1-1 ALGEBRA 2 LESSON 1-1 Simplify. 1.–(–7.2)2.1 – (–3) 3.–9 + (–4.5) 4.(–3.4)(–2) (For help, go to Skills Handbook page 845.) Properties of Real Numbers

1-1

ALGEBRA 2 LESSON 1-1ALGEBRA 2 LESSON 1-1

Simplify.

1. –(–7.2) 2. 1 – (–3)

3. –9 + (–4.5) 4. (–3.4)(–2)

(For help, go to Skills Handbook page 845.)

Properties of Real NumbersProperties of Real Numbers

5. –15 ÷ 3 6. – + (– )25

35



Solutions

1. –(–7.2) = 7.2

3. –9 + (–4.5) = –13.5

5. –15 ÷ 3 = –5

6. – + (– ) = = – = –125

35

–2 + (–3)5

55

2. 1 – (–3) = 1 + 3 = 4

4. (–3.4)(–2) = 6.8

1-1

A pilot uses the formula d = a2 + 7290a to find the number

of miles d that can be seen when flying at an altitude of a miles.

Which set of numbers best describes the values for each variable?



The pilot’s instrument panel will display the altitude a as a rational number in the form of a terminating decimal.

The value of d obtained by replacing a with a rational number can be either irrational or rational (consider a = 1.47).

So, the values of d are best described as real numbers.

1-1

Graph the numbers – , 7 , and 3.6 on anumber line.



34

1-1

Use a calculator to find that 7 2.65.

– is between –1 and 0.34

Compare –9 and – 9. Use the symbols < and >.



Since –9 < –3, it follows that

9 = 3, so – 9 = –3.

–9 < – 9.

1-1

Find the opposite and the reciprocal of each number.



a. –317

Opposite: –4

1-1

Opposite: –(–3 ) = 317

17

Reciprocal: = = – 7 2222

7

1

–317

1

–Reciprocal: 1

4

b. 4

Which property is illustrated?



a. (–7)(2 • 5) = (–7)(5 • 2) b. 3 • (8 + 0) = 3 • 8

The given equation is true because 2 • 5 = 5 • 2.

So, the equation uses the Commutative Propertyof Multiplication.

The given equation is true because 8 + 0 = 8.

This is an instance of the Identity Property of Addition.

1-1

Simplify | 4 |, |–9.2|, and |3 – 8|.



13

–9.2 is 9.2 units from 0, so |–9.2| = 9.2.

|3 – 8| = |–5| and –5 is 5 units from 0. So, |–5| = 5, and hence |3 – 8| = 5.

1-1

4 is 4 units from 0, so | 4 | = 4 .13

13

13

13

Properties of Real NumbersProperties of Real NumbersALGEBRA 2 LESSON 1-1ALGEBRA 2 LESSON 1-1

Pages 8–10 Exercises

1. natural numbers, whole numbers, integers, rational numbers, real numbers

2. irrational numbers, real numbers


4. integers, rational numbers, real numbers

5. whole numbers, integers, rational numbers, real numbers

6. rational numbers, real numbers



9. whole numbers, rational numbers

10. natural numbers, rational numbers

11. real numbers

12.

13.

14.

15.

1-1


16.

17. >

18. =

19. >

20. <

21. <

22. =

23. >

24. <

25. – > – , – < –26. 0.075 < 0.39,

0.39 > 0.075

27. –2.3 < 2.1, 2.1 > –2.3

28. –5.2 < –4.8, –4.8 > –5.2

29. 3.04 < 3.4, 3.4 > 3.04

14

13

13

14

30. 0.4 < 0.4, 0.4 > 0.4

31. –4 < – 4,– 4 > – 4

32. 5 < 7, 7 > 5

33. – 3 > – 5– 5 < – 3

34. –200,

35. –3 ,

36. 0.01, –100

1200

35

518

1-1


37.

38. – 3,

39. –2 ,

40. 2.34, –

41. 3 – ,

42. Identity Prop. Of Mult.

43. Dist. Prop.

44. Comm. Prop. of Add.

45. Assoc. Prop. of Mult.

46. Comm. Prop. of Mult.

47. Identity Prop. of Add.

48. Inverse Prop. of Add.

49. Inverse Prop. of Mult.

50. Dist. Prop.


52. Assoc. Prop. of Add.

53. 10.3

54. 0.06

55. –25

56. 1.6

57.

