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Solution Manual for College Algebra, 3rd Edition

Solution Manual for College Algebra, 3rd Edition simplifies the toughest textbook questions, providing easy-to-follow solutions for every chapter.

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    CONTENTS Chapter P Basic Concepts of Algebra P.1 The Real Numbers and Their Properties...................................................... 1 P.2 Integer Exponents and Scientific Notation .................................................. 6 P.3 Polynomials................................................................................................ 10 P.4 Factoring Polynomials ............................................................................... 14 P.5 Rational Expressions.................................................................................. 19 P.6 Rational Exponents and Radicals............................................................... 27 Chapter P Review Exercises ................................................................................... 35 Chapter P Practice Test ........................................................................................... 40 Chapter 1 Equations and Inequalities 1.1 Linear Equations in One Variable.............................................................. 41 1.2 Quadratic Equations................................................................................... 55 1.3 Complex Numbers: Quadratic Equations with Complex Solutions........... 69 1.4 Solving Other Types of Equations ............................................................. 77 1.5 Inequalities................................................................................................. 97 1.6 Equations and Inequalities Involving Absolute Value............................. 113 Chapter 1 Review Exercises.................................................................................. 129 Chapter 1 Practice Test A ..................................................................................... 140 Chapter 1 Practice Test B...................................................................................... 141 Chapter 2 Graphs and Functions 2.1 The Coordinate Plane............................................................................... 144 2.2 Graphs of Equations................................................................................. 154 2.3 Lines......................................................................................................... 167 2.4 Functions.................................................................................................. 180 2.5 Properties of Functions ............................................................................ 189 2.6 A Library of Functions............................................................................. 198 2.7 Transformations of Functions .................................................................. 209 2.8 Combining Functions; Composite Functions........................................... 222 2.9 Inverse Functions ..................................................................................... 238 Chapter 2 Review Exercises.................................................................................. 250 Chapter 2 Practice Test A ..................................................................................... 260 Chapter 2 Practice Test B...................................................................................... 261 Cumulative Review Exercises (Chapters P−2) ..................................................... 262 Chapter 3 Polynomial and Rational Functions 3.1 Quadratic Functions ................................................................................. 266 3.2 Polynomial Functions .............................................................................. 285 3.3 Dividing Polynomials .............................................................................. 298 3.4 The Real Zeros of a Polynomial Function ............................................... 309 3.5 The Complex Zeros of a Polynomial Function ........................................ 328 3.6 Rational Functions ................................................................................... 337 3.7 Variation .................................................................................................. 355 Chapter 3 Review Exercises.................................................................................. 360 Chapter 3 Practice Test A ..................................................................................... 376 Chapter 3 Practice Test B...................................................................................... 378 Cumulative Review Exercises (Chapters P−3) ..................................................... 379

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    Chapter 4 Exponential and Logarithmic Functions 4.1 Exponential Functions.............................................................................. 382 4.2 Logarithmic Functions ............................................................................. 394 4.3 Rules of Logarithms................................................................................. 404 4.4 Exponential and Logarithmic Equations and Inequalities........................ 415 4.5 Logarithmic Scales................................................................................... 427 Chapter 4 Review Exercises.................................................................................. 438 Chapter 4 Practice Test A ..................................................................................... 446 Chapter 4 Practice Test B...................................................................................... 447 Cumulative Review Exercises (Chapters P−4) ..................................................... 448 Chapter 5 Systems of Equations and Inequalities 5.1 Systems of Linear Equations in Two Variables....................................... 451 5.2 Systems of Linear Equations in Three Variables..................................... 469 5.3 Partial-Fraction Decomposition ............................................................... 490 5.4 Systems of Nonlinear Equations .............................................................. 507 5.5 Systems of Inequalities ............................................................................ 521 5.6 Linear Programming ................................................................................ 534 Chapter 5 Review Exercises.................................................................................. 549 Chapter 5 Practice Test A ..................................................................................... 562 Chapter 5 Practice Test B...................................................................................... 565 Cumulative Review Exercises (Chapters P−5) ..................................................... 566 Chapter 6 Matrices and Determinants 6.1 Matrices and Systems of Equations ......................................................... 569 6.2 Matrix Algebra......................................................................................... 586 6.3 The Matrix Inverse................................................................................... 604 6.4 Determinants and Cramer’s Rule............................................................. 625 Chapter 6 Review Exercises.................................................................................. 637 Chapter 6 Practice Test A ..................................................................................... 647 Chapter 6 Practice Test B...................................................................................... 649 Cumulative Review Exercises (Chapters P−6) ..................................................... 650 Chapter 7 The Conic Sections 7.2 The Parabola ............................................................................................ 653 7.3 The Ellipse ............................................................................................... 668 7.4 The Hyperbola.......................................................................................... 685 Chapter 7 Review Exercises.................................................................................. 708 Chapter 7 Practice Test A ..................................................................................... 717 Chapter 7 Practice Test B...................................................................................... 719 Cumulative Review Exercises (Chapters P−7) ..................................................... 720 Chapter 8 Further Topics in Algebra 8.1 Sequences and Series ............................................................................... 723 8.2 Arithmetic Sequences; Partial Sums ........................................................ 731 8.3 Geometric Sequences and Series ............................................................. 737 8.4 Mathematical Induction ........................................................................... 744 8.5 The Binomial Theorem ............................................................................ 752 8.6 Counting Principles.................................................................................. 759 8.7 Probability................................................................................................ 766 Chapter 8 Review Exercises.................................................................................. 770 Chapter 8 Practice Test A ..................................................................................... 774 Chapter 8 Practice Test B...................................................................................... 775 Cumulative Review Exercises (Chapters P−8) ..................................................... 776

