complete each statement with. 6.–3 –57.4.29 4.88.(–3)(–4) 12 9.–1 – 2 6 – 910.11. + +...

Complete each statement with <, =, or >.

6. –3 –5 7. 4.29 4.8 8. (–3)(–4) 12

9. –1 – 2 6 – 9 10. 11. + +

Graph the numbers on the same number line.

1. 4 2. –3 3. 4. 0 5. 1.5

ALGEBRA 1 LESSON 3-1ALGEBRA 1 LESSON 3-1

93

34– 4

5–13

13

12

12

(For help, go to Lesson 1-3.)

Inequalities and Their GraphsInequalities and Their Graphs

3-1


Solutions

1.–5.

6. –3 > –5 7. 4.29 < 4.8

8. (–3)(–4) 12

12 = 12

9. –1 – 2 6 – 9

–3 = –3

34– 4

5–10.

–0.75 > –0.8

13

13

12

1211. + +

< 123


3-1

Is each number a solution of x 5?


>

Yes, 5 5 is true. >No, –2 5 is not true. > Yes, 10 5 is true.>

a. –2 b. 10 c. 255


3-1


Is each number a solution of 3 + 2x < 8?

a. –2 b. 3

3 + 2x < 8

–2 is a solution.

3 + 2(–2) < 8 Substitute for x.

3 – 4 < 8 Simplify.

–1 < 8 Compare.

3 + 2x < 8

3 is not a solution.

3 + 2(3) < 8 Substitute for x.

3 + 6 < 8 Simplify.

9 < 8 Compare.


3-1

a. Graph d < 3.


b. Graph –3 ≥ g.

The solutions of d < 3 are all the points to the left of 3.

The solutions of –3 g are –3 and all the points to the left of –3.

>


3-1

Write an inequality for each graph.


x < 2 Numbers less than 2 are graphed.

x > –2 Numbers greater than –2 are graphed.

x –3 Numbers less than or equal to –3 are graphed.<


3-1

>x Numbers greater than or equal to are graphed.

12

12


Define a variable and write an inequality for each situation.

a. A speed that violates the law when the speed limit is 55 miles per hour.

b. A job that pays at least $500 a month.

Let v = an illegal speed.

The speed limit is 55, so v > 55.

Let p = pay per month.

The job pays $500 or more,

so p 500.>


3-1



pages 136–139 Exercises

1. yes

2. no

3. yes

4. yes

5. no

6. yes

7. no

8. yes

9. a. no b. no c. yes

10. a. yes b. no c. yes

11. a. no b. no c. no

12. a. yes b. no c. no

13. a. no b. yes c. no

14. a. no b. no c. yes

15. C

16. B

17. D

18. A

19.

20.

21.

22.

3-1

24.

25.

26.

27–37. Choice of variable may vary.

27. x > –3

28. x 7

29. x 1

30. x < –6

31. x 4.5

32. x < –0.5

23.

>–

<–

>–



33. Let s = number of students. s 48

34. Let a = age. a 16

35. Let w = number of watts. w 60

36. Let s = number of students. s 350

37. Let a = number of appearances. a > 75

38. n is less than 5.

39. b is greater than 0.

40. 7 is greater than or equal to x, or x is less than or equal to 7.

41. z is greater than or equal to –5.6.

42. q is less than 4, or 4 is greater than q.

43. –1 is greater than or equal to m, or m is less than or equal to –1.

44. 35 is greater than or equal to w, or w is less than or equal to 35.

45. g minus 2 is less than 7.

46. a is less than or equal to 3.

47. 6 plus r is greater than –2.

48. 8 is less than or equal to h, or h is greater than or equal to 8.

49. 1.2 is greater than k, or k is less than 1.2.

50. Use an open dot for < or >. Use a closed dot for or .

51. Answers may vary. Sample: For x = –1, 3(–1) + 1 = –3 + 1 = –2. –2 > 0. /

3-1

<–

>–

<–

>–

>–<–



52. Put an open dot at 3 and color the rest of the number line.

53. Answers may vary. Sample: Every class has at least 18 students.

54. x > 2

55. b –5

56. r 0

57. a < 5

58.

59.

60.

61.

62.

63.

64. “At least” is translated as .“At most” is translated as .

65. x 2451

66. Option A < Option B

67. “4 greater than x” means x + 4; “4 > x” means 4 is greater than x.

68. C; the inequality is true for x = 3 but not true for x = 5, so C is correct.

69. Answers may vary. Sample: a = –1, b = 12

3-1

>–

<–

>–<–

>–

70. a is negative, and a and b are opposites.

71. 5a + 4s 4000

72.

73.

74. C

75. H

76. B

77. H

78. B

79. [2]

Since 6 is half, m > 6 .

