Chapter 2 Algebra Basics, Equations, and Inequalitiescollege.cengage.com/mathematics/hubbard/elementary_algebra/2e/students/... · SSM: Elementary Algebra Chapter 2: Algebra Basics,

1 Copyright © Houghton Mifflin Company. All rights reserved.

Chapter 2

Algebra Basics, Equations, and Inequalities

Section 2.1

1. A numerical expression is an algebraic expression that does not contain a variable.

3. 2x − 7 = 2 ⋅3− 7 Replace x with 3= 6 − 7= −1

5. −3(1− x) = −3[1− (−1)] Replace x with − 1= −3(1+1) = −3(2)= −6

7. x − (1− 3x) = −2− [1− 3(−2)] Replace x with − 2= −2− (1+ 6) = −2− 7=−9

9. 2x2 + 3x + 2 = 2 ⋅32 + 3⋅ 3+ 2 Replace x with 3= 18 + 9+ 2 = 29

11. 3x − y = 3⋅2 − (−4) Replace x with 2 and y with − 4= 6+ 4 = 10

13. x2 + xy + 3y2 = (−3)2 + (−3) ⋅1+ 3 ⋅12 Replace x with − 3 and y with 1 = 9 − 3+ 3= 9

15.5− y3− x

= 5− (−1)3− 2

Replace x with 2 and y with −1

= 61

= 6

17.a + b

c= 3+ (−2)

4 Replace a with 3, b with − 2, and c with 4

= 14

19. (a + b)2 − 3c = [3+(−2)]2 − 3 ⋅4 Replace a with 3, b with − 2, and c with 4=12 −12 =1−12 =−11

21. For x = –3.92, x – 2(x – 1) = 5.92.

23. For x = –1.2, x + 52x −1

≈ −1.12 .

2 Chapter 2: Algebra Basics, Equations, and Inequalities SSM: Elementary Algebra

Copyright © Houghton Mifflin Company. All rights reserved.

25. There are two terms in 2x + 1 but onlyone term in 2(x + 1). In 2(x + 1) both 2and x + 1 are factors.

27. Terms: 3y, x, –8Coefficients: 3, 1, –8

29. Terms: x2, 3x, –6Coefficients: 1, 3, –6

31. Terms: 7b, − c , 3a7

Coefficients: 7, −1, 37

33. 3x + 8x = (3 + 8)x = 11x

35. –y + 3y = (–1 + 3)y = 2y

37. –3x – 5x = (–3 – 5)x = –8x

39. 4t2 − t2 = 4 t2 −1 ⋅ t2

= (4 −1)t2 = 3t2

41. 6x – 7x – 3 = –x + 3

43. 9x – 12 – 3x + 5 = 6x – 7

45. 2m – 7n – 3m + 6n = –m – n

47. 10 – 2b + 3b – 9 = b + 1

49. 2a2 + a3 − a2 − 5a3 =− 4a3 + a2

51. ab + 3ac – 2ab + 5ac = –ab + 8ac

53.13

a+ 12

b − a + 23

b = − 23

a + 76

b

55. 2.1x – 4.2y + 0.3y + 1.2x = 3.3x –3.9y

57. Remove parentheses and combine liketerms.

59. −5(3x) = (−5 ⋅3)x = −15x

61. − 14

(4y) = − 14

⋅ 4

y = −y

63. − 56

− 910

x

= − 56

⋅ −910

x = 34

x

65. 5(4 + 3b) = 5(4)+ 5(3b)= 20 +15b= 15b+ 20

67. −3(5 − 2x) = −3(5) − 3(−2x)=−15 + 6x= 6x −15

69. 2(3x + y) = 2(3x) + 2(y) = 6x + 2y

71. −3(−x + y − 5) = −3(−x) − 3(y)− 3(−5)= 3x − 3y + 15

73. − 35

(5x + 20) = − 35

(5x)− 35

(20)

= −3x −12

75. – (x – 4) = –x + 4

77. –(3 + n) = –3 – n

79. – (2x + 5y – 3) = –2x – 5y + 3

81. 2(x −3) + 4 = 2 x − 6+ 4 Distributive Property= 2x − 2 Combine like terms

83. 5− 3(x −1) = 5 − 3x + 3 Distributive Property= −3x + 8 Combine like terms

85. −2(x − 3y) + y + 2x = −2x + 6y + y + 2x Distributive Property= 7y Combine like terms

87. (2a + 7) + (5 – a) = a + 12 Combine like terms

89. 2 + (2a + b) + a – b = 3a + 2 Combine like terms

SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 3


91. 4(1−3x)− (3x + 4) = 4− 12x −3x − 4 Distributive Property= −15x Combine like terms

93. 2 5 − (2x − 3)[ ]+ 3x = 2 5 − 2x +3[ ] + 3x= 2 −2x + 8[ ]+ 3x = −4x +16 + 3x = 16 − x

Distributive PropertySimplify inside bracketsDistributive propertyCombine like terms

95. 4 + 5 x −3(x + 2)[ ]= 4 + 5 x −3x − 6[ ] = 4 + 5 −2x − 6[ ] = 4 −10x − 30 = −10x − 26

Distributive PropertySimplify inside bracketsDistributive propertyCombine like terms

97. The student will study for 8 hours, so t = 8.8t + 20 = 8 ⋅8 + 20

= 64 + 20= 84

The predicted grade is an 84.

99. (a) t = 1985 – 1980 = 5t = 2005 – 1980 = 25

(b)152

(5)3 = 937.5

152

(25)3 = 117,187.5

The model is most accurate for year 2005.

101. 2(3n + 5) − 7 − 3(n+ 1)− 2(n − 4)[ ]= 2(3n + 5) − 7− 3n + 3− 2n + 8[ ]= 2(3n + 5) − 7 − [n+11]= 6n +10 − 7 − n − 11= 5n −8

103.x + y

2 x + 3y= 1.2 + (−3.1)

2(1.2)+ 3(−3.1) Replace x with 1.2 and y with − 3.1

= 1.2− 3.12.4 − 9.3

= −1.9−6.9

≈ 0.28

105. 2a2 + (c − 4)2 = 2(−5.3)2 + (2.15− 4)2 Replace a with − 5.3 and c with 2.15= 2(−5.3)2 + (−1.85)2 = 2(28.09) + 3.4225= 56.18 + 3.4225= 59.6025

Section 2.2

1. The number lines are called the x-axisand y-axis. Their point of intersection iscalled the origin.

3. Quadrant I

5. Quadrant III

7. Quadrant II



9. Quadrant IV

11. x-axis

13. y-axis

15. A(5, 0)B(–7, 2)C(2, 6)D(0, –4)E(–3, –5)F(4, –4)

17. A(0, 6)B(–5, 0)C(7, 3)D(–6, –3)E(–2, 8)F(3, –3)

19. The order of the coordinates isdifferent.

21.

