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Solution Manual for College Algebra, 6th Edition

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    ’ S S OLUTIONS M ANUAL E DGAR R EYES Southeastern Louisiana University C OLLEGE A LGEBRA S IXTH E DITION Mark Dugopolski Southeastern Louisiana University Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo

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    Table of Contents Chapter P...……………………………………………………………………………………1 Chapter 1……………………………………………………………………………………..34 Chapter 2…………………………………………………………………………………….121 Chapter 3…………………………………………………………………………………….179 Chapter 4…………………………………………………………………………………….255 Chapter 5…………………………………………………………………………………….306 Chapter 6…………………………………………………………………………………….388 Chapter 7…………………………………………………………………………………….458 Chapter 8…………………………………………………………………………………….501

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    34 Chapter 1 Equations, Inequalities, and Modeling For Thought 1. True, since 5(1) = 6 − 1 . 2. True, since x = 3 is the solution to both equations. 3. False, − 2 is not a solution of the first equation since √ − 2 is not a real number. 4. True, since x − x = 0. 5. False, x = 0 is the solution. 6. True 7. False, since | x | = − 8 has no solution. 8. False, x x − 5 is undefined at x = 5. 9. False, since we should multiply by − 3 2 . 10. False, 0 · x + 1 = 0 has no solution. 1.1 Exercises 1. equation 2. linear 3. equivalent 4. solution set 5. identity 6. inconsistent equation 7. conditional equation 8. extraneous root 9. No, since 2(3) − 4 = 2 6 = 9 . 10. Yes 11. Yes, since ( − 4) 2 = 16. 12. No, since √ 16 6 = − 4 . 13. Since 3 x = 5, the solution set is { 5 3 } . 14. Since − 2 x = − 3, the solution set is { 3 2 } . 15. Since − 3 x = 6, the solution set is {− 2 } . 16. Since 5 x = − 10, the solution set is {− 2 } . 17. Since 14 x = 7, the solution set is { 1 2 } . 18. Since − 2 x = 2, the solution set is {− 1 } . 19. Since 7 + 3 x = 4 x − 4, the solution set is { 11 } . 20. Since − 3 x + 15 = 4 − 2 x , the solution set is { 11 } . 21. Since x = − 4 3 · 18, the solution set is {− 24 } . 22. Since x = 3 2 · ( − 9), the solution set is { − 27 2 } . 23. Multiplying by 6 we get 3 x − 30 = − 72 − 4 x 7 x = − 42 . The solution set is {− 6 } . 24. Multiplying by 4 we obtain x − 12 = 2 x + 12 − 24 = x. The solution set is {− 24 } . 25. Multiply both sides of the equation by 12. 18 x + 4 = 3 x − 2 15 x = − 6 x = − 2 5 . The solution set is { − 2 5 } . 26. Multiply both sides of the equation by 30. 15 x + 6 x = 5 x − 10 16 x = − 10 x = − 5 8 . The solution set is { − 5 8 } . 27. Note, 3( x − 6) = 3 x − 18 is true by the distributive law. It is an identity and the solution set is R .

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    1.1 Linear, Rational, and Absolute Value Equations 35 28. Subtract 5 a from both sides of 5 a = 6 a to get 0 = a . The latter equation is conditional whose solution set is { 0 } . 29. Note, 5 x = 4 x is equivalent to x = 0. The latter equation is conditional whose solution set is { 0 } . 30. Note, 4( y − 1) = 4 y − 4 is true by the distributive law. The equation is an identity and the solution set is R . 31. Equivalently, we get 2 x + 6 = 3 x − 3 or 9 = x . The latter equation is conditional whose solution set is { 9 } . 32. Equivalently, we obtain 2 x + 2 = 3 x + 2 or 0 = x . The latter equation is conditional whose solution set is { 0 } . 33. Using the distributive property, we find 3 x − 18 = 3 x + 18 − 18 = 18 . The equation is inconsistent and the solution set is ∅ . 34. Since 5 x = 5 x + 1 or 0 = 1, the equation is inconsistent and the solution set is ∅ . 35. An identity and the solution set is { x | x 6 = 0 } . 36. An identity and the solution set is { x | x 6 = − 2 } . 37. Multiplying by 2( w − 1), we get 1 w − 1 − 1 2 w − 2 = 1 2 w − 2 2 − 1 = 1 . An identity and the solution set is { w | w 6 = 1 } 38. Multiply by x ( x − 3). ( x − 3) + x = 9 2 x = 12 A conditional equation with solution set { 6 } . 39. Multiply by 6 x . 6 − 2 = 3 + 1 4 = 4 An identity with solution set { x | x 6 = 0 } . 40. Multiply by 60 x . 12 − 15 + 20 = − 17 x 17 = − 17 x A conditional equation with solution set {− 1 } . 41. Multiply by 3( z − 3). 3( z + 2) = − 5( z − 3) 3 z + 6 = − 5 z + 15 8 z = 9 A conditional equation with solution set { 9 8 } . 42. Multiply by ( x − 4). 2 x − 3 = 5 2 x = 8 x = 4 Since division by zero is not allowed, x = 4 does not satisfy the original equation. We have an inconsistent equation and so the solution set is ∅ . 43. Multiplying by ( x − 3)( x + 3). ( x + 3) − ( x − 3) = 6 6 = 6 An identity with solution set { x | x 6 = 3 , x 6 = − 3 } . 44. Multiply by ( x + 1)( x − 1). 4( x + 1) − 9( x − 1) = 3 4 x + 4 − 9 x + 9 = 3 − 5 x = − 10 A conditional equation with solution set { 2 } . 45. Multiply by ( y − 3). 4( y − 3) + 6 = 2 y 4 y − 6 = 2 y y = 3 Since division by zero is not allowed, y = 3 does not satisfy the original equation. We have an inconsistent equation and so the solution set is ∅ .