[1] incorrect inequality OR no diagram



80. y = – x + 2

81. y = 5x + 4

82. y = x – 3

83. y = – x – 5

84. 6.27

85. 5

86. 9

87. I =

88. = or = – w

89. b = P – a – c

90. Associative Property of Multiplication

91. Commutative Property of Multiplication

92. Commutative Property of Addition

23

45

12

VRP – 2w

2P2

3-1

>–

310

310


1. Is each number a solution of x > –1?

a. 3 b. –5

2. Is 2 a solution of 3x – 4 < 2?

3. Graph x > 4.

4. Write an inequality for the graph.

5. Graph each inequality.

a. t is at most 2. b. w is at least 1.

yes no

no

p –2<


3-1


(For help, go to Lessons 1-4, 1-5, and 2-1.)


1. –3 + 4 –5 + 4

2. –3 – 6 4 + 6

3. –3.4 + 2 –3.45 + 2

Solve each equation.

4. x – 4 = 5 5. n – 3 = –5

6. t + 4 = –5 7. k + = 23

56

Solving Inequalities Using Addition and SubtractionSolving Inequalities Using Addition and Subtraction

3-2


1. –3 + 4 –5 + 4

1 > –1

2. –3 – 6 4 – 6

–9 < –2

3. –3.4 + 2 –3.45 + 2

–1.4 > –1.45

4. x – 4 = 5x = 9

5. n – 3 = –5n = –2

6. t + 4 = –5t = –9

Solutions


3-2

7. k + =

k = – = – =

23

56

23

56

56

46

16


Solve p – 4 < 1. Graph the solutions.

p – 4 + 4 < 1 + 4 Add 4 to each side.

p < 5 Simplify.


3-2


Solve 8 d – 2. Graph and check your solution.>

8 + 2 d – 2 + 2 Add 2 to each side.>

10 d, or d 10 Simplify.> <

Check: 8 = d – 2 Check the computation.8 10 – 2 Substitute 10 for d.8 = 88 d – 2 Check the direction of the inequality.8 9 – 2 Substitute 9 for d.8 7

>>>


3-2


Solve c + 4 > 7. Graph the solutions.

c + 4 – 4 > 7 – 4 Subtract 4 from each side.

c > 3 Simplify.


3-2


In order to receive a B in your literature class, you must earn more than 350 points of reading credits. Last week you earned 120 points. This week you earned 90 points. How many more points must you earn to receive a B?


3-2

points pointsearned required

points needed

Relate: plusis more

than

Define: Let = the number of points needed.p

Write: 120 + 90+ 350p >


(continued)

You must earn 141 more points.

210 + p > 350 Combine like terms.

120 + 90 + p > 350

210 + p – 210 > 350 – 210 Subtract 210 from each side.

p > 140 Simplify.


3-2

12

12. d > 3

13. y –4

14. q < 3

15. x 2.5

16. r < 4.5

17. m < –1.6

18. b <

19. n > 3

20. 2

21.

22. 3.1

23. w 5;

12




1. 5

2. 8

3. 4.3

4. x > 11

5. t < 1

6. b < –4

7. d 10

8. s –4

9. r 9

10. n > 10

11. w –2

3-2

13

>–

<–

<–

53

>–

<–

>–

<–



36. h – ;

37. d > –1.5;

38. t < – ;

39. s + 637 2000, $1363

40. s + 6.50 + 5 15, $3.50

41. r + 17 + 12 50, 21 reflectors

42. Add 4 to each side.

43. Subtract 9 from each side.

44. Add to each side.

45. w 11

46. c 3

47. y < 3.1

24. m > –8;

25. b > –7;

26. a –6;

27. r –6;

28. k 1;

29. x < –1;

30. p > –6;

31. z –1;

32. y < 5.5;

33. m > –1 ;

34. a –0.3;

35. p > –7;

12

12

48. n < –5

49. z < –9.7

50. y < –3.2

51. t >

52. v 16

53. k –8.1

54. b < 6

55. m –3.5

56. k 4

57. h –

58. x –5.7

59. w < –14

16

12

45

34

3-2

12

12

>–

<–

<–

<–

>–

<–

>–

<–

>–

<–

<–

>–

>–

<–

<–

>–

<–



60. w –1

61. x < 16

62. m > 5.5

63. t 12.9

64. k < 2

65. n < 6

66. m 5

67. s > 10.3

68. a. yesb. noc. Answers may vary. Sample:

The = sign indicates that each side is equal, so the two sides may be interchanged.

16

35

The < sign does not indicate equality. One side cannot be both greater than and less than the other side.

69. a. Let n = points scored on floor exercises. 8.8 + 7.9 + 8.2 + n 34.0; n 9.1

b. Your sister must score 9.1 points or more to qualify for regional gymnastics competition.

c. Answers may vary. Sample: 9.1, 9.2, 9.3

70. nearly 51.2 MB

71. at least $15.50

72. 40 or more points

73. a-b. Check students’ work.

3-2

>–

<–

<–

>– >–



74. a. No; the solution is z 13.8, so z 14 is not correct.

b. Answers may vary. Sample: Substituting values does not work because there is always the possibility that the solution lies between values that make the inequality true and a value that does not.