23.

25.

The next ordered pair is (2, 4).

27.

The next ordered pair is (6, –12).

29.

The next ordered pair is (7, 10).

31.

The next ordered pair is (10, –7).

33. (a) +, –

(b) +, +

(c) –, +

(d) –, –

35. The point is either on the x-axis, or inquadrant I or IV.

37. The point is either on the y-axis, or inquadrant III or IV.

39. The point is in quadrant I or III.

41. The point is on the y-axis.

43. The point is in quadrant II or IV.

45. (a) The point is on the x-axis.



(b) The point is on the y-axis.

(c) The point is in quadrant IV.

(d) The point is in quadrant I or III.

47. P(0, 13)

49. P(9, 9)

51. P(–12, 6)

53.

55.

57. B(–4, 6)

59. B(17, 0)

61. The length is 5 and the width is 3. Thearea is 15 and the perimeter is 16.

63. The length of each side is 3. The area is9 and the perimeter is 12.

65. (a) The second coordinates are:–1, 1, 3, 5, 7

(b) y = x + 2

67. (a) The second coordinates are:3, 1, 0, –1, –3

(b) y = –x

69. (a) The fourth vertex is (–40, –10).

(b) The length is 50 feet, the width is20feet, so the area is 1000 square feet.

71. (a) Sales/marketing and Writtencommunications would be the samedistance above the x-axis becausethey are both 35%

(b) Basic computer would be thegreatest distance above the x-axisbecause 68% is the largest.

73. They would all lie in the first quadrant.

75. (a) The years in which the number ofresolutions exceed the average forthe period

(b) The years in which the number ofresolutions was less than theaverage for the period

77. The midpoint is (2, 3).

79. The midpoint is (–4, 2).

81. The other two vertices are 8 units aboveor below the two given vertices. Sinceone of these vertices must be inquadrant I, it must be 8 units above thevertex in quadrant IV, (7, –3). Thecoordinates of P are (7, 5).



Section 2.3

1. It is not necessary to enter the expression each time it is to be evaluated.

3. x –1 0 4 11

x – 6 –7 –6 –2 5

5. x –8 3 6 12

–2.1x + 7 23.8 0.7 –5.6 –18.2

7. x –9 –4 3 5

x2 + 4x – 21 24 –21 0 24

9. x –8 –1 2 10

2(x + 8) 0 14 20 36

11. x –5 –1 –0.5 8

2x +1 9 1 0 17

13. x –5 1 25 50

x +12x

0.4 1 0.52 0.51

15. x –2 1 4 7 10

x – 3 –5 –2 1 4 7

(x, x – 3) (–2, –5) (1, –2) (4, 1) (7, 4) (10, 7)

17. x 3 7 0 –4 –5

2x + 5 11 19 5 –3 –5

(x, 2x + 5) (3, 11) (7, 19) (0, 5) (–4, –3) (–5, –5)

19. x 1 4 6 –3 –5

x2 – 3x – 4 –6 0 14 14 36

(x, x2 – 3x – 4) (1, –6) (4, 0) (6, 14) (–3, 14) (–5, 36)

21. You obtain an error message for x = 1 because division by 0 is not defined.

23. When x = 3, y = 4.

25. When x = –1, y = 1.



27.

x 2 5 7

2x –10

–6 0 4

29.

x –30 –9 12

13

x + 5 –5 2 9

31.

x –6 0 5

–(3x + 1) 17 –1 –16

33. The first coordinate, 3, is the value ofthe variable, and the second coordinate,7, is the value of the expression forx = 3.

35.

x 2 4 0 5 –1 –2

x – 4 –2 0 –4 1 –5 –6

37.

x –32 –8 –12 16 6 28

14

x + 3 –5 1 0 7 4.5 10

39.

(a) x = 3

(b) x = 9

(c) x = –5

(d) x = –10

41.

(a) x = –1

(b) x = 2

(c) x = 5

(d) x = 8

43.

(a) x = –23

(b) x = 6

(c) x = 20



(d) x = 33

45.

(a) x = –2, 11

(b) x = –1, 10

(c) x = 1, 8

(d) x = 2, 7

47.

(a) x = –12, 22

(b) x = –3, 13

(c) x = 5

(d) x = 1, 9

49. 2x + 1 = 4 – x when x = 1.

51. 3x = 8 – x when x = 2.

53.

When x = 28, there are 3 pizzas left.

55. Year ReportedNumber

ModeledNumber

Difference

1983 620 13(3) + 588 = 627 –7

1985 660 13(5) + 588 = 653 7

1987 680 13(7) + 588 = 679 1

1989 700 13(9) + 588 = 705 –5

1991 740 13(11) + 588 = 731 9

1993 750 13(13) + 588 = 757 –7The model is most accurate for 1987.

57. Year xValue of the expression

1965 0 3241970 5 17871975 10 32501980 15 47141985 20 61771990 25 7640

59. For the year 2000, x = 35, and the valueof the expression is 10,567.

61. The y-coordinate is one more than twicethe x -coordinate.

63. The y-coordinate is twice the sum of thex-coordinate and 1.

65. (a) For c = 0.5 and k = 6.1,c2 + 5k = 30.75.

(b) For c = –0.5 and k = 6.1,c2 + 5k = 30.75.



Section 2.4

1. An equation has an = symbol, but anexpression does not.

3. Expression

5. Equation

7. Equation

9. Expression

11. A solution of an equation is a value ofthe variable that makes the equationtrue.

13. x − 5 = 914 − 5 = 9 Replace x with 14

9 = 9 TrueYes

15. 5− 3x = 115 − 3 ⋅2 = 11 Replace x with 2

5− 6 = 11−1 = 11 False

No

17. 10x − 3 = 5x +110 ⋅1− 3 = 5⋅1+1 Replace x with 1

10 − 3 = 5+17 = 6 False

No

19. x2 + 3x + 2 = 0(−1)2 + 3(−1)+ 2 = 0 Replace x

with −11 − 3+ 2 = 0

0 = 0 TrueYes

21. Yes

23. Yes

25. Yes

27. No

29.

x = 14

31.

x = 12

33.

x = 4

35.

x = –14

37.

x = –32

39.

x = –20

41. Graph each side of the equation andtrace to the point of intersection. Thex-coordinate of that point is theestimated solution.

