x + 2). x+1 2x + 1 1 x+1 2x + 1 1 + - = + x 2 + 2x 6x + 12 6 x(x + 2) 6(x + 2) 6 = 6(x + 1) x(2x + 1) 1x(x + 2) + 6x(x + 2) 6x(x + 2) 6x(x + 2) Factor denominators. Build up to the LCD. dug84356_ch06a.qxd 412 9/14/10 12:39 PM Page 412 6-32 Chapter 6 Rational Expressions = 6x + 6 2x2 + x x2 + 2x + 6x(x + 2) 6x(x + 2) 6x(x + 2) Simplify numerators. = 6x + 6 + 2x2 + x - x2 - 2x 6x(x + 2) Combine the numerators. = x2 + 5x + 6 6x(x + 2) = (x + 3)(x + 2) 6x(x + 2) Factor. = x+3 6x Reduce. Combine like terms. Now do Exercises 65–70 U3V Applications We have seen how rational expressions can occur in problems involving rates. In Example 7, we see an applied situation in which we add rational expressions. E X A M P L E 7 Adding work Harry takes twice as long as Lucy to proofread a manuscript. Write a rational expression for the amount of work they do in 3 hours working together on a manuscript. Solution Let x = the number of hours it would take Lucy to complete the manuscript alone and 2x = the number of hours it would take Harry to complete the manuscript alone. Make a table showing rate, time, and work completed: Rate Time Work Lucy 1 msp x hr 3 hr 3 msp x Harry 1 msp 2x hr 3 hr 3 msp 2x Now find the sum of each person’s work. 3 3 2 3 3 + = + x 2x 2 x 2x = 6 3 + 2x 2x 9 2x So in 3 hours working together they will complete = 9 2x of the manuscript. Now do Exercises 81–86 dug84356_ch06a.qxd 9/17/10 8:08 PM Page 413 6-33 6.4 413 ▼ Fill in the blank. 1. We can rational expressions only if they have identical denominators. 2. We can any two rational expressions so that their denominators are identical. 3 4 29 5. -- + -- = -5 3 15 4 5 3 6. -- - -- = -5 7 35 5 3 7. -- + -- = 1 20 4 2 3 8. For any nonzero value of x, -- + 1 = --. x x 1 a+1 9. For any nonzero value of a, 1 + -- = --. a a 1 4a - 1 10. For any value of a, a - -- = --. 4 4 True or false? 1 1 2 3. -- + -- = -2 3 5 1 7 1 4. -- - -- = -2 12 12 Exercises U Study Tips V • When studying for a midterm or final, review the material in the order it was originally presented. This strategy will help you to see connections between the ideas. • Studying the oldest material first will give top priority to material that you might have forgotten. U1V Addition and Subtraction of Rational Numbers Perform the indicated operation. Reduce each answer to lowest terms. See Example 1. 1 1 1. -- + -10 10 1 3 2. -- + -8 8 7 1 3. -- - -8 8 4 1 4. -- - -9 9 1 5 5. -- - -6 6 3 7 6. --- - -8 8 7 1 7. --- + -8 8 9 3 8. --- + --20 20 ( ) Perform the indicated operation. Reduce each answer to lowest terms. See Example 2. 1 2 9. -- + -3 9 1 5 10. -- + -4 6 7 5 -11. -10 + 6 5 3 12. -6- + -1-0 7 5 13. -- + -16 18 7 4 14. -- + -6 15 1 9 15. -- - -8 10 2 5 16. -- - -15 12 ( ) 3 1 17. --- - --6 8 ( ) 1 1 18. --- - --5 7 6.4 Warm-Ups Addition and Subtraction dug84356_ch06a.qxd 414 9/14/10 12:39 PM Page 414 6-34 Chapter 6 Rational Expressions Perform the indicated operation. Reduce each answer to lowest terms. See Examples 5 and 6. U2V Addition and Subtraction of Rational Expressions Perform the indicated operation. Reduce each answer to lowest terms. See Example 3. 47. 1 1 + x x+2 19. 1 1 + 2x 2x 20. 1 2 + 3y 3y 48. 1 2 + y y+1 21. 3 7 + 2w 2w 22. 5x 7x + 3y 3y 49. 2 3 x+1 x 23. 15 3a + a+5 a+5 24. a + 7 9 - 5a + a-4 a-4 50. 1 2 a-1 a 25. q - 1 3q - 9 q-4 q-4 51. 2 1 + a-b a+b 26. 3-a a-5 3 3 52. 3 2 + x+1 x-1 4h - 3 h-6 27. h(h + 1) h(h + 1) 53. 3 4 x 2 + x 5x + 5 2t - 9 t-9 28. t(t - 3) t(t - 3) 54. 3 2 a + 3a 5a + 15 55. 2a a + a2 - 9 a - 3 56. x 3 + x 1 x -1 57. 4 4 + a-b b-a 58. 2 3 + x-3 3-x x2 - x - 5 1 - 2x 29. + (x + 1)(x + 2) (x + 1)(x + 2) 2x - 5 x - 2x + 1 + (x - 2)(x + 6) (x - 2)(x + 6) 2 30. Perform the indicated operation. Reduce each answer to lowest terms. See Example 4. 2 2 31. 1 1 + a 2a 32. 1 2 + 3w w 59. 3 2 2a - 2 1 - a 33. x x + 3 2 34. y y + 4 2 60. 5 3 2x - 4 2 - x 35. m +m 5 36. y + 2y 4 61. 1 3 x 2 - 4 x2 - 3x - 10 37. 1 2 + x y 38. 2 3 + a b 62. 2x 3x + 2 x2 - 9 x + 4x + 3 39. 3 1 + 2a 5a 40. 3 5 6y 8y 63. 3 4 + 2 x2 + x - 2 x + 2x - 3 41. w-3 w-4 9 12 42. y+4 y-2 10 14 64. x+4 x-1 + 2 x - x - 12 x + 5x + 6 43. b2 -c 4a 44. y + 3 7b 65. 1 1 1 + + a b c 45. 2 3 + wz2 w2z 46. 1 5 a5b ab3 66. 1 1 1 + 2+ 3 x x x 2 dug84356_ch06a.qxd 9/14/10 12:39 PM Page 415 6-35 6.4 415 Addition and Subtraction 2 1 1 + x x-1 x+2 U3V Applications 68. 1 2 3 + a a+1 a-1 69. 5 3 4 - + 2 3a - 9 2a a - 3a 81. Perimeter of a rectangle. Suppose that the length of a rectangle is 3 feet and its width is 5 feet. Find a rational x 2x expression for the perimeter of the rectangle. 70. 5 3 c-4 - 2 6c 4c + 2 2c + c 67. Solve each problem. See Example 7. 82. Perimeter of a triangle. The lengths of the sides of a triangle are 1, 1 , and Match each expression in (a)–(f) with the equivalent expression in (A)–(F). 71. a) d) A) D) 72. a) d) A) D) 1 +2 y 1 1 + y 2y 3 y y+2 y 1 -x x 1 -x x2 1 - x3 x2 1-x x2 1 2 + y y 2 e) +1 y 3 B) 2y y+2 E) 2y 1 1 b) - 2 x x 1 e) x x 1-x B) x x2 - 1 E) x b) 1 1 + y 2 y f) + 1 2 y+2 C) 2 2y + 1 F) y 1 c) -1 x 1 1 f) 2 x x 1 - x2 C) x x-1 F) x2 x 2x c) 1 — x Figure for Exercise 82 83. Traveling time. Janet drove 120 miles at x mph before 6:00 A.M. After 6:00 A.M., she increased her speed by 5 mph and drove 195 additional miles. Use the fact that T = D to complete the following table. R Rate 1 3 73. 