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    36 Chapter 1 Equations, Inequalities, and Modeling 46. Multiply by x + 6. x − 3( x + 6) = ( x + 6) − 6 − 2 x − 18 = x − 6 = x Since division by zero is not allowed, x = − 6 does not satisfy the original equation. We have an inconsistent equation and so the solution set is ∅ . 47. Multiply by t + 3. t + 4 t + 12 = 2 5 t = − 10 A conditional equation with solution set {− 2 } . 48. Multiply by x + 1. 3 x − 5( x + 1) = x − 11 3 x − 5 x − 5 = x − 11 6 = 3 x A conditional equation with solution set { 2 } . 49. Since − 4 . 19 = 0 . 21 x and − 4 . 19 0 . 21 ≈ − 19 . 952, the solution set is approximately {− 19 . 952 } . 50. Since 0 . 92 x = 5 . 9, we get x = 5 . 9 0 . 92 ≈ 6 . 413. The solution set is approximately { 6 . 413 } . 51. Divide by 0 . 06. x − 3 . 78 = 1 . 95 0 . 06 x = 32 . 5 + 3 . 78 x = 36 . 28 The solution set is { 36 . 28 } . 52. Divide by 0 . 86. 3 . 7 − 2 . 3 x = 4 . 9 0 . 86 − 2 . 3 x = 4 . 9 0 . 86 − 3 . 7 x = 4 . 9 0 . 86 − 3 . 7 − 2 . 3 x ≈ − 0 . 869 The solution set is approximately {− 0 . 869 } . 53. 2 a = − 1 − √ 17 a = − 1 − √ 17 2 a ≈ − 1 − 4 . 1231 2 a ≈ − 2 . 562 The solution set is approximately {− 2 . 562 } . 54. 3 c = √ 38 − 4 c = √ 38 − 4 3 x ≈ 6 . 1644 − 4 3 x ≈ 0 . 721 The solution set is approximately { 0 . 721 } . 55. 0 . 001 = 3( y − 0 . 333) 0 . 001 = 3 y − 0 . 999 1 = 3 y 1 3 = y The solution set is { 1 3 } . 56. Multiply by t − 1. ( t − 1) + 0 . 001 = 0 t − 0 . 999 = 0 The solution set is { 0 . 999 } . 57. Factoring x , we get x ( 1 0 . 376 + 1 0 . 135 ) = 2 x (2 . 6596 + 7 . 4074) ≈ 2 10 . 067 x ≈ 2 x ≈ 0 . 199 The solution set is approximately { 0 . 199 } .