43.

x = –1.25



45.

x = 0.14

47.

x = –1.71

49.

x = –7

51.

x = –12

53.

x = –12

55.

x = –20

57.

x = 8

59.

x = –9

61.

x = 15

63.

x = –9

65.

x = 15

67. (a) The graphs of the two sides of theequation do not intersect. Thesolution set is Ø.

(b) The graphs of the two sides of theequation coincide. The solution setis ℜ.



69.

x = 0

71.

Ø

73.

ℜ

75.

Ø

77.

ℜ

79.

x = 8

81.

x = 24

83.

Ø

85.

ℜ

87. (a) 2.1x + 22.8 = 37.50; 7 (1997)

(b) 2.1x + 20.8 = 39.70; 9 (1999)

89.

x = 110

91.

x = –6, 30

93.

Ø



Section 2.5

1. Equivalent equations are equations thathave exactly the same solutions.

3. 7x − 6x = 9x = 9 Combine like terms

5. 1.8x − 0.8x = 3.5x = 3.5 Combine like terms

7. (i)

9. (ii)

11. x + 5 = 9x + 5 − 5 = 9 − 5 Subtract 5 from both sides

x = 4

13. −4 = x − 6x − 6 = −4 Exchange sides

x − 6 + 6 = −4+ 6 Add 6 to both sidesx = 2

15. 25 + x = 3925 + x − 25 = 39 − 25 Subtract 25 from both sides

x = 14

17. x + 27.6 = 8.9 x + 27.6 − 27.6 = 8.9− 27.6 Subtract 27.6 from both sides

x = −18.7

19. 12 = x +12x + 12 = 12 Exchange sides

x +12 − 12 = 12 −12 Subtract 12 from both sidesx = 0

21. 3x = 2x −1 3x − 2 x = 2x −1 − 2x Subtract 2 x from both sides

x =−1

23. 7x = 5 + 6x7x − 6x = 5 + 6x − 6x Subtract 6x from both sides

x = 5

25. −4x + 2 = −3x−4x + 2 + 4x = −3x + 4x

2 = x x = 2

Add 4 x to both sides

Exchange sides

27. 9x − 6 = 10x9x − 6− 9 x = 10x − 9x

−6 = x x = −6

Subtract 9x from both sides

Exchange sides

29. −3a + 5a = 9+ a + 62a = a + 15

2a − a = a + 15− aa = 15

Combine like termsSubtract a from both sides



31. 3x + 3− 3x + 7= 6x − 6− 5x10 = x − 6 Combine like terms

10 +6 = x − 6 +6 Add 6 to both sides16 = x

x = 16 Exchange sides

33. 7x − 9 = 8 + 6x7x − 9− 6 x = 8+ 6x − 6x

x −9 = 8x − 9+ 9 = 8 + 9

x = 17


Add 9 to both sides

35. 1− 3y = 4 − 4y1 − 3y + 4y = 4− 4 y + 4y Add 4y to both sides

1 + y = 4 1 + y − 1= 4 −1 Subtract 1 from both sides

y = 3

37. 5x + 3= 6x +15x + 3 − 5x = 6x +1− 5x

3 = x +13 −1= x +1 −1

2 = xx = 2


Subtract 1 from both sides

Exchange sides

39. 1− 5x = 1 − 4x1 − 5x + 5x = 1− 4x + 5x

1 = 1+ x1 −1= 1 + x −1

0 = xx = 0

Add 5x to both sides


Exchange sides

41. When we add 3 to both sides of x – 3 = 12, the left side becomes x – 3 + 3 = x + 0 = x. When

we divide both sides of 3x = 12 by 3, the left side becomes 3x3

= x . In both cases, the variable

is isolated.

43. (ii)

45. (i)

47. All of the given steps are correct. In each case, the result is 1x = 20 because

(i)43

and 34

are reciprocals whose product is 1.

(ii)34

divided by 34

is 1.

(iii) Multiplying by 4 and dividing by 3 is equivalent to multiplying by 43

.

49. 5x = 205x5

= 205

Divide both sides by 5

x = 4



51. 48 = −16x−16x = 48−16x−16

= 48−16

x = −3

Exchange sidesDivide both sides by −16

53. −y = −3−y−1

= −3−1

Divide both sides by −1

y = 3

55. −3x = 0−3x−3

= 0−3

Divide both sides by − 3

x = 0

57.x9

= −4

9 ⋅ x9

= 9(−4) Multiply both sides by 9

x = −36

59.−t4

= −5

−4 ⋅ −t4

= −4(−5) Multiply both sides by − 4

t = 20

61.34

x = 9

43

⋅ 34

x = 43

⋅9 Multiply both sides by 43

x = 12

63. 2x − 7x = 0−5x = 0 Combine like terms−5x−5

= 0−5


x = 0

65. y − 6y =−25−5y =−25−5y−5

= −25−5

y = 5

Combine like terms


67. 8 = 3t − 4t8 =− t

− t = 8−t−1

= 8−1

t =− 8

Combine like termsExchange sidesDivide both sides by −1

69. 9x = 8x9x − 8x = 8x − 8x Subtract 8x from both sides

x = 0



71. −4x =−28−4 x−4

= −28−4


x = 7

73. 5− 7x = −6x5− 7x + 7x = −6x + 7x

5 = xx = 5

Add 7 x to both sides

Exchange sides

75. −3+ x = 0−3 + x + 3= 0+ 3 Add 3 to both sides

x = 3

77. 1

4=−2y

−2y = 1

4−2y−2

=14−2

y = − 18

Exchange sides


79. Distributive Property

81. Multiplication Property of Equations

83. Addition Property of Equations

85. 0.7x = 3.50.7x0.7

= 3.50.7

Divide both sides by 0.7

x = 5The sales in 1991 were at $3.5 billion.

87. 0.7x = 5.60.7x0.7

= 5.60.7


x = 8The solution is 8, which means that sales for 1994 were projected to be $5.6 billion.

89. x + a = bx + a − a = b − a Subtract a from both sides

x = b − a

91. 5x − a = b5x − a + a = b + a

5x = a + b5x5

= a + b5

x = a + b

5

Add a to both sides




93. x − k = 0−4 − k = 0 Replace x with − 4

−4 − k + k = 0+ k Add k to both sides−4 = k

k = −4 Exchange sides

95. kx + 1 = 7k ⋅ 2+ 1= 7 Replace x with 22k + 1 = 7

2k +1− 1 = 7−1 Subtract 1 from both sides2k = 6 2k2

= 62


k = 3

97. 2x −1= 52x −1 +1= 5 +1

2x = 62x2

= 62

x = 3

Solve for xAdd 1 to both sides


3x − 4 = 3⋅3 − 4= 9 − 4= 5

Replace x with 3

Section 2.6

1. Dividing both sides by 2 would create fractions and make the equation harder to solve.The best first step is to add 5 to both sides.

