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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition

Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition is the ultimate guide for understanding and solving textbook problems.

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    I NSTRUCTOR ’ S S OLUTIONS M ANUAL E LEMENTARY AND I NTERMEDIATE A LGEBRA G RAPHS AND M ODELS F IFTH E DITION Marvin L. Bittinger Indiana University P u rdue University Indianapolis David J. Ellenbogen Community College of Vermont Barbara L. Johnson Ivy Tech Community College of Indiana

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    CONTENTS Chapter 1 Introduction to Algebraic Expressions ................................ ................................ ... 1 Chapter 2 Equations, Inequalities, and Problem Solving ................................ .................. 5 1 Chapter 3 Introduction to Graphing and Functions ................................ ........................... 157 Chapters 1 – 3 Cumulative Review ................................ ................................ ........................... 281 Chapter 4 Systems of Equations in Two Variables ................................ ........................... 289 Chapter 5 Polynomials ................................ ................................ ................................ ................... 377 Chapter 6 Polynomial Factorizations and Equations ................................ ........................ 465 Chapters 1 – 6 Cumulative Review ................................ ................................ ........................... 551 Chapter 7 Rational Expressions, Equations, and Functions ................................ .......... 559 Chapter 8 Inequalities ................................ ................................ ................................ ..................... 683 Chapter 9 More on Systems ................................ ................................ ................................ ......... 751 Chapters 1 – 9 Cumulative Review ................................ ................................ ........................... 829 Chapter 10 Exponents and Radical Functions ................................ ................................ ..... 837 Chapter 11 Quadratic Functions and Equations ................................ ................................ . 921 Chapter 12 Exponential Functions and Logarithmic Functions ............................... 1015 Chapters 1 – 12 Cumulative Review ................................ ................................ ..................... 1109 Chapter 13 Conic Sections ................................ ................................ ................................ ........ 1119 Chapter 14 Sequences, Series, and the Binomial Theorem ................................ ........ 1179 Chapters 1 – 14 Cumulative Review ................................ ................................ ..................... 1223 Chapter R Elementary Algebra Review ................................ ................................ .............. 1233

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    Chapter 1 Introduction to Algebraic Expressions Exercise Set 1.1 1. Expression 2. Equation 3. Equation 4. Expression 5. Equation 6. Equation 7. Expression 8. Equation 9. Equation 10. Expression 11. Expression 12. Expression 13. Substitute 9 for a and multiply. 3 3 9 27 a =  = 14. Substitute 7 for x and multiply. 8 8 7 56 x =  = 15. Substitute 2 for t and add. 6 2 6 8 t + = + = 16. Substitute 9 for r and subtract. 13 13 9 4 r − = − = 17. Substitute 2 for x , 14 for y . Add and then divide the sum by 4. 2 14 16 4 4 4 4 x y + + = = = 18. Substitute 15 for c , 20 for d , add and then divide the sum by 7. 15 20 35 5 7 7 7 c d + + = = = 19. Substitute 20 for m , 6 for n , subtract and then divide the difference by 2. 20 6 2 2 14 7 2 m n − − = = = 20. Substitute 23 for x , 5 for y , subtract and divide the difference by 6. 23 5 18 3 6 6 6 x y − − = = = 21. Substitute 6 for m and 18 for q . 9 9 6 54 3 18 18 m q  = = = 22. Substitute 9 for z and 15 for y . 5 5 9 45 3 15 15 z y × = = = 23. Enter the expression in the graphing calculator, replacing a with 136 and b with 13. We see that 27 18 3438 a b − = for 136 and 13. a b = = 24. Enter the expression in the graphing calculator, replacing x with 87 and y with 29. We see that 19 9 13 47,531 xy x y − + = for 87 and 29. x y = =

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    2 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 25. 2 1 2 1 2 (6ft)(3 ft) (6)(3 )(ft)(ft) 21ft , or 21 square feet A bh = = = = 26. 1400 t v = 1400 mi 3.5hr 400 mph = = 27. 1 2 A bh =  2 1 (5cm)(6 cm) 2 1 (5)(6)(cm)(cm) 2 15cm , or 15 square centimeters = = = 28. a. 30 sec. By substitution, t = 2 2 30 sec 60 sec t = × = b. 35min. By substitution, t = 2 2 35 min 70 min t = × = c. 1 2 2 hr. By substitution, t = 1 2 2 2 2 hr 5 hr t = × = 29. A bh = 2 (67 ft)(12 ft) (67)(12)(ft)(ft) 804 ft , or 804 square feet = = = 30. Average h a = 8 0.364 22 =  31. Let r represent Ron’s age. 5 more than Ron’s age, or Ron’s age plus 5 can be expressed as 5 , or 5 r r + + 32. Let s represent Luke’s speed. Times means multiplication, so we have 8 s × , or 8 s 33. Product is the result of multiplication, so we have 4, or 4 a a  34. Let l represent Lou’s weight; 7 l + or 7 l + 35. 9 less than c ; 9 c − 36. 4 less than d ; 4 d - 37. 6 increased by q ; to increase is to add; 6 , or 6 q q + + 38. 11 increased by z ; to increase is to add; 11 , z + or 11 z + 39. The difference of m and n ; m n - 40. t subtracted from p ; p t - 41. x less than y , or decrease y by x ; y – x 42. Let a represent Lorrie’s age; a – 2 43. x divided by w (Note: x is the dividend, or numerator , and w is the divisor, or denominator); , or x w x w  44. Let s and t represent the two numbers; quotient is the result of division; , or s t s t  45. Let l and h represent the box’s length and height, respectively. We have the sum l h + or . h l + 46. Sum of d and f ; , or d f f d + + 47. Let p represent Panya’s speed and let w represent the speed of the wind. 2 p w − 48 . The product of 9 and twice m . Twice m is 2 , m  or 2 . m We have 9 2 , or 2 9 m m   49 . L et y represent “some numbe r”; 1 4 4 12, or 12 y y − − 50. Let n represent the number; 6 20 n - 51. Let a and b represent the numbers; 8( ) a b - 52. One third of the sum is 1 3 × sum; let x and y represent the two numbers. We have 1 1 3 3 . sum ( ) x y × = +

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    Exercise Set 1. 1 3 53. 64% 0.64, = so 64% of w , where w = the number of women, is 0.64 w . (Note: replace “of ” with multiplication) 54. Let y = the number; 38% 0.38, = so we have 0.38 y . 55. | 17 32 15 17 ? 32 32 32 x + = + Writing the equation Substituting 15 for x 32 = 32 is TRUE Since the left - hand and right - hand sides are the same, 15 is a solution. 56. | 28 93 75 28 ? 93 103 93 y + = + Writing the equation Substituting 75 for y 103 = 93 is FALSE No ; since the left - hand and right - hand sides are not the same, 75 is not a solution. 57. | 28 75 93 28 ?75 65 75 a − = − Writing the equation Substituting 93 for a 65 = 75 is FALSE Since the left - hand and right - hand sides are not the same, 93 is not a solution. 58. | 8 96 8 12 ?96 96 96 t =  Writing the equation Substituting 12 for t 96 = 96 is TRUE Since the left - hand and right - hand sides are the same, 12 is a solution. 59. 7 63 7 9 | ? 9 9 9 t = 9 = 9 is TRUE Since the left - hand and right - hand sides are the same, 63 is a solution. 60. 8 52 8 6 | ? 6 6.5 6 x = 6.5 = 6 is FALSE Since the left - hand and right - hand sides are not the same, 52 is not a solution. 61. 108 108 3 36 | ? 36 36 36 x = 36 = 36 is TRUE Since the left - hand and right - hand sides are the same, 3 is a solution. 62. 94 94 7 12 | ? 12 3 13 12 7 y = 3 13 12 is FALSE. 7 = Since the left - hand and right - hand sides are not the same, 7 is not a solution. 63. Let x represent the number. 7 times what number is 1596? Translating: 7 1596 7 1596 x x       = = 64. Let x represent the number. What number added to 73 is 201? Translating: 73 201 73 201 x x ¯ ¯ ¯ ¯ ¯ + = + = 65. Let x represent the number. Rewording: 42 times what number is 2352? Translating: 42 2352 42 2352 x x ¯ ¯ ¯ ¯ ¯ × = = 66. Let x represent the number. Rewording: what number plus 345 is 987? Translating: 345 987 345 987 x x      + = + = 67. Let s represent the number of unoccupied squares. Number number Rewording: plus of occupied of unoccupied Translating: 19 64 19 64 is 64. s s ¯ ¯ ¯ ¯ ¯ + = + =