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    41 Chapter 1 Equations and Inequalities 1.1 Linear Equations in One Variable 1.1 Practice Problems 1. a. Both sides of the equation 7 5 3 x − = are defined for all real numbers, so the domain is ( , ) −∞ ∞ . b. The left side of the equation 2 4 2 x = − is not defined if x = 2. The right side of the equation is defined for all real numbers, so the domain is ( ) ( ) , 2 2, . −∞ ∞ ∪ c. The left side of the equation 1 0 x − = is not defined if x < 1. The right side of the equation is defined for all real numbers, so the domain is [ ) 1, . ∞ 2. 2 3 1 7 3 2 6 3 x x − = − To clear the fractions, multiply both sides of the equation by the LCD, 6. 4 9 1 14 4 9 14 1 14 14 4 5 1 4 5 4 1 4 5 3 5 3 5 5 3 5 x x x x x x x x x x x − = − − + = − + + = + − = − = − − = = − Solution set: { } 3 5 − 3. ( ) ( ) ( ) 3 2 6 1 7 1 3 2 6 6 7 1 3 4 6 7 1 3 4 6 7 1 7 6 7 1 7 6 7 7 1 7 6 1 x x x x x x x x x x x x x x x x x x x x − ⎡ − + ⎤ = − ⎣ ⎦ − − − = − − − − = − + + = − + = − + − = − − = − Since 6 = −1 is false, no number satisfies this equation. Thus, the equation is inconsistent, and the solution set is . ∅ 4. ( ) ( ) 2 3 6 5 12 19 6 6 12 5 12 19 6 6 7 7 6 6 7 6 7 6 6 7 7 7 7 7 7 0 0 x x x x x x x x x x − + = − − − + = − + − = − + − − = − + − − = − − + = − + = The equation 0 = 0 is always true. Therefore, the original equation is an identity, and the solution set is ( ) , . −∞ ∞ 5. 9 32 5 9 50 32 5 9 50 32 32 32 5 9 18 5 5 5 9 18 9 9 5 10 F C C C C C C = + = + − = + − = ⋅ = ⋅ = Thus, 50ºF converts to 10º C . 6. 2 2 P l w = + Subtract 2 l from both sides. 2 2 P l w − = Now, divide both sides by 2. 2 2 P l w − = 7. Let w = the width of the rectangle. Then 2 w + 5 = the length of the rectangle. P = 2 l + 2 w , so we have 28 2(2 5) 2 28 4 10 2 28 6 10 18 6 3 w w w w w w w = + + = + + = + = = The width of the rectangle is 3 m and the length is 2(3) + 5 = 11 m 8. Let x = the amount invested in stocks. Then 15,000 − x = the amount invested in bonds. ( ) 3 15, 000 45, 000 3 4 45, 000 11, 250 x x x x x x = − = − = = Tyrick invested $11,250 in stocks and $15,000 − $11,250 = $3,750 in bonds .

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    42 Chapter 1 Equations and Inequalities 9. Let x = the amount of capital. Then 5 x = the amount invested at 5%, 6 x = the amount invested at 8%, and 19 5 6 30 x x x x ⎛ ⎞ − + = = ⎜ ⎟ ⎝ ⎠ the amount invested at 10%. Principal Rate Time Interest 5 x 0.05 1 0.05 5 x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 6 x 0.08 1 0.08 6 x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 19 30 x 0.1 1 0.1 19 30 x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ The total interest is $130, so 19 0.05 0.08 0.1 130 5 6 30 Multiply by the LCD, 30. 0.3 0.4 1.9 3900 2.6 3900 1500 x x x x x x x x ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + + = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ + + = = = The total capital is $1500. 10. Let x = the length of the bridge. Then x + 130 = the distance the train travels. rt = d , so ( ) 25 21 130 525 130 395 x x x = + ⇒ = + ⇒ = The bridge is 395 m long. 11. Following the reasoning in example 10, we have x + 2 x = 3 x is the maximum extended length (in feet) of the cord. 3 7 10 120 3 17 120 3 17 17 120 17 3 103 3 103 34.3 3 3 x x x x x x + + = + = + − = − = = ⇒ ≈ The cord should be no longer than 34.3 feet. 1.1 Basic Concepts and Skills 1. The domain of the variable in an equation is the set of all real number for which both sides of the equation are defined. 2. Standard form for a linear equation in x is of the form ax + b = 0. 3. Two equations with the same solution sets are called equivalent. 4. A conditional equation is one that is not true for some values of the variables. 5. False. The interest I = (100)(0.05)(3). 6. False. Since the rate is given in feet per second, the time must also be converted to seconds. 15 minutes = 15(60) = 900 seconds Therefore, d = 60(900) feet. 7. a. Substitute 0 for x in the equation 2 5 6 x x − = + : 0 2 5(0) 6 2 6 − = + ⇒ − ≠ So, 0 is not a solution of the equation. b . Substitute –2 for x in the equation 2 5 6 x x − = + : 2 2 5( 2) 6 4 10 6 4 4 − − = − + ⇒ − = − + ⇒ − = − So, –2 is a solution of the equation. 8. a. Substitute –1 for x in the equation 8 3 14 1 x x + = − : 8( 1) 3 14( 1) 1 8 3 14 1 5 15 − + = − − ⇒ − + = − − ⇒ − ≠ − So, –1 is not a solution of the equation. b. Substitute 2 3 for x in the equation 8 3 14 1 x x + = − : 2 2 16 28 8 3 14 1 3 1 3 3 3 3 25 25 3 3 ⎛ ⎞ ⎛ ⎞ + = − ⇒ + = − ⇒ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = So, 2 3 is a solution of the equation. 9. a. Substitute 4 for x in the equation 2 1 1 3 2 x x = + + : 2 1 1 1 1 1 1 1 4 3 4 2 2 3 6 2 2 = + ⇒ = + ⇒ = + So, 4 is a solution of the equation. b. Substitute 1 for x in the equation 2 1 1 3 2 x x = + + : 2 1 1 1 1 2 2 2 1 3 1 2 3 3 3 = + ⇒ = + ⇒ ≠ + So, 1 is not a solution of the equation. 10. a. Substitute 1 2 for x in the equation ( 3)(2 1) 0 x x − + = : 1 1 5 3 2 1 0 (2) 0 2 2 2 5 0 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − ⋅ + = ⇒ − = ⇒ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ − ≠ So, 1 2 is not a solution of the equation.