3. 5x +14 = 45x + 14 −14 = 4 −14

5x =−105x5

= −105

x =−2



The solution is –2.

5. 2 −3x = 02 − 3x − 2 = 0 − 2 Subtract 2 from both sides

−3x = −2 −3x−3

= −2−3


x = 23

The solution is 23

.

7. 0.2 − 0.37t = −0.91 0.2 − 0.37t − 0.2 = −0.91 − 0.2 Subtract 0.2 from both sides

−0.37t = −1.11 −0.37t−0.37

= −1.11−0.37

Divide both sides by − 0.37

t = 3 The solution is 3.



9. 7x = 2x − 57x − 2x = 2x − 5 − 2x Subtract 2x from both sides

5x = −5 5x5

= −55


x = −1 The solution is –1.

11. 3x = 5x +143x − 5x = 5x +14 − 5x Subtract 5x from both sides

−2x = 14−2x−2

= 14−2


x = −7 The solution is –7.

13. 4x − 7 = 6 x + 34 x − 7 − 6x = 6x + 3 − 6x Subtract 6x from both sides

−2x −7 = 3−2 x − 7 + 7 = 3+ 7 Add 7 to both sides

−2 x =10−2x−2

= 10−2


x = −5The solution is –5.

15. 5x + 2 = 2 − 3x5x + 2 + 3x = 2 − 3x + 3x Add 3x to both sides

8x + 2 = 2 8x + 2 − 2 = 2 − 2 Subtract 2 from both sides

8x =0 8x8

= 08


x = 0 The solution is 0.

17. x + 6 = 5x −10x + 6 − 5x = 5x −10 − 5x Subtract 5 x from both sides

−4x + 6 = −10−4 x + 6 − 6 = −10 − 6 Subtract 6 from both sides

−4x = −16−4x

−4= −16

−4Divide both sides by –4

x = 4The solution is 4.

19. 5 − 9x = 8x + 55 − 9x − 8x = 8x + 5 −8x

5 −17x = 55 −17x − 5 = 5 − 5

−17x =0−17x−17

= 0−17

x = 0




The solution is 0.



21. 9 − x = x +159 − x − x = x +15 − x

9 − 2x = 159 − 2x −9 = 15 − 9

−2 x = 6−2 x−2

= 6−2

x = −3

Subtract x from both sides




23. 3x + 8− 5x = 2− x + 2x − 3−2x +8 = x −1−3x +8 = −1

−3x = −9x = 3

Combine like termsSubtract x from both sidesSubtract 8 from both sidesDivide by − 3

The solution is 3.

25. 3+ 5y − 24 + 3y = y −1+ y8y − 21 = 2y −16y − 21 =−1

6y = 20

y = 206

= 103

Combine like terms Subtract 2y from both sides Add 21 to both sides

Divide by 6

The solution is 103

.

27. 5− 5x +8 − 4 = −9+ x9− 5x = −9+ x Combine like terms9− 6x = −9 Subtract x from both sides

−6x = −18 Subtract 9 from both sidesx = 3 Divide by − 6

The solution is 3.

29. 3(x +1) = 13x + 3 = 1

3x = −2

x = − 2

3

Remove grouping symbolsSubtract 3 from both sides

Divide by 3

The solution is − 23

.

31. 5(2x −1) = 3(3x −1)10x − 5 = 9 x − 3 Remove grouping symbols

x − 5 = −3 Subtract 9x from both sidesx = 2 Add 5 to both sides

The solution is 2.

33. −10 = 3x + 8+ 4(x − 1)−10 = 3x + 8 + 4x − 4 Remove grouping symbols−10 = 7x + 4 Combine like terms−14 = 7x Subtract 4 from both sides

−2 = x Divide by 7 The solution is –2.



35. 4(x −1) = 5 +3(x − 6)4x − 4 = 5 +3x −184x − 4 = 3x − 13

x − 4 = −13x = −9

Remove grouping symbolsCombine like termsSubtract 3 x from both sidesAdd 4 to both sides


37. 6x −(5x −3) = 06x − 5x + 3= 0 Remove grouping symbols

x + 3 = 0 Combine like termsx = −3 Subtract 3 from both sides


39. 2 − 2(3x − 4) = 8− 2(5 + x )2 − 6x + 8= 8 −10 − 2x

10 − 6x = −2 − 2x12 − 6x = −2x

12 = 4x3= x

Remove grouping symbolsCombine like termsAdd 2 to both sidesAdd 6 x to both sidesDivide by 4

The solution is 3.

41. The number n is the LCD of all fractions in the equation.

43.2

3x − 2

3= 5

3 The LCD is 3

3 ⋅ 2

3x − 3 ⋅ 2

3= 3 ⋅ 5

3 Multiply by 3 to clear denominators

2x − 2 = 5 2x = 7 Add 2 to both sides

x = 72

Divide by 2

The solution is 72

.

45. 1+ 3t4

= 14

The LCD is 4

4 ⋅1 + 4 ⋅ 3t4

= 4 ⋅ 14

Multiply by 4 to clear denominators

4+ 3t = 13t = −3 Subtract 4 from both sidest = −1 Divide by 3


47.12

x − 2 = 1+ 23

x + 16

6 ⋅ 12

x − 6 ⋅2 = 6 ⋅1 + 6 ⋅ 23

x + 6 ⋅ 16

3x −12 = 6 + 4 x +13x −12 = 7+ 4 x

−12 = 7+ x−19 = x

The LCD is 6

Multiply by 6 to clear denominators

Combine like termsSubtract 3x from both sidesSubtract 7 from both sides




49.x + 5

3= − x

2+ 5 The LCD is 6

6 ⋅ x + 5

3= 6 ⋅ −x

2+ 6 ⋅ 5 Multiply by 6 to clear denominators

2(x + 5)= −3x + 30

2x +10 = −3x + 30 Remove grouping symbols

5x +10 = 30 Add 3 x to both sides

5x = 20 Subtract 10 from both sides

x = 4 Divide by 5The solution is 4.