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    4 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 68. Let h represent the number of hours. Rewording: $35 times the number of hours is $3150 Translating: $35 = $3150 $35 $3150 h h       = 69. Let w represent the total amount of waste generated, in millions of tons. the total amount of million 87 Rewording: 34.5% of waste is tons. Translating:34.5% = 34.5% 87, or 0.345 87 87 w w w       = = 70. Let t represent the average commute in Fort Bliss in minutes. The average The average commuting commuting time in time in Rewording: is min. Fort Bliss minus 51.2 Indian Wells Translating: 59.8 51.2 t ¯ ¯ ¯ ¯ ¯ = - 71. f 72. h 73. d 74. c 75. g 76. b 77. e 78. a. 79. Look for a pattern in the data. Observe that the daily recommended number of grams of dietary fiber for each age is 5 more than that age. We reword and translate. Grams of fiber is 5 more than the age 5 f a      = + We have 5 . f a = + This equation could also be written as 5. f a = + 80. Observe that the tuition is 100 times the number of class hours. Reword and translate. Tuition is 100 times number of class hours 100 c h      =  We have 100 . c h = 81. Look for a pattern in the data. Observe that it costs $2.42 more to process nonmachinable packages than to process machinable packages . We reword and translate. Cost of Cost of nonmachinable machinable is packages $2.42 packages more than $2.42 n m      = + We have $2.42 . n m = + This equation could also be written as 2.42. n m = + 82. Observe that in each case the amount received is $3.00 less than the amount removed from her account . Reword and tran slate. Amount amount $3.00 minus is removed received $3.00 r s      = − We have $3.00, r s = − or r = s – 3. 83. Observe that the number of vehicle miles traveled is 10,000 times the number of drivers. Reword and translate. Number of Number of is Vehicle miles drivers 10,000 times 10, 000 v d      =  We have 10,000 . v d =

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    Exercise Set 1. 1 5 84. Observe that the depth of the water is one - tenth the depth of the snow. Reword and translate. 1 10 Depth Depth is one-tenth of of Water of snow w s      =  1 10 , w s = or, we could write this as 10. w s =  85. Thinking and Writing Exercise. A variable is a letter that is used to stand for any number chosen from a set of numbers. An algebraic expression is an expression that consists of variables, constants, operation signs, and/or grouping symbols. A variable expression is an algebraic expression that contains a variable. An equation is a number sentence with the verb = The symbol ‘ = ’ is used to indicate that the algebraic expressions on either side of the symbol represent the same quantity. 86. Thinking and Writing Exercise. To evaluate an algebraic expression means to find the value of the expression when its variables are given values. 87. Thinking and Writing Exercise. No; for a square with side s , the area is given by A = s  s . The area of a square with side 2 s is given by (2 )(2 ) 4 4 2 . s s s s A A =   =  88. Thinking and Writing Exercise. Answer may vary. Juliet was born in 1998. How old will she be in 2006? 89. The sign is a triangle whose area is 2 1 1 , or (3ft)(2.5ft) 3.75ft 2 2 b h × × = The cost is $ 120 for each square foot. Thus, $120(3.75) $450 C = = 90. The shaded area is the area of a rectangle with dimensions 20 cm by 10 cm less the area of a triangle with base 20 cm 4 − cm – 5 cm, or 11 cm, and height 7.5 cm. We perform the computation : 1 (20 cm)(10 cm) (11 cm)(7.5) 2 − 2 2 2 200 cm 41.25 cm square 158.75 cm , or 158.75 centimeters = − = 91. When y is twice x and 6, x = 2 6 12 6 12 2 2 18 9 2 y x y =  = + + = = = 92. When a is three times b , then b is one - third a , so 18 6 3 18 6 3 3 12 4 3 b a b = = − − = = = 93. The next whole number is one more than 3: w + 3 1 4 w w + + = + 94 . The preceding odd number is 2 less than 2: d + 2 2 d d + − = 9 5 . Adding the length of the sides, we have: , or 2 2 l w l w l w + + + + 9 6 . Adding the length of the sides, we have: , or 4 s s s s s + + + 9 7 . Ella’s time is 5 more than Kyle’s time, or 5 . E K = + Kyle’s time is 3 more than Ava’s time ( t ), or 3 . K t = + Substituting 3 for into 5 We have: 5 (3 ), or 8 , t K E K t t + = + + + + or 8. t + Also: Kyle is 3 more than Ava, or 3 t + and Ella is 5 more than Kyle, or ( 3) 5 8 t t + + = + 98 . Ray is currently 2 years older than Monique, or 2. a + In 7 years, he will be 2 7, or 9. a a + + +

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    6 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 99 . Thinking and Writing Exercise. Yes; the area of a triangle with base b and height h is given by 1 2 . A bh = The area of a triangle with base b and height 2 h is given by 1 1 (2 ) 2 2 . 2 2 b h bh A   = =     Exercise Set 1.2 1. commutative 2. associative 3. associative 4. commutative 5. distributive 6. associative 7. associative 8. commutative 9. commutative 10. distributive For exercises 11 – 18, change the order of the addends . 11. 7 x + 12. 2 a + 13 . 3 y x + 14 . 3 9 y x + 15. c ab + 16. xy uv + 17. 5(1 ) a + 18. 9(5 ) x + For exercises 19 – 26, change the order of the factors. 19. 2 a  20. yx 21. ts 22. 4 x  23. 5 ba + 24. 3 x y +  25. ( 1)5 a + 26. ( 5)9 x + For exercises 27 – 32, regroup the addends. 27. (5 ) a b + + 28. 5 ( ) m r + + 29. ( ) 7 r t + + 30. ( 2) x y + + 31. ( ) ab c d + + 32. ( ) m np r + + For exercises 33 – 38, regroup the factors . 33. 7( ) mn 34. 13( ) xy 35. (2 ) a b 36. (9 ) r p 37. (3 2)( ) a b  + 38. (5 )(2 ) x y +

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    Exercise Set 1.2 7 39. 2 ( 6) 2 (6 ) Commutative law of addition (2 6) Associative law of addition 8 Simplification t t t t + + = + + = + + = + A nswer s may vary. 40. (11 ) 4 ( 11) 4 Commutative law of addition (11 4) Associative law of addition 15 Simplification v v v v + + = + + = + + = + Answers may vary. 41. (3 ) 7 ( 3) 7 Commutative law of multiplication (3 7) Associative law of multiplication 21 or 21 Simplification a a a a a  =   =  =  Answers may vary. 42. 5( 8) 5(8 ) Commutative law of multiplication (5 8) Associative law of multiplication 40 Simplification x x x x  = =  = Answers may vary. 43. (5 ) 2 x + + ( 5) 2 Commutative law (5 2) Associative law 7 Simplifying x x x = + + = + + = + 44. (2 )4 4(2 ) Commutative law a a = (4 2) Associative law 8 Simplifying a a =  = 45. ( 3)7 (3 7) Associative law (3 7) Commutative law 21 Simplification m m m m × = × = × = 46. 4 (9 ) x + + (4 9) Associative law (4 9) Commutative law 13 Simplification x x x = + + = + + = + For exercises 47 – 62, use the distributive law. 47. 4( 3) 4 4 3 4 12 a a a + =  +  = + 48. 3( 5) 3 3 5 3 15 x x x + =  +  = + 49. 6(1 ) 6 1 6 6 6 x x x + =  +  = + 50. 6( 4) 6 6 4 6 24 v v v + =  +  = + 51. ( 5)2 2 5 2 2 10 n n n + =  +  = + 52. (1 )3 1 3 3 3 3 t t t + =  +  = + 53. 8(3 5 ) 8 3 8 5 24 40 x y x y x y + =  +  = + 54. 7(4 5 ) 7 4 7 5 28 35 x y x y x y + = × + × = + 55. 9(2 6) 9 2 9 6 x x + =  +  18 54 x = + 56. 9(6 7) 9 6 9 7 m m + =  +  54 63 m = + 57. 5( 2 3 ) 5 5 2 5 3 r t r t + + =  +  +  5 10 15 r t = + + 58. 4(5 8 3 ) 4 5 4 8 4 3 x p x p + + =  +  +  20 32 12 x p = + + 59. ( )2 2 2 2 2 a b a b a b + =  +  = + 60. ( 2)7 7 2 7 x x + =  +  7 14 x = + 61. ( 2) 5 5 5 2 5 x y x y + + =  +  +  5 5 10 x y = + +