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    1.1 Linear, Rational, and Absolute Value Equations 37 58. 1 x = 10 . 379 − 5 6 . 72 1 = x ( 10 . 379 − 5 6 . 72 ) 1 10 . 379 − 5 6 . 72 = x 0 . 104 ≈ x The solution set is approximately { 0 . 104 } . 59. x 2 + 6 . 5 x + 3 . 25 2 = x 2 − 8 . 2 x + 4 . 1 2 14 . 7 x = 4 . 1 2 − 3 . 25 2 14 . 7 x = 16 . 81 − 10 . 5625 14 . 7 x = 6 . 2475 x = 0 . 425 The solution set is { 0 . 425 } . 60. 0 . 25(4 x 2 − 6 . 4 x + 2 . 56) = x 2 − 1 . 8 x + 0 . 81 x 2 − 1 . 6 x + 0 . 64 = x 2 − 1 . 8 x + 0 . 81 0 . 2 x = 0 . 17 x = 0 . 85 The solution set is { 0 . 85 } . 61. (2 . 3 × 10 6 ) x = 1 . 63 × 10 4 − 8 . 9 × 10 5 x = 1 . 63 × 10 4 − 8 . 9 × 10 5 2 . 3 × 10 6 x ≈ − 0 . 380 The solution set is approximately {− 0 . 380 } . 62. Note, 3 . 45 × 10 − 8 ≈ 0 . x = 1 . 63 × 10 4 − 3 . 45 × 10 − 8 − 3 . 4 × 10 − 9 x ≈ − 4 . 794 × 10 12 The solution set is approximately {− 4 . 794 × 10 12 } . 63. Solution set is {± 8 } . 64. Solution set is {± 2 . 6 } . 65. Since x − 4 = ± 8, we get x = 4 ± 8. The solution set is {− 4 , 12 } . 66. Since x − 5 = 3 . 6 or x − 5 = − 3 . 6, we get x = 8 . 6 or x = 1 . 4. The solution set is { 1 . 4 , 8 . 6 } . 67. Since x − 6 = 0, we get x = 6. The solution set is { 6 } . 68. Since x − 7 = 0, we get x = 7. The solution set is { 7 } . 69. Since the absolute value of a real number is not a negative number, the equation | x + 8 | = − 3 has no solution. The solution set is ∅ . 70. Since the absolute value of a real number is not a negative number, the equation | x + 9 | = − 6 has no solution. The solution set is ∅ . 71. Since 2 x − 3 = 7 or 2 x − 3 = − 7, we get 2 x = 10 or 2 x = − 4. The solution set is {− 2 , 5 } . 72. Since 3 x + 4 = 12 or 3 x + 4 = − 12, we find 3 x = 8 or 3 x = − 16. The solution set is {− 16 / 3 , 8 / 3 } . 73. Multiplying 1 2 | x − 9 | = 16 by 2 we obtain | x − 9 | = 32. Then x − 9 = 32 or x − 9 = − 32. The solution set is {− 23 , 41 } . 74. Multiplying 2 3 | x + 4 | = 8 by 3 2 we obtain | x + 4 | = 12. Then x + 4 = 12 or x + 4 = − 12. The solution set is {− 16 , 8 } . 75. Since 2 | x + 5 | = 10, we find | x + 5 | = 5. Then x + 5 = ± 5 or x = ± 5 − 5. The solution set is {− 10 , 0 } . 76. Since 8 = 4 | x + 3 | , we obtain 2 = | x + 3 | . Then x + 3 = ± 2 or x = ± 2 − 3. The solution set is {− 5 , − 1 } . 77. Dividing 8 | 3 x − 2 | = 0 by 8, we obtain | 3 x − 2 | = 0. Then 3 x − 2 = 0 and the solution set is { 2 / 3 } .

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    38 Chapter 1 Equations, Inequalities, and Modeling 78. Dividing 5 | 6 − 3 x | = 0 by 5, we obtain | 6 − 3 x | = 0. Then 6 − 3 x = 0 and the solution set is { 2 } . 79. Subtracting 7, we find 2 | x | = − 1 and | x | = − 1 2 . Since an absolute value is not equal to a negative number, the solution set is ∅ . 80. Subtracting 5, we obtain 3 | x − 4 | = − 5 and | x − 4 | = − 5 3 . Since an absolute value is not a negative number, the solution set is ∅ . 81. Since 0 . 95 x = 190, the solution set is { 200 } . 82. Since 1 . 1 x = 121, the solution set is { 110 } . 83. 0 . 1 x − 0 . 05 x + 1 = 1 . 2 0 . 05 x = 0 . 2 The solution set is { 4 } . 84. 0 . 03 x − 0 . 2 = 0 . 2 x + 0 . 006 − 0 . 206 = 0 . 17 x − 0 . 206 0 . 17 = x − 0 . 206 0 . 17 · 1000 1000 = x − 206 170 = x The solution set is { − 103 85 } . 85. Simplifying x 2 + 4 x + 4 = x 2 + 4, we obtain 4 x = 0. The solution set is { 0 } . 86. Simplifying x 2 − 6 x +9 = x 2 − 9, we get 18 = 6 x . The solution set is { 3 } . 87. Since | 2 x − 3 | = | 2 x + 5 | , we get 2 x − 3 = 2 x + 5 or 2 x − 3 = − 2 x − 5. Solving for x , we find − 3 = 5 (an inconsistent equation) or 4 x = − 2. The solution set is {− 1 / 2 } . 88. Squaring the terms, we find (9 x 2 − 24 x + 16) + (16 x 2 + 8 x + 1) = 25 x 2 + 20 x + 4. Setting the left side to zero, we obtain 0 = 36 x − 13. The solution set is { 13 / 36 } . 89. Multiply by 4. 2 x + 4 = x − 6 x = − 10 The solution set is {− 10 } . 90. Multiply by 12. − 2( x + 3) = 3(3 − x ) − 2 x − 6 = 9 − 3 x x = 15 The solution set is { 15 } . 91. Multiply by 30. 15( y − 3) + 6 y = 90 − 5( y + 1) 15 y − 45 + 6 y = 90 − 5 y − 5 26 y = 130 The solution set is { 5 } . 92. Multiply by 10. 2( y − 3) − 5( y − 4) = 50 2 y − 6 − 5 y + 20 = 50 − 36 = 3 y The solution set is {− 12 } . 93. Since 7 | x + 6 | = 14, | x + 6 | = 2. Then x + 6 = 2 or x + 6 = − 2. The solution set is {− 4 , − 8 } . 94. From 3 = | 2 x − 3 | , it follows that 2 x − 3 = 3 or 2 x − 3 = − 3 2 x = 6 or 2 x = 0 . The solution set is { 3 , 0 } . 95. Since − 4 | 2 x − 3 | = 0, we get | 2 x − 3 | = 0. Then 2 x − 3 = 0 and the solution set is { 3 / 2 } . 96. Since −| 3 x +1 | = | 3 x +1 | , we get 0 = 2 | 3 x +1 | . Then | 3 x + 1 | = 0 or 3 x + 1 = 0. The solution set is {− 1 / 3 } .