51.35

(x −1) = 9 The LCD is 5

53

⋅ 35

(x −1) = 53

⋅9 Multiply by 53

to clear denominators and isolate x

x −1 =15x =16 Add 1 to both sides

The solution is 16.

53.x3

+ 2 = 32

(x + 3) − 6 The LCD is 6

6 ⋅ x3

+ 6 ⋅ 2 = 6 ⋅ 32

(x + 3) − 6 ⋅6 Multiply by 6 to clear denominators

2x +12 = 9(x + 3) −362x +12 = 9x + 27 − 36 Remove grouping symbols2x +12 = 9x − 9 Combine like terms

12 = 7x − 9 Subtract 2 x from both sides21 = 7x Add 9 to both sides3 = x Divide by 7

The solution is 3.

55.37

x = 514

(x +1)+ 17

(8 − x) The LCD is 14

14 ⋅ 37

x = 14 ⋅ 514

(x +1) +14 ⋅ 17

(8 − x) Multiply by 14 to clear denominators

6x = 5(x + 1)+2(8 − x)6x = 5x + 5 +16 − 2x Remove grouping symbols6x =3x + 21 Combine like terms3x = 21 Subtract 3x from both sides

x = 7 Divide by 3The solution is 7.

57. 7.35x + 5.1 = 2.6x − 0.67.35x = 2.6 x − 5.74.75x = −5.7

x = −1.2

Subtract 5.1 from both sidesSubtract 2.6 x from both sidesDivide by 4.75

The solution is –1.2.



59. 2.7(3x −1.5) + 6.3 = 9.1x + 6.558.1x − 4.05 + 6.3 = 9.1x + 6.55

8.1x + 2.25 = 9.1x + 6.558.1x = 9.1x + 4.3

−x = 4.3x = −4.3

Remove grouping symbolsCombine like termsSubtract 2.25 from both sidesSubtract 9.1x from both sidesMultiply by −1

The solution is –4.3.

61. x + 7 = 7+ xx + 7 − x = 7+ x − x Subtract x from both sides

7 = 7 TrueIdentitySolution set: ℜ

63. x = x + 3x − x = x + 3− x Subtract x from both sides

0 = 3 FalseContradictionSolution Set: Ø

65. 2x = 3x2x − 2x = 3x − 2x Subtract 2x from both sides

0 = xConditionalThe solution is 0.

67. 6t + 7 = 76t = 0 Subtract 7 from both sidest = 0 Divide by 6

ConditionalThe solution is 0.

69. 0x = 30 = 3 False

ContradictionThe solution set is Ø.

71. 6x − 3− x = 6x + 55x − 3= 6x + 5

−3 = x + 5−8 = x

Combine like termsSubtract 5x from both sidesSubtract 5 from both sides


73. 2 + 3x + 5 = 6x + 7 − 6x3x + 7 = 7 Combine like terms

3x = 0 Subtract 7 from both sidesx = 0 Divide by 3

The solution is 0.



75.3

4x + 5

2= 1

2(5 + x)+ 1

4x The LCD is 4

4 ⋅ 34

x + 4 ⋅ 52

= 4⋅ 12

(5 + x) + 4 ⋅ 14

x Multiply by 4 to clear denominators

3x +10 = 2(5 + x )+ x

3x +10 =10 + 2x + x Remove grouping symbols

3x +10 = 3x +10 Combine like terms

10 =10 True Subtract 3x from both sidesSolution set: ℜ

77. 4(x − 2) +8 = 6(x +1)− 2x4x −8 +8 = 6x + 6 − 2x Remove grouping symbols

4x = 4x + 6 Combine like terms0 = 6 False Subtract 4x from both sides

Solution set: Ø

79. 2(3x − 4) = 2 − 6(1− x)6x − 8 = 2 − 6 + 6x Remove grouping symbols6x − 8 = −4 + 6x Combine like terms

−8 = −4 False Subtract 6x from both sidesSolution set: Ø

81. 3x −1− 2(x − 4) = 2x + 7 − x3x −1− 2x + 8 = 2 x + 7− x Remove grouping symbols

x + 7 = x +7 Combine like terms7 = 7 True Subtract x from both sides

Solution set: ℜ

83. (a) (i) 5.88x + 40.58 = 150 The cost at a private college is $150,000.(ii) 2.28x + 15.7 = 2(31.637) 2.28(7) + 15.7 = 31.66 The cost at a public college is double the 1997 cost.

(b) (i) 5.88x + 40.58 =1505.88x =109.42

x =18.61 1990 + 19 = 2009(ii) 2.28x +15.7 = 2(31.637)

2.28x = 47.574x = 20.87

1990 + 21 = 2011

85. For 1976, t = 20Men: (−0.015)(20) +10.5 = −0.3+10.5

= 10.2 secondsWomen: ( −0.018)(20)+11.5 = −0.36 +11.5

= 11.14 seconds

87. The two times will be equal in the year 2289. For both men and women, the winning times are5.51 seconds.



89. x − 5 x − 5 x − 5(x − 5)[ ]{ } = 1x − 5 x − 5 x − 5x + 25[ ]{ } =1 Remove parentheses

x − 5 x − 5 −4x + 25[ ]{ } = 1 Combine like termsx − 5 x + 20x −125{ } = 1 Remove grouping symbols

x − 5 21x −125{ }= 1 Combine like termsx −105x + 625 = 1 Remove grouping symbols

−104x + 625 = 1 Combine like terms−104x = −624 Subtract 625 from both sides

x = 6 Divide by − 104 The solution is 6.

91. x − 2(x − 3) − (x + 1)[ ] = −3(x −1)+ 4 + 3xx − 2x − 6 − x −1[ ] = −3x + 3 + 4+ 3x Remove parentheses

x − 2x + 6 + x +1 = −3x + 3 + 4+ 3x Remove brackets7 = 7 True Combine like terms

Solution set: ℜ

93. ax + b = cax = c − b Subtract b from both sides

x = c − ba

Divide by a

95. 2t2 −15 = t(t + 5)+ t2

2 t2 −15 = t2 + 5t + t2 Remove grouping symbols2 t2 −15 = 2t2 + 5t Combine like terms

−15 = 5t Subtract 2 t2 from both sides−3 = t Divide by 5


Section 2.7

1. The value of the expression x + 3 is less than or equal to 7.

3. x = 3 x = 5 x = 9 x = 11

2 + x > 7 2 + x > 7 2 + x > 7 2 + x > 7

2 + 3 > 7 2 + 5 > 7 2 + 9 > 7 2 + 11 > 7

5 > 7 7 > 7 11 > 7 13 > 7

False False True True

The solutions are x = 9 and x = 11.