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    8 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 62. (2 )6 2 6 6 6 a b a b + + =  +  +  12 6 6 a b = + + For exercises 63 – 66, the terms are separated by the addition/plus signs. 63. , , 19 x xyz 64. 9, 17 , a abc 65. 2 , , 5 a a b b 66. 4 3 , 20, a xy b 67. 2 2 2 2 The common factor is 2 a b a b + =  +  2( ) Using the distributive law a b = + Check: 2( ) 2 2 2 2 a b a b a b + =  +  = + 68. 5 5 5 5 The common factor is 5. y z y z + =  +  5( ) Using the distributive law y z = + Check: 5( ) 5 5 5 5 y z y z y z + =  +  = + 69. 7 7 7 1 7 The common factor is 7 y y + =  +  7(1 ) Using the distributive law y = + Check: 7(1 ) 7 1 7 7 7 y y y + =  + = + 70. 13 13 13 1 13 The common factor is 13. x x + =  +  13(1 ) Using the distributive law x = + Check: 13(1 ) 13 1 13 13 13 x x x + =  +  = + 71. 18 3 3 6 3 1 The common factor is 3. x x + =  +  3(6 1) Using the distributive law x = + Check: 3(6 1) 3 6 3 1 18 3 x x x + =  +  = + 72. 20 5 5 4 5 1 The common factor is 5. a a + =  +  5(4 1) Using the distributive law a = + Check: 5(4 1) 5 4 5 1 20 5 a a a + = × + × = + 73. 5 10 15 5 5 2 5 3 5( 2 3 ) x y x y x y + + =  +  +  = + + Check: 5( 2 3 ) 5 5 2 5 3 5 10 15 x y x y x y + + =  +  +  = + + 74. 3 27 6 3 1 3 9 3 2 3(1 9 2 ) b c b c b c + + = × + × + × = + + Check: 3(1 9 2 ) 3 1 3 9 3 2 3 27 6 b c b c b c + + = × + × + × = + + 75. 12 9 3 4 3 3 3(4 3) x x x + =  +  = + Check: 3(4 3) 3 4 3 3 12 9 x x x + =  +  = + 76. 25 30 5 5 5 6 5(5 6) y y y + =  +  = + Check: 5(5 6) 5 5 5 6. 25 30 y y y + = × + × = + 77. 3 9 3 3 3 3( 3 ) a b a b a b + =  +  = + Check: 3( 3 ) 3 3 3 3 9 a b a b a b + = + × = + 78. 5 15 5 5 3 5( 3 ) a b a b a b + = × + × × = + Check: 5( 3 ) 5 5 3 5 15 a b a b a b + =  +  = +

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    Exercise Set 1.2 9 79. 44 88 66 22 2 22 4 22 3 22(2 4 3 ) x y z x y z x y z + + = × + × + × = + + Check: 22(2 4 3 ) 22 2 22 4 22 3 44 88 66 x y z x y z x y z + + = × + × + × = + + 80. 24 48 60 12 2 12 4 12 5 12(2 4 5) a b a b a b + + = × + × + × = + + Check: 12(2 4 5) 12 2 12 4 12 5 24 48 60 a b a b a b + + = × + × + × = + + 81. and s t 82. 5 and x 83. 3 and ( x + y ) 84. ( ) and 6 a b + 85. 7, a , and b 86. m , n, and 2 87. ( ) and ( ) a b x y − − 88. (3 ) and ( ) a b c − + 89. Terms are separated by addition and subtraction signs. A single term may include plus or minus signs only if they are inside parentheses, brackets, or other grouping symbols. Factors are numbers, variables, or expressions that are multiplied together. 90. 2(3 4 ) 2 3 2 4 Distributive law 6 8 Simplification x y x y x y + =  +  = + 91. Let k represent Kylie’s salary. Then we have 1 2 2 , or k k . 92. To multiply a sum by a number, put the sum in parentheses; 2( 7) m + 93 . Thinking and Writing Exercise. No; in general, when subtracting, the result depends on the order in which the operation is performed. 94 . Thinking and Writing Exercise. No; in general, when dividing, the result depends on the grouping. 95. The expressions are equivalent by the distributive law. 8 4( ) 8 4 4 4 2 4 4 4(2 ) a b a b a b a b + + = +  +  =  +  +  = + + 96. The expressions are not equal. Distribution applies only to sums and differences. 5(a ) 5 , but 5 5 5 5 25 . b ab a b a b ab  =    =    = 97 . The expressions are not equal. For example, let 1. m = Then we have 7 7 7 3 1 1 , but 3 3 3 3 1 7 3 7 . 7 ¸ × = × = × ¸ = ¸ = 98 . The expressions are equivalent by the commutative law of multiplication and the distributive law . ( )5 5( ) 5 ( ) 5 ( ) rt st t r t s t r s t r s + =  +  =  + = + 99. The expressions are not equivalent. 5[2( 3 )] 5 2( 3 ) 10( 3 ) 10 10 3 10 30 30 10 x y x y x y x y x y y x + =  + = + =  +  = + = + 100. The expressions are equivalent by the commutative law of multiplication and the distributive law . [ (2 3 )]5 5[ (2 3 )] 5 (2 3 ) 5 (2 3 ) 5 2 5 3 10 15 . c b c b c b c b c c b c bc + = + =  + = + =  +  = +

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    10 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 101. Thinking and Writing Exercise. 3(2 ) 3(2 0) x + = + 3 2 6; 6 6 0 6 x =  = + = + = The result indicates that 3(2 ) x + and 6 x + are equivalent when 0. x = (By the distributive law, we know they are not equivalent for all values of x .) 102. Writing Exercise . 15 40 5(3 8) 15 4 40 60 40 100 5(3 4 8) 5(12 8) 5 20 100 x x + = +  + = + =  + = + =  = Although the expressions 15 40 x + and 5(3 8) x + are equivalent for 4, x = this result does not guarantee that the factorizatio n is correct for all values of x . (See Exercise 10 1 . ) Exercise Set 1.3 1. b 2. c 3. d 4. a 5. The paired factors of 50 are: 1 50, 2 25, and 5 10    T he factors of 50 are: 1, 2, 5, 10, 25, 50. 6. The paired factors of 70 are: 1 70, 2 35, 5 14, and 7 10     The factors of 70 are: 1, 2, 5, 7, 10, 14, 35, 70. 7. The paired factors of 42 are: 1 42, 2 21, 3 14, and 6 7     The factors of 42 are: 1, 2, 3, 6, 7, 14, 21, 42 8. The paired factors of 60 are: 1 60, 2 30, 3 20, 4 15, 5 12, and 6 10       The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 9. Factors of 21 are 1, 3, 7, 21 . Composite. 10. Factors of 15 are 1, 3, 5, 15. Composite. 11. The factors of 31 are: 1 and 31. Prime. 12. The factors of 35 are: 1, 5, 7, 35. Composite. 13. The factors of 25 are: 1, 5, 25. Composite. 14. The factors of 37 are: 1 and 37. Prime. 15 . Th e factors of 2 are : l and 2. Prime. 16. 1 is the only factor of 1 . It is neither prime nor composite. (It is a unit .) 17. 0 is neither prime nor composite. (Note: The number must be a counting number greater than 1). 18 . The factors of 4 are: 1, 2, 4. Composite. 19. The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40. Composite . 20. The factors of 75 are: 1, 3, 5, 15, 25, 75. Composite . 21 . 26 2 13 =  22 . 15 3 5 =  23 . 30 2 15 2 3 5 =  =   24 . 55 5 11 =  25 . 27 3 9 3 3 3 =  =   26 . 98 2 49 2 7 7 =  =   27. 40 2 20 2 2 10 2 2 2 5 =  =   =    28. 54 2 27 2 3 9 2 3 3 3 =  =   =    29. 43; Prime 30. 120 2 60 2 2 30 2 2 2 15 2 2 2 3 5 =  =   =    =     31. 210 2 105 2 3 35 2 3 5 7 =  =   =    32. 79 ; Prime 33. 115 5 23 = 