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    Section 1.1 Linear Equations in One Variable 43 b. Substitute 3 for x in the equation ( 3)(2 1) 0 x x − + = : ( ) ( ) 3 3 2 3 1 0 (0)(7) 0 0 0 − ⋅ + = ⇒ = ⇒ = So, 3 is a solution of the equation. 11. a. The equation 2 3 5 x x x + = is an identity, so every real number is a solution of the equation. Thus 157 is a solution of the equation. This can be checked by substituting 157 for x in the equation: 2(157) 3(157) 5(157) 314 471 785 785 785 + = ⇒ + = ⇒ = b. The equation 2 3 5 x x x + = is an identity, so every real number is a solution of the equation. Thus 2046 − is a solution of the equation. This can be checked by substituting 2046 − for x in the equation: 2( 2046) 3( 2046) 5( 2046) 4092 6138 10, 230 10, 230 10, 230 − + − = − − − = − − = − 12. Both sides of the equation (2 ) 4 7 3( 4) x x x − − = − + are defined for all real numbers, so the domain is ( , ) −∞ ∞ . 13. The left side of the equation 3 1 2 y y y = − + is not defined if 1 y = , and the right side of the equation is not defined if 2 y = − . The domain is ( , 2) ( 2, 1) (1, ) −∞ − − ∞ ∪ ∪ . 14. The left side of the equation 1 2 y y = + is not defined if 0 y = . The right side of the equation is not defined if 0 y < , so the domain is (0, ) ∞ . 15. The left side of the equation 3 2 9 ( 3)( 4) x x x x = + − − is not defined if 3 x = or 4 x = . The right side is defined for all real numbers. So, the domain is ( , 3) (3, 4) (4, ) −∞ ∞ ∪ ∪ . 16. The left side of the equation 2 1 1 x x = − is not defined if 0 x ≤ . The right side of the equation is defined for all real numbers. So the domain is (0, ) ∞ . 17. Substitute 0 for x in 2 3 5 1 x x + = + . Because 3 1 ≠ , the equation is not an identity. 18. When the like terms on the right side of the equation 3 4 6 2 (3 2) x x x + = + − − are collected, the equation becomes 3 4 3 4 x x + = + , which is an identity. 19. When the terms on the left side of the equation 1 1 2 2 2 x x x + + = are collected, the equation becomes 2 2 2 2 x x x x + + = , which is an identity. 20. The right side of the equation 1 1 1 3 3 x x = + + is not defined for 0 x = , while the left side is defined for 0 x = . Therefore, the equation is not an identity. In exercises 21–46, solve the equations using the procedures listed on page 79 in your text: eliminate fractions, simplify, isolate the variable term, combine terms, isolate the variable term, and check the solution. 21. 3 5 14 3 5 5 14 5 3 9 3 9 3 3 3 x x x x x + = + − = − = = = Solution set: {3} 22. 2 17 7 2 17 17 7 17 2 24 2 24 2 2 12 x x x x x − = − + = + = = = Solution set: {12} 23. 10 12 32 10 12 12 32 12 10 20 10 20 10 10 2 x x x x x − + = − + − = − − = − = − − = − Solution set: {−2} 24. 2 5 6 2 5 5 6 5 2 1 2 1 1 2 2 2 x x x x x − + = − + − = − − = − = ⇒ = − − − Solution set: 1 2 ⎧ ⎫ − ⎨ ⎬ ⎩ ⎭

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    44 Chapter 1 Equations and Inequalities 25. 3 4 3 3 4 3 7 7 y y y y − = − − − = − − − = − ⇒ = Solution set: {7} 26. 2 7 23 2 7 2 23 2 7 21 7 21 3 7 7 y y y y y − = − − = − − = − = ⇒ = − − − Solution set: {−3} 27. 7 7 2( 1) 7 7 2 2 7 7 7 2 2 7 7 2 5 7 2 2 5 2 5 5 5 5 1 5 5 x x x x x x x x x x x x x x x + = + + = + + − = + − = − − = − − = − − = ⇒ = − Solution set: {−1} 28. 3( 2) 4 3 6 4 3 6 4 4 6 4 4 6 6 4 6 4 2 4 2 1 4 4 2 x x x x x x x x x x x x x + = − + = − + + = − + + = + − = − = − − = ⇒ = − Solution set: 1 2 ⎧ ⎫ − ⎨ ⎬ ⎩ ⎭ 29. 3(2 ) 5 3 6 3 5 3 6 2 3 6 2 2 3 2 6 y y y y y y y y y y y y y − + = − + = + = + − = − ⇒ = Solution set: {6} 30. 9 3( 1) 6 9 3 3 6 6 3 6 6 3 6 5 3 6 5 3 3 6 3 5 3 5 3 3 5 5 5 y y y y y y y y y y y y y y y y y − − = + − + = + + = + + − = + − + = + − = − = = ⇒ = Solution set: 3 5 ⎧ ⎫ ⎨ ⎬ ⎩ ⎭ 31. 4 3 7 2 (7 ) Distribute 1 to clear the parentheses. 7 2 7 7 5 7 5 5 5 12 y y y y y y y y − + − = − − − = − + = − + + = − + + ⇒ = Solution set: {12} 32. 3( 1) 6 4 2 4 3 3 4 4 3 3 3 4 4 3 3 4 1 3 4 4 1 4 1 1 y y y y y y y y y y y y y y y y − = − + − − = − − + = − + = − − = − − − = − ⇒ = Solution set: {1} 33. 3( 2) 2(3 ) 1 3 6 6 2 1 1 x x x x x − + − = − + − = ⇒ = Solution set: {1} 34. 2 3 (3 1) 6 Distribute 1 to clear the parentheses. 2 3 3 1 6 2 6 2 2 6 2 8 8 x x x x x x x x − − − = − − − + = − − = − − + = + − = ⇒ = − Solution set: {−8} 35. 2 3( 4) 7 10 2 3 12 7 10 5 12 7 10 5 12 12 7 10 12 5 7 22 5 7 7 22 7 2 22 2 22 11 2 2 x x x x x x x x x x x x x x x x x x x + − = + + − = + − = + − + = + + = + − = + − − = − = ⇒ = − − − Solution set: {−11} 36. 3(2 3 ) 4 3 10 6 9 4 3 10 6 13 3 10 6 13 6 3 10 6 13 3 16 13 3 3 16 3 16 16 16 16 1 16 16 x x x x x x x x x x x x x x x x x x x − − = − − − = − − = − − − = − − − = − − − = − − − = − − − = ⇒ = − − Solution set: {1} 37. 4[ 2(3 )] 2 1 Distribute 2 to clear the inner parentheses. 4[ 6 2 ] 2 1 Combine like terms within the brackets. 4[6 ] 2 1 Distribute 4 to clear the brackets. 24 4 2 1 x x x x x x x x x x + − = + + − = + − = + − = + 24 4 24 2 1 24 4 2 23 4 2 23 6 23 6 23 23 6 6 6 x x x x x x x x x − − = + − − = − − − = − − = − − − = ⇒ = − − Solution set: 23 6 ⎧ ⎫ ⎨ ⎬ ⎩ ⎭