5. x = –4 x = –2 x = –1 x = 0

4 ≤ 1 – 3x 4 ≤ 1 – 3x 4 ≤ 1 – 3x 4 ≤ 1 – 3x

4 ≤ 1 – 3(–4) 4 ≤ 1 – 3(–2) 4 ≤ 1 – 3(–1) 4 ≤ 1 – 3 ? 0

4 ≤ 1 + 12 4 ≤ 1 + 6 4 ≤ 1 + 3 4 ≤ 1 – 0

4 ≤ 13 4 ≤ 7 4 ≤ 4 4 ≤ 1

True True True False

The solutions are x = –4, –2, and –1.

7.x = –1 x = 2 x = 5 x = 6 x = 9

–3 ≤ 2x – 7 < 5 –3 ≤ 2x – 7 < 5 –3 ≤ 2x – 7 < 5 –3 ≤ 2x – 7 < 5 –3 ≤ 2x – 7 < 5

–3 ≤ 2(–1) – 7 < 5 –3 ≤ 2 ? 2 – 7 < 5 –3 ≤ 2 ? 5 – 7 < 5 –3 ≤ 2 ? 6 – 7 < 5 –3 ≤ 2 ? 9 – 7 < 5

–3 ≤ –2 – 7 < 5 –3 ≤ 4 – 7 < 5 –3 ≤ 10 – 7 < 5 –3 ≤ 12 – 7 < 5 –3 ≤ 18 – 7 < 5

–3 ≤ –9 < 5 –3 ≤ –3 < 5 –3 ≤ 3 < 5 –3 ≤ 5 < 5 –3 ≤ 11 < 5

False True True False False

The solutions are x = 2 and x = 5.

9. The symbols [ and ] indicate that anendpoint is included. The symbols ( and) indicate that an endpoint is notincluded.

11.

13.

15.

17.

19.

21.

23. n ≥ 0

25. x ≤ 3

27. n < 0

29. x ≤ 6

31. x < –3

33. x ≥ 5

35. –4 < x ≤ 7

37. The point of intersection represents asolution if the inequality symbol is ≤ or≥.

39. x > –15

41. x ≤ 0

43.

x > –17



45.

x ≥ –5

47.

x < –3

49.

x ≤ 7

51.

x < 0

53.

x < 2

55.

x ≥ –8

57.

x < 12

59. If the inequality is y1 > y2 or y1 ≥ y2the solution set is ℜ. If the inequality isy1 < y2or y1 ≤ y2 , the solution set is Ø.

61. The solution set is ℜ.

63. (a)

The solution set is ℜ.

(b)

The solution set is Ø.65.

–11 ≤ x ≤ 23

67.

–19 < x ≤ 13



69.

–13 < x ≤ 9

71.

–6 < x < 8

73.

x ≤ 9

75.

The solution set is Ø.

77.

x > –9

79.


81.

x < 0

83.


85. x + 6 ≥ 3

x ≥ –3

87. 8 – x > –2

x < 10



89. (a) –0.85x + 26.6 ≥ 20

(b) –0.85x + 26.6 ≥ 20

x ≤ 7.8; before 1988

91. (a) –0.575x + 22.6 > –0.85x + 26.6

(b)

x > 14.5; after 1995

93.


95.

(a) The solution set is Ø.

(b) The solution set is ℜ.

97.


99.


Section 2.8

1. The properties allow us to add the samenumber to both sides of an equation oran inequality.

3. –4(–1) > 1(–1)

5. 3 + a > 3 + b

7. –4a > –4b

9.a

−5< b

−5

11. x + 5 < 1x + 5 − 5 < 1− 5 Subtract 5 from

both sidesx <−4

13. −3 ≤ 2 + x−3 −2 ≤ 2 + x −2 Subtract 2

from both sides−5 ≤ x

x ≥ −5



15. 5x −1≥ −1 +4x5x −1− 4 x ≥ −1+ 4x − 4x Subtract 4 x from both sides

x −1≥ −1x −1+1 ≥ −1+1 Add 1 to both sides

x ≥ 0

17. x − 2 > 2x + 6x − 2 − x > 2x + 6 − x Subtract x from both sides

−2 > x + 6−2 − 6 > x + 6 − 6 Subtract 6 from both sides

−8> xx < −8

19. 3x > 123x3

> 123


x > 4

21. −5x <15−5x−5

> 15−5

Divide both sides by − 5 and reverse the inequality

x > −3

23. 0 ≤ −x−1(0)≥ −1(−x) Multiply by −1 and reverse the inequality

0 ≥ xx ≤ 0

25.34

x ≤12

43

⋅ 34

x ≤ 43

⋅12 Multiply both sides by 43

x ≤16

27.x

−5≤ 1

−5x

−5

≥ −5 ⋅1 Multiply by − 5 and reverse the inequality

x ≥ −5

29. No. Because we multiplied both sides by a negative number, the inequality symbol must bereversed: 32x > 89.

31. 5+ 3x >−43x >−9 Subtract 5 from both sides3x

3> −9

3 Divide both sides by 3

x >−3

33. 3− x ≤ 5−x ≤ 2 Subtract 3 from both sides−x

−1≥ 2

−1 Divide by −1 and reverse the inequality

x ≥ −2



35. 3− 4 x ≤ 19−4 x ≤ 16−4x−4

≥ 16−4

x ≥ −4

Subtract 3 from both sidesDivide by − 4 and reverse the inequality

37. 5x + 5 ≤ 4x − 2x + 5 ≤ −2 Subtract 4x from both sides

x ≤ −7 Subtract 5 from both sides

39. 6− 3x > x − 26 − 4x >−2 Subtract x from both sides

−4x >−8 Subtract 6 from both sides−4x−4

< −8−4

Divide by − 4 and reverse the inequality

x < 2

41. 7+ x ≤ 3(1− x)

7 + x ≤ 3− 3x

7 + 4x ≤ 3

4x ≤−44x4

≤ −44

x ≤−1

Remove grouping symbols

Add 3x to both sides



43. 6(2 − 3x) > 11(3− x)12 −18x > 33 −11x Remove grouping symbols12 − 7x > 33 Add 11x to both sides

−7x > 21 Subtract 12 from both sides−7x−7

< 21−7

Divide by − 7 and reverse the inequality

x < −3

45. 3y + 2(2y +1) >11+ y3y + 4 y + 2 >11+ y

7y + 2 >11+ y6y + 2 >11

6y > 96y6

> 96

y > 32

Remove grouping symbolsCombine like termsSubtract y from both sidesSubtract 2 from both sidesDivide both sides by 6