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    Exercise Set 1.3 11 34. 143 11 13 =  35. Factoring numerator and 14 2 7 denominator 21 3 7  =  Rewriting as a product 2 7 of two fractions 3 7 2 7 1 1 3 7 2 Using the identity property of 1 3 = × = × = = 36. Factoring numerator and 20 2 10 denominator 26 2 13  =  Rewriting as a product 2 10 of two fractions 2 13 10 2 1 1 13 2 10 Using the identity property of 1 13 =  =  = = 37. 16 2 2 2 2 56 2 2 2 7    =    2 2 2 2 2 2 2 2 7 7 2 2 1 1 1 7 7 2 7 =    = =    = = 38. 72 2 2 2 3 3 27 3 3 3     =   2 2 2 3 3 3 3 3 8 3     =   = 39. 6 6 1 48 6 8 8 = =  40. 18 6 3 3 84 6 14 14  = =  41. 52 13 4 4 13 13 1 × = = × 42. 132 11 11 = 12 11  12 = 43. 19 19 76 = 4 19  1 4 = 44. 17 17 51 = 3 17  1 3 = 45. 150 6 25 25  = 25 6 = 46. 170 5 34 34  = 34 5 = 47. 42 2 21 21 50 2 25 25  = =  48. 75 5 15 15 80 5 16 16  = =  49. 120 2 60 60 82 2 41 41  = =  50. 75 5 15 45  = 3 15  5 3 = 51. 210 2 98 = 3 5 7 2    15 7 7 7 =   52. 140 2 70 350  = 5 70  2 5 = 53. 1 3 1 3 3 2 7 2 7 14   = =  54. 9 3 9 3 27 4 8 4 8 32   = =  55. 12 10 12 10 3 4 5 2 4 2 8 5 9 5 9 5 3 3 3 3       = = = =    56. 11 12 11 12 11  = 12  12 11  1 =

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    12 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 57. 1 3 1 3 4 4 8 8 8 8 + + = = = 4 1 2 2 =  58. Use 8 as the common 1 1 4 1 1 denominator 2 8 4 2 8 + =  + 4 1 4 1 8 8 8 5 8 + = + = = 59. Use 18 as the common 4 13 2 4 13 denominator 9 18 2 9 18 + = × + 8 13 21 3 7 18 18 18 3 6 7 6  = + = =  = 60. 4 8 3 4 8 5 15 3 5 15 + = × + 12 8 20 15 15 15 4 5 4 3 5 3 = + =   = =   61. 3 3 3 7 7 7 b b b a a a   = =  62. 5 5 5 x y x y xy z z z   = =  63. 4 3 4 3 7 a a a a + + = = 64. 7 5 7 5 2 a a a a − − = = 65. Use 30 as the common denominator 3 8 3 3 2 8 10 15 3 10 2 15 9 16 9 16 25 30 30 30 30 5 5 5 5 6 6 + =  +  + = + = =  = =  66. 7 5 3 7 2 5 Use 24 as the common denominator 8 12 3 8 2 12 + =  +  21 10 31 24 24 24 = + = 67. 9 2 9 2 7 1 7 7 7 7 − − = = = 68. 12 2 12 2 10 2 5 2 5 5 5 5 5 −  − = = = = 69. 13 4 13 2 4 Use 18 as the common denominator 18 9 18 2 9 13 8 18 18 13 8 5 18 18 - = - × = - - = = 70. 13 8 3 13 8 Use 45 as the common denominator 15 45 3 15 45 39 8 45 45 39 8 31 45 45 − =  − = − − = = 71. 20 2 2 10 30 3  − = 3 10  2 3 − 2 2 0 3 3 = − = 72. 5 5 3 5 5 7 21 3 7 21 − =  − 15 5 21 10 21 − = = 73. Multiply by the reciprocal 7 3 7 5 of the divisor 6 5 6 3  =  7 5 6 3 35 18  =  =

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    Exercise Set 1.3 13 74. Multiply by the reciprocal 7 3 7 4 of the divisor 5 4 5 3  =  7 4 5 3 28 15  =  = 75. 8 4 8 15 9 15 9 4  =  8 15 9 4 2 4 3 5 3 3 4 10 3  =     =   = 76. 1 1 1 4 8 4 8 1 4 1 1 4 2 2  =   = =  77. 3 7 12 12 7 3  =  3 4 7 28 3   = = 78. 10 10 1 1 10 9 9 10 9  =  = 79. 7 7 7 13 7 13 13 13 7 ¸ = × = 13 × 13 7 × 1 = 80. 17 5 17 6 17 2 3 51 8 6 8 5 2 4 5 20    =  = =   81. 2 7 5 3 2 5 2 3 6 7 3 7 5 35 =  =  = 82. 3 8 1 5 3 1 3 5 15 8 5 8 1 8 =  =  = 83. 1 2 9 1 9 9 2 18 2 = ¸ = × = 84. 3 7 3 3 1 3 1 1 6 6 7 7 6 7 3 2 14  =  =  = =   85. Thinking and Writing Exercise. There is an infinite number of even numbers. There is an infinite number of whole numbers , and each of these can be multiplied by 2, producing an even number. 86. Thinking and Writing Exercise. If the fractions have the same denominators, it would probably be easier to compute the ir sum than their product . 87. 5( 3) ( 3)5 x x + = + Commutative Law of M ultiplication 5( 3) 5(3 ) x x + = + Commutative Law of Addition 88. 7 ( ) 7 ( ) or a b b a + + = + + 7 ( ) ( ) 7 a b a b + + = + + 89. Thinking and Writing Exercise. Bryce is canceling incorrectly. The number 2 is not a common factor of both terms in the numerator, so it cannot be canceled. For example, let 1. x = Then (2 1) / 8 3 / 8 + = but (1 1) / 4 2 / 4 1/ 2. + = = The expressions are not equivalent. 90. 0 is not a natural number, and we consider only natural numbers greater than 1. 91. Product 56 63 36 72 140 96 168 Factor 7 7 2 36 14 8 8 Factor 8 9 18 2 10 12 21 Sum 15 16 20 38 24 20 29 92. We need to find the least number that has both 6 and 8 as factors. Starting with 6 we list some numbers with a factor of 6, and starting with 8 we also list some numbers with a factor of 8. Then we find the first number that is on both lists. 6,12,18, 24,30,36,... 8,16, 24,32, 40, 48,... Since 24 is the least number that is on both lists, the carton should be 24 in. long.

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    14 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 93. 16 9 4 4 4 3 3 2 2 2 15 8 12 3 5 2 4 3 4 5        = =        94. 9 2 4 1 9 8 1 2 36 2 4 9 x y xy xy x y       = =      95. 3 9 27 3 9 9 p q r s pqrs q prst p r s t t      = =     96. 247 13 19 323  = 17 19  13 17 = 97. 15 15 4 9 6 25 15 xy x y   =   2 2 3 3 2 3 25 15 x y x           6 25 y =  98 . 10 10 12 25 2 30 20 x y z x y × × = × × 2 2 3 x × × × × 5 5 2 3 10 y z × × × × × × 2 2 5 5 2 x y z × × × × × = 99. 27 15 18 25 27 18 27 25 27 25 15 25 15 18 15 18 3 9 5 5 5 3 5 2 9 2 ab mn bc np ab bc ab np ab np mn np mn bc mn bc a b n p ap m n b c cm  =  =  =         = =        100. 45 24 30 32 45 32 24 30 xyz ab xz ac xyz ac ab xz =  3 = 15 × 2 2 8 x y z × × × × × × 3 8 a c × × × 2 15 a b × × × × 2 x z cy b × × = 101. 3 4 1 2 5 3 1 5 4 4 4 2 rs rs st st =  23 9 23 2 4 2 4 9 23 2 23 2 2 9 18 rs st rs st rs r s t t =  =     = =      3 3 4 5 3 Note: 5 5 4 4 4 1 4 20 3 23 4 4 = + =  + + = = 1 2 4 1 8 1 9 Similarly: 4 2 2 1 2 2 2 + =  + = = 102. 5 7 4 5 3 5 4 3 2 2 7 5 mn mn np np =  26 14 7 5 26 5 7 14 2 13 5 7 2 7 65 49 mn np mn np m n np m p = ¸ = × / × × × / = ×/ × / = 103. 4 7 m m 5 9 A lw     = =         2 4 7 (m)(m) 5 9 28 28 m , or square meters 45 45     =         = 104. 1 1 10 5 m m 2 2 7 4 A bh     = =         2 2 2 1 10 5 (m)(m) 2 7 4 1 10 5 m 2 7 4 1 2 5 5 m 2 7 4 25 25 m , or square meters 28 28     =           =      =   = 105. 5 4 4 3 m 9 32 4 m 9 128 2 m, or 14 m 9 9 P s   = =     =  =

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    Exercise Set 1.4 15 106. 4 7 2 2 2 m 2 m 5 9 P l w     = + = +         8 14 m m 5 9 8 14 m 5 9 8 9 14 5 m 5 9 9 5 72 70 m 45 45 142 7 m, or 3 m 45 45 = + æ ö = + ç ÷ è ø æ ö = × + × ç ÷ è ø æ ö = + ç ÷ è ø = 107. The cube has 12 edges, all with length of 3 2 cm. 10 3 12 2 10 23 2 6 23 12 10 2 5 138 3 27 cm 5 5 TL =    =  =  = = Exercise Set 1.4 1. repeating 2. terminating 3. integer 4. whole number 5. rational number 6. irrational number 7. natural number 8. absolute value 9. 100 corresponds to the highest temperature recorded in Alaska. – 80 corresponds to the lowest. 10. – 2 50 corresponds to burning 2 50 calories, and 65 corresponds to consuming 65 calories. 11. – 777.68 corresponds to falling 777.68 points , and 936.42 corresponds to gaining 936.42 points . 12. – 1340 corresponds to 1340 feet below sea level, and 29,035 corresponds to 29,035 feet above sea level. 13. – 12 ,500 corresponds to taking on a debt of $12,500, and 5000 corresponds to receiving $5000 as a scholarship. 14. 18 . 7 corresponds to the birth rate of 18 . 7 births per 1000 people, and – 7 . 89 corresponds to 7 . 89 deaths per 1000 people. 15. 8 corresponds to the 8 yard gain, and – 5 corresponds to the 5 yard loss. 16. 6 corresponds to being 6 over par , and – 17 corresponds to being 17 under par . 17 . 18 . 19 . 20 . 21 . 22 . 23. 24.