75. x 1

76. n < 5

77. t –3

78. k > 10

79. r < –2

80. r –13

81. a –25

82. 5 m

83. d > –8

84. y < 6

85. a 24

86. a < –8

87. a. a > c – bb. b > a – cc. c > b – ad. The length of the third side

must be greater than the difference of the lengths of the other two sides.

88. not true; sample counter example: for a = 5 and b = –6, 5 –(–6) < 5 +(–6)

89. true

90. true

91. not true; sample counter example: for a = 0, b = 1, and c = –2, 0 < 1 and 0 < 1 + (–2)

92. Answers may vary. Sample: For x = 2, y = 1,z = 4, and w = 3, 2 > 1 and 4 > 3, but 2 – 4 > 1 – 3.

/

/

/

3-2

>–

<–

>–

>–

>–

>–<–

<–



93. B

94. G

95. A

96. I

97. D

98. [2] at least 33 points;24(19.5) + x > 25(20)

468 + x > 500 x > 32

(OR equivalent explanation)[1] incorrect answer OR no work or explanation

99. Let c = length of octopus in feet. c 10

110. 22

111. 64

112. 98

113. 7

114. 0.56

115. 31

116. 48

117. 73

118. 58

119. 82

120. 26

100. Let h = distance in miles a hummingbird migrates. h > 1850

101. Let a = average. a 90

102. Let p = number of pages to read. p 25

103. 13

104. –1

105. –12

106. 4

107. 13

108. –18

109. –28

3-2

>–

<–

>–


Solve each inequality. Graph the solutions.

1. p – 7 –5 2. w – 3 < –9

3. x + 6 > 4 4. 13 9 + h

>

>

p 2> w < –6

x > –2 <4 h, or h 4>


3-2

(For help, go to Lessons 2-1 and 3-1.)


Solve each equation.

1. 8 = t 2. 14 = –21x 3. = –1

4. 5d = 32 5. x = –12 6. 0.5n = 9

Write an inequality for each graph.

7. 8.

12

x6

23

Solving Inequalities Using Multiplication and DivisionSolving Inequalities Using Multiplication and Division

3-3


Solutions

1. 8 = t

t = 8

t = 8(2) = 16

12

12

2. 14 = –21x

–21x = 14

x = – = –1421

23

3. = –1

x = –1(6) = –6

x6 4. 5d = 32

d = = 6.4325

5. x = –12

• x = • –12

x = –18

32

23

23

32

6. 0.5n = 9

=

n = 18

0.5n0.5

90.5

7. x –1 8. x > 3<


3-3


Solve > –2. Graph and check the solutions.z3

z > –6 Simplify each side.

3 > 3(–2) Multiply each side by 3. Do not reverse the inequality symbol.

z3( )

z3 > –2 Check the direction of the inequality.

Check: = –2 Check the computation.z3

– = –2 Substitute –6 for z.63

–2 = –2 Simplify.

– > –2 Substitute –3 for z.33

–1 > –2 Simplify.


3-3


Solve 3 – x. Graph and check the solutions.< 35

( )53–( )5

3– (3) > ( )35

xMultiply each side by the reciprocal of – , which

is – , and reverse the inequality symbol.

35

53

–5 x, or x –5 Simplify.<>


3-3


(continued)

Check: 3 = – x Check the computation.

3 = – (–5) Substitute –5 for x.

3 = 3

3 – x Check the direction of the inequality.

3 – (–10) Substitute –10 for x.

3 6

35

35

<

<

<

35

35


3-3


Solve –4c < 24. Graph the solutions.

c > –6 Simplify.

Divide each side by –4. Reverse the inequality symbol.–4c–4 >

24–4


3-3


Your family budgets $160 to spend on fuel for a trip. How many times can they fill the car’s gas tank if it cost $25 each time?

Your family can fill the car’s tank at most 6 times.

cost per total fueltank budget

Relate: timesnumberof tanks

is atmost

Define: Let = the number of tanks of gas.t

Write: 25 • 160t <

25t 160<


3-3

Divide each side by 25.<25t25

16025

t 6.4 Simplify.<




1. t –4;

2. s < 6;

3. w –2;

4. p < –8;

5. y > –4;

6. v –1.5;

7. x < 6;

8. k –2;

9. x < 0;

10. y 0;

11. x < 7;

12. d –4;

13. c > – ;

14. b –1 ;

15. u < – ;

16. n > ;

17. t < –3;

18. m ≥ 2;

19. w –5;

20. c > 4;

21. z –9;

22. b < –6;

23. d < – ;

23

12

12

34

23

3-3

<–

<–

>–

<–

<–

>–

>–

>–

>–



34. , 0, –1, –2

35. –6, –7, –8, –9

36. –2, –3, –4, –5

37. 5, 4, 3, 2

38. –10.4, –11, –12, –13

39. Multiply each side by –4 and reverse the inequality symbol.

40. Multiply each side by 5.

41. Divide each side by 5.

42. Multiply each side by .

43. Divide each side by 4

12

44. Multiply each side by –1 and reverse the inequality symbol.

45. –2

46. –5

47. 4

48. 18

49. –10

50. 7.5

51. x and y are equal.