2p + 8 2p After x+5 3 3 + 2 a2 + 3a + 2 a + 5a + 6 76. 4 12 + 2 w2 + w w - 3w 77. 2 1 - 2 b2 + 4b + 3 b + 5b + 6 78. 9 6 - 2 m -m-2 m -1 79. 3 3 2 - 2 t + 2t 2t t+2 80. 4 2 2 + + 2 3n n+1 n +n Time mi hr x 75. 2 — 3x 1 — 2x Before 3 3 2y 2y + 4 meters. Find a rational expression for the perimeter of the triangle. Perform the indicated operation. Reduce each answer to lowest terms. 74. 2 3x Distance 120 mi mi hr 195 mi Write a rational expression for her total traveling time. Evaluate the expression for x = 60. 84. Traveling time. After leaving Moose Jaw, Hanson drove 200 kilometers at x km/hr and then decreased his speed by 20 km/hr and drove 240 additional kilometers. Make a table like the one in Exercise 83. Write a rational expression for his total traveling time. Evaluate the expression for x = 100. 2 85. House painting. Kent can paint a certain house by himself in x days. His helper Keith can paint the same house by himself in x + 3 days. Suppose that they work together on the job for 2 days. To complete the table on the next page, use the fact that the work completed is the product of the dug84356_ch06a.qxd 416 9/14/10 12:39 PM Page 416 6-36 Chapter 6 Rational Expressions Rate Time Kent 1 job x day 2 days Keith 1 job 2 days x + 3 day Work rate and the time. Write a rational expression for the fraction of the house that they complete by working together for 2 days. Evaluate the expression for x = 6. Photo for Exercise 86 Getting More Involved 86. Barn painting. Melanie can paint a certain barn by herself in x days. Her helper Melissa can paint the same barn by herself in 2x days. Write a rational expression for the frac­ tion of the barn that they complete in one day by working together. Evaluate the expression for x = 5. 87. Writing Write a step-by-step procedure for adding rational expressions. 88. Writing Explain why fractions must have the same denominator to be added. Use real-life examples. Math at Work Gravity on the Moon Hundreds of years before humans even considered traveling beyond the earth, Isaac Newton established the laws of gravity. So when Neil Armstrong made the first human step onto the moon in 1969, he knew what amount of gravitational force to expect. Let’s see how he knew. Newton’s equation for the force of gravity between two objects is 2 F = G m1m , where m1 and m2 are the masses of the objects (in kilograms), d is d2 the distance (in meters) between the centers of the two objects, and G is the gravitational constant 6.67 � 10-11. To find the force of gravity for Armstrong on earth, use 5.98 � 1024 kg for the mass of the earth, 6.378 � 106 m for the radius of the earth, and 80 kg for Armstrong’s mass. We get F = 6.67 � 10-11 5.98 � 1024 kg 80 kg � 784 Newtons. (6.378 � 106 m)2 To find the force of gravity for Armstrong on the moon, use 7.34 � 1022 kg for the mass of the moon and 1.737 � 106 m for the radius of the moon. We get F = 6.67 � 10-11 7.34 � 1022 kg 80 kg � 130 Newtons. (1.737 � 106 m)2 So the force of gravity for Armstrong on the moon was about one-sixth of the force of gravity for Armstrong on earth. Fortunately, the moon is smaller than the earth. Walking on a planet much larger than the earth would present a real problem in terms of gravitational force. dug84356_ch06a.qxd 9/14/10 12:39 PM Page 417 6-37 Mid-Chapter Quiz Reduce to lowest terms. 36 1. 84 3. w2 - 1 2w + 2 2. 8x - 2 8 4. 2a2 - 10a + 12 6 - 3a Perform the indicated operation. 6 21 3xy2 5. 6. 7 10 5z 7. 9. a2 - 9 2a + 4 5a + 10 2a - 6 5 25 � 9 33 s2 s2 11. + 21 3 Sections 6.1 through 6.4 8. 10. 8x2z3 8y4 m2 - 8m + 7 12. � (m - 7) 2m 6.5 In This Section U1V Complex Fractions U2V Using the LCD to Simplify Complex Fractions 417 Chapter 6 13. 5 5 6 21 14. 5 4 3 + 2 ab ab 15. x 3x + x + 1 x2 + 2x + 1 16. y y y+5 y+2 17. 1 1 1 + + a b c b2 b6 � 21 3 3x - 9 x2 - 6x + 9 � 8 12 6.5 Miscellaneous. 3x - 6 18. What numbers(s) can’t be used in place of x in ? 2x + 1 19. Find the value of 3x - 6 when x = -2. 2x + 1 20. Find R(-1) if R(x) = 6x2 + 3 . 5x - 1 Complex Fractions In this section, we will use the idea of least common denominator to simplify complex fractions. Also we will see how complex fractions can arise in applications. Complex Fractions U3V Applications U1V Complex Fractions A complex fraction is a fraction having rational expressions in the numerator, denominator, or both. Consider the following complex fraction: 1 2 + 2 3 1 5 4 8 ← Numerator of complex fraction ← Denominator of complex fraction Since the fraction bar is a grouping symbol, we can compute the value of the numer­ ator, the value of the denominator, and then divide them, as shown in Example 1. dug84356_ch06b.qxd 418 9/14/10 12:44 PM Page 418 6-38 Chapter 6 Rational Expressions E X A M P L E 1 Simplifying complex fractions Simplify. 