58. 3

72

, – 27

1 3

3 3or

12

50117

1 – 3 1

3

1-1


59. 3

60. –2

61–68. Answers may vary. Samples are given.

61. –5

62. –3

63. –1

64.

65. 1

66. 3

67. 4

68. 4.8

69. natural numbers, whole numbers, integers, rational numbers, real numbers






75. >

76. >

77. <

78. >

12

14

12

23

13

1-1


79. <

80. <

81. <

82. <

83. Answers may vary. Sample: 4 is a whole

number, but is not

a whole number.

84. Answers may vary. Sample: 7 is a natural number, but –7 is not a natural number.

85. 0 is a whole number, and since –0 = 0, the opposite of 0 is a whole number.

86. Answers may vary. Sample: The integer –1 has –1 as its reciprocal, so –1 is an integer whose reciprocal is an integer.

87. Answers may vary. Sample: 2 and 2 are irrational numbers, but their product (2) is a rational number.

88. Check students’ work.

89. All except the Identity Prop. of Add. (since 0 is not in the set of natural numbers) and the inverse properties

14

1-1


90. all except the Inverse Prop. of Mult. and the Inverse Prop. of Add.

91. all except the Inverse Prop. of Mult.

92. all the properties

93. Comm., Assoc., and Distr. Prop.

94. No; explanations may vary. Sample: The only pairs of integers that have a product of –12 are –1 and 12, –2 and 6, –3 and 4, –4 and 3, –6 and 2, and –12 and 1. None of these pairs has a sum of –3.

95. D

96. H

97. C

98. I

99. A

1-1


102. 1.9

103. –3.8

104. 27

105. 0

106. –0.4

107. 7

108. , or 2

109. , or 11

110. ,or 1

100.[2] The opposite of the reciprocal of 5 is the

opposite of , or – , and the reciprocal of the

opposite of 5 is the reciprocal of –5, or – .

[1] includes a statement or example that a reciprocal is a multiplicative inverse or

that a number times its reciprocal is 1, OR a statement or example that an opposite is

an additive inverse or that a number plus its opposite is 0

101.[2] 1 and –1; for a number n to be equal to its

reciprocal, n = , which means n2 = 1. So

n = 1 or –1.

[1] only includes answer 1 and –1 with no explanation

15

15

15

1n

94

14

353

23

32

12

1-1


111. 5

112. 38

113. 45

1-1


1. Graph –2.3, 12, and on a number line.

2. Replace each with the symbol <, >, or = to make the sentence true.

a. 2.03 2.8

b. 1 1.2

3. Fit the opposite and the reciprocal of 1 .

4. Name the property of real numbers illustrated by the equation.

(–2)(3 + ) = (–2)( + 3)

5. Simplify |9 – (–5)|.

29

75

<

=

58

Comm. Prop. of Add.

14

1-1

–1 , 58

8 13

1-2


Use the order of operations to simplify each expression.

1. 8 • 3 – 2 • 4 2. 8 – 4 + 6 ÷ 3

3. 24 ÷ 12 • 4 ÷ 3 4. 3 • 82 + 12 ÷ 4

5. 27 + 18 ÷ 9 – 32 + 1 6. (40 + 24) ÷ 8 – (23 + 1)


Algebraic ExpressionsAlgebraic Expressions

Solutions



1. 8 • 3 – 2 • 4 = (8 • 3) – (2 • 4) = 24 – 8 = 16

2. 8 – 4 + 6 ÷ 3 = 8 – 4 + (6 ÷ 3) = 8 – 4 + 2 = (8 – 4) + 2 = 4 + 2 = 6

3. 24 ÷ 12 • 4 ÷ 3 = (24 ÷ 12) • 4 ÷ 3 = 2 • 4 ÷ 3 = (2 • 4) ÷ 3 = 8 ÷ 3 =

or 2

4. 3 • 82 + 12 ÷ 4 = 3 • 64 + 12 ÷ 4 + (3 • 64) + (12 ÷ 4) = 192 + 3 = 195

5. 27 + 18 ÷ 9 – 32 + 1 = 27 + 18 ÷ 9 – 9 + 1 = 27 + (18 ÷ 9) – 9 + 1 = 27 + 2 – 9 + 1 = (27 + 2) – 9 + 1 = 29 – 9 + 1 = (29 – 9) + 1 = 20 + 1 = 21

6. (40 + 24) ÷ 8 – (23 + 1) = (40 + 24) ÷ 8 – (8 + 1) = 64 ÷ 8 – 9 = (64 ÷ 8) – 9 = 8 – 9 = –1

83

23

1-2



Evaluate 7x – 3xy for x = –2 and y = 5.