47. 6(t − 3) + 3≤ 2(4 t + 3) + 5t6t −18 + 3≤ 8t + 6 + 5t

6t −15 ≤13t + 6−7t −15 ≤ 6

−7t ≤ 21−7t−7

≥ 21−7

t ≥ −3

Clear grouping symbolsCombine like termsSubtract 13t from both sidesAdd 15 to both sidesDivide by − 7 and reverse the inequality



49. 2x + 12

≥ 13

The LCD is 6

6 ⋅2x + 6 ⋅ 12

≥ 6 ⋅ 13

Clear denominators

12x + 3 ≥ 212x ≥ −1 Subtract 3 from both sides12x12

≥ −112


x ≥ − 112

51.1

2t − 3

4< − 1

3t The LCD is 12

12 ⋅ 1

2t −12 ⋅ 3

4<12 − 1

3t

Clear denominators

6 t − 9 < −4t10t − 9 < 0 Add 4 t to both sides

10t < 9 Add 9 to both sides10t10

< 910


t < 910

53. −1.6t −1.4 ≥ 4.2 − 3t1.4t −1.4 ≥ 4.2 Add 3t to both sides

1.4t ≥ 5.6 Add 1.4 to both sides1.4t1.4

≥ 5.61.4


t ≥ 4

55. If the resulting inequality is true, the solution set is ℜ. If the resulting inequality is false, thesolution set is Ø.

57. 3(x − 2) ≥ 2x + x3x − 6 ≥ 2x + x Remove grouping symbols3x − 6 ≥ 3x Combine like terms

−6 ≥ 0 Subtract 3x from both sidesThe solution set is Ø.

59. 2(3− x) + x ≤ 6− x6 − 2x + x ≤ 6− x Remove grouping symbols

6 −x ≤ 6− x Combine like terms6 ≤ 6 Add x to both sides


61. 2x − 5+ x ≤−3(2 − x )2x − 5 + x ≤−6 + 3x

3x − 5 ≤ 3x − 6−5 ≤−6

Remove grouping symbolsCombine like termsSubtract 3x from both sides


63. 4 < x + 2 < 114 −2 < x + 2 −2 < 11− 2 Subtract 2 from all three parts

2 < x < 9



65. 10 ≤ −5x < 3510

−5≥ −5x

−5> 35

−5 Divide by − 5 and reverse the inequalities

−2 ≥ x >−7−7 < x ≤−2

67.0 ≤ y

3≤ 2 The LCD is 3

3 ⋅ 0 ≤ 3⋅ y3

≤ 3⋅ 2 Multiply all three parts by 3

0 ≤ y ≤ 6

69. −5 < 2 x −1 <−1−5 +1 < 2x −1 +1< −1+ 1 Add 1 to all three parts

−4 < 2x < 0 −42

< 2x2

< 02

Divide all three parts by 2

−2 < x < 0

71. −2 ≤ 4 − 3x ≤ 4−2 − 4 ≤ 4− 3x − 4 ≤ 4 − 4

−6 ≤ −3x ≤ 0−6−3

≥ −3x−3

≥ 0−3

2 ≥ x ≥ 00 ≤ x ≤ 2

Subtract 4 from all three parts

Divide by − 3 and reverse the inequalities

73. −3x + 5≥ x −15≥ 4x −1 Add 3x to both sides6 ≥ 4x Add 1 to both sides6

4≥ 4x

4 Divide both sides by 4

3

2≥ x

x ≤ 3

2

75. 5x − 6 ≥ 5x −1−6 ≥ −1 Subtract 5x from both sides

False, so the solution set is Ø.

77. x + 5 ≤ x + 75 ≤ 7 Subtract x from both sides

True, so the solution set is ℜ.

79. 1− 3(x + 2) < 71− 3x − 6< 7

−3x − 5 < 7−3x < 12−3x−3

> 12−3

x > −4

Remove grouping symbolsCombine like termsAdd 5 to both sidesDivide by − 3 and reverse the inequality



81. 2x +1> 3x −12

The LCD is 2

2 ⋅2x + 2 ⋅1 > 2 ⋅ 3x −12

Clear denominator

4 x + 2 > 3x −1x + 2 > −1 Subtract 3x from both sides

x > −3 Subtract 2 from both sides

83. 2(x + 3) < 2x +12x + 6 < 2x +1 Remove grouping symbols

6 < 1 Subtract 2 x from both sides

False, so the solution set is Ø.

85. 2.3y + 0.75 ≥ 4.22.3y ≥ 3.452.3y

2.3≥ 3.45

2.3y ≥1.5

Subtract 0.75 from both sides


87. y2 =100 − (0.58x + 27.4)y2 =100 − 0.58x − 27.4y2 =−0.58x + 72.6

89. Until 1979 the number of employed women was less than the number of women who were notemployed.

91. x + a < bx < b − a Subtract a from both sides

93. ax + b ≤ 0

ax ≤ −baxa

≥ −ba

x ≥ − ba

Subtract b from both sides

Divide by a and reverse the inequality since a < 0

95. x(2x − 5) > 2x2 + 5x + 202x2 − 5x > 2x2 + 5x + 20 Remove grouping symbols

−5x > 5x + 20 Subtract 2x2 from both sides−10x > 20 Subtract 5x from both sides−10x−10

< 20−10

Divide by −10 and reverse the inequality

x < −2

97. 3+ x < 2x − 4 < x + 63 + x − x < 2x − 4 − x < x + 6 − x Subtract x from all three parts

3 < x − 4 < 63 + 4 < x − 4 + 4 < 6 + 4 Add 4 to all three parts

7 < x < 10

Chapter 2 Review Exercises

1. (a)−4x + 7 = −4(−6) + 7 Replace x with − 6= 24 + 7= 31



(b) −x2 +1= −(1)2 +1 Replace x with 1= −1+1= 0

3. For x = –3, 7 + 2x − x2 = −8.

5. (a) 3xy + x – 5xy + 4x = –2xy + 5x

(b) 8c2 − 6c2 + 5c2 = 7c2

7. −3(2a − b+ 4) = −3(2a)− 3(−b) − 3(4) Distributive property= −6a + 3b −12

9. −(x + 2y) − 4(3x − y +1) = −x − 2y − 4(3x) − 4(−y)− 4(1)= −x − 2y − 12x + 4y − 4= −13x + 2y − 4

Distributive property

Combine like terms

11. R(4, –1)

13. Point Q has coordinates (–3, 2).

15. The set of all points with a y-coordinate of 0 is the x-axis.

17. The point (0, 0) is called the origin.

19. x –4 –1 3 7

2x – 3 –11 –5 3 11

21.

x = 18

23. The y-coordinate will be –3 because they-coordinate is the value of the expression when x = 5.

25. Equation (b) is a conditional equation.

27. Yes, –3 is a solution.

29.