1 2 + 2 3 a) — 1 5 4 8 2 4 5 b) — 1 +3 10 Solution a) Combine the fractions in the numerator: 1 2 1 3 2 2 3 4 + + + 2 3 2 3 3 2 6 6 Combine the fractions in the denominator as follows: 1 5 1 2 5 2 5 4 8 4 2 8 8 8 Now divide the numerator by the denominator: 2 4 5 b) — 1 +3 10 7 1 2 + 2 3 6 7 — — 3 6 1 5 8 4 8 20 2 18 18 5 5 5 — — 30 31 1 5 + 10 10 10 3 8 7 6 31 10 18 10 5 31 7 6 3 8 8 3 56 18 28 9 36 31 Now do Exercises 1–12 U2V Using the LCD to Simplify Complex Fractions A complex fraction can be simplified by performing the operations in the numerator and denominator, and then dividing the results, as shown in Example 1. However, there is a better method. All of the fractions in the complex fraction can be eliminated in one step by multiplying by the LCD of all of the single fractions. The strategy for this method is detailed in the following box and illustrated in Example 2. Strategy for Simplifying a Complex Fraction 1. Find the LCD for all the denominators in the complex fraction. 2. Multiply both the numerator and the denominator of the complex fraction by the LCD. Use the distributive property if necessary. 3. Combine like terms if possible. 4. Reduce to lowest terms when possible. E X A M P L E 2 Using the LCD to simplify a complex fraction Use the LCD to simplify 1 2 + 2 3 —. 1 5 4 8 dug84356_ch06b.qxd 9/14/10 12:44 PM Page 419 6-39 6.5 Complex Fractions 419 U Calculator Close-Up V Solution You can check Example 2 with a calculator as shown here. The LCD of 2, 3, 4, and 8 is 24. Now multiply the numerator and denominator of the complex fraction by the LCD: 1 2 + 2 3 — 1 5 4 8 1 2 + 24 2 3 —— 1 4 Multiply the numerator and denominator by the LCD. 5 24 8 1 2 24 + 24 2 3 —— 1 5 24 24 8 4 Distributive property 12 + 16 6 15 Simplify. 28 9 28 9 Now do Exercises 13–20 CAUTION We simplify a complex fraction by multiplying the numerator and denomi­ nator of the complex fraction by the LCD. Do not multiply the numerator and denominator of each fraction in the complex fraction by the LCD. In Example 3 we simplify a complex fraction involving variables. E X A M P L E 3 A complex fraction with variables Simplify 2 1 x2 1 x . 1 2 U Helpful Hint V Solution When students see addition or subtraction in a complex fraction, they often convert all fractions to the same denominator. This is not wrong, but it is not necessary. Simply multiplying every fraction by the LCD eliminates the denominators of the original fractions. The LCD of the denominators x, x 2, and 2 is 2x 2: 1 2 x — 1 1 x2 2 2 1 (2x2) x —— 1 x2 Multiply the numerator and denominator by 2x2. 1 (2x2) 2 1 2 2x2 2x2 x —— 1 1 2x2 2x2 x2 2 Distributive property dug84356_ch06b.qxd 420 9/14/10 12:44 PM Page 420 6-40 Chapter 6 Rational Expressions 4x2 2x Simplify. 2 x2 The numerator of this answer can be factored, but the rational expression cannot be reduced. Now do Exercises 21–30 E X A M P L E 4 Simplifying a complex fraction Simplify 1 2 x 2 x+2 —— . 3 4 + 2 x x+2 Solution Because x 2 and 2 x are opposites, we can use (x 2)(x + 2) as the LCD. Multiply the numerator and denominator by (x 2)(x + 2): 1 2 1 2 (x 2)(x + 2) (x 2)(x + 2) x 2 x+2 x 2 x+2 —— ————— 3 4 3 4 (x 2)(x + 2) + (x 2)(x + 2) + 2 x x+2 2 x x+2 x + 2 2(x 2) 3( 1)(x + 2) + 4(x 2) x x 2 2 x 1 x + 2 2x + 4 3x 6 + 4x 8 Distributive property x+6 14 Combine like terms. Now do Exercises 31–46 U3V Applications As their name suggests, complex fractions arise in some fairly complex situations. E X A M P L E 5 Fast-food workers A survey of college students found that 1 of the female students had jobs and 2 of the male 3 2 students had jobs. It was also found that 1 of the female students worked in fast-food 4 restaurants and 1 of the male students worked in fast-food restaurants. If equal numbers of 6 male and female students were surveyed, then what fraction of the working students worked in fast-food restaurants? Solution Let x represent the number of males surveyed. The number of females surveyed is also x. The total number of students working in fast-food restaurants is 1 1 x + x. 4 6 dug84356_ch06b.qxd 9/14/10 12:44 PM Page 421 6-41 6.5 Complex Fractions 421 The total number of working students in the survey is 1 2 x + x. 2 3 So the fraction of working students who work in fast-food restaurants is 1 1 x+ x 4 6 2 1 x+ x 3 2 . The LCD of the denominators 2, 3, 4, and 6 is 12. Multiply the numerator and denominator by 12 to eliminate the fractions as follows: 1 1 x+ x 4 6 — 2 1 x+ x 3 2 1 1 x + x 12 6 4 —— Multiply numerator and denominator by 12. 2 1 x + x 12 3 2 3x + 2x 6x + 8x 5x 14x 5 14 So 5 14 Distributive property Combine like terms. Reduce. (or about 36%) of the working students work in fast-food restaurants. Now do Exercises 61–62 Warm-Ups ▼ Fill in the blank. 1. A fraction has fractions in its numerator, denominator, or both. 2. To simplify a complex fraction, you can multiply its and by the LCD of all of the fractions. True or false? 3. The LCD for the denominator 4, x, 6, and x2 is 12x3. 4. The LCD for the denominator a 6a 6b. b, 2b 2a, and 6 is 1 1 + 2 3 5. To simplify — , we multiply the numerator and 1 1 4 6 denominator by 12. 1 1 + 12 2 3 6+4 3 2 6. —— 1 1 12 4 6 1 1 + 2 3 7. — 1 1 4 6 5 6 — 1 12 1 x 2 8. For any real number x, — 1 x+ 3 2x 1 . 3x + 1 6.5 dug84356_ch06b.qxd 9/17/10 8:11 PM Page 422 Exercises U Study Tips V • Stay calm and confident. Take breaks when you study. Get 6 to 8 hours of sleep every night. • Keep reminding yourself that working hard throughout the semester will really pay off in the end. U1V Complex Fractions Simplify each complex fraction. See Example 1. 