3-3

43

24. x –5 ;

25. q > –3 ;

26. h < 5;

27. d > 4;

28. m –4.5;

29. 4.5c 300, 67 cars

30. 6.15h 100, 17 hours

31–38. Answers may vary. Samples are given.

31. –2, –3, –4, –5

32. –12, –10, –8, 0

33. –3, –4, –5, –6

1312

>–

<–

>–

>–

59. d –7

60. u > 35

61. s < –

62. k –30

63. y < 9

64. t –2.35

65. h –4

66. x > 2

67. x < –9

68. b < 0

69. p > –2

70. m –47

71. g < –

72. n 2

73. m < 3.5

74. z –2

75. Yes; in each case, y is greater than 6.

76. < r; r > 0.15

77. a. She should have divided each side by –15.

b. 150 satisfies the original inequality –15q 135.



14

3-3

12

2314

3 20

12

52–55. Estimates may vary.

52. r > –2

53. j > –6

54. p 42

55. s 28

56. 104 concrete blocks

57. For both the equation and the inequality, you multiply each side by –3. For the inequality, you must reverse the inequality symbol.

58. Answers may vary. Sample:

2x > 6, x > 4, –x < –3, – < –43

x5

35

>–

<–

<–

>–

<–

<–

<–

>–

>–

<–



77. (continued)c. Answers may vary.

Sample: –15

78. a > 0

79. a > 0

80. a 0

81. a < 0

82. d > 15, 5-in. box or 6-in. box

83. x 18(15), 480 tiles

84. 40

85. 50,400

86. 54

87. 14

88. 4

89. 13339

90. x –7

91. w < –

92. t 9.2

93. d > 4

94. –3.9 m

95. y < 8

96. 1 > a

97. k 3

98. t = 2

99. n = –2

100. x = 1

101. Prop. of Opposites

102. Prop. of Reciprocals

103. Ident. Prop. of Add.

104. Ident. Prop. of Mult.

105. Assoc. Prop. of Mult.

106. Comm. Prop. of Mult.

3-3

18/=

916

>–

>–

<–

>–

<–


Solve each inequality. Graph the solution.

1. –3

2. – < –1

3. 6x < 30

4. 48 –12h

y2 >

p3

>

y –6>

p > 3

x < 5

><–4 h, or h –4


3-3



Solve each equation, if possible. If the equation is an identity or if it has no

solution, write identity or no solution.

1. 3(c + 4) = 6 2. 3t + 6 = 3(t – 2)

3. 5p + 9 = 2p – 1 4. 7n + 4 – 5n = 2(n + 2)

5. k – + k = 6. 2t – 32 = 5t + 1

Find the missing dimension of each rectangle.

7. 8.

12

23

76

Solving Multi-Step InequalitiesSolving Multi-Step Inequalities

3-4


1. 3(c + 4) = 6 2. 3t + 6 = 3(t – 2)

c + 4 = 2 3t + 6 = 3t – 6

c = –2 6 = –6

no solution

3. 5p + 9 = 2p – 1 4. 7n + 4 – 5n = 2(n + 2)

3p = –10 2n + 4 = 2n + 4

p = –3 identity13


Solutions

3-4


5. k – + k = 6. 2t – 32 = 5t + 1

k = –3t = 33

k = = 1 t = –11

7. P = 2( + w) 8. P = 2( + w)

110 = 2( + 15) 78 = 2(26 + w)

55 = + 15 39 = 26 + w

40 = 13 = w

length = 40 cm width = 13 in.

12

23

76

32

116

116

29

Solutions


3-4


Solve 5 + 4b < 21.

5 + 4b – 5 < 21 – 5 Subtract 5 from each side.

4b < 16 Simplify.

b < 4 Simplify.

< Divide each side by 4.4b4

164

Check: 5 + 4b = 21 Check the computation.

5 + 4b < 21 Check the direction of the inequality.

5 + 4(3) < 21 Substitute 3 for b.

5 + 4(4) 21 Substitute 4 for b.

21 = 21

17 < 21


3-4


The band is making a rectangular banner that is 20 feet long

with trim around the edges. What are the possible widths the banner

can be if there is no more than 48 feet of trim?

twice the the lengthlength of trim

Relate: plustwice the

widthcan be nomore than

Write: 2(20) + 2w 48<


3-4


(continued)

The banner’s width must be 4 feet or less.

2(20) + 2w 48<

40 + 2w 48 Simplify 2(20).<

40 + 2w – 40 48 – 40 Subtract 40 from each side.<

2w 8 Simplify.<

w 4 Simplify.<

Divide each side by 2.<2w2

82


3-4


Solve 3x + 4(6 – x) < 2.