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    Section 1.1 Linear Equations in One Variable 45 38. 3 [ 3( 2)] 4 Distribute 3 to clear the parentheses. 3 [ 3 6] 4 Combine like terms in the brackets. 3 [ 2 6] 4 Distribute 1 to clear the brackets. 3 2 6 4 2 9 4 2 9 9 4 9 2 5 2 5 5 2 2 2 x x x x x x x x x x x − − + = − − − − = − − − = − + + = + = + − = − = − − = ⇒ = − Solution set: 5 2 ⎧ ⎫ − ⎨ ⎬ ⎩ ⎭ 39. 3(4 3) 4[ (4 3)] Distribute 3 on the left side and 1 on the right side to clear parentheses. 12 9 4[ 4 3] Combine like terms in the brackets. 12 9 4[ 3 3] Distribute 4 to clear the brackets. 12 9 1 y y y y y y y y y − = − − − − = − + − = − + − = − 2 12 12 9 9 12 12 9 12 12 21 12 12 12 21 12 24 21 24 21 21 7 24 24 24 8 y y y y y y y y y y y y + − + = − + + = − + + = − + + = = ⇒ = = Solution set: 7 8 ⎧ ⎫ ⎨ ⎬ ⎩ ⎭ 40. 5 (6 9) 2 2( 1) Distribute 1 on the left and 2 on the right to clear the parentheses. 5 6 9 2 2 2 4 4 2 2 4 4 4 2 2 4 4 2 6 4 2 2 6 2 6 6 6 6 6 6 1 y y y y y y y y y y y y y y y y y y y − + + = + − − − + = + − − = + − − + = + + − = + − − = + − − = − = − − = − Solution set: {−1} 41. 2 3(2 ) ( 3) 2 1 Distribute 3 on the left to clear the parentheses. 2 6 3 3 2 1 5 6 3 2 x x x x x x x x x x − − = − + + − − + = − + + − = − 5 6 6 3 2 6 5 3 4 5 3 3 4 3 2 4 2 4 2 2 2 x x x x x x x x x x x − + = − + = + − = + − = = ⇒ = Solution set: {2} 42. 5( 3) 6( 4) 5 Distribute 5 to clear the first set of parentheses. Distribute 6 to clear the second set of parentheses. 5 15 6 24 5 9 5 9 9 5 9 14 14 x x x x x x x x − − − = − − − − + = − − + = − − + − = − − − = − ⇒ = Solution set: {14} 43. 2 1 4 1 9 6 To clear the fractions, multiply both sides of the equation by the least common denominator, 36. 2 1 4 36 36(1) 9 6 4(2 1) 6( 4) 36 8 4 6 24 36 2 20 36 2 20 20 36 20 2 56 2 56 2 2 2 x x x x x x x x x x x x x + + − = + + ⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ + − + = + − − = − = − + = + = = ⇒ = 8 Solution set: {28} 44. 2 3 1 3 7 3 7 To clear the fractions, multiply both sides of the equation by the least common denominator, 21. 2 3 1 3 21 21 7 3 7 3(2 3 ) 7( 1) 3(3 ) 6 9 7 7 9 1 2 9 1 2 2 9 2 1 11 1 x x x x x x x x x x x x x x x x x x x − − + = − − ⎛ ⎞ ⎛ ⎞ + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ − + − = − + − = − − = − − + = + − = − 11 1 11 11 11 x x = ⇒ = − Solution set: 1 11 ⎧ ⎫ − ⎨ ⎬ ⎩ ⎭