7x – 3xy = 7(–2) – 3(–2) (5) Substitute –2 for x and 5 for y.

= –14 – (–30) Multiply first.

= –14 + 30 To subtract, add the opposite.

= 16 Add.

1-2


Evaluate (k – 18)2 – 4k for k = 6.


(k – 18)2 – 4k = (6 – 18)2 – 4(6) Substitute 6 for k.

= (–12)2 – 4(6) Subtract within parentheses.

= 144 – 4(6) Simplify the power.

= 144 – 24 Multiply.

= 120 Subtract.

1-2

Algebraic ExpressionsAlgebraic ExpressionsALGEBRA 2 LESSON 1-2ALGEBRA 2 LESSON 1-2

The expression –0.08y 2 + 3y models the percent increase of

Hispanic voters in a town from 1990 to 2000. In the expression, y

represents the number of years since 1990. Find the approximate

percent of increase of Hispanic voters by 1998.

Since 1998 – 1990 = 8, y = 8 represents the year 1998.

–0.08y2 + 3y = –0.08(8)2 + 3(8) Substitute 8 for y

19

The number of Hispanic voters had increased by about 19%.

1-2


Simplify by combining like terms.

2h – 3k + 7(2h – 3k)

2h – 3k + 7(2h – 3k) = 2h – 3k + 14h – 21k Distributive Property

= 2h + 14h – 3k – 21k Commutative Property

= (2 + 14)h – (3 + 21)k Distributive Property

= 16h – 24k

1-2


Find the perimeter of this figure. Simplify the answer.

= 2c + 4d

= c + c + 4d

1-2

P = c + + d + (d – c) + d + + c + dc2

c2

= c + + d + d – c + d + + c + d c2

c2

= + + c + 4dc2

c2

= + c + 4d2c2



1. –30

2. 26

3. 368

4. –16

5. –16

6. 28

7. –70

8. –12

9. 1 ft

10. 4 ft

11. 64 ft

12. 1600 ft

13. 0.013 mm

14. 0.032 mm

15. 0.4 mm

16. 1.4 mm

17. $1210

18. $1331

19. $1464.10

20. $1610.51

21. 4a

22. 2s + 5

23. –9a + b

24. 6a + 3b

1-2


25. 10r + 5s

26. 6w + 5z

27. 2x2 + x

28. xy +3x

29. –0.5x

30. 6x – 5

31. 10y + x – 4

32. –y – 6x

33. –3a + 15b

34. 4g – 2

35. –3x + 4y – z

36. 4a

37. 4a

38. 5

39. 17

40. 41

41. 66

42. –1

43.

44. 10

45. –765

46. a. about 180 million voters

b. about 242 million voters; about 263 million voters

c. –0.0078y2 + 1.265y +

65.27

d. about 87 million

414

1-2


47. – a2 + 2b2

48.

49. +

50. y

51. 3x + 6y

52. 4x + 2y

53. –2x2 + 2y2

54. F

55. C

56. A

57. G

58. B

59. H

60. E

61. I

62. D

63. Assoc. Prop. of Add., Comm. Prop. of Add., Assoc. Prop. of Add., Identity Prop. of Mult., Distr. Prop. of Add.

64. Distr. Prop.; addition; Distr. Prop.

65. Answers may vary. Sample: x3 + x2 + x – x3 – 2x, –x2 + 2x2 + 3x – 3x – x, x2 – 5x + 4x – x5 + x5, 3x2 – x2 – x2 + 7x – 8x

34

5x2

2

7y 2

122y15

1-2


66. Answers may vary. Sample: 2(b – a) + 5(b – a) = (2 + 5) (b – a) Dist. Prop.

= 7 (b – a) Addition= 7b – 7a Dist.

Prop.

67. a. 18

b. 2x2, 18

c. Properties of operations were used to simplify the original expression. Those properties convert one expression into an equivalent

expression, and equivalent expressions have equal values for all replacements of their variables.

d. No; explanations may vary. Sample: Students should check each step in the simplification; they also should substitute more than one value for x in the original and simplified expressions.