3x + 24 – 4x < 2 Use the Distributive Property.

–x + 24 < 2 Combine like terms.

–x + 24 – 24 < 2 – 24 Subtract 24 from each side.

–x < –22 Simplify.

x > 22 Simplify.

> Divide each side by –1. Reverse the inequality symbol.

–x–1

–22–1


3-4

Solve 8z – 6 < 3z + 12.


8z – 6 – 3z < 3z + 12 – 3z Subtract 3z from each side.

5z – 6 < 12 Combine like terms.

5z – 6 + 6 < 12 + 6 Add 6 to each side.

5z < 18 Simplify.

< Divide each side by 5.5z5

185

z < 3 Simplify.35


3-4


Solve 5(–3 + d) 3(3d – 2).<

–15 – 4d + 15 –6 + 15 Add 15 to each side. <

–4d 9 Simplify.<

–15 + 5d – 9d 9d – 6 – 9d Subtract 9d from each side. <

–15 – 4d –6 Combine like terms.<

–15 + 5d 9d – 6 Use the Distributive Property.<

Divide each side by –4. Reverse the inequality symbol.

–4d–4

9 –4>

d –2 Simplify.>14


3-4



10. 8t 420 and t 52.5, so the average rate of speed must be at least 52.5 mi/h.

11. 27 2s + 8 and s 9.5, so the two equal sides must be no longer than 9.5 cm.

12. –6x + 9

13. j 1

14. b <

15. h > 5

16. x 3

17. y > –8

18. w 4

19. c < –7

20. r 2

13

12

21. n 9

22. w < 3

23. t –1

24. d 6

25. n 2

26. k –4

27. s < 10

28. p > 3

29. x > 2

30. m > –

31. d < –1

32. y –1

14

3-4


1. d 4

2. m > –3

3. x > –2

4. n 3

5. q 4

6. h –2

7. a 3

8. b > 6

9. c < 2

12

1c

12

>–

<–

>–

<–

<–

>– >–

>– <–

>–

>–

<–

<–

<–

<–

<–

>–

>–

<–



3-4

33. k

34. v –4

35. q –2

36. r 1

37. x < 1

38. m > –8

39. v 2



42. Subtract y from each side and add 5 to each side.

23

4513

43. Add 2 to each side, then multiply each side by –5, and reverse the inequality sign.

44. Subtract 3j from each side and subtract 5 from each side.

45. Answers may vary. Sample: Multiply q – 3 by 2, add 3q to each side, add 6 to each side, and divide each side by 5.

46. a. –3t –9, t 3b. 9 3t, t 3c. The results are the same.

47. 6 – (r + 3) < 15, r > –12

48. (t – 6) 4, t 1412

>–

<– <–

<–>–<–

>–

<–

>–

>–

>–



49. 3(z + 2) > 12, z > 2

50. Answers may vary. Sample: To solve 2.5(p – 4) > 3(p + 2), first use the Distributive Property to simplify each side. The result is 2.5p – 10 > 3p + 6. Then use the Subtraction Property of Inequality. Subtract 6 from each side and 2.5p from each side. The result is –16 > 0.5p. Then use the Division Property of Inequality. Divide each side by 0.5. The result is –32 > p. So the solution is –32 > p or p < –32.

51. a. i. always trueii. always trueiii. never true

b. If the coefficients of the only variable on each side of an inequality are the same, then the inequality will either be always true or never true.

52. For x = the number of guests, (0.75)200 + 1.25x 250; x 80; at least 80 guests must attend.

53. a. maximumb. no more than 135

54. B

55. E

56. F

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>–>–



57. A

58. D

59. C

60. Answers may vary. Samples:

– x – 5 > 0, x < –15;

– x – 5 10, x –45

61. r < 5

62. m –2

63. s 4.4

64. n –2

65. s < 8

66. k 1.4

1313

58

12

77. For x = number of hours of work each month, 15x – (490 + 45 + 65) 600, so to make a profit he must work at least 80 h per month.

78. For x = amount of sales, 250 + 0.03x 460, so to reach her goal, she must have at least $7000 in sales.

79. Add 2x to each side rather than subtract 2x, so x .

80. Distribute 4 to 2 as well as n, so n > –7.

81. a. x >

b. x <

25

67. n > –2

68. r 1

69. x > 1

70. a

71. m 1

72. k 2

73. n –23

74. x 0

75. a < –6

76. c 4

34

c – ba

c – ba

3-4

12

16

23

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82. x is an integer between –3 and 7, inclusive.

83. 8.8 in. 14.2 in.