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    46 Chapter 1 Equations and Inequalities 45. 1 5 1 2( 1) 3 4 2 8 To clear the fractions, multiply both sides by the least common denominator, 8. 1 5 1 2( 1) 8 8 3 4 2 8 Distribute the 8 on both sides. 1 5 1 2( 1) 8 8 8(3) 8 4 2 x x x x x x x x x − + + + = − − + + ⎛ ⎞ ⎛ ⎞ + = − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ − + + ⎛ ⎞ ⎛ ⎞ + = − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 8 2(1 ) 4(5 1) 8(3) 2( 1) Simplify by collecting like terms and combining constants. 2 2 20 4 24 2 2 18 6 22 2 18 6 2 22 2 2 20 6 22 20 6 6 22 6 20 16 20 16 20 20 16 4 20 5 x x x x x x x x x x x x x x x x x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ − + + = − + − + + = − − + = − + + = − + + = + − = − = = = = Solution set: 4 5 ⎧ ⎫ ⎨ ⎬ ⎩ ⎭ 46. 4 1 3 2 2 3 2 6 x x x + + + − = To clear the fractions, multiply both sides of the equation by the least common denominator, 6. 4 1 3 2 6 2 6 3 2 6 x x x + + ⎛ ⎞ ⎛ ⎞ + − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Distribute the 6 on both sides. ( ) ( ) 4 1 3 2 6 6 2 6 6 3 2 6 2 4 12 3 3 2 x x x x x x + + ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + − = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ + + − = + Simplify by collecting like terms and combining constants. 2 8 12 3 3 2 14 5 3 2 14 5 3 3 2 3 11 5 2 11 5 5 2 5 11 3 11 3 3 11 11 11 x x x x x x x x x x x x x x + + − = + + = + + − = + − + = + − = − = − − = ⇒ = − Solution set: 3 11 ⎧ ⎫ − ⎨ ⎬ ⎩ ⎭ 47. To solve d rt = for r , divide both sides of the equation by t . d r t = . 48. To solve F ma = for a, divide both sides of the equation by m. F a m = . 49. To solve 2 C r π = for r, divide both sides of the equation by 2 π . 2 C r π = . 50. 2 2 2 2 2 2 2 2 To solve 2 for , subtract from both sides. 2 2 Divide both sides by 2 . 2 2 2 2 A rx r x r A r rx r r A r rx r A r rx r r A r x r π π π π π π π π π π π π π π π π = + − = + − − = − = − = 51. To solve for , multiply both sides by . E I R R R E RI R RI E R = ⎛ ⎞ = ⇒ = ⎜ ⎟ ⎝ ⎠ Divide both sides by . I RI E E R I I I = ⇒ = 52. To solve (1 ) for , distribute . Subtract from both sides. Divide both sides by . A P rt t P A P Prt P A P P Prt P A P Prt Pr A P Prt Pr Pr A P t Pr = + = + − = + − − = − = − = 53. ( ) To solve for , multiply both 2 sides by 2. 2 ( ) Divide both sides by ( ). 2 ( ) 2 a b h A h A a b h a b A a b h A h a b a b a b + = = + + + = ⇒ = + + +

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    Section 1.1 Linear Equations in One Variable 47 54. To solve ( 1) for , subtract from both sides. ( 1) ( 1) Divide both sides by ( 1). ( 1) 1 1 1 T a n d d a T a a n d a T a n d n T a n d T a d n n n = + − − = + − − − = − − − − − = ⇒ = − − − 55. 1 1 1 To solve for , clear the fractions by multiplying both sides by the least common denominator, . 1 1 1 u f u v fuv fuv fuv f u v = + ⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 1 1 1 Simplify. Subtract from both sides. Factor the left side. ( ) Divide both sides by . ( ) fuv fuv fuv f u v uv fv fu fu uv fu fv fu fu uv fu fv u v f fv v f u v f fv fv u v f v f v f ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = + − = + − − = − = − − = ⇒ = − − − 56. 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 2 2 1 1 1 To solve for , clear the fractions by multiplying both sides by the least common denominator, . 1 1 1 1 1 1 Si R R R R R R R R R R R R R R R R R R R R R R R R R R R R = + ⎛ ⎞ ⎛ ⎞ ⋅ ⋅ = ⋅ ⋅ + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⋅ ⋅ = ⋅ ⋅ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ + ⎜ ⎟ ⎝ ⎠ i i i i 1 2 2 1 mplify. R R RR RR = + 2 1 2 2 2 1 2 1 2 2 1 Subtract from both sides. RR R R RR RR RR RR R R RR RR − = + − − = 2 1 1 1 2 1 1 1 2 1 1 1 Factor the left side. ( ) Divide both sides by ( ). ( ) R R R RR R R R R R RR RR R R R R R R R − = − − = ⇒ = − − − 57. To solve for , subtract from both sides. Divide both sides by . y mx b m b y b mx b b y b mx x y b mx y b m x x x = + − = + − ⇒ − = − − = ⇒ = 58. To solve for , subtract from both sides. Divide both sides by . ax by c y ax ax by ax c ax by c ax b by c ax c ax y b b b + = + − = − ⇒ = − − − = ⇒ = 59. 0.065 x 60. 5 6 x 61. $22, 000 x − 62. 229.50 72 x − 1.1 Applying the Concepts 63. The formula for volume is V lwh = . Substitute 2808 for V , 18 for l , and 12 for h . Solve for w . 2808 18 12 2808 216 2808 216 216 216 13 w w w w = ⋅ ⋅ = = = The width of the pool is 13 ft. 64. The formula for volume is V lwh = . Substitute 168 for V , 7 for l , and 3 for w . Solve for h . 168 7 3 168 21 168 21 21 21 8 h h h h = ⋅ ⋅ = = = The hole must be 8 ft deep. 65. Let w = the width of the rectangle. Then 2 w − 5 = the length of the rectangle. ( ) 2 2 2 5 80 2 4 10 80 6 10 80 6 90 15, 2 5 25 w w w w w w w w + − = + − = − = = = − = The width of the rectangle is 15 ft and its length is 25 feet.