68. B

69. G

1-2


70. [2] x = 0 or x = 3

since for 6x(x – 3) = 0

either 6x or x – 3 must be equal

to 0.

[1] x = 0 and x = 3

with no

explanation

71. A

72. D

73. C

74. –1.5, – 2,–1.4, –0.5

75. –4.3, –|3.4|,|–3.4|, |–4.3|

76. – , – , – ,

77. – , – ,

,

78. >

79. >

56

34

38

12

18

110

116

14

80. >

81. <

82. <

83. =

1-2


1. Evaluate each expression for the given values of the variables.

a. 4p – 3q + 8; p = –6, q = –10

b. –3m2 – (m + n)2; m = 2, n = –4

2. Simplify by combining like terms.

a. 7 – 4(x + y) + 2(x – 3y)

b. 3y2 + 8 – (2y2 + 12)

c. –12(x2 + y) + 3(y + x2)

3. Find the perimeter. Simplify the answer.

14

–16

7 – 2x – 10y

y2 – 4

–9x2 – 9y

8c + 4d

1-2

1-3


Simplify each expression.

1. 5x – 9x + 3 2. 2y + 7x + y – 1

3. 10h + 12g – 8h – 4g 4. + + – y

5. (x + y) – (x – y) 6. –(3 – c) – 4(c – 1)

(For help, go to Lesson 1-2.)

Solving EquationsSolving Equations

y3

2y3

x3


Solving EquationsSolving Equations

Solutions

1. 5x – 9x + 3 = (5 – 9)x + 3 = –4x + 3

2. 2y + 7x + y – 1 = 7x + (2 + 1)y – 1 = 7x + 3y – 1

3. 10h + 12g – 8h – 4g = (12 – 4)g + (10 – 8)h = 8g + 2h

4. + + – y = + + – = =

= =

5. (x + y) – (x – y) = (x + y) + (y – x) = (x – x) + (y + y) = 0 + 2y = 2y

6. –(3 – c) – 4(c – 1) = (c – 3) – 4c – 4(–1) = c – 3 – 4c + 4 = (1 – 4)c

+ (–3 + 4) = –3c + 1

1-3

x3

y3

2y3

x3

y3

2y3

3y3

x + y + 2y – 3y3

x + (1 + 2 – 3)y3

x + 0y3

x3

Solving EquationsSolving EquationsALGEBRA 2 LESSON 1-3ALGEBRA 2 LESSON 1-3

Solve 7x + 3 = 2x – 12.

7x + 3 = 2x – 12

5x + 3 = –12 Subtract 2x from each side.

5x = –15 Subtract 3 from each side.

x = –3 Divide each side by 5.

1-3

Check: 7x + 3 = 2x – 12

7(–3) + 3 2(–3) – 12

–18 = –18


Solve 4(m + 9) = –3(m – 4).

4(m + 9) = –3(m – 4)

4m + 36 = –3m + 12 Distributive Property

7m + 36 = 12 Add 3m to each side.

7m = –24 Subtract 36 from each side.

1-3

m = – Divide each side by 7.247


The formula for the surface area of a rectangular prism

units long, w units wide, and h units high is A = 2( w + h + wh).

Solve the formula for w.

A = 2( w + h + wh)

A = 2 w + 2 h + 2wh Distributive Property

A – 2 h = 2 w + 2wh Subtract 2 h from each side.

A – 2 h = (2 + 2h)w Distributive Property

1-3

= w Divide both sides by 2 + 2h.A – 2 h2 + 2h


Solve + 8 = b for x. Find any restrictions on a and b.xa

x + 8a = ab Simplify.

1-3

+ 8

= b

xa

a( ) + a(8) = ab Multiply each side by the least common denominator (LCD), a.

xa

x = ab – 8a Subtract 8a from each side.

=/The denominator cannot be zero, so a 0.


Adrian will use part of a garage wall as one of the long sides

of a rectangular rabbit pen. He wants the pen to be 3 times as long as

it is wide. He plans to use 68 ft of fencing. Find the dimensions of the

pen.

Relate: 2 • width + length = perimeter

Define: Let w = the width.

Then 3w = the length.