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    16 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 25. 7 8 means 7 8,  so we divide. 0.8 7 5 8 7.000 6 4 6 0 5 6 4 0 4 0 0 7 8 0.875. = 26. 1 8 means (1 8), so we divide. − −  0.125 8 1.00 0 8 2 0 16 4 0 4 0 0 1 8 0.125. − = − 27. We first find decimal notation for 3 3 4 4 . Since means 3 4, we divide.  0.7 5 4 3.0 0 2 8 2 0 2 0 0 3 4 0.75. − = − 28. 11 6 means 11 6,so we divide.  1.833... 6 11.000 6 50 48 20 18 2 11 6 1.83. = 29. 7 6 means 7 6, so we divide. - - ¸ 1.16 6 6 7.0 0 0 6 10 6 4 0 36 4 0 36 4 7 6 1.16. - = - 30. 5 12 means 5 12, so we divide. − −  0.4166 12 5.000 0 4 8 2 0 12 80 7 2 80 7 2 8 5 12 0.416. - = - 31. 2 3 means 2 3,  so we divide. 0.6 6 6 3 2.0 0 0 1 8 2 0 18 2 0 18 2 2 3 0.6. = 32. 1 4 means 1 4,  so we divide. 0.2 5 4 1.0 0 8 2 0 2 0 0 1 4 0.25. =

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    Exercise Set 1.4 17 33. 1 2 means (1 2), so − −  we divide 0.5 2 1.0 1 0 0 1 1 2 2 0.5, so 0.5. = − = − 34. 2 9 − means 2 9, −  so we divide. 0.2 9 2.0 0 18 2 2 9 0.2. - = - 35. Since the denominator is 100, we know that 13 100 0.13. = We could also divide 13 by 100 to find this result. 36. 7 20 means (7 20), − −  so we divide. 0.35 20 7.0 0 6 0 1 0 0 1 0 0 0 7 20 0.35. − = − 37. 38. 39. 40. 41 . 7 > 0, since 7 is to the right of 0. 42 . 8 > – 8, since 8 is to the right of – 8. 43 . – 6 < 6, since – 6 is to the left of 6. 44 . 0 > – 7, since 0 is to the right of – 7. 4 5 . – 8 < – 5, since – 8 is to the left of – 5. 46 . – 4 < – 3, since – 4 is to the left of – 3. 47 . – 5 > – 11, since – 5 is to the right of – 11. 48 . – 3 > – 4, since – 3 is to the right of – 4. 49 . – 12.5 < – 9.4, since – 12.5 is to the left of – 9.4. 50 . – 10.3 > – 14.5, since – 10.3 is to the right of – 14.5. 51 . 5 11 12 25 , −  − since 5 12 − is to the right of 11 25 . − 52 . 27 14 17 35 , − −  since 14 17 − is to the left of 27 35 . - 53 . – 7 > x has the same meaning as x < – 7. 54 . a > 9 has the same meaning as 9 < a. 55 . – 10  y has the same meaning as 10. y  − 56 . 12 t  has the same meaning as 12. t  57 . 3 11 −  − is true. 58 . 5 5  − is false. 59 . 0 8  is false . 60 . 5 7 −  is true . 61 . 8 8 −  − is true . 62 . 8 8  is true . 63 . 58 58 - = 64 . 47 47 - = 65. 5.6 5.6 = 66. 2 2 5 5 − = 67. 2 2 = 68. 456 456 - = 69. 9 9 7 7 − =

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    18 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 70. 3 3 - = 71. 0 0 = 72. 4.3 4.3 = 73. 8 8 x = - = 74. 5 5 a = - = 75. 5 18, 4.7, 0, , 2.16, 37 9 − − − 76. 18 77. 18 , 0, 37 − 78. , 17  79. All of them: 5 18, 4.7, 0, , , 17, 2.16, 37 9  − − − 80. 18, 0 81. Thinking and Writing Exercise. Yes; every integer can be written as n /1, a quotient of the form a / b where 0. b  82. Thinking and Writing Exercise. No; for instance, – 2 is an integer and – 2 is not a natural number. 83. 3 3 2 7 42 xy =   = 84. 5 5 , or ab ab + = + 5 5. ab ba + = + 85. Thinking and Writing Exercise. According to the three given measures, the Spurs are the best team overall, with the highest number of points per game, the greatest points per game differential, and the third - greatest field goal percentage differential. 86. Thinking and Writing Exercise. There are infinitely many rational numbers between 0 and 1. Consider rational numbers of the form 1 , n where n is an integer greater than 1. There are infinitely many integers greater than 1, so there are infinitely many numbers 1 , n all between 0 and 1. (These numbers are a subset of the rational numbers between 0 and 1.) 87. Thinking and Writing Exercise. No; every positive number is nonnegative, but zero is nonnegative and zero is not positive. 88. List the numbers as they occur on the number line, from left to right: – 17, – 12, 5, 13 89. List the numbers as they occur on the number line, from left to right: – 23, – 17, 0, 4 90. 2 1 3 5 3 1 , , , , , 3 2 4 6 8 6 − − − can be written in decimal notation as 0.666,0.5, 0.75, 0.833,0.375,0.166, − − − respectively. Listing from least to greatest (in fractional form), we have 5 3 2 1 3 1 , , , , , . 6 4 3 6 8 2 − − − 91. Converting to decimal notation, we can write 4 4 4 4 4 4 4 , , , , , , as 5 3 8 6 9 2 3 − 0.8, 1.33, 0.5, 0.66, 0.44, 2, 1.33, − respectively. List the numbers (in fraction form) as they occur on the number line, from left to right: 4 4 4 4 4 4 4 , , , , , , 3 9 8 6 5 3 2 − 92. 5 5 - = and 2 2, - = so 5 2 . - > - 93. 4 4 = and 7 7, - = so 4 7 . < - 94. 8 8 - = and 8 8, = so 8 8 . - = 95. 23 23 and 23 23, so 23 23 . = - = = - 96. 7 x = x represents a number whose distance from 0 is 7. Thus, 7 x = or 7. x = − 97 . 3 x < x represents an integer whose distance from 0 is less than 3 units. Thus, 2, x = − – 1, 0, 1, 2.

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    Mid - Chapter Review 19 98 . 2 5 x < < x represents an integer whose distance from 0 is greater than 2 and also less than 5. Thus, = − 4, x – 3, 3, 4 99. 1 3 0.3 1 1 1 0.1 1 3 3 3 3 9 = = =  = 10 0 . ( ) 1 3 0.99 3 0.3 3 3 3 = =  = 10 1 . ( ) 1 50 5.55 50 0. 1 50 9 9 = =  = (See Exercise 99 ). 10 2 . ( ) 1 70 7.77 70 0. 1 70 9 9 = =  = (See Exercise 99 ). 103. 0 a  104. 0 x  105. 10 x  106. 20 t  10 7 . Thinking and Writing Exercise. The number entered by hand is an approximation of 2 while the value that is squared immediately after being calculated is actually regarded by the calculator as 2. 108 . Thinking and Writing Exercise. Yes. We have three possibilities: a is negative, 0, a = or a is positive. We will show an example of each possibility. ( ) 2 2 2 2; 2 4 2; | 2| 2 0; 0 0 0; | 0 | 0 7; 7 49 7; |7| 7. a a a = - - = = - = = = = = = = = = Mid - Chapter Review Guided Solutions 1. 22 10 12 4 3 3 3 x y − − = = = 2. 40 8 8 5 8 1 8(5 1) x x x + =  +  = + Mixed Review 1. x + y = 3 + 12 = 15 2. 2 2 10 20 4 5 5 5 a  = = = 3. 10 d − 4. Let h be the number of hours worked. Then eight times the number of hours work is: 8 h . 5. Let n be the number of students that originally enrolled in Janine’s class. Then reword “five fewer than the number that originally enrolled” as “the number that originally enrolled minus five” and write the e quation: The number that originally Twenty-seven is minus five enrolled 27 5 n      = − 27 5 n = − 6. 13 8 104 94,  =  so 8 t = is not a solution to 13 94. t = 7. 7 10 10 7 x x + = + 8. 3( ) (3 ) ab a b = 9. 4(2 8) 4 2 4 8 8 32 x x x + =  +  = + 10. 3(2 5 10) 3 2 3 5 3 10 6 15 30 m n m n m n + + = × + × + × = + + 11. 18 9 9 2 9 1 9(2 1) x x x + = × + × = + 12. 8 24 20 4 2 4 6 4 5 4(2 6 5) a y a y a y + + = × + × + × = + + 13. 84 2 42 2 2 21 2 2 3 7 = × = × × = × × × 14. 8 6 48 6 40 8 5 5 × = = × 15. 135 3 45 3 3 15 3 3 3 5 315 3 105 3 3 35 3 3 5 7 3 3 3 5 135 3 315 3 3 5 7 7 = × = × × = × × × = × = × × = × × × × × × = = × × ×