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    1.1 Linear, Rational, and Absolute Value Equations 39 97. Since − 5 | 5 x + 1 | = 4, we find | 5 x + 1 | = − 4 / 5. Since the absolute value is not a negative num- ber, the solution set is ∅ . 98. Since | 7 − 3 x | = − 3 and the absolute value is not a negative number, the solution set is ∅ . 99. Multiply by ( x − 2)( x + 2). 3( x + 2) + 4( x − 2) = 7 x − 2 3 x + 6 + 4 x − 8 = 7 x − 2 7 x − 2 = 7 x − 2 An identity with solution set { x | x 6 = 2 , x 6 = − 2 } . 100. Multiply by ( x − 1)( x + 2). 2( x + 2) − 3( x − 1) = 8 − x 2 x + 4 − 3 x + 3 = 8 − x 7 − x = 8 − x An inconsistent equation and the solution set is ∅ . 101. Multiply ( x + 3)( x − 2) to both sides of 4 x + 3 + 3 x − 2 = 7 x + 1 ( x + 3)( x − 2) . Then we find 4( x − 2) + 3( x + 3) = 7 x + 1 4 x − 8 + 3 x + 9 = 7 x + 1 7 x + 1 = 7 x + 1 . An identity and the solution set is { x | x 6 = 2 and x 6 = − 3 } . 102. Multiply by x ( x − 1) to both sides of 3 x + 4 x − 1 = 7 x − 3 x ( x − 1) . Then we get 3( x − 1) + 4 x = 7 x − 3 3 x − 3 + 4 x = 7 x − 3 7 x − 3 = 7 x − 3 . An identity and the solution set is { x | x 6 = 0 and x 6 = 1 } . 103. Multiply by ( x − 3)( x − 4). ( x − 4)( x − 2) = ( x − 3) 2 x 2 − 6 x + 8 = x 2 − 6 x + 9 8 = 9 An inconsistent equation and so the solution set is ∅ . 104. Multiply by ( y + 4)( y − 2). ( y − 2)( y − 1) = ( y + 4)( y + 1) y 2 − 3 y + 2 = y 2 + 5 y + 4 − 2 = 8 y. A conditional equation and the solution set is {− 1 / 4 } . 105. a) About 1995 b) Increasing c) Let y = 0 . 95. Solving for x , we find 0 . 95 = 0 . 0102 x + 0 . 644 0 . 95 − 0 . 644 0 . 0102 = x 30 = x. In the year 2020 (= 1990 + 30), 95% of mothers will be in the labor force. 106. Let y = 0 . 644. Solving for x , we find 0 . 644 = 0 . 0102 x + 0 . 644 0 = 0 . 0102 x 0 = x. In the year 1990 (= 1990 + 0), 64 . 4% of mothers were in the labor force. 107. Since B = 21 , 000 − 0 . 15 B , we obtain 1 . 15 B = 21 , 000 and the bonus is B = 21 , 000 1 . 15 = $18 , 260 . 87. 108. Since 0 . 30(200 , 000) = 60 , 000, we find S = 0 . 06(140 , 000 + 0 . 3 S ) S = 8400 + 0 . 018 S 0 . 982 S = 8400 . The state tax is S = 8400 0 . 982 = $8553 . 97 and the federal tax is F = 0 . 30(200 , 000 − 8553 . 97) = $57 , 433 . 81.