Write: 2 w + 3w = 68

1-3


(continued)

5w = 68 Add.

Check: Is the answer reasonable? Since the dimensions are about 14 ft by 41 ft and 14 + 14 + 41 = 69, the perimeter is about 69 ft. The answer is reasonable.

1-3

w = 13 Divide each side by 5.35

3w = 40 Find the length.45

35The width is 13 ft and the length is 40 ft.4

5


The sides of a quadrilateral are in the ratio 1 : 2 : 3 : 6. The

perimeter is 138 cm. Find the lengths of the sides.

Relate: Perimeter equals the sum of the lengths of the four sides.

Define: Let x = the length of the shortest side.

Then 2x = the length of the second side.

Then 3x = the length of the third side.

Then 6x = the length of the fourth side.

1-3


(continued)

Write: 138= x + 2x + 3x + 6x138 = 12xCombine like terms.11.5 = x2x = 2(11.5) 3x = 3(11.5) 6x = 6(11.5) Find the length of

= 23 = 34.5 = 69 each side.

Check: Is the answer reasonable? Since 12 + 23 + 35 + 69 = 139, the answer is reasonable.

The lengths of the sides are 11.5 cm, 23 cm, 34.5 cm, and 69 cm.

1-3


A plane takes off from an airport and flies east at a speed of

350 mi/h. Thirty-five minutes later, a second plane takes off from the

same airport and flies east at a higher altitude at a speed of 400 mi/h.

How long does it take the second plane to overtake the first plane?

1-3

Relate: distance first plane travels = distance second plane travels.

Define: Let t = the time in hours for the second plane.

Then t + = the time in hours for the first plane.3560

Write: 400t = 350 (t + ) 7 12

400t = 350t + Distributive Property 12256


(continued)

50t = Solve for t.12256

1-3

t = 4 h or about 4 h 5 min 1 12

Check: Is the answer reasonable? In 4 h, the second plane travels 1600

mi. In 4 h, the first plane travels about 1600 mi. The answer is

reasonable.

23


1-3


1. 23

2. 8

3.

4. –5

5.

6. –

7.

8.

9. 8

10. –6

11. 2

12. 0

13. –

14. –6

15. 2

16.

172

72

19

177

32

23

12

17. h =

18. g =

19. w =

20. r =

21. r =

22. h =

23. x = , a –b

24. x = , b c

2Ab

2st2

Vlh

Ipt

S2 h

V r

2

ca + b

cc – b

=/

=/


25. x = a(c – b) or ac – ab, a 0

26. x = a(b + 5) or ab + 5a, a 0

27. x = 2(m + n) + 2 or2m + 2n + 2

28. x = – 1

29. 4 h

30. 300 mi/h, 600 mi/h

31. width = 4.5 cm, length = 7.5 cm

32. 4 in., 5 in., 6 in.

33. 11 cm, 11 cm, 16.5 cm, 16.5 cm

34. 7.5 cm, 10 cm, 12.5 cm

35. a. x + (x + 1) + (x + 2) = 90; 29, 30, 31

b. (x – 1) + x + (x + 1) = 90; 29, 30, 31

36. , or 1

37. 3

38. , or 5

39.

40. 30

41. , or 0.9875

42. R =

43. r2 =

44. h =

45. v =

23

23

23

4639

739

275

25

32

7980

r1r2

r1 + r2

Rr1

r1 – R

h + 5t2

t

1-3

5g2

=/

=/

S – 2 r 2

2 r


1-3

46. h =

47. b2 = – b1

48. 40°, 140°

49. 34°, 56°

50. about 169.4 mi

51. 2.98 m

52. $360; $746.4053. 43, 45, 47, 49

54. 34, 36, 38, 40

55. x = ab – b2 – a, b 0

56. x = , b d

57. x = , a c

58. x = , a b

59. x = , b c

60. x = , 3at cd

61. x = , 5bp aq

62. x = + 6, a, b, d 0

63. x = , a 0

64. x = + a, m 0,

x a

65. a. t =

b. about 40.9°F

c. C = (F – 32)

d. about 4.9°C

2(v – s2)s

2Ah

c – ab – d

b + dc – a

3a – b – 8a – b

3b + 2c – 5b – c

2ab – 2c3at – cd

4a – 3bcaq – 5bp

cb2da

10ca

a – cm

59

s – 10551.1

=/

=/

=/

=/

=/

=/

=/

=/

=/

=/

=/


66. a. 10 cows, 30 chickens; Sample equation: 4c + 2(40 – c) = 100, where c is the number of cows.

b. 80 legs; 20 legs; 2 legs; 10 cows

c. Answers may vary. Sample: In all, a repair shop has 11 bicycles and tricycles to repair.