1 5 1 1 1 1 �� � �� �� � �� �� � �� 3 6 2 3 2 4 2. — 3. — 1. — 1 3 2 1 1 1 �� � �� �� � �� �� � �� 2 4 3 6 4 2 1 1 �� � �� 3 4 4. — 1 1 �� � �� 3 6 2 5 1 �� � �� � �� 5 6 2 5. —— 1 1 1 �� � �� � �� 2 3 15 2 2 1 �� � �� � �� 5 9 3 6. —— 1 1 2 �� � �� � �� 3 5 15 1 1 � �� 2 7. — 1 2 � �� 4 1 �� � 1 3 8. — 1 �� � 2 6 1 3 � �� 2 9. — 3 5 � �� 4 1 1 � �� 12 10. — 1 1 � �� 12 1 2 1 � �� � �� 6 3 11. —— 1 3 1 � �� � �� 15 10 2 1 3 � �� � �� 9 6 12. —— 5 1 �� � �� � 2 18 3 U2V Using the LCD to Simplify Complex Fractions Simplify each complex fraction. See Examples 2 and 3. See the Strategy for Simplifying a Complex Fraction box on page 418. 1 1 �� � �� 2 3 — 13. 1 1 �� � �� 2 4 1 1 �� � �� 4 3 — 14. 1 1 �� � �� 4 6 2 1 �� � �� 5 10 15. — 1 1 �� � �� 5 4 3 4 �� � �� 10 5 16. — 1 3 �� � �� 2 4 2 1 1 � �� � �� 3 2 17. —— 1 3 2 � �� � �� 3 2 1 3 3 � �� � �� 5 10 18. —— 6 3 2 � �� � �� 5 10 2 5 1 �� � �� � �� 3 6 2 19. —— 1 1 1 �� � �� � �� 6 3 2 2 3 7 �� � �� � �� 5 2 10 20. —— 1 1 1 �� � �� � �� 5 2 10 1 1 �� � �� a b 21. — 2 2 �� � �� a b 1 1 �� � �� x y 22. — 3 3 �� � �� x y 1 3 �� � �� a b 23. — 1 3 �� � �� b a 1 3 �� � �� x 2 24. — 3 1 �� � �� 4 x 3 5 � �� a 25. — 1 3 � �� a 3 4 � ��y 26. — 2 1 � �� y 1 2 �� � �� 2 x — 27. 1 3 � ��2 x 2 5 �� � �� a 3 — 28. 3 3 �� � ��2 a a 3 1 �� � �� 2b b 29. � 3 1 �� � ��2 4 b 3 4 �� � �� 2w 3w 30. � 1 5 �� � �� 4w 9w Simplify each complex fraction. See Example 4. 1 �� � 1 x�1 31. —— 3 �� � 3 x�1 2 �� � 1 x�3 32. —— 4 �� � 2 x�3 3 1 � �� y�1 33. —— 1 3 � �� y�1 1 2 � �� a�3 34. —— 1 3 � �� a�3 dug84356_ch06b.qxd 9/14/10 12:44 PM Page 423 6-43 6.5 4 x+ x 2 35. —— x+1 x x 2 x 6 x x 1 36. —— x + 15 x x 1 2x 9 6 53. — 2x 3 9 2x 4y 2 1 2 5 3 x 37. —— 1 2 x 3 x x 5 38. —— 3x 1 5 x 5 1 a 1 39. —— 2 3 1 a 1 2 3 9 x 40. —— 1 1 6 x 9 4 m m 3 41. 3 1 + m 3 m 1 y+3 1 2 42. 1 y 4 y 2 y+3 3 w 1 w+1 43. —— 4 5 + w+1 w 1 1 3 x+2 x+3 44. —— 2 3 + x+3 x+2 1 1 a b a+b 45. —— 1 1 + b a b+a 1 1 2+x 2 x 46. —— 1 1 x+2 x 2 Simplify each complex fraction. Reduce each answer to lowest terms. 1 1 4 + 1 3 y a2 48. — 47. —— y 3 2 8 1+ 3 y a a2 1 1 + 2 4x 49. — x 1 3 12x 1 1 + 9 3x 50. — x 1 9 x 1 5 2 + 3 3x x2 51. —— 1 3 3 x2 1 3 1 + 2 2x x2 52. —— 1 1 2 2x2 xy 55. — 3x 6y x3y a2 + 2a 24 a+1 57. —— a2 a 12 (a + 1)2 x x+1 59. —— 1 1 x2 1 x 1 Complex Fractions 423 a 5 12 54. — a+2 15 ab + b2 4ab5 56. — a+b 6a2b4 y2 3y 18 y2 4 58. —— y2 + 5y + 6 y 2 a 2 a b2 60. —— 1 1 + a+b a b U3V Applications Solve each problem. See Example 5. 61. Sophomore math. A survey of college sophomores showed that 5 of the males were taking a mathematics class and 3 of 6 4 the females were taking a mathematics class. One-third of the males were enrolled in calculus, and 1 of the females were 5 enrolled in calculus. If just as many males as females were surveyed, then what fraction of the surveyed students taking mathematics were enrolled in calculus? Rework this prob­ lem assuming that the number of females in the survey was twice the number of males. 62. Commuting students. At a well-known university, 1 of the 4 undergraduate students commute, and 1 of the graduate 3 students commute. One-tenth of the undergraduate students drive more than 40 miles daily, and 1 of the graduate 6 students drive more than 40 miles daily. If there are twice as many undergraduate students as there are graduate students, then what fraction of the commuters drive more than 40 miles daily? Photo for Exercise 62 dug84356_ch06b.qxd 424 9/14/10 12:44 PM Page 424 6-44 Chapter 6 Rational Expressions Getting More Involved 64. Discussion A complex fraction can be simplified by writing the numerator and denominator as single fractions and then dividing them or by multiplying the numerator and denominator by the LCD. Simplify the complex fraction 63. Exploration Simplify 1 1 1 —, , and . 1 1 1 1+2 1+ 1+ 1 1 1+ 1+2 1 1+ 2 a) Are these fractions getting larger or smaller as the fractions become more complex? b) Continuing the pattern, find the next two complex 4 6 xy2 xy — 2 4 2+ 2 x xy by using each of these methods. Compare the number of steps used in each method, and determine which method requires fewer steps. fractions and simplify them. c) Now what can you say about the values of all five complex fractions? 6.6 In This Section U1V Equations with Rational Expressions 2 U V Extraneous Solutions Solving Equations with Rational Expressions Many problems in algebra can be solved by using equations involving rational expressions. In this section you will learn how to solve equations that involve rational expressions, and in Sections 6.7 and 6.8 you will solve problems using these equations. U1V Equations with Rational Expressions We solved some equations involving fractions in Section 2.3. In that section, the equations had only integers in the denominators. Our first step in solving those equations was to multiply by the LCD to eliminate all of the denominators. E X A M P L E 1 Integers in the denominators Solve 1 x 2 2 3 1 . 6 Solution U Helpful Hint V Note that it is not necessary to convert each fraction into an equiva­ lent fraction with a common denominator here. Since we can multiply both sides of an equation by any expression we choose, we choose to multiply by the LCD. This tactic eliminates the fractions in one step. The LCD for 2, 3, and 6 is 6. Multiply each side of the equation by 6: 1 x 2 2 3 1 x 2 6 2 3 6 2 1 2 3 6� Original equation 6 1 6 Multiply each side by 6. 