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    48 Chapter 1 Equations and Inequalities 66. Let l = the length of the rectangle. Then 1 3 2 l + = the width of the rectangle. 1 2 2 3 36 2 2 6 36 3 6 36 3 30 1 10, 3 8 2 l l l l l l l l ⎛ ⎞ + + = ⎜ ⎟ ⎝ ⎠ + + = + = = = + = The length of the rectangle is 10 ft and its width is 8 ft. 67. The formula for circumference of a circle is 2 C r π = . Substitute 114 π for C . Solve for r . 114 2 114 2 57 2 2 r r r π π π π π π = ⇒ = ⇒ = The radius is 57 cm. 68. The formula for perimeter of a rectangle is 2 2 P l w = + . Substitute 28 for P and 5 for w . Solve for l . 28 2 2(5) 28 2 10 28 10 2 10 10 18 2 18 2 2 2 9 l l l l l l = + = + − = + − = = = The length is 9 m. 69. The formula for surface area of a cylinder is 2 2 2 S rh r π π = + . Substitute 6 π for S and 1 for r . Solve for h . 2 6 2 (1) 2 (1 ) 6 2 2 6 2 2 2 2 4 2 4 2 2 2 2 h h h h h h π π π π π π π π π π π π π π π π π = + = + − = + − = = ⇒ = The height is 2 m. 70. The formula for volume of a cylinder is 2 V r h π = . Substitute 148 π for V and 2 for r . Solve for h . 2 148 2 148 4 148 4 4 4 37 h h h h π π π π π π π π = ⋅ ⋅ = = = The height of the can is 37 cm. 71. The formula for area of a trapezoid is ( ) 1 2 1 2 A h b b = + . Substitute 66 for A , 6 for h , and 3 for 1 b . Solve for 2 b . ( ) ( ) 2 2 2 2 2 2 2 1 66 6 3 2 66 3 3 66 9 3 66 9 9 3 9 57 3 3 57 3 3 19 b b b b b b b = ⋅ + = + = + − = + − = = = The length of the second base is 19 ft. 72. The formula for area of a trapezoid is ( ) 1 2 1 2 A h b b = + . Substitute 35 for A , 9 for 1 b , and 11 for 2 b . Solve for h . ( ) ( ) 1 35 9 11 2 1 35 20 2 35 10 35 10 3.5 10 10 h h h h h = + = = = ⇒ = The height of the trapezoid is 3.5 cm. 73. Let x = the cost of the less expensive land. Then x + 23,000 = the cost of the more expensive land. Together they cost $147,000, so ( 23, 000) 147, 000 2 23, 000 147, 000 2 124, 000 62, 000 x x x x x + + = + = = ⇒ = The less expensive piece of land costs $62,000 and the more expensive piece of land costs $62,000 + $23,000 = $85,000. 74. Let x = the amount the assistant manager earns. Then x + 450 = the amount the manager earns. Together they earn $3700, so ( 450) 3700 2 450 3700 2 3250 1625 x x x x x + + = + = = ⇒ = The assistant manager earns $1625, and the manager earns $1625 + $450 = $2075.

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    Section 1.1 Linear Equations in One Variable 49 75. Let x = the lottery ticket sales in July. Then 1.10 x = the lottery ticket sales in August. A total of 1113 tickets were sold, so 1.10 1113 2.10 1113 530 x x x x + = = ⇒ = 530 tickets were sold in July, and 1.10(530) = 583 tickets were sold in August. 76. Let x = Jan’s commission in March. Then 15 + 0.5 x = Jan’s commission in February. She earned a total of $633, so (15 0.5 ) 633 1.5 15 633 1.5 618 412 x x x x x + + = + = ⇒ = ⇒ = Jan’s commission was $412 in March and 15 + 0.5(412) = $221 in February. 77. Let x = the amount the younger son receives. Then 4 x = the amount the older son receives. Together they receive $225,000, so 4 225, 000 5 225, 000 45, 000 x x x x + = ⇒ = ⇒ = The younger son will received $45,000, and the older son will receive 4($45,000)  = $180,000. 78. Let x  = the amount Kevin kept for himself. Then 2 x = the amount he gave his daughter, and 4 x = the amount he gave his dad. He won $735,000, so 735, 000 2 4 4 4(735, 000) 2 4 4 2 2, 940, 000 7 2, 940, 000 420, 000 x x x x x x x x x x x + + = ⎛ ⎞ + + = ⎜ ⎟ ⎝ ⎠ + + = = ⇒ = Kevin kept $420,000 for himself. He gave $420, 000 2 $210, 000 = to his daughter and $420, 000 4 $105, 000 = to his dad. 79. a. Let x = the number of points needed to average 75. 87 59 73 75 4 219 300 81 x x x + + + = + = = You need to score 81 in order to average 75. b. 87 59 73 2 75 5 219 2 375 2 156 78 x x x x + + + = + = = = You need to score 78 in order to average 75 if the final carries double weight. 80. Let x = the amount invested in real estate. Then 4200 –   x = the amount invested in a savings and loan. Investment Principal Rate Time Interest Real estate x 0.15 1 0.15 x Savings 4200 – x 0.08 1 0.08(4200 – x ) The total income was $448, so 0.15 0.08(4200 ) 448 0.15 336 0.08 448 0.07 336 448 0.07 112 1600 x x x x x x x + − = + − = + = = ⇒ = So, the real estate agent invested $1600 in real estate and 4200 – 1600 = $2600 in a savings and loan. 81. Let x = the amount invested in a tax shelter. Then 7000 – x = the amount invested in a bank. Investment Principal Rate Time Interest Tax shelter x 0.09 1 0.09 x Bank 7000 – x 0.06 1 0.06(7000 – x ) The total interest was $540, so 0.09 0.06(7000 ) 540 0.09 420 0.06 540 0.03 420 540 0.03 120 4000 x x x x x x x + − = + − = + = = ⇒ = Mr. Mostafa invested $4000 in a tax shelter and 7000 – 4000 = $3000 in a bank. 82. Let x = the amount invested at 6%. Then 4900 – x = the amount invested at 8% Principal Rate Time Interest x 0.06 1 0.06 x 4900 – x 0.08 1 0.08(4900 – x ) The amount of interest for each investment is equal, so 0.06 0.08(4900 ) 0.06 392 0.08 0.14 392 2800 x x x x x x = − = − = ⇒ = Ms. Jordan invested $2800 at 6% and $2100 at 8%. The amount of interest she earned on each investment is $168, so she earned $336 in all.