84. –2

85. 10 h

86. a. 74 boxesb. 5 trips

87. C

88. F

89. A

90. F

91. [2] Maxwell should continue ordering from the current

supplier. Let x = number of bottles Maxwell orders per month. The cost from the current supplier is 3x + 25. The cost from the competitor is 4x + 5. The inequality 4x + 5 3x + 25 represents the number of bottles of shampoo for which the competitor will be less expensive. Since x = 20, the competitor will not be less expensive for orders of 30 bottles or more. (OR equivalent explanation)

[1] incorrect answer OR insufficient explanation

92. m –4

93. y –8

94. x > –12

95. t –3

96. b < 27

97. w > –98

98. p > 1

99. d 7

100. 6 h

101. –16

102. 16

103. 24

104. –16

3-4

13

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<–

>–

<–

>–


Solve each inequality.

1. 8 + 5a 23 2. – p < p – 6

3. 3(x – 4) > 4x + 7 4. 3(3c + 2) 2(3c – 2)

>

<

13

12

a 3> p > 7 15

x < –19 <c –313


3-4



Graph each pair of inequalities on one number line.

1. c < 8; c 10 2. t –2; t –5 3. m 7; m > 12

Use the given value of the variable to evaluate each expression.

4. 3n – 6; 4 5. 7 – 2b; 5

6. ; 17 7. ; 9

> <> <

12 + 13 + y3

2d – 35

Compound InequalitiesCompound Inequalities

3-5


1. c < 8; c 10

2. t –2; t –5

3. m 7; m > 12

4. 3n – 6 for n = 4: 3(4) – 6 = 12 – 6 = 6

5. 7 – 2b for b = 5: 7 – 2(5) = 7 – 10 = –3

6. for y = 17: = = 14

7. for d = 9: = = = 3

>

> <

<

12 + 13 + y3

2d – 35

12 + 13 + 173

423

2(9) – 35

18 – 35

155


Solutions

3-5


Write two compound inequalities that represent each situation.

Graph the solutions.

a. all real numbers that are at least –1 and at most 3.

b –1 and b 3 –1 b 3 > < < <

b. all real numbers that are at less than 31, but greater than 25.

31 > n and n > 25 31 > n > 25


3-5

Solve 5 > 5 – f > 2. Graph the solutions.


5 – f – 5 > 2 – 5

–f > –3

–3 –1

–f–1 <

5 – 5 > 5 – f – 5

0 > –f

0 –1

–f–1<

5 – f > 25 > 5 – f and

f < 30 < f and

0 < f < 3

Write the compound inequality as two inequalities joined by and .


3-5


Your test grades in science so far are 83 and 87. What

possible grades g can you make on your next test to have an average

between 85 and 90?

Relate: the averagewhich is less

than or equal tois less thanor equal to

85 90

Write: 85 90< 83 + 87 + g 3

g<


3-5


(continued)

85 < 83 + 87 + g 3

< 90

Multiply by 3.3(85) < < 3(90)83 + 87 + g 33( )

< <255 170 + g 270 Simplify.

< <255 – 170 170 + g – 170 270 – 170 Subtract 170.

< <85 g 100 Simplify.

The third test grade must be between 85 and 100, inclusive.


3-5


Write an inequality that represents each situation.Graph the solutions.

a. all real numbers that are less than 0 or greater than 3.

n < 0 or n > 3

b. Discounted tickets are available to children under 7 years old or to adults 65 and older.

a < 7 or a 65; because age cannot be negative, a > 0.

>


3-5


Solve the compound inequality 3x + 2 < –7 or –4x + 5 < 1.

Graph the solution.

3x + 2 – 2 < –7 – 2

3x < –9

<3x3

–93

–4x + 5 – 5 < 1 – 5

–4x < –4

>–4x–4

–4–4

3x + 2 < –7 –4x + 5 < 1or

x < –3 x > 1or


3-5




1. –4 < x and x < 6 or –4 < x < 6;

2. 2 n and n 9 or 2 n 9;

3. 23 < c < 23.5;

4. 40 w 74;

5. –5 < j < 5;

6. 1 w 5;

7. 2 < n 6;

8. –4 < p –2;

9. 1 < x < 3;

10. –1.5 < w 3.5;

11. 5 < x < 8;

12. –2 < m 1;

13. 3 s < 4;

14. 1 < t < 3 ;

15. 1 r 2;

16. –2 < r 1.5;

17. 5 a 17;

18. 1 < x < 7;

19. –1 x 11;

20. n –3 or n 5,

21. x < 3 or x > 7;

22. h < 1 or h > 3;

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12

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23. b < 100 or b > 300;

24. b < –2 or b > 2;

25. k < –5 or k > –1;

26. c < 2 or c 3;

27. a 4 or a > 5;

28. c 2 or c 3;

29. y –2 or y 5;

30. d –3 or d 0;

31. z < –1 or z > 2;

32. x < –3 or x 5;

33. n < –5 or n > 3;

34. –2 < x < 3

35. x < –3 or x 2

36. x ≤ 0 or x > 2

37. –4 x 3

38. q ≤ –2 or q > 4

39. h < –7 or h > 4

40. 4 ≤ t ≤14

41. r < 16 or r > 25

42. –6 ≤ t < 1

43. x < –6 or x > 9

44. 20 ≤ d ≤ 32

45. all real numbers except 5

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23

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46. The word “and” means both statements must be true. The word “or” means that at least one of the statements must be true.