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    40 Chapter 1 Equations, Inequalities, and Modeling 109. Rewrite the left-hand side as a sum. 10 , 000 + 500 , 000 , 000 x = 12 , 000 500 , 000 , 000 x = 2 , 000 500 , 000 , 000 = 2000 x 250 , 000 = x Thus, 250 , 000 vehicles must be sold. 110. (a) The harmonic mean is 5 1 15 . 1 + 1 7 . 3 + 1 5 . 9 + 1 3 . 6 + 1 2 . 8 which is about $4.96 trillion. (b) Let x be the GDP of Brazil, and A = 1 15 . 1 + 1 7 . 3 + 1 5 . 9 + 1 3 . 6 + 1 2 . 8 . Applying the harmonic mean formula, we get 6 A + 1 x = 4 . 26 6 = 4 . 26 A + 4 . 26 x 6 − 4 . 26 A = 4 . 26 x x = 4 . 26 6 − 4 . 26 A x = $2 . 49 trillion 111. The third side of the triangle is √ 3 by the Pythagorean Theorem. Then draw radial lines from the center of the circle to each of the three sides. Consider the square with side r that is formed with the 90 ◦ angle of the triangle. Then the side of length √ 3 is divided into two segments of length r and √ 3 − r . Similarly, the side of length 1 is divided into segments of length r and 1 − r . Note, the center of the circle lies on the bi- sectors of the angles of the triangle. Using congruent triangles, the hypotenuse consists of line segments of length √ 3 − r and 1 − r . Since the hypotenuse is 2, we have ( √ 3 − r ) + (1 − r ) = 2 √ 3 − 1 = 2 r. Thus, the radius is r = √ 3 − 1 2 . 112. The hypotenuse is √ 2 by the Pythagorean Theorem. Then draw radial lines from the center of the circle to each of the three sides. Consider the square with side r that is formed with the 90 ◦ angle of the triangle. Then each side of length 1 consists of line segments of length r and 1 − r . Note, the center of the circle lies on the bi- sectors of the angles of the triangle. Using congruent triangles, the hypotenuse consists of line segments of length 1 − r and 1 − r . Since the hypotenuse is √ 2, we have (1 − r ) + (1 − r ) = √ 2 2 − √ 2 = 2 r. Thus, the radius is r = 2 − √ 2 2 . 115. − 3 , 0 , 7 116. 20 − 24 = − 4 117. − 5 x + 20 − 24 + 12 x = 7 x − 4 118. 1 4 − 1 3 1 5 − 1 4 = − 1 12 − 1 20 = 20 12 = 5 3 119. √ 5 √ 3 · √ 3 √ 3 = √ 15 3 120. Since x 6 = 0, domain is ( −∞ , 0) ∪ (0 , ∞ ). 121. $9, $99, $999, $9,999, $99,999, $999,999, $9,999,999 $99,999,999, $999,999,999 122. Notice, 5! + 6! + · · · + 1776! is a multiple of 5! = 120. Then the units digit of 1!+ · · · +1776! is the same as the units digit of 1!+2!+3!+4! = 33. Thus, the units digit of the sum is 3.

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    1.2 Constructing Models to Solve Problems 41 1.1 Pop Quiz 1. Since 7 x = 6, we get x = 6 / 7. A conditional equation and the solution set is { 6 / 7 } . 2. Since 1 4 x = 1 3 + 1 6 = 1 2 , we get x = 4 · 1 2 = 2 . A conditional equation, the solution set is { 2 } . 3. Since 3 x − 27 = 3 x − 27 is an identity, the solution set is R . 4. Since w − 1 = 6 or w − 1 = − 6, we get w = 7 or w = − 5. A conditional equation and the solution set is {− 5 , 7 } . 5. Since 2 x + 12 = 2 x + 6, we obtain 12 = 6 which is an inconsistent equation. The solution set is ∅ . 6. Since x 2 + 2 x + 1 = x 2 + 1, we obtain 2 x = 0. This is a conditional equation and the solution set is { 0 } . 1.1 Linking Concepts (a) The power expenditures for runners with masses 60 kg, 65 kg, and 70 kg are 60( a · 400 − b ) ≈ 22 . 9 kcal/min, 65( a · 400 − b ) ≈ 24 . 8 kcal/min, and 70( a · 400 − b ) ≈ 26 . 7 kcal/min, respectively. (b) Power expenditure increases as the mass of the runner increases (assuming constant velocity). (c) Since v = 1 a ( P M + b ) , the velocities are v = 1 a ( 38 . 7 80 + b ) ≈ 500 m/min, v = 1 a ( 38 . 7 84 + b ) ≈ 477 m/min, and v = 1 a ( 38 . 7 90 + b ) ≈ 447 m/min. (d) The velocity decreases as the mass increases (assuming constant power expenditure). (e) Note, P = M va − M b . Solving for v , we find 52(480) a − 52 b = 50 va − 50 b 52(480) a − 2 b = 50 va 52(480) a − 2 b 50 a = v 498 . 2 ≈ v. The velocity is approximately 498 . 2 m/min. (f ) With weights removed and constant power expenditure, a runner’s velocity increases. (g) The first graph shows P versus M (with v = 400) 20 40 60 M 25 50 p and the second graph shows v versus M (with P = 40) 25 50 75 M 500 1000 1500 v For Thought 1. False, P (1 + rt ) = S implies P = S 1 + rt . 2. False, since the perimeter is twice the sum of the length and width. 3. False, since n + 1 and n + 3 are even integers if n is odd. 4. True 5. True, since x + ( − 3 − x ) = − 3 . 6. False, since P = 2 S