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    50 Chapter 1 Equations and Inequalities 83. Let x = the amount to be invested at 8%. Principal Rate Time Interest 5000 0.05 1 250 x 0.08 1 0.08 x 5000 + x 0.06 1 0.06(5000 + x ) The amount of interest for the total investment is the sum of the interest earned on the individual investments, so 0.06(5000 ) 250 0.08 300 0.06 250 0.08 50 0.06 0.08 50 0.02 2500 x x x x x x x x + = + + = + + = = ⇒ = So, $2500 must be invested at 8%. 84. Let x = the selling price. Then x – 480 = the profit. So 480 0.2 x x − = ⇒ 480 0.8 x − = − ⇒ 600 x = . The selling price is $600. 85. There is a profit of $2 on each shaving set. They want to earn $40,000 + $30,000 = $70,000. Let x = the number of shaving sets to be sold. Then 2 x = the amount of profit for x shaving sets. So, 2 70, 000 35, 000 x x = ⇒ = They must sell 35,000 shaving sets. 86. Let t = the time each traveled. Then 100 t = Angelina’s rate and 150 t = Harry’s rate. Rate Time Distance Angelina 100 t t 100 Harry 150 t t 150 Harry’s rate is 15 meters per minute faster than Angelina’s, so we have 100 150 15 100 15 150 15 50 10 min 3 t t t t t + = + = = = So, Angelina jogged at 100 30 10 3 = meters per minute. Harry biked at 150 45 10 3 = meters per minute. 87. Let x = the time the second car travels. Then 1 + x = the time the first car travels. So, Rate Time Distance First car 50 1 + x 50(1 ) x + Second car 70 x 70 x The distances are equal, so 50(1 ) 70 50 50 70 50 20 2.5 x x x x x x + = + = = ⇒ = So, it will take the second car 2.5 hours to overtake the first car. 88. Let x = the time the planes travel. So, Rate Time Distance First plane 470 x 470 x Second plane 430 x 430 x The planes are 2250 km apart, so 470 430 2250 900 2250 2.5 x x x x + = ⇒ = ⇒ = So, the planes will be 2250 km apart at 2.5 hours. 89. At 20 miles per hour, it will take Lucas two minutes to bike the remaining 2/3 of a mile. 20 mi 20 mi 1mi 1 hr 60 min 3 min ⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ So his brother will have to bike 1 mile in 2 minutes: 1mi 30 mi 30 mi 2 min 60 min 1 hr = = 90. Driving at 40 miles per hour, it will take Karen’s husband 45 40 hours or 1 hour and 7.5 minutes to get to the airport. Driving at 60 miles per hour, it will take Karen 45 minutes to get to the airport. Her husband has already driven for 15 minutes, so it will take him an additional 52.5 minutes to get to the airport. Karen will get there before he does. 91. Let x = the rate the slower car travels. Then x + 5 = the rate the faster car travels. So, Rate Time Distance First car x 3 3 x Second car x + 5 3 ( ) 3 5 x + ( continued on next page )

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    Section 1.1 Linear Equations in One Variable 51 ( continued ) The cars are 405 miles apart, so ( ) 3 3 5 405 3 3 15 405 6 15 405 6 390 65 x x x x x x x + + = + + = + = ⇒ = ⇒ = One car is traveling at 65 miles per hour, and the other car is traveling at 70 miles per hour. 92. Let x = the distance to Aya’s friend’s house. Rate Distance Time go 16 x 16 x return 80 x 80 x She traveled for a total of 3 hours, so 3 16 80 80 80(3) 16 80 5 240 6 240 40 x x x x x x x x + = ⎛ ⎞ + = ⎜ ⎟ ⎝ ⎠ + = ⇒ = ⇒ = So, her friend lives 40 km away. 93. Substitute 170 for P into the formula 200 0.02 P q = − . Solve for q . 170 200 0.02 170 200 200 0.02 200 30 0.02 30 0.02 1500 0.02 0.02 q q q q q = − − = − − − = − − − = ⇒ = − − Note that the solution must fall between 100 and 2000 cameras. 1500 cameras must be ordered. 94. Substitute 37,000 for Q, 1500 for L , and 3200 for I into the formula A I Q L − = . Solve for A . 3200 37, 000 1500 3200 1500(37, 000) 1500 1500 55, 500, 000 3200 55,500, 000 3200 3200 3200 55,503, 200 A A A A A − = − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = − + = − + = The current assets are $55,503,200. 95. Let x = one number. Then 3 x = the other number. 3 28 4 28 7, 3 3(7) 21 x x x x x + = ⇒ = ⇒ = = = The numbers are 7 and 21. 96. Let x = the first even integer. Then x + 2 = the second even integer, and x + 4 = the third even integer. ( ) ( ) 2 4 42 3 6 42 3 36 12 2 14, 4 16 x x x x x x x x + + + + = + = = ⇒ = + = + = The numbers are 12, 14, and 16. 97. Let x = one number. Then 2 x is the second number. 2 14 14 x x x − = = The numbers are 14 and 28. 98. Let x = one number. Then x + 5 = the second number. Note that the second number is the larger number. ( ) 2 5 49 2 10 49 3 10 49 3 39 13 x x x x x x x + + = + + = + = = = The numbers are 13 and 18. 1.1 Beyond the Basics 99. a. The solution set of 2 x x = is {0,1} , while the solution set of 1 x = is {1}. Therefore, the equations are not equivalent. b. The solution set of 2 9 x = is { 3, 3} − , while the solution set of 3 x = is {3}. Therefore, the equations are not equivalent. c. The solution set of 2 1 1 x x − = − is {0, 1}, while the solution set of 0 x = is {0}. Therefore, the equations are not equivalent. d. The equation 2 2 2 x x x = − − is an inconsistent equation, so is solution set is ∅ . The solution set of 2 x = is {2}. Therefore, the equations are not equivalent. 100. First, solve 7 2 16 x + = . Subtracting 2 from both sides, we have 7 14 x = . Then divide both sides by 7; we obtain 2 x = . Now substitute 2 for x in 3 1 x k − = . This becomes 3(2) 1 k − = , so 5 k = .