47. 2.5 < x < 7.5

48. 6 < x < 30

49. 7 < x < 49

50. 11 < x < 21

51. 66 C 88

52. 15 D 30

53. Charlotte: 29 C 90Detroit: 15 D 83

54. Answers may vary. Sample: Elevation near a coastline varies between 2 m below and 8 m above sea level.

55.

56. –2 < x < 0 or 0 < x < 3

57. n = 0 or n 3

58. a. 130 R 157b. 20 years old

59. 16, 18, 20

60. 10, 12, 14

61. B

62. G

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<– <–<– <–

<– <–



63. [2] 35 10.4 + 0.0059g 5024.6 0.0059g 39.64169 < g < 6712minimum consumption: 4169 galmaximum consumption: 6712 gal(OR equivalent explanation)

[1] incorrect answer OR insufficient explanation

64. b >

65. n 3

66. r 7

67. x = –1

68. w = 2

69. identity

13

3-5

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<–

<–


1. Write two compound inequalities that represent the given situation. Graph the solution.all real numbers that are at least 2 and at most 5

2. Write an inequality that represents the given situation. Graph the solution.all real numbers that are less than –3 or greater than –1

3. Solve –2 2x – 4 < 6. Graph the solution.

4. Solve 3x – 2 < –8 or –2x + 5 3. Graph the solution.

<

<

b 2 and b 5, 2 b 5> < < <

n < –3 or n > –1

1 x < 5<

>x < –2 or x 1


3-5

11. |–4 | + 2 2 12. |–3 – 4 | 7



Simplify.

1. |15| 2. |–3| 3. |18 – 12|

4. –|–7| 5. |12 – (–12)| 6. |–10 + 8|


7. |3 – 7| 4 8. |–5| + 2 6

9. |7| – 1 8 10. |6 – 2 | 314

23

13

12

18

12

Absolute Value Equations and InequalitiesAbsolute Value Equations and Inequalities

3-6

58

58


1. |15| = 15 2. |–3| = 3

3. |18 – 12| = |6| = 6 4. –|–7| = –(7) = –7

5. |12 – (–12)| = |12 + 12| = |24| = 24

6. |–10 + 8| = |–2| = 2

7. |3 – 7| 4

|–4| 4

4 = 4

8. |–5| + 2 6

5 + 2 6

7 > 6


Solutions

3-6


Solutions

10. |6 – 2 | 3

|3 | 3

3 > 3

1434

58

58

34

58

9. |7| – 1 8

7 – 1 8

6 < 8

11. |–4 | + 2 2

4 + 2 2

7 > 2

23

13

12

23

13

12

12

12. |–3 – 4 | 7

|–7 | 7

7 = 7

18

12

58

58

58

58

58


3-6


Solve and check |a| – 3 = 5.

|a| – 3 + 3 = 5 + 3 Add 3 to each side.

|a| = 8 Simplify.

a = 8 or a = –8 Definition of absolute value.

Check: |a| – 3 = 5

|8| – 3 5 Substitute 8 and –8 for a. |–8| – 3 5

8 – 3 = 5 8 – 3 = 5


3-6


Solve |3c – 6| = 9.

The value of c is 5 or –1.


3-6

3c – 6 = 9 Write two equations. 3c – 6 = –9

3c – 6 + 6 = 9 + 6 Add 6 to each side. 3c – 6 + 6 = –9 + 6

3c = 15 3c = –3

Divide each side by 3.3c3 =

153

3c3 =

–33

c = 5 c = –1


Solve |y – 5| 2. Graph the solutions.<

3 y 7<<

Write a compound inequality.y – 5 –2> y – 5 2<and

Add 5 to each side.y – 5 + 5 –2 + 5> y – 5 + 5 2 + 5<

Simplify.y 3> y 7<and


3-6

The ideal diameter of a piston for one type of car is 88.000 mm. The actual diameter can vary from the ideal diameter by at most 0.007 mm. Find the range of acceptable diameters for the piston.


greatest difference betweenactual and ideal

Relate: 0.007 mmis at most

Define: Let d = actual diameter in millimeters of the piston.

Write: | d – 88.000| 0.007<


3-6

(continued)


The actual diameter must be between 87.993 mm and 88.007 mm, inclusive.