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    42 Chapter 1 Equations, Inequalities, and Modeling 7. False, for if the house sells for x dollars then x − 0 . 09 x = 100 , 000 0 . 91 x = 100 , 000 x = $109 , 890 . 11 . 8. True 9. False, a correct equation is 4( x − 2) = 3 x − 5. 10. False, since 9 and x + 9 differ by x. 1.2 Exercises 1. formula 2. function 3. uniform 4. rate, time 5. r = I P t 6. R = D T 7. Since F − 32 = 9 5 C , C = 5 9 ( F − 32) . 8. Since 9 5 C = F − 32, 9 5 C + 32 = F or F = 9 5 C + 32. 9. Since 2 A = bh , we get b = 2 A h . 10. Since 2 A = bh , we have h = 2 A b . 11. Since By = C − Ax , we obtain y = C − Ax B . 12. Since Ax = C − By , we get x = C − By A . 13. Multiplying by RR 1 R 2 R 3 , we find R 1 R 2 R 3 = RR 2 R 3 + RR 1 R 3 + RR 1 R 2 R 1 R 2 R 3 − RR 1 R 3 − RR 1 R 2 = RR 2 R 3 R 1 ( R 2 R 3 − RR 3 − RR 2 ) = RR 2 R 3 . Then R 1 = RR 2 R 3 R 2 R 3 − RR 3 − RR 2 . 14. Multiplying by RR 1 R 2 R 3 , we find R 1 R 2 R 3 = RR 2 R 3 + RR 1 R 3 + RR 1 R 2 R 1 R 2 R 3 − RR 2 R 3 − RR 1 R 2 = RR 1 R 3 R 2 ( R 1 R 3 − RR 3 − RR 1 ) = RR 1 R 3 . Then R 2 = RR 1 R 3 R 1 R 3 − RR 3 − RR 1 . 15. Since a n − a 1 = ( n − 1) d , we obtain n − 1 = a n − a 1 d n = a n − a 1 d + 1 n = a n − a 1 + d d . 16. Multiplying by 2, we get 2 S n = n ( a 1 + a n ) 2 S n = na 1 + na n 2 S n − na n = na 1 . Then a 1 = 2 S n − na n n . 17. Since S = a 1 (1 − r n ) 1 − r , we obtain a 1 (1 − r n ) = S (1 − r ) a 1 = S (1 − r ) 1 − r n . 18. Since S = 2 LW + H (2 L + 2 W ), we obtain H (2 L + 2 W ) = S − 2 LW H = S − 2 LW 2 L + 2 W . 19. Multiplying by 2.37, one finds 2 . 4(2 . 37) = L + 2 D − F √ S 5 . 688 − L + F √ S = 2 D and D = 5 . 688 − L + F √ S 2 . 20. Multiplying by 2.37, one finds 2 . 4(2 . 37) = L + 2 D − F √ S F √ S = L + 2 D − 5 . 688 F = L + 2 D − 5 . 688 √ S .