2 �6 1 �6 Distributive property 2) 1 2x + 4 1 x �3 2(x 3 1 6 2x x Simplify. Distributive property 6 3 Subtract 7 from each side. Divide each side by 2. dug84356_ch06b.qxd 9/14/10 12:44 PM Page 425 6-45 6.6 U Helpful Hint V Check x Always check your solution in the original equation by calculating the value of the left-hand side and the value of the right-hand side. If they are the same, your solution is correct. Solving Equations with Rational Expressions 425 3 in the original equation: 1 2 3 2 1 2 3 1 3 3 6 2 6 1 6 Since the right-hand side of the equation is 1, you can be sure that the solution to the 6 equation is 3. Now do Exercises 1–12 CAUTION When a numerator contains a binomial, as in Example 1, the numer­ ator must be enclosed in parentheses when the denominator is eliminated. To solve an equation involving rational expressions, we usually multiply each side of the equation by the LCD for all the denominators involved, just as we do for an equation with fractions. E X A M P L E 2 Variables in the denominators Solve 1 + 1 x 6 1 . 4 Solution We multiply each side of the equation by 12x, the LCD for 4, 6, and x : 1 1 + x 6 12x 1 1 + x 6 2 1 1 +� 12x 6� �x 12x� 12 + 2x 12 1 4 Original equation 12x 3 � 12x 1 4 Multiply each side by 12x. 1 Distributive property 4� 3x Simplify. x Subtract 2x from each side. Check that 12 satisfies the original equation: 1 1 + 12 6 1 2 + 12 12 3 12 1 4 The solution to the equation is 12. Now do Exercises 13–24 E X A M P L E 3 An equation with two solutions Solve the equation 100 + x 100 x+5 9. dug84356_ch06b.qxd 426 9/14/10 12:44 PM Page 426 6-46 Chapter 6 Rational Expressions Solution The LCD for the denominators x and x + 5 is x(x + 5): 100 100 + 9 x x+5 100 100 x(x + 5) + x(x + 5) x(x + 5)9 +5 x x 9x + 25 (x + 5)100 + x(100) (x2 + 5x)9 100x + 500 + 100x 9x2 + 45x 500 + 200x 9x2 + 45x 0 or 25 9 x x 25 9 A check will show that both Multiply each side by x(x + 5). All denominators are eliminated. Simplify. 0 9x2 155x 500 Get 0 on one side. 0 (9x + 25)(x 20) Factor. 20 0 x 20 or Original equation Zero factor property and 20 satisfy the original equation. Now do Exercises 25–32 U2V Extraneous Solutions In a rational expression, we can replace the variable only by real numbers that do not cause the denominator to be 0. When solving equations involving rational expressions, we must check every solution to see whether it causes 0 to appear in a denominator. If a number causes the denominator to be 0, then it cannot be a solution to the equa­ tion. A number that appears to be a solution but causes 0 in a denominator is called an extraneous solution. Since a solution to an equation is sometimes called a root to the equation, an extraneous solution is also called an extraneous root. E X A M P L E 4 An equation with an extraneous solution Solve the equation 1 x x 2 2x 4 + 1. Solution Because the denominator 2x 2(x 2) 1 x 2 2(x 4 factors as 2(x 2) 2 x + 2x 2 3x 6 3x 2 x x 2(x 4 2) + 2(x 2), the LCD is 2(x 2) 1 2). Multiply each side of the original equation by 2(x 2). Simplify. 4 Check 2 in the original equation: 1 2 +1 2 2 2 2 4 The denominator 2 2 is 0. So 2 does not satisfy the equation, and it is an extraneous solution. The equation has no solutions. Now do Exercises 33–36 dug84356_ch06b.qxd 9/14/10 12:44 PM Page 427 6-47 6.6 427 Solving Equations with Rational Expressions If the denominators of the rational expressions in an equation are not too compli­ cated, you can tell at a glance which numbers cannot be solutions. For example, the 2 3 x 2 equation x + x 1 x + 5 could not have 0, 1, or 5 as a solution. Any solution to this equation must be in the domain of all three of the rational expressions in the equation. E X A M P L E 5 Another extraneous solution Solve the equation 1 + x 1 x x x 3 2 . 3 Solution The LCD for the denominators x and x 3 is x(x 1 x 2 1 + x x 3 x 3 1 x 1 x(x 3) + x(x 3) x(x 3) x x x 3 x 3+x 2x x 3 0 or x 3 or x 3 x(x 2) 2 2x x 0 x 4x + 3 0 (x 3)(x 1 0 x 1 2 3): Original equation 2 3 Multiply each side by x(x 3). 1) If x 3, then the denominator x 3 has a value of 0. If x satisfied. The only solution to the equation is 1. 1, the original equation is Now do Exercises 37–40 CAUTION Always be sure to check your answers in the original equation to determine whether they are extraneous solutions. Warm-Ups ▼ Fill in the blank. 1. The usual first step in solving an equation involving rational expressions is to multiply by the . 2. An solution is a number that appears to be a solution but does not check in the original equation. 6. To solve 7. To solve 3 5 + x x 2 1 x 1 2 , multiply each side by 3x2 3 1 , multiply each side by x+1 +2 x2 – 1. True or false? 2 3. To solve x 8x, we divide each side by x. 4. An extraneous solution is an irrational number. 5. Both 0 and 2 satisfy 3 5 + x x 2 2 . 