87.993 d 88.007 Simplify.<<

Add 88.000.–0.007 + 88.000 d – 88.000 + 88.000 0.007 + 88.000 <<

|d – 88.000| 0.007<

–0.007 d – 88.000 0.007 Write a compound inequality.<<


3-6




1. –2, 2

2. –4, 4

3. – ,

4. –6, 6

5. –3, 3

6. –7, 7

7. –5, 5

8. –2, 2

9. 0

10. no solution

11. –3, 3

12

12

12. –4, 4

13. 3, 13

14. –8, 4

15. –3, 1

16. –5, 1

17. –5, 9

18. no solution

19. –7, 1

20. –1, 1

21. –0.6, 0.6

22. a. less thanb. greater than

3-6

23. k < –2.5 or k > 2.5;

24. –2 < w < 2;

25. –8 < x < 2;

26. n –11 or n –5;

27. 1 y 3;

28. 1 p 7;

29. –2 < c < 7;

30. y –2 or y 5;

31. t < –3 or t > 2 ;

32. x < –3 or x > 2.5;

13

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<– <–

<– <–

<– >–



33. t –2.4 or t 4;

34. –2 < r < 8;

35. between 12.18 mm and 12.30 mm, inclusive

36. between 49 in. and 50 in., inclusive

37. –9, 9

38. –3, 3

39. –1 , 1

40. –1.8, 1.8

41. –5, 3

42. 0, 8

43. d –2 or d 2

44. n < –10 or n > 10

2532

7 32

12

12

45. c < –16 or c > 2

46. –12.6, 12.6

47. –6 < x < 6

48. –8, 8

49. –8, 8

50. –3, 3

51. –8 < m < 4

52. |n| < 3

53. |n| > 7.5

54. |n – 6| > 2

55. |n + 1| 3

56. |w – 16| 0.05, 15.95 w 16.05

3-6

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57. 39%, 45%

58. a. |w – 454| 5b. between 449 g and

459 g, inclusive

59. |g – 6.27| 0.02

60. The absolute value of a number cannot be less than zero.

61. Sample: |5x – 12| = 3; 1 , 3

62. |x – 4| = 2

63. |x – 2| = 4

64. |x – 3| = 6

65. |x – 12 | = 3

45

66. |x – 3| = 4

67. |x – 5 | = 2

68. |x + 9| = 6

69. |x – 6| = 4

70. a. between 7.075 oz and 7.105 oz, inclusive

b. No; the excess weight of some coins may be balanced by the lower weight of other coins.

71. a. 11°F, 39°Fb. |t – 25| = 14

72. 2

73. , 312

12

12

12

23

74. 3

75.

76.

77. =

78. =

79. |x + 1| > 3

80. |x – 2| 4

81. A

82. G

83. D

84. I

85. A

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86. [2] a.

b. The overlap of the three graphs is from 20 to 25.

[1] no graph OR insufficient explanation

87. –282 e 20,320

88. Let t = body temperature (°C), 36.0 t 37.2.

89. –3

90. 3

91. 5

92. –8

93.



13

94. 7

95. –2.5, –2, 0, 3,

96. –2, –1.5, – , 7,

97. 0.001, 0.009, 0.01, 0.011

98. – , –3, –2.5, 3, 2

43

152

3-6

<– <–

<– <–

Solve.1. |a| + 6 = 9 2. |2x + 3| = 7

3. |p + 6| 1 4. 3|x + 4| > 15<

a = 3 or a = –3 x = 2 or x = –5

–7 p –5< < x > 1 or x < –9



3-6

ALGEBRA 1 CHAPTER 3ALGEBRA 1 CHAPTER 3

Solving InequalitiesSolving Inequalities

1. a. no b. yes c. yes

2. a. yes b. no c. no

3. c 7

4. r 5

5. g 40

6. h > 32

7. n < –7

8. n 4.5

9. n –5

10. n > 1.5

11. z 2;

12. –4 y;

13. x > –6;

14. u < 4;

15. t 2;

16. w 1;

17. m > –4;

18. y < 2.5;

19. x 2 or x 8;

20. –3 < h < 2;

21. –6 < b –3;

22. –2.5 < q < 3;

23. n < –5 or n –1;

24. k 4 and k < 7.5;

3-A

>–

<–

>–

<–

<–

<–

>–

<–

>–

<– >–

<–

>–

>–

Solving InequalitiesSolving Inequalities

25. d > –

26. b < 6

27. m

28. x –0.8

29. Answers may vary. Sample: x –6 or x > 2

30. Answers may vary. Sample: –3 < x < 2

31. –2.25 or 3.25

32. –33 or 13

33. –6 or 5

34. 0 or 10

13

23

13

35. Answers may vary. Sample: Depending on the sign of a, the inequality symbol may need to be reversed.

a = 2: 2x – 1 < 32x < 4x < 2

a = –2: –2x – 1 < 3–2x < 4

x > –2

36. Answers may vary. Sample: |x| < 6

37. Let c = number of cansc + 702 + 470 + 492 + 547 3000c 789

38. 33 boxes

39. Let = lengthw = width| – 11.125| 0.005|w – 7.625| 0.00511.120 11.1307.620 w 7.630

ALGEBRA 1 CHAPTER 3ALGEBRA 1 CHAPTER 3

3-A

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<– <–<– <–

complete each statement with. 6.–3 –57.4.29 4.88.(–3)(–4) 12 9.–1 – 2 6 – 910.11. + +...

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