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    20 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 16. 12 2 6 2 2 3 and 8 2 2 2, so the LCM of 12 and 8 is 2 2 2 3 24. 11 11 2 22 3 3 3 9 and , so 12 12 2 24 8 8 3 24 11 3 22 9 22 9 13 12 8 24 24 24 24 = × = × × = × × × × × = = × = = × = - - = - = = 17. 8 6 8 11 2 2 2 11 44 15 11 15 6 3 5 2 3 45 × × ¸ = × = × = × × 18. 2.5 is 2.5 units to the left of zero on the number line: - 19. 3 (3 20) and 20 0.15 3 20 20 3.00 20 10 100 100 0 - = - ¸ ¸ = So, 3 0.15 20 − = − 20. 16 24, because 16 is to the right of 24 on the number line. - > - - - 21. 22 2 11 and 15 3 5, so the LCM of 22 and 15 is 2 3 5 11 330. 3 3 15 45 22 22 15 330 2 2 22 44 15 15 22 330 3 45 44 2 Therefore 22 330 330 15 since 45 is to the left of 44 on the number line. = × = × × × × = - - - = × = - - - = × = - - - - = < = - - 22. 9 is equivalent to 9 . x x ³ £ 23. 6 is to the left of 5 on the number line, so the statement 6 5 is true. - - - £ - 24. | 5.6 | 5.6 - = 25. | 0 | 0 = Exercise Set 1.5 1. f 2. d 3. e 4. a 5. b 6. c 7. Start at 5. Move 8 units to the left. 5 ( 8) 3 + − = − 8. Start at 2. Move 5 units to the left. 2 ( 5) 3 + − = − 9. Start at – 5. Move 9 units to the right. 5 9 4 − + = 10. Start at – 3. Move 8 units to the right. 3 8 5 − + = 11. Start at 4 − . Move 0 units. 4 0 4 − + = − 12. Start at 6 − . Move 0 units. 6 0 6 − + = −

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    Exercise Set 1.5 21 13. Start at – 3. Move 5 units to the left. 3 ( 5) 8 − + − = − 14. Start at – 4. Move 6 units to the left. 4 ( 6) 10 − + − = − 15. 6 ( 5). − + − Two negatives. Add their absolute values . Make the answer negative. 6 ( 5) (6 5) 11 − + − = − + = − 16. 8 ( 12). − + − Two negatives. Add their absolute values. Make the answer negative. 8 ( 12) (8 12) 20 − + − = − + = − 17. 10 ( 15). + − The absolute values are 10 and 15. Their difference is 15 10 5. − = The negative number has the larger absolute value, so the answer is negative. 10 ( 15) (15 10) 5 + − = − − = − 18. 12 ( 22). + − The absolute values are 12 and 22. Their difference is 22 12 10. − = The negative number has the larger absolute value, so the answer is negative. 12 ( 22) (22 12) 10 + − = − − = − 19. 12 ( 12). + − The numbers have the same absolute value and their difference is 0. 12 ( 12) 0 + − = 20. 17 ( 17). + − The numbers have the same absolute value and their difference is 0. 17 ( 17) 0 + - = 21. 24 ( 17). − + − Two negatives. Add their absolute values. Make the answer negative. 24 ( 17) (24 17) 41 − + − = − + = − 22. 17 ( 25). − + − Two negatives. Add their absolute values. Make the answer negative. 17 ( 25) (17 25) 42 − + − = − + = − 23. 13 13. − + The numbers have the same absolute value and their difference is 0. 13 13 0 − + = 24. 18 18. − + The numbers have the same absolute value and their difference is 0. 18 18 0 − + = 25. 18 ( 11). + − The absolute values are 18 and 11. Their difference is 18 11 7. − = The positive number has the greater absolute value, so the answer is positive. 18 ( 11) 18 11 7 + − = − = 26. 8 ( 5). + − The absolute values are 8 and 5. Their difference is 8 5 3. − = The positive number has the greater absolute value, so the answer is positive. 8 ( 5) 8 5 3 + − = − = 27. 36 0. − + 0 is the additive identity. 36 0 36 − + = − 28. 0 ( 74). + − 0 is the additive identity. 0 ( 74) 74 + − = − 29. 3 14. − + The absolute values are 3 and 14. Their difference is 14 3 11. − = The positive number has the larger absolute value, so the answer is positive. 3 14 14 3 11 − + = − = 30. 13 ( 6). + − The absolute values are 13 and 6. Their difference is 13 6 7. − = The positive number has the larger absolute value, so the answer is positive. 13 ( 6) 13 6 7 + − = − = 31. 14 ( 19). − + − Two negatives. Add their absolute values. Make the answer negative. 14 ( 19) (14 19) 33 − + − = − + = − 32. 11 ( 9). + − The absolute values are 11 and 9. Their difference is 11 9 2. − = The positive number has the larger absolute value, so the answer is positive. 11 ( 9) 11 9 2 + − = − =

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    22 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 33. 19 ( 19). + − The numbers have the same absolute value and their difference is 0. 19 ( 19) 0 + − = 34. 20 ( 6). − + − Two negatives. Add their absolute values. Make the answer negative. 20 ( 6) (20 6) 26 − + − = − + = − 35. 23 ( 5). + − The absolute values are 23 and 5. Their difference is 23 5 − or 18. The positive number has the larger absolute value, so the answer is positive. 23 ( 5) 23 5 18 + − = − = 36. 15 ( 7). − + − Two negatives. Add their absolute values. Make the answer negative. 15 ( 7) (5 17) 22 − + − = − + = − 37. 31 ( 14). − + − Two negatives. Add their absolute values. Make the answer negative. 31 ( 14) (31 14) 45 − + − = − + = − 38. 40 ( 8). + − The absolute values are 40 and 8. Their difference is 40 8 32. − = . The positive number has the larger absolute value, so the answer is positive. 40 ( 8) 40 8 32 + − = − = 39. 40 ( 40). + − The numbers have the same absolute value and their difference is 0. 40 ( 40) 0 + − = 40. 25 25. − + The numbers have the same absolute value and their difference is 0. 25 25 0 − + = 41. 85 ( 65). + − The absolute values are 85 and 65. Their difference is 85 65 20. − = The positive number has the larger absolute value, so the answer is positive. 85 ( 65) 85 65 20 + − = − = 42. 63 ( 18). + − The absolute values are 63 and 18. Their difference is 63 18 45. − = The positive number has the larger absolute value, so the answer is positive. 63 ( 18) 63 18 45 + − = − = 43. 3.6 1.9. − + The absolute values are 3.6 and 1.9. Their difference is 3.6 1.9 1.7. − = The negative number has the larger absolute value, so the answer is negative. 3.6 1.9 (3.6 1.9) 1.7 − + = − − = − 44. 6.5 4.7. − + The absolute values are 6.5 and 4.7. Their difference is 6.5 4.7 1.8. − = The negative number has the larger absolute value, so the answer is negative. 6.5 4.7 (6.5 4.7) 1.8 − + = − − = − 45. 5.4 ( 3.7). − + − Two negatives. Add their absolute values. Make the answer negative. 5.4 ( 3.7) (5.4 3.7) 9.1 − + − = − + = − 46. 3.8 ( 9.4). − + − Two negatives. Add their absolute values. Make the answer negative. 3.8 ( 9.4) (3.8 9.4) 13.2 − + − = − + = − 47. 3 4 5 5 . - + The absolute values are 3 5 and 4 5 . Their difference is 3 4 1 5 5 5 . − = The positive number has the larger absolute value, so the answer is positive. 3 4 4 3 1 5 5 5 5 5 − + = − = 48. 3 2 7 7 . − + The absolute values are 2 7 and 3 7 . Their difference is 3 2 1 7 7 7 . − = The positive number has the large r absolute value, so the answer is positive. 2 3 3 2 1 7 7 7 7 7 − + = − = 49. 4 2 7 7 . − − + Two negatives. Add their absolute values. Make the answer negative. 4 2 4 2 6 7 7 7 7 7 − −   + = − + = −     50. 5 2 9 9 . − − + Two negatives. Add their absolute values. Make the answer negative. 5 2 5 2 7 9 9 9 9 9 − −   + = − + = −    