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    52 Chapter 1 Equations and Inequalities 101. Let x = the average speed for the second half of the trip. Rate Distance Time 1 st half 75 D 75 D 2 nd half x D D x Entire trip 60 2 D 2 60 D 2 75 60 2 300 300 75 60 4 300 5 (2 ) 4 300 10 300 6 50 D D D x D D D x x x Dx D x D Dx D Dx D Dx x + = ⎛ ⎞ ⎛ ⎞ + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ + = + = = ⇒ = The average speed for the second half of the drive is 50 mph. 102. First we need to compute how much time it will take for Davinder and Mikhael to meet. Let t = the time it will take for them to meet. So, Rate Time Distancee Davinder 3.7 t 3.7 t Mikhail 4.3 t 4.3 t 3.7 4.3 2 8 2 0.25 t t t t + = ⇒ = ⇒ = They will be walking for 0.25 hour until they meet. The dog starts with Davinder. Let 1 d t = the amount of time it takes for the dog to meet Mikhail. So, Rate Time Distance dog 6 1 d t 1 6 d t Mikhail 4.3 1 d t 1 4.3 d t 1 1 1 6 4.3 2 10.3 2 0.19 d d d t t t t + = = = The dog meets Mikhail for the first time when they have walked for 0.19 hour. The dog will have traveled 1.14 mi. While the dog has been running towards Mikhail, Davinder has continued to walk. During the 0.19 hour, he walked 0.70 mi, so now he and the dog are 1.14 – 0.70 = 0.44 mi apart. Let 2 d t = the time it takes the dog to meet Davinder. Rate Time Distance dog 6 2 d t 2 6 d t Davinder 3.7 2 d t 2 3.7 d t 2 2 2 6 3.7 0.44 9.7 0.44 0.05 d d d t t t t + = = = So, the dog meets Davinder when they have walked for another 0.05 hour. The dog will have traveled 0.3 mi. They have now walked for 0.19 + 0.05 = 0.24 hr. Since Davinder and Mikhail don’t meet until they have walked for 0.25 hours, the dog must walk for 0.01 hr more. In that time, the dog will travel 0.06 mi. So in total the dog will travel 1.14 + 0.3 + 0.06 = 1.5 mi. 103. Let x = the number of liters of water in the original mixture. Then 5 x = the number of liters of alcohol in the original mixture, and 6 x = the total number of liters in the original mixture. x + 5 = the number of liters of water in the new mixture. Then 6 x + 5 = the total number of liters in the new mixture. Since the ratio of alcohol to water in the new mixture is 5:2, then the amount of alcohol in the new mixture is 5/7 of the total mixture or 5 (6 5) 7 x + . There was no alcohol added, so the amount of alcohol in the original mixture equals the amount of alcohol in the new mixture. This gives 5 (6 5) 5 7 5(6 5) 35 30 25 35 25 5 5 x x x x x x x x + = + = + = ⇒ = ⇒ = So, there were 5 liters of water in the original mixture and 25 liters of alcohol. 104. Let x = the amount of each alloy. There are 13 parts in the first alloy and 8 parts in the second alloy. We can use the following table to organize the information: Total Zinc Copper Alloy 1 x 5 13 x 8 13 x Alloy 2 x 5 8 x 3 8 x Total 2 x 5 5 13 8 x x + 8 3 13 8 x x + ( continued on next page )

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    Section 1.1 Linear Equations in One Variable 53 ( continued ) The amount of zinc in the new mixture is 5 5 105 13 8 104 x x x + = , and the amount of copper in the new mixture is 8 3 103 13 8 104 x x x + = . So, the ratio of zinc to copper in the new mixture is 105 103 : 104 104 x x or 105:103. 105. Let x = Democratus’ age now. Then 6 x = the number of years as a boy, 8 x = the number of years as a youth, and 2 x = the number of years as a man. He has spent 15 years as a mature adult. So, 15 6 8 2 24 15 24 6 8 2 4 3 12 360 24 19 360 24 360 5 72 x x x x x x x x x x x x x x x x + + + = ⎛ ⎞ + + + = ⎜ ⎟ ⎝ ⎠ + + + = + = = ⇒ = Democratus is 72 years old. 106. Let x = the man’s age now. When the woman is x years old, the man will be 119 – x years old. So the difference in their ages is (119 ) 119 2 x x x − − = − years. So the woman’s age now is (119 2 ) 3 119 x x x − − = − . When the man was 3 119 x − years old, she was 3 119 2 x − years old. Since the difference in their ages is 119 2 x − , we have ( ) 3 119 3 119 119 2 2 6 238 3 119 238 4 3 119 238 4 7 119 238 7 357 51 x x x x x x x x x x x − − − = − − − + = − − = − − = = ⇒ = So the man is now 51 years old. Check by verifying the facts in the problem. When she is 51 years old, he will be 119 – 51 = 68 years old. The difference in their ages is 68 – 51 = 17 years. So she is 51 – 17 = 34 years old now. When he was 34 years old, she was 17 years old, which is 1/2 of 34. 107. There are 180 minutes from 3 p.m. to 6 p.m. So, the number of minutes before 6 p.m. plus 50 minutes plus 4 × the number of minutes before 6 p.m. equals 180 minutes. Let x = the number of minutes before 6 p.m. So, 50 4 180 x x + + = ⇒ 5 50 180 x + = ⇒ 5 130 26 x x = ⇒ = So it is 26 minutes before 6 p.m. or 5:34 p.m. Check this by verifying that 26 + 50 = 76 minutes before 6 p.m. is the same time as 4(26) = 104 minutes after 3 p.m. Seventy-six minutes before 6 p.m. is 4:44 p.m., while 104 minutes after 3 p.m. is also 4:44 p.m. 108. Let x = the number of minutes pipe B is open. Pipe A is open for 18 minutes, so it fills 18/24 or 3/4 of the tank. Pipe B fills x /32 of the tank. So, 3 3 1 32 32(1) 4 32 4 32 x x ⎛ ⎞ + = ⇒ + = ⇒ ⎜ ⎟ ⎝ ⎠ 24 32 8 x x + = ⇒ = Pipe B should be turned off after 8 minutes. 109. a. Because of the head wind, the plane flies at 140 mph from Atlanta to Washington and 160 mph from Washington to Atlanta. Let x = the distance the plane flew before turning back. So, Rate Distance Time to 140 x 140 x from 160 x 160 x 1.5 140 160 160 140 1.5(140)(160) 300 33, 600 112 x x x x x x + = + = = ⇒ = The plane flew 112 miles before turning back. b. The plane traveled 224 miles in 1.5 hours, so the average speed is 224 149.33 1.5 = mph.

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