3 8. The solution set to 9. The solution set to 1 x 1 1 1 + x 2 +2 1 is { 1, 1}. x +1 3 is {4}. x 6x. 6.6 dug84356_ch06b.qxd 9/17/10 8:12 PM Page 428 Exercises U Study Tips V • The last couple of weeks of the semester is not the time to slack off. This is the time to double your efforts. • Make a schedule and plan every hour of your time. U1V Equations with Rational Expressions Solve each equation. See Example 3. Solve each equation. See Example 1. x 5 25. -- = -2 x+3 x 4 26. -- = -3 x+1 x 6 27. -- = -x+1 x+7 x 2 28. -- = -x+3 x-3 2 1 1 29. -- = -- + -x+1 x 6 1 1 3 30. -- - -- = -w + 1 2w 40 x x 1. -- + 1 = -2 4 x x 2. -- + 2 = -3 6 x x 3. -- - 5 = -- - 7 3 2 x x x 4. -- - -- = -- - 11 3 2 5 y 2 y 1 5. -- - -- = -- + -5 3 6 3 z 5 z 3 6. -- + -- = -- - -6 4 2 4 t 3 t-4 7. -- - -- = -4 3 12 4 v-1 v-5 8. -- - -- = -5 10 30 x x + 1 x + 15 9. -- + -- = -3 4 12 x x + 4 6x + 5 10. -- + -- = -8 12 24 1 1 w + 10 w+1 11. -- - -- = -- - -10 5 15 6 a-1 1 a+4 31. -+ -- = -2 a -4 a-2 a+2 1 b + 17 b-2 32. -- -- = -b2 - 1 b + 1 b - 1 U2V Extraneous Solutions q q - 1 13 q + 1 12. -- - -- = -- - -5 2 20 4 Solve each equation. Watch for extraneous solutions. See Examples 4 and 5. Solve each equation. See Example 2. 1 x 2 33. -- + -- = -x-1 x x-1 1 1 13. -- + -- = 3 x 2 2 3 14. -- + -- = 5 x 4 1 2 15. -- + -- = 7 x x 5 6 16. -- + -- = 12 x x 1 1 3 17. -- + -- = -x 2 4 3 1 5 18. -- + -- = -x 4 8 2 1 7 19. -- + -- = -3x 2x 24 1 1 1 20. -- - -- = -6x 8x 72 1 a-2 a+2 21. -- + -- = -2 a 2a 1 1 b-1 3 22. -- + -- = -- + -b 5 5b 10 4 3 x 1 34. -- + -- = -- - -x x-3 x-3 3 5 2 x-1 35. -- + -- = -x+2 x-3 x-3 6 7 y-1 36. -- + -- = -y-2 y-8 y-8 3y 6 37. 1 + -- = -y-2 y-2 5 y+7 38. -- = -- + 1 y - 3 2y - 6 1 k+3 1 k-1 23. -- - -- = -- - -3 6k 3k 2k z 1 2z + 5 39. -- - -- = -z + 1 z + 2 z2 + 3z + 2 3 p + 3 2p - 1 5 24. -- - -- = -- - -6 p 3p 2p 1 z 7 40. -- - -- = z - 2 z + 5 z2 + 3z - 10 dug84356_ch06b.qxd 9/14/10 12:44 PM Page 429 6-49 6.7 Applications of Ratios and Proportions Miscellaneous Applications Solve each equation. Solve each problem. a 4 w 43. 6 5 45. x x 47. 41. 49. 51. 53. 55. 56. 57. 58. 59. 60. 61. 62. 5 2 3w 11 x 5 3 x 3 5 x 1 x x+2 x+2 1 1 3 + 2x 4 x 2 2 3 2 a2 a 6 a2 4 4 1 25 c 2 2 c c+6 3 1 10 x + 1 1 x x2 1 1 3 4 + x2 9 x + 3 x 3 3 5 1 x 2 x + 3 x2 + x 6 3 1 1 2x + 4 x + 2 3x + 1 5 1 1 2m + 6 m + 1 m + 3 2t 1 3t 1 t + 3t + 3 6t + 6 t + 1 4w 1 w 1 w 1 3w + 6 3 w+2 42. 44. 46. 48. 50. 52. 54. y 6 3 5 2m 3m 3 2 3 x x 3 a+4 a+4 2 a 3 w w+2 w+2 7 1 4 3x 9 x 3 3 8 6 a2 + a 6 a2 9 U1V Ratios U2V Proportions 63. Lens equation. The focal length f for a camera lens is related to the object distance o and the image distance i by the formula 1 1 1 + . f o i See the accompanying figure. The image is in focus at distance i from the lens. For an object that is 600 mm from a 50-mm lens, use f 50 mm and o 600 mm to find i. o i Figure for Exercise 63 64. Telephoto lens. Use the formula from Exercise 63 to find the image distance i for an object that is 2,000,000 mm from a 250-mm telephoto lens. Photo for Exercise 64 6.7 In This Section 429 Applications of Ratios and Proportions In this section, we will use the ideas of rational expressions in ratio and proportion problems. We will solve proportions in the same way we solved equations in Section 6.6. U1V Ratios In Chapter 1 we defined a rational number as the ratio of two integers. We will now give a more general definition of ratio. If a and b are any real numbers (not just integers), with b � 0, then the expression a is called the ratio of a and b or the ratio of a to b. b dug84356_ch06b.qxd 430 9/14/10 12:44 PM Page 430 6-50 Chapter 6 Rational Expressions The ratio of a to b is also written as a: b. A ratio is a comparison of two numbers. Some examples of ratios are 1 3 4.2 3.6 100 4 , , —, , and . 1 4 2.1 5 1 2 Ratios are treated just like fractions. We can reduce ratios, and we can build them up. We generally express ratios as ratios of integers. When possible, we will convert a ratio into an equivalent ratio of integers in lowest terms. E X A M P L E 1 Finding equivalent ratios Find an equivalent ratio of integers in lowest terms for each ratio. 1 4 b) — 1 2 4.2 a) 2.1 c) 3.6 5 Solution a) Because both the numerator and the denominator have one decimal place, we will multiply the numerator and denominator by 10 to eliminate the decimals: 4.2 2.1 4.2(10) 2.1(10) 42 21 21 2 21 1 2 Do not omit the 1 in a ratio. 1 So the ratio of 4.2 to 2.1 is equivalent to the ratio 2 to 1. b) This ratio is a complex fraction. We can simplify this expression using the LCD method as shown in Section 6.5. Multiply the numerator and denominator of this ratio by 4: 1 1 4 4 4 1 1 1 2 4 2 2 c) We can get a ratio of integers if we multiply the numerator and denominator by 10. 3.6 5 3.6(10) 5(10) 18 25 36 50 Reduce to lowest terms. Now do Exercises 1–16 In Example 2, a ratio is used to compare quantities. E X A M P L E 2 Nitrogen to potash In a 50-pound bag of lawn fertilizer there are 8 pounds of nitrogen and 12 pounds of potash. What is the ratio of nitrogen to potash? dug84356_ch06b.qxd 9/14/10 12:44 PM Page 431 6-51 6.7 Applications of Ratios and Proportions 431 Solution The nitrogen and potash occur in this fertilizer in the ratio of 8 pounds to 12 pounds: 8 12 2 4� 3 4� 2 3 So the ratio of nitrogen to potash is 2 to 3. Now do Exercises 17–18 E X A M P L E 3 Males to females In a class of 50 students, there were exactly 20 male students. What was the ratio of males to females in this class? Solution Because there were 20 males in the class of 50, there were 30 females. The ratio of males to females was 20 to 30, or 2 to 3. Now do Exercises 19–20 Ratios give us a means of comparing