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    Exercise Set 1.5 23 51. 2 1 5 3 . − + The absolute values are 2 5 and 1 3 . Their difference is 6 5 1 15 15 15 . − = The negative number has the larger absolute value, so the answer is negative. 2 1 6 5 6 5 1 5 3 15 15 15 15 15 - - æ ö + = + = - - = - ç ÷ è ø 52. 4 1 13 2 . − + The absolute values are 4 13 and 1 2 . Their difference is 13 8 5 26 26 26 . − = The positive number has the larger absolute value, so the answer is positive. 4 1 8 13 13 8 5 13 2 26 26 26 26 26 − − + = + = − = 53. 4 2 9 3 . − + The absolute values are 4 9 and 2 3 . Their difference is 6 4 2 9 9 9 . − = The positive number has the larger absolute value, so the answer is positive. 4 2 4 6 6 4 2 9 3 9 9 9 9 9 − − + = + = − = 54. 1 1 6 3 . − + The absolute values are 1 6 and 1 3 . The difference is 2 1 1 6 6 6 . − = The positive number has the larger absolute values, so the answer is positive. 1 1 1 2 2 1 1 6 3 6 6 6 6 6 − − + = + = − = 55. 35 ( 14) ( 19) ( 5) + − + − + − Using the associative law of addition Adding the negatives Adding a positive and a negative 35 [( 14) ( 19) ( 5)] 35 ( 38) 3 = + − + − + − = + − = − 56. 28 ( 44) 17 31 ( 94) + − + + + − Using the commutative and associative laws of addition Adding the positives and adding the negatives Adding a positive and a negative [28 17 31] [( 44) ( 94)] 76 ( 138) 62 = + + + − + − = + − = − 57. 4.9 8.5 4.9 ( 8.5) − + + + − Note that we have two pairs of numbers with different signs and the same absolute value: 4.9 − and 4.9, 8. 5 and 8.5. − The sum of each pair is 0, so the result is 0 0, + or 0. 58. 24 3.1 ( 44) ( 8.2) 63 + + − + − + Using the commutative and associative laws of addition Adding the positives and adding the negatives Adding a positive and a negative [24 3.1 63] [( 44)] ( 8.2)] 90.1 ( 52.2) 37.9 = + + + − + − = + − = 59. Rewording: First second plus plus change change Translating: 0.05 + ( 0.03) + third total is change change total 0.07 change     − −    = Since 0.05 ( 0.03) 0.07 0.01, − + − + = − the pric e dropped 1¢, or the cost changed $0.01. − 60. Rewording: First second change change plus plus Translating: ( 0.06) + 0.12 + third total change change is total ( 0.04) change     −    − = Since 0.06 (0.12) ( 0.04) 0.02, − + + − = the price rose 2¢, or the cost changed $0.02.

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    24 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 61. ( ) ( ) ( ) 7 15 8 8 19 7 10 10 Rewording: First second change plus change plus + ( ) third fourth level change plus is change change = total change     + −      + − Since 15 7 19 7 8 8 10 10 75 35 76 28 40 40 40 40 75 35 76 28 88 11 40 40 5         − + + − +                         = − + + − +                 − + − +   = = − = −     the lake level dropped by 11 5 f ee t. 62. Rewording: 2014 2015 loss plus loss plus Translating: 26,500 + ( 10, 200) 2016 total profit is profit or loss 32,400 total profit or loss     − − +   = Since 26,500 ( 10, 200) 32, 400 36, 700 32, 400 4300, − + − + = − + = − the loss was $4300, or the profit was – $4300. 63. Rewording: First second third Total try plus try plus try is gain Translating: 13 + ( 12) 21 Total gain ¯ ¯ ¯ ¯ ¯ ¯ ¯ - + = Since 13 ( 12) 21 22, + − + = the total gain wa s 22 yd. 64. Rewording: Original change from balance plus writing first check plus Translating: 350 + ( 530) new change from deposit plus is balance writing second check 75 = new ( 90) balan     − +      + − ce Since 350 ( 530) (75) ( 90) + − + + − (350 75) [ 530 ( 90)] 425 ( 620) 195 = + + − + − = + − = − The balance is $195. − 65. Rewording: Original change balance plus from check plus Translating: 82 + ( 50) Change from is New balance August charges New balance 63     − +    = Since 82 ( 50) 63 95, + − + = his new balance is $95. 6 6. Rewording: Original change balance plus from check plus Translating: 470 + ( 45) new change total plus is charges from check change 160 + ( 500) new balance     − +      − = Since 470 ( 45) 160 ( 500) 85, + − + + − = Ian still owes $85. 67. Rewording: base plus rise is elevation Translating:( 19,684) 33, 480 elevation      − + = Since ( 19,684) 33, 480 13,796, − + = the ele vation of the peak is 13,796 ft a bove sea level.

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    Exercise Set 1.5 25 68. Rewording: Change from change from withdrawals plus additions Translating: 5 + 8 change from change in original size is plus "no shows" = change in ( 4) original size ¯ ¯ ¯ - ¯ ¯ ¯ ¯ + - Since 5 8 ( 4) − + + − 3 ( 4) 1, = + − = − the class lost 1 student , or the class size changed by – 1 . 69. 7 10 (7 10) 17 a a a a + = + = 70. 3 8 (3 8) 11 x x x x + = + = 71. 3 12 (12 3) 9 x x x x − + = − = 72. 2 ( 7 ) (7 2) 5 m m m m + − = − − = − 73. 5 ( 8 ) (8 5) 3 a a a a + − = − − = − 74. 3 8 (8 3) 5 x x x x − + = − = 75 . 5 2 (5 2) 7 a a a a − + − = − + = − 76 . 10 ( 17 ) (17 10) 7 n n n n + − = − − = − 77 . 3 8 4 ( 10 ) x x − + + + − 3 4 8 ( 10 ) Using the commutative law of addition ( 3 4) [8 ( 10)] Using the distributive law 1 2 Adding x x x x = − + + + − = − + + + − = − 78 . 8 5 ( ) ( 3) a a + + − + − 8 ( ) 5 ( 3) Using the commutative law of addition [8 ( 1)] [5 ( 3)] Using the distributive law 7 2 Adding a a a a = + − + + − = + − + + − = + 79 . 6 9 ( 9 ) ( 10 ) (10 6) (9 9) 4 m n n m m n m + + − + − = − − + − = − 80 . 11 ( 8 ) ( 3 ) 8 [ 11 ( 3)] ( 8 8) 14 s t s t s t s − + − + − + = − + − + − + = − 81 . 4 6.3 ( ) ( 10.2) [( 4) ( 1)] (10.2 6.3) 5 3.9 x x x x − + + − + − = − + − − − = − − 82 . 7 10.5 13 ( 11.5 ) (11.5 10.5) (13 7) 1 6 6, or 6 y y y y y y − + + + − = − − + − = − + = − + − 83 . Perimeter 8 5 9 7 x x = + + + 8 9 5 7 (8 9) (5 7) 12 17 x x x x = + + + = + + + = + 84 . Perimeter 8 4 5 6 a a = + + + (8 5) (4 6 ) (8 5) (4 6) 10 13 a a a a = + + + = + + + = + 85 . Perimeter 3 3 7 5 9 4 t r t r = + + + + + (3 5 ) (3 4 ) (7 9) (3 5) (3 4) (7 9) 8 7 16 t t r r t r t r = + + + + + = + + + + + = + + 86 . Perimeter 2 7 5 3 8 4 x z x z = + + + + + (2 3 ) (5 4 ) (7 8) (2 3) (5 4) (7 8) 5 9 15 x x z z x z x z = + + + + + = + + + + + = + + 87 . Perimeter 9 6 7 8 4 n n n = + + + + 9 7 6 8 4 (9 7) (6 8 4) 18 16 n n n n n = + + + + = + + + + = + 88 . Perimeter 2 6 5 7 3 7 n n n = + + + + + (5 7 7 ) (2 6 3) (5 7 7) (2 6 3) 19 11 n n n n n = + + + + + = + + + + + = +