These have a total of 26 wheels. How many bicycles and how many tricycles are there? 7 bicycles, 4 tricycles

67. a. If you solve ax – b = c for x, you

get x = . Since b and c are

integers, b + c is an integer. But

a is a nonzero integer. So

is the quotient of two integers and hence, by the definition of a

rational number, is a

rational number.

b + ca

b + ca

b + ca

1-3


1-3

b. Solutions are rational when 0, a 0,

and is a perfect square (a whole number

perfect square or, in simplest form, a fraction whose numerator and denominator are whole number perfect squares).

68. about 269.4 ft

69. 18.5

70. 20

71. 19.38°

72. 560 cm2

c – ba

c – ba

73. –7

74. – , or –5

75. –20

76. – , or –5

77. 7x2 – 2x

78. ab – 6a

79. 2y – 7x

80. 11x – 6

81. –6x – 5

82. –2r – 5s

163

13

112

12

=/

67. (continued)>–


1. Solve 16x – 15 = –5x + 48.

2. Solve 5(1 – 3m) = 30 – 2(4m + 7).

3. Solve s = for b.

4. Mrs. Chern drove at a rate of 45 mi/h from her home to her sister’s house. She spent 1.5 hours having lunch with her sister. She then drove back home at a rate of 55 mi/h. The entire trip, including lunch, took 4 hours. How far does Mrs. Chern live from her sister?

5. Find three consecutive odd integers whose sum is 111.

a + b + c2

3

b = 2s – a – c

35, 37, 39

1-3

– 117

61 mi78

1-4


State whether each inequality is true or false.

1. 5 < 12 2. 5 < –12

3. 5 12 4. 5 –12

5. 5 5 6. 5 5

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

Solving InequalitiesSolving Inequalities

Solve each equation.

7. 3x + 3 = 2x – 3 8. 5x = 9(x – 8) + 12

>– <–

<– >–

2. 5 < –12, false

4. 5 –12, false

6. 5 5, true

8. 5x = 9(x – 8) + 125x = 9x – 72 + 12–4x = –60x = 15

Solutions


1. 5 < 12, true

3. 5 12, false

5. 5 5, true

7. 3x + 3 = 2x – 33x – 2x = –3 – 3x = –6


1-4

>–<–

<–>–

Solving InequalitiesSolving InequalitiesALGEBRA 2 LESSON 1-4ALGEBRA 2 LESSON 1-4

Solve –2x < 3(x – 5). Graph the solution.

–2x < 3(x – 5)

–2x < 3x – 15 Distributive Property

–5x < –15 Subtract 3x from both sides.

x > 3 Divide each side by –5 and reverse the inequality.

1-4


Solve 7x 7(2 + x). Graph the solution.

1-4

7x 7(2 + x)>–

7x 14 + 7x Distributive Property>–

0 14 Subtract 7x from both sides.>–

The last inequality is always false, so 7x 7(2 + x) is always false. It has no solution.

>–

>–


A real estate agent earns a salary of $2000 per month plus

4% of the sales. What must the sales be if the salesperson is to have

a monthly income of at least $5000?

Define: Let x = sales (in dollars).

The sales must be greater than or equal to $75,000.

1-4

Relate: $2000 + 4% of sales $5000>–

Write: 2000 + 0.04x 5000>–

0.04x 3000 Subtract 2000 from each side.>–

x 75,000 Divide each side by 0.04.>–


Graph the solution of 2x – 1 3x and x > 4x – 9.

1-4

<–

2x – 1 3x and x > 4x – 9<–

–1 x 9 > 3x<–

–1 x and 3 > x<–

This compound inequality can be written as –1 x < 3.<–


Graph the solution of 3x + 9 < –3 or –2x + 1 < 5.

3x + 9 < –3 or –2x + 1 < 5

3x < –12 –2x < 4

x < –4 or x > –2

1-4


A strip of wood is to be 17 cm long with a tolerance of ± 0.15

cm. How much should be trimmed from a strip 18 cm long to allow it to

meet specifications?