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    1.2 Constructing Models to Solve Problems 43 21. R = D/T 22. T = D/R 23. Since LW = A , we have W = A/L . 24. Since P = 2 L + 2 W , we have 2 W = P − 2 L and W = P/ 2 − L . 25. r = d/ 2 26. d = 2 r 27. By using the formula I = P rt , one gets 51 . 30 = 950 r · 1 0 . 054 = r. The simple interest rate is 5 . 4%. 28. Note, I = P rt and the interest is $5. 5 = 100 r · 1 12 60 = 100 r 0 . 6 = r Simple interest rate is 60%. 29. Since D = RT , we find 5570 = 2228 · T 2 . 5 = T. The surveillance takes 2 . 5 hours. 30. Since C = 2 πr , the radius is r = 72 π 2 π = 36 in. 31. Note, C = 5 9 ( F − 32). If F = 23 o F , then C = 5 9 (23 − 32) = − 5 o C. 32. Since F = 9 5 C + 32 and C = 30 o C, we get F = 9 5 · 30 + 32 = 54 + 32 = 86 o F. 33. If x is the cost of the car before taxes, then 1 . 08 x = 40 , 230 x = $37 , 250 . 34. If x is the minimum selling price, then x − 0 . 06 x − 780 = 128 , 000 0 . 94 x = 128 , 780 x = $137 , 000 . 35. Let S be the saddle height and let L be the inside measurement. S = 1 . 09 L 37 = 1 . 09 L 37 1 . 09 = L 33 . 9 ≈ L The inside leg measurement is 33.9 inches. 36. Let h , a , and r denote the target heart rate, age, and resting heart rate, respectively. Substituting we obtain 144 = 0 . 6[220 − ( a + r )] + r 144 = 0 . 6[220 − (30 + r )] + r 144 = 0 . 6[190 − r ] + r 144 = 114 + 0 . 4 r 30 = 0 . 4 r. The resting heart rate is r = 30 0 . 4 = 75. 37. Let x be the sale price. 1 . 1 x = 50 , 600 x = 50 , 600 1 . 1 x = $46 , 000 38. Let x be the winning bid. 1 . 05 x = 2 . 835 x = 2 . 835 1 . 05 x = 2 . 7 The winning bid is 2.7 million pounds.

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    44 Chapter 1 Equations, Inequalities, and Modeling 39. Let x be the amount of her game-show winnings. 0 . 14 x 3 + 0 . 12 x 6 = 4000 6 ( 0 . 14 x 3 + 0 . 12 x 6 ) = 24 , 000 0 . 28 x + 0 . 12 x = 24 , 000 0 . 40 x = 24 , 000 x = $60 , 000 Her winnings is $60,000. 40. Let x and 4 . 7 − x be the amounts of the high school and stadium contracts, respectively, in millions of dollars. 0 . 05 x + 0 . 04(4 . 7 − x ) = 0 . 223 0 . 05 x + 0 . 188 − 0 . 04 x = 0 . 223 0 . 01 x = 0 . 035 x = 3 . 5 The high school contract and stadium contract are are $3 . 5 million and $1 . 2 million, respec- tively. 41. If x is the length of the shorter piece in feet, then the length of the longer side is 2 x + 2. Then we obtain x + x + (2 x + 2) = 30 4 x = 28 x = 7 . The length of each shorter piece is 7 ft and the longer piece is (2 · 7 + 2) or 16 ft. 42. If x is the width of the old field, then ( x + 3) x is the area of the old field. The larger field has an area of ( x + 5)( x + 2) and so ( x + 5)( x + 2) = ( x + 3) x + 46 x 2 + 7 x + 10 = x 2 + 3 x + 46 4 x = 36 x = 9 . Since x + 3 = 9 + 3 = 12, the dimension of the old field is 9 m by 12 m. 43. If x is the length of the side of the larger square lot then 2 x is the amount of fencing needed to divide the square lot into four smaller lots. The solution to 4 x + 2 x = 480 is x = 80 . The side of the larger square lot is 80 feet and its area is 6400 ft 2 . 44. If x is the length of a shorter side, then the longer side is 3 x and the amount of fencing for the three long sides is 9 x . Since there are four shorter sides and there is 65 feet of fencing, we have 4 x + 9 x = 65. The solution to this equation is x = 5 and the dimension of each pen is 5 ft by 15 ft. 45. Let d denote the distance from Fairbanks to Coldfoot, which is the same distance from Coldfoot to Deadhorse. d 50 + d 40 = 11 . 25 hr 90 d = 11 . 25(2000) d = 250 miles 46. Let x be the distance from Dawson City to Inuvik. x 23 + 3 = x 20 x = − 3 1 23 − 1 20 x = 460 miles 47. Note, Bobby will complete the remaining 8 laps in 8 90 of an hour. If Ricky is to finish at the same time as Bobby, then Ricky’s average speed s over 10 laps must satisfy 10 s = 8 90 . ( Note: time = distance speed ) . The equation is equivalent to 900 = 8 s . Thus, Ricky’s average speed must be 112.5 mph.