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    26 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 89 . Thinking and Writing Exercise. Answer may vary. One possible explanation follows. Consider performing the addition on a number line. We start to the left of 0 and then move farther left, so the result must be a negative number. 90 . Thinking and Writing Exercise. Each nonzero integer from – 10 to 10 can be added to its opposite with the sum of each pair being 0. When this is added to the rem aining integer sums, the total is 0. 91 . 7(3 2) 7 3 7 7 2 z y z y + + =  +  +  21 7 14 z y = + + 92 . 7 3 7 8 2 8 2 3 7 8 2 3 7 2 4 2 3 28 3  =   =    =  = 9 3 . Thinking and Writing Exercise. The sum will be positive when the positive number is greater than the sum of the absolute values of the negative numbers. 9 4 . Thinking and Writing Exercise. No; when we add real numbers with different signs, we must know how to subtract absolute values. That is, we must be able to calculate a b − where a and b are both nonnegative with . a b  9 5 . We’re looking for the difference between the amount in Travis’s account after the deposit and his eventual overdrawn amount . Rewording: Original deposit plus minus amount Translating: 257.33 + (152) overdrawn check is amount amount check 42.37 amount ¯ ¯ ¯ ¯ - ¯ ¯ ¯ - = Since 257.33 152 ( 42.37) 451.70, + − − = the check’s amount $ 451.70 . 9 6 . Rewording: Final opposite value of drop plus plus Translating: 61 + 12 + opposite original of rise is value original ( 17.50) value        − = Since 61 12 ( 17.50) 55.50, + + − = the original value was $55.50. 9 7 . 4 __ ( 9 ) ( 2 ) 5 7 x x y x y + + − + − = − − Simplify the left side of the equation. 4 ___ ( 9 ) ( 2 ) x x y + + − + − [4 ( 9 )] ___ ( 2 ) [4 ( 9)] ___ ( 2 ) 5 ___ ( 2 ) x x y x y x y = + − + + − = + − + + − = − + + − We now have: 5 __ ( 2 ) 5 7 5 5 , so__+( 2 ) 7 x y x y x x y y − + + − = − − − = − − = − The missing term is – 5 y , since 5 2 7 y y y − + − = − 98. 3 9 __ 5 2 6 a b a a b − + + + = − Simplify the left side of the equation. 3 9 __ 5 ( 3 5 ) 9 __ ( 3 5) 9 __ 2 9 __ a b a a a b a b a b − + + + = − + + + = − + + + = + + We now have : 2 9 __ 2 6 2 2 ,so 9 __ 6 a b a b a a b b + + = − = + = − The missing term is – 15 b , since 9 ( 15 ) 6 b b b + − = −

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    Exercise Set 1.6 27 99. 3 2 __ ( 2 ) 2 ( 6 ) m n m n m + + + − = + − Simplify the left side of the equation. 3 2 __ ( 2 ) [3 ( 2 )] 2 __ [3 ( 2)] 2 __ 2 __ 2 __ m n m m m n m n m n n m + + + − = + − + + = + − + + = + + = + + We now have: 2 __ 2 ( 6 ) 2 2 ,so __ 6 n m n m n n m m + + = + − = + = − The missing term is – 7 m , since ( 7 ) 6 m m m + − = − 100 . __ 9 ( 4 ) 10 7 x y x x y + + − + = − (9 ) ( 4 ) __ 10 7 10 ( 4 ) __ 10 7 10 10 ,so ( 4 ) __ 7 x x y x y x y x y x x y y + + − + = − + − + = − = − + = − The missing term is – 3 y , since 4 ( 3 ) 7 y y y − + − = − 101. Note that, in order for the sum to be 0, the two missing terms must be the opposites of the given terms. Thus, the missing terms are – 7 t and – 23. 102 . 2 2 7 10 p l w x = + = + We know 2 2 5 10, l =  = so 2 w is 7 x . Then the width is a number which yields 7 x when added to itself. Since 3.5 3.5 7 , x x x + = the width is 3.5 x , or 7 2 . x 103 . 3 ( 3) 2 ( 2) 1 5 − + − + + − + = − Since the total is 5 under par after the five rounds and 5 1 ( 1) ( 1) ( 1) ( 1), − = − + − + − + − + − the golfer was 1 under par on average. Exercise Set 1.6 1. d 2. g 3. f 4. h 5. a 6. c 7. b 8. e 9. four minus ten 10. five minus thirteen 11. two minus negative nine 12. four minus negative one 1 3 . the negative/opposite of x minus y 1 4 . the negative/opposite of a minus b 1 5 . negative three minus the negative/opposite of n 1 6 . negative seven minus the negative/opposite of m 17 . The additive inverse of 39 is – 39 because 39 ( 39) 0 + − = 18 . The additive inverse of – 17 is 17 because 17 17 0 − + = 19 . The additive inverse of 11 2 − is 11 2 , because 11 11 0 2 2 − + = 20 . The additive inverse of 7 2 is 7 2 − because ( ) 7 7 2 2 0 + − = 21 . The additive inverse of 3.14 − is 3.14 because 3.14 3.14 0 − + = 22 . The additive inverse of 48.2 is 48.2 − because 48.2 48.2 0 + − = 23 . If 45, x = − then ( 45) 45. x − = − − = 24 . If 13, x = then 13. x − = − 25 . If 14 3 , x = − then ( ) 14 14 3 3 . x − = − − =

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    28 Chapter 1: Elemen tary and Intermediate Algebra: Graphs and Models 26 . If 1 328 , x = then 1 328 . x − = − 27 . If 0.101, x = then 0.101. x − = − 28 . If 0, x = then 0 0. x − = − = 29 . If 72, x = then ( ) 72. x x − − = = 30 . If 29, x = then ( ) 29. x x − − = = 31 . If 2 5 , x = − then 2 5 ( ) . x x − − = = − 32 . If 9.1, x = − then ( ) 9.1. x x − − = = − 33 . ( 1) 1 − − = 34 . ( 7) 7 − − = 3 5 . (7) 7 − = − 36 . (10) 10 − = − 37 . 6 8 (8 6) 2 − = − − = − 38 . 4 13 (13 4) 9 − = − − = − 39 . 0 5 0 ( 5) 5 − = + − = − 40 . 0 8 0 ( 8) 8 − = + − = − 4 1 . 4 3 (4 3) 7 − − = − + = − 4 2 . 5 6 (5 6) 11 − − = − + = − 4 3. 9 ( 3) 9 3 6 − − − = − + = − 44 . 9 ( 5) 9 5 4 − − − = − + = − 45 . Note that we are subtracting a number from itself. The result is 0. We could also do this exercise as follows: 8 ( 8) 8 8 0 − − − = − + = 46 . 10 ( 10) 10 10 0 − − − = − + = See exercise 45 . 47 . 30 40 30 ( 40) 10 − = + − = − 48 . 20 27 20 ( 27) 7 − = + − = − 49 . 7 ( 9) 7 9 2 − − − = − + = 50 . 8 ( 3) 8 3 5 − − − = − + = − 51 . 9 ( 9) 9 9 0 − − − = − + = 52 . 40 ( 40) 40 40 0 − − − = − + = 53 . 5 5 5 ( 5) 0 − = + − = 54 . 7 7 7 ( 7) 0 − = + − = 55 . 4 ( 4) 4 4 8 − − = + = 56 . 6 ( 6) 6 6 12 − − = + = 57 . 7 4 7 ( 4) 11 − − = − + − = − 58 . 6 8 6 ( 8) 14 − − = − + − = − 59 . 6 ( 10) 6 10 16 − − = + = 6 0 . 3 ( 12) 3 12 15 − − = + = 6 1 . 6 ( 5) 6 5 1 − − − = − + = − 62 . 4 ( 7) 4 7 3 − − − = − + = 63 . 5 ( 12) 5 12 17 − − = + = 64 . 5 ( 6) 5 6 11 − − = + = 65 . 0 ( 10) 0 10 10 − − = + = 66 . 0 ( 1) 0 1 1 − − = + = 67 . 5 ( 2) 5 2 3 − − − = − + = − 68 . 3 ( 1) 3 1 2 − − − = − + = − 69 . 7 14 7 ( 14) 21 − − = − + − = − 70 . 9 16 9 ( 16) 25 − − = − + − = − 71 . 8 0 8 0 8 − − = − + = −

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