Yes, 4 is a solution.

31. The point is (–2, 20).

33. (a) The lines are parallel.



(b) The lines coincide.

35. Step (iii) would not produce an equivalent equation.

37. The next in solving equation (ii) would be to isolate the variable term.

39. 7 = 9x − 8x7 = x Combine like terms

The solution is 7.

41. −5x + 8= −4x−5x +8 + 5x = −4x + 5x Add 5x to both sides

8 = xThe solution is 8.

43. − x3

= −6

−3 − x3

= −3(−6) Multiply both sides by − 3

x =18The solution is 18.

45. 9 = 2x − 615 = 2x Add 6 to both sides152

= x Divide both sides by 2

The solution is 152

.

47. 6 + 3x − 4 − 7x − 2 = 6− 5x − 10−4x = −5x − 4 Combine like terms

x = −4 Add 5 x to both sidesThe solution is –4.

49.43

x + 149

= x9

+ 13

The LCD is 9

9 ⋅ 43

x + 9 ⋅149

= 9 ⋅ x9

+ 9 ⋅ 13

Clear denominators

12x +14 = x + 311x +14 =3 Subtract x from both sides

11x =−11 Subtract 14 from both sidesx =−1 Divide both sides by 11


51. 2(x −3) − 3(x +1)= 02x − 6− 3x −3 = 0

−x − 9 = 0−9 = x

Remove grouping symbolsCombine like termsAdd x to both sides


53. 3x + 7 − x = 4 + 2x + 32x + 7 = 2x + 7 Combine like terms

7 = 7 Subtract 2x from both sidesThe solution set is ℜ.



55. −3(x − 4) = 4x − (7x −13)−3x + 12 = 4x − 7x + 13−3x + 12 = −3x + 13

12 = 13

Remove grouping symbolsCombine like termsAdd 3 x to both sides


57. (iii) is a correct translation of x ≤ 5.

59. The point of intersection represents a solution for ≤ or ≥ inequalities.

61. (a) x ≤ 2

(b) –1 < x ≤ 3

63. (a) 8 + x > x – 6

(b) 8 + x < x – 6

65. Inequality (iii) is equivalent to x < 7.

67. −8 ≥ 3 − x−11 ≥ −x Subtract 3 from both sides−11−1

≤ −x−1

Divide by −1 and reverse the inequality

11 ≤ xx ≥ 11

69. − 34

x ≤ 15

− 43

⋅ − 34

x

≥ − 43

⋅15 Multiply by − 43

and reverse the inequality

x ≥ −20

71. −3(x + 1) < −(x −1)−3x − 3 < −x +1

−3x < −x + 4−2 x < 4−2x−2

> 4−2

x > −2

Remove grouping symbolsAdd 3 to both sidesAdd x to both sidesDivide by − 2 and reverse the inequality

73. −12 < −4x ≤16−12

−4> −4x

−4≥ 16

−4 Divide by − 4 and reverse all inequalities

3> x ≥−4−4 ≤ x < 3

75. 3(x −1)− (x +1) ≤ 2(x + 4)3x − 3 − x −1≤ 2 x + 8

2x − 4 ≤ 2 x + 8−4 ≤ 8

Remove grouping symbolsCombine like termsSubtract 2 x from both sides


Looking Ahead

1. 9 + 2(–6)



3.34

(−28)

5. 2 – 6

7.95

C + 32 = 95

(30) + 32

= 54 + 32= 86

9. Answers will vary.

11. x − 0.32 x = 5781x − 0.32 x = 578(1− 0.32)x = 578

0.68x = 5780.68x0.68

= 5780.68

x = 850

Chapter 2 Test

1.2xy

− (x + y) = 2 ⋅ 2−4

− [2 + (−4)] Replace x with 2 and y with − 4

= 4−4

− (−2)

= −1 +2= 1

3. −(x − 1)+ 4(2 x + 3) =− x +1 +8x +12= 7x +13

Remove grouping symbolsCombine like terms

5. Quadrant II

7. The y-coordinate of a point of the graph of an expression corresponds to the value of theexpression. If x is replaced with –4, the value of the expression 3x – 2 is –14. Thus if thex-coordinate is –4, the y-coordinate is –14.

9.

x = 10

11. (a) The equation is an identity.

(b) The equation is a contradiction.

13. 14x − 20 = 1+ 13xx − 20 = 1 Subtract 13x from both sides

x = 21 Add 20 to both sidesThe solution is 21.



15.23

(x − 5) = x + 2 The LCD is 3

3 ⋅ 23

(x − 5) = 3(x + 2) Clear denominator

2(x − 5) = 3(x + 2)2x − 10 = 3x + 6 Remove grouping symbols

−10 = x + 6 Subtract 2x from both sides−16 = x Subtract 6 from both sides


17. 5− 5(1− x) = 3 + 5x5 − 5+ 5x = 3+ 5x Remove grouping symbols

5x = 3 + 5x Combine like terms0 = 3 Subtract 5x from both sides


19. x – 3 ≥ –2

21. (a) The solution set is Ø.

(b) The solution set is ℜ.

The graph of 9 – x is always above the graph of –11 – x. Thus the solution set for part (a) is Øand for part (b) is ℜ.

23. x + 14

≥ − 4x5

−1.55 The LCD is 20

20 ⋅x + 20 ⋅ 14

≥ 20 − 4x5

− 20(1.55) Clear denominators

20x + 5 ≥ −16x − 3136x + 5 ≥−31 Add 16x to both sides

36x ≥ −36 Subtract 5 from both sides36x36

≥ −3636


x ≥ −1

25. −4 ≤ 2(x − 7) < 0−42

≤ 2(x − 7)2

< 02

Divide all three parts by 2

−2 ≤ x − 7 < 0−2 + 7 ≤ x − 7 + 7 < 0 + 7 Add 7 to all three parts

5 ≤ x < 7

Chapter 2 Algebra Basics, Equations, and Inequalitiescollege.cengage.com/mathematics/hubbard/elementary_algebra/2e/students/... · SSM: Elementary Algebra Chapter 2: Algebra Basics,

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