the size of two quantities. For this reason the numbers compared in a ratio should be expressed in the same units. For example, if one dog is 24 inches high and another is 1 foot high, then the ratio of their heights is 2 to 1, not 24 to 1. E X A M P L E 4 Quantities with different units What is the ratio of length to width for a poster with a length of 30 inches and a width of 2 feet? Solution Because the width is 2 feet, or 24 inches, the ratio of length to width is 30 to 24. Reduce as follows: 30 5 6 5 24 4 6 4 So the ratio of length to width is 5 to 4. Now do Exercises 21–24 U2V Proportions A proportion is any statement expressing the equality of two ratios. The statement a c or a:b c:d b d is a proportion. In any proportion the numbers in the positions of a and d shown here are called the extremes. The numbers in the positions of b and c as shown are called the means. In the proportion 30 5 , 24 4 the means are 24 and 5, and the extremes are 30 and 4. dug84356_ch06b.qxd 432 9/14/10 12:44 PM Page 432 6-52 Chapter 6 Rational Expressions If we multiply each side of the proportion a c b d by the LCD, bd, we get U Helpful Hint V The extremes-means property or cross-multiplying is nothing new. You can accomplish the same thing by multiplying each side of the equa­ tion by the LCD. a bd b c bd d a d b c. or We can express this result by saying that the product of the extremes is equal to the product of the means. We call this fact the extremes-means property or crossmultiplying. Extremes-Means Property (Cross-Multiplying) Suppose a, b, c, and d are real numbers with b � 0 and d � 0. If a b c , then d ad bc. We use the extremes-means property to solve proportions. E X A M P L E 5 Using the extremes-means property Solve the proportion 3 x 5 x+5 for x. Solution Instead of multiplying each side by the LCD, we use the extremes-means property: 3 x 3(x + 5) 5 x+5 5x 3x + 15 5x 15 2x 15 2 x Original proportion Extremes-means property Distributive property Check: 3 15 2 3 2 15 2 5 5 5 2 2 5 15 25 5 25 +5 2 2 So 15 is the solution to the equation or the solution to the proportion. 2 Now do Exercises 25–38 dug84356_ch06b.qxd 9/14/10 12:45 PM Page 433 6-53 E X A M P L E 6.7 6 Applications of Ratios and Proportions 433 Solving a proportion The ratio of men to women at Brighton City College is 2 to 3. If there are 894 men, then how many women are there? Solution Because the ratio of men to women is 2 to 3, we have Number of men Number of women 2 . 3 If x represents the number of women, then we have the following proportion: 894 x 2 3 2x 2682 Extremes-means property x 1341 The number of women is 1341. Now do Exercises 39–42 Note that any proportion can be solved by multiplying each side by the LCD as we did when we solved other equations involving rational expressions. The extremes-means property gives us a shortcut for solving proportions. E X A M P L E 7 Solving a proportion In a conservative portfolio the ratio of the amount invested in bonds to the amount invested in stocks should be 3 to 1. A conservative investor invested $2850 more in bonds than she did in stocks. How much did she invest in each category? Solution Because the ratio of the amount invested in bonds to the amount invested in stocks is 3 to 1, we have Amount invested in bonds Amount invested in stocks 3 . 1 If x represents the amount invested in stocks and x + 2850 represents the amount invested in bonds, then we can write and solve the following proportion: x + 2850 x 3 1 3x x + 2850 Extremes-means property 2x 2850 x 1425 x + 2850 4275 So she invested $4275 in bonds and $1425 in stocks. As a check, note that these amounts are in the ratio of 3 to 1. Now do Exercises 43–46 dug84356_ch06b.qxd 434 9/14/10 12:45 PM Page 434 6-54 Chapter 6 Rational Expressions Example 8 shows how conversions from one unit of measurement to another can be done by using proportions. E X A M P L E 8 Converting measurements There are 3 feet in 1 yard. How many feet are there in 12 yards? Solution Let x represent the number of feet in 12 yards. There are two proportions that we can write to solve the problem: 3 feet x feet 1 yard 12 yards 3 feet 1 yard x feet 12 yards The ratios in the second proportion violate the rule of comparing only measurements that are expressed in the same units. Note that each side of the second proportion is actually the ratio 1 to 1, since 3 feet 1 yard and x feet 12 yards. For doing conversions we can use ratios like this to compare measurements in different units. Applying the extremesmeans property to either proportion gives 3 12 x 1, or x 36. So there are 36 feet in 12 yards. Now do Exercises 47–50 Warm-Ups ▼ Fill in the blank. 1. A is a comparison of two numbers. 2. A is an equation that expresses the equality of two ratios. a c 3. In , b and c are the . b d a c 4. In , a and d are the . b d a c 5. The property says that if , then b d ad bc. True or false? 6. The ratio of 40 men to 30 women can be expressed as the ratio 4 to 3. 7. The ratio of 3 feet to 2 yards can be expressed as the ratio 3 to 2. 8. The ratio of 1.5 to 2 is equivalent to the ratio 3 to 4. 9. The product of the extremes is equal to the product of the means. 