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    1.2 Constructing Models to Solve Problems 45 48. Let d be the distance from the camp to the site of the crab traps. Note, rate = distance ÷ time . against d 1/2 2 d going d 1/6 6 d distance time rate Since the speed going with the tide is increased by 2 mph, his normal speed is 6 d − 2. Similarly, since his speed against the tide is decreased by 2 mph, his normal speed is 2 d + 2. Then 6 d − 2 = 2 d + 2 for the normal speeds are the same. Solving for d , we get d = 1 mile which is the distance between the camp and the site of the crab traps. 49. Let d be the halfway distance between San Antonio and El Paso, and let s be the speed in the last half of the trip. Junior took d 80 hours to get to the halfway point and the last half took d s hours to drive. Since the total distance is 2 d and distance = rate × time , 2 d = 60 ( d 80 + d s ) 160 sd = 60 ( sd + 80 d ) 160 sd = 60 sd + 4800 d 100 sd = 4800 d 100 d ( s − 48) = 0 . Since d 6 = 0, the speed for the last half of the trip was s = 48 mph. 50. Let A be Parker’s average for the remaining games and let n be the number of games in the entire season. The number of points she scored through two-thirds of the season is 18 ( 2 n 3 ) points. If she must have an average of 22 points per game for the entire season, then 18 ( 2 n 3 ) + A ( n 3 ) n = 22 36 3 + A 3 = 22 36 + A = 66 A = 30 . For the remaining games, she must average 30 points per game. 51. If x is the part of the start-up capital invested at 5% and x + 10 , 000 is the part invested at 6%, then 0 . 05 x + 0 . 06( x + 10 , 000) = 5880 0 . 11 x + 600 = 5880 0 . 11 x = 5280 x = 48 , 000 . Norma invested $48 , 000 at 5% and $58 , 000 at 6% for a total start-up capital of $106 , 000. 52. If x is the amount in cents Bob borrowed at 8% and x 2 is the amount borrowed by Betty at 16%, then 0 . 08 x + 0 . 16 ( x 2 ) = 24 0 . 08 x + 0 . 08 x = 24 0 . 16 x = 24 x = 150 cents . Bob borrowed $1 . 50 and Betty borrowed $0 . 75. 53. Let x and 1500 − x be the number of employees from the Northside and Southside, respectively. Then (0 . 05) x + 0 . 80(1500 − x ) = 0 . 5(1500) 0 . 05 x + 1200 − 0 . 80 x = 750 450 = 0 . 75 x 600 = x. There were 600 and 900 employees at the Northside and Southside, respectively. 54. Let x be the number of ounces of a 30% solution. 70% soln 40 + x 0 . 7(40 + x ) 80% soln 40 0.8(40) 30% soln x 0 . 3 x amt of soln amt of alcohol

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    46 Chapter 1 Equations, Inequalities, and Modeling Adding the amounts of pure alcohol we get 0 . 3 x + 0 . 8(40) = 0 . 7(40 + x ) 0 . 3 x + 32 = 28 + 0 . 7 x 4 = 0 . 4 x 10 = x. Then 10 ounces of the 30% solution are needed. 55. Let x be the number of hours it takes both combines working together to harvest an entire wheat crop. combined 1 /x new 1 / 48 old 1 / 72 rate Then 1 72 + 1 48 = 1 x . Multiply both sides by 144 x and get 2 x + 3 x = 144. The solution is x = 28 . 8 hr, which is the time it takes both combines to harvest the entire wheat crop. 56. Let x be the number of hours it takes Rita and Eduardo, working together, to process a batch of claims. together 1 /x Eduardo 1 / 2 Rita 1 / 4 rate It follows that 1 2 + 1 4 = 1 x . Multiplying both sides by 4 x , we get 2 x + x = 4 . The solution is x = 4 3 hr, which is how long it will take both of them to process a batch of claims. 57. Let t be the number of hours since 8:00 a.m. Robin 1 / 12 t t/ 12 Batman 1 / 8 t − 2 ( t − 2) / 8 rate time work completed t − 2 8 + t 12 = 1 24 ( t − 2 8 + t 12 ) = 24 3( t − 2) + 2 t = 24 5 t − 6 = 24 t = 6 At 2 p.m., all the crimes have been cleaned up. 58. Let t be the number of hours since noon. Don 1 / 15 t t/ 15 Della 1 / 10 t − 3 ( t − 3) / 10 rate time work completed Since the sum of the works completed is 1, t − 3 10 + t 15 = 1 30 ( t − 3 10 + t 15 ) = 30 3( t − 3) + 2 t = 30 5 t − 9 = 30 t = 7 . 8 . They will finish the job in 7.8 hrs or at 7:48 p.m. 59. Since there are 5280 feet to a mile and the circumference of a circle is C = 2 πr , the radius r of the race track is r = 5280 2 π . Since the length of a side of the square plot is twice the radius, the area of the plot is ( 2 · 5280 2 π ) 2 ≈ 2 , 824 , 672 . 5 ft 2 . Dividing this number by 43 , 560 results to 64 . 85 acres which is the acreage of the square lot. 60. Since the volume of a circular cylinder is V = πr 2 h , we have 12(1 . 8) = π ( 2 . 375 2 ) 2 · h . Solving for h , we get h ≈ 4 . 876 in., the height of a can of Coke.

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