2 3 10. If , then 5x 6. x 5 11. If 4 of the 12 members of the supreme council are women, then the ratio of men to women is 1 to 3. Exercises U Study Tips V • Get an early start studying for your final exams. • If you have several final exams, it can be difficult to find the time to prepare for all of them in the last couple of days. U1V Ratios For each ratio, find an equivalent ratio of integers in lowest terms. See Example 1. 4 1.  6 10 2.  20 200 3.  150 1000 4.  200 2.5 5.  3.5 4.8 6.  1.2 0.32 7.  0.6 0.05 8.  0.8 35 9.  10 3 88 4.5 10.  11.  12.  33 7 2.5 1 2   2 3 5 13.  14.  15.  1 3 1    5 4 3 4 16.  1  4 Find a ratio for each of the following, and write it as a ratio of integers in lowest terms. See Examples 2–4. 17. Men and women. Find the ratio of men to women in a bowling league containing 12 men and 8 women. 19. Smokers. A life insurance company found that among its last 200 claims, there were six dozen smokers. What is the ratio of smokers to nonsmokers in this group of claimants? 20. Hits and misses. A woman threw 60 darts and hit the target a dozen times. What is her ratio of hits to misses? 21. Violence and kindness. While watching television for one week, a consumer group counted 1240 acts of violence and 40 acts of kindness. What is the violence to kindness ratio for television, according to this group? 22. Length to width. What is the ratio of length to width for the rectangle shown? L W 2.5 ft 48 in. Figure for Exercise 22 23. Rise to run. What is the ratio of rise to run for the stairway shown in the figure? 18. Coffee drinkers. Among 100 coffee drinkers, 36 said that they preferred their coffee black and the rest did not prefer their coffee black. Find the ratio of those who prefer black coffee to those who prefer nonblack coffee. Rise Run 8 in. 1 ft Figure for Exercise 23 24. Rise and run. If the rise is 3 and the run is 5, then what is 2 the ratio of the rise to the run? U2V Proportions Solve each proportion. See Example 5. Photo for Exercise 18 4 2 25.    x 3 9 3 26.    x 2 6.7 dug84356_ch06b.qxd 9/17/10 8:12 PM Page 435 dug84356_ch06b.qxd 436 27. a 2 Page 436 1 5 28. 3 x 30. 5 9 x 2 x 7 5 33. 10 x 35. a a+1 m m 37. 12:45 PM 6-56 Chapter 6 Rational Expressions 29. 31. 9/14/10 43. Basketball blowout. As the final buzzer signaled the end of the basketball game, the Lions were 34 points ahead of the Tigers. If the Lions scored 5 points for every 3 scored by the Tigers, then what was the final score? 3 4 3 4 5 x 4 32. x+1 x 3 a+3 a 36. c+3 c 1 m 3 m+4 38. c+2 c 3 h h 44. The golden ratio. The ancient Greeks thought that the most pleasing shape for a rectangle was one for which the ratio of the length to the width was approximately 8 to 5, the golden ratio. If the length of a rectangular painting is 2 ft longer than its width, then for what dimensions would the length and width have the golden ratio? 2 x x+1 2 34. 34 x + 12 1 2 b 3 h 3 h 9 Use a proportion to solve each problem. See Examples 6–8. 39. New shows and reruns. The ratio of new shows to reruns on cable TV is 2 to 27. If Frank counted only eight new shows one evening, then how many reruns were there? 45. Automobile sales. The ratio of sports cars to luxury cars sold in Wentworth one month was 3 to 2. If 20 more sports cars were sold than luxury cars, then how many of each were sold that month? 46. Foxes and rabbits. The ratio of foxes to rabbits in the Deerfield Forest Preserve is 2 to 9. If there are 35 fewer foxes than rabbits, then how many of each are there? 47. Inches and feet. If there are 12 inches in 1 foot, then how many inches are there in 7 feet? 40. Fast food. If four out of five doctors prefer fast food, then at a convention of 445 doctors, how many prefer fast food? 48. Feet and yards. If there are 3 feet in 1 yard, then how 41. Voting. If 220 out of 500 voters surveyed said that they would vote for the incumbent, then how many votes could the incumbent expect out of the 400,000 voters in the state? 49. Minutes and hours. If there are 60 minutes in 1 hour, then how many minutes are there in 0.25 hour? many yards are there in 28 feet? 50. Meters and kilometers. If there are 1000 meters in 1 kilometer, then how many meters are there in 2.33 kilometers? 51. Miles and hours. If Alonzo travels 230 miles in 3 hours, then how many miles does he travel in 7 hours? 52. Hiking time. If Evangelica can hike 19 miles in 2 days on the Appalachian Trail, then how many days will it take her to hike 63 miles? Photo for Exercise 41 42. New product. A taste test with 200 randomly selected people found that only three of them said that they would buy a box of new Sweet Wheats cereal. How many boxes could the manufacturer expect to sell in a country of 280 million people? 53. Force on basketball shoes. The force exerted on shoe soles in a jump shot is proportional to the weight of the person jumping. If a 70-pound boy exerts a force of 980 pounds on his shoe soles when he returns to the court after a jump, then what force does a 6 ft 8 in. professional ball player weighing 280 pounds exert on the soles of his shoes when he returns to the court after a jump? Use the accompanying graph to estimate the force for a 150-pound player. dug84356_ch06b.qxd 9/14/10 12:45 PM Page 437 6-57 6.7 Applications of Ratios and Proportions 437 Force (thousands of pounds) WASTE GENERATION AT A FAST-FOOD RESTAUR...