At least 0.85 cm and no more than 1.15 cm should be trimmed off to meet specifications.

1-4

Relate: minimum length final length maximum length

Define: Let x = number of centimeters to remove.

<– <–

Write: 17 – 0.15 18 – x 17 + 0.15<– <–

16.85 18 – x 17.15 Simplify.<– <–

–1.15 – x –0.85 Subtract 18.<– <–

1.15 x 0.85 Multiply by –1.>– >–

5. y –14

6. y –6

7. x 8

8. m < 10


1. x –

2. k > –9

3. a > 11

4. t 11

1-4


9. n > 8

10. All real numbers are solutions.


12

<–

<–

<–

<–

<–



13. no solutions

14. The width is less than 11.5 in., and the length is 3 in. greater than the width.

15. The longest side is less than 21 cm.

16. The smaller number is an integer greater than or equal to 8.

17. 4546 or more chips

18. –5 < x < 2

19. –4 x 2

20. –4 x < 6

1-4

<– <–

<–


21. –5 < x 6

22. x < 4 or x > 12


24. x –8

25. x –3 or x 9

26. between about 36.4 lb and 45.5 lb flour

27. between 4 and 5 days

28. between 0.26 cm and 0.30 cm

29. z 6

30. y

12

12

313

1-4

>–

<–

<–

>–

>–

<–


31. x –48

32. x < 42

33. no solutions


35. Answers may vary. Sample: Mario has a coin collection that consists of dimes and nickels. There are half as many dimes as nickels. There are no more than 60 coins in the collection. Describe the collection.

36. 2 < AB < 6

37. between $204,000 and $254,000

38. a. 0 makes y 20 true, but it does

not make (y – 16)

y + 2 true.

b. y –20

39. Dist. Prop.; arithmetic; Sub. Prop. of Inequality; Mult. Prop. of Inequality

12

1-4

>– <–

>–

<–


40. Mult. Prop. of Inequality; Dist. Prop.; Add. Prop. of Inequality; Subt. Prop. of Inequality; Div. Prop. of Inequality

41. –1 < x < 8

42. 3 < y < 6

43. no solutions

44. –7 z < 4


46. b < 2 or b > 5


25

1-4

<–


48. no solutions


50. no solutions

51. Answers may vary. Sample: 2x – 7 –11

52. Answers may vary. Sample: –3x + 1 > 4

53. Answers may vary. Sample: –9 < 5x + 1 < 6

1-4

>–

54. Answers may vary. Sample: 2x + 4 0 or –3x – 3 0

55. a. no

b. yes; values of a that are 8 or

greater

c. (a) yes; values of a that are less than 8, (b) no

56. C

57. B

58. H

59. D

60. [2] The maximum number of songs can be recorded when the

songs are short, so the maximum number of songs is

= 30

songs; the minimum number of songs can be recorded when the songs are long, so the minimum number of songs is

= 18 songs.

90 min3 min per song

90 min5 min per song

<–<–

The solutions to the two simple inequalities overlap. By writing or in the box, the solution is any value on either of the two arrows, or all real numbers.

x + 5 < 0 x + 5 > 0

x < –5 x > –5

The solutions to the two simple inequalities do not overlap. By writing and in the box, the solution is any point on both arrows, or no real numbers.


[1] provides 30 and 18 but not the explanation

61. [4] Answer includes all of the following six parts of the explanation (or equivalent statements), which

consist of solving each compound inequality, graphing each

compound inequality, and explaining why the choices of or and and result in solutions of all real numbers and no real

numbers.

x + 5 > 0 x – 3 < 0

x > –5 x < 3

1-4


[3] omits one or two parts of the six parts of the explanation

[2] omits three or four of the six parts of the explanation

[1] includes at least one of the six parts of the explanation

62. 4

63. no solution

64.

65. –20

66. 7a + 5

67. –2x + 14y

68.

69. 1.61 – 0.1k

910

b + 1212

1-4

1. Solve –2(x – 3) 4. Graph the solution.

2. Solve –5(4 – x) < 5x. Graph the solution.

3. Graph the solution of 3x + 4 1 and –2x + 7 5.

4. A copper wire is to have a length of 16 cm with a tolerance of ±0.02 cm. How much must be trimmed from a wire that is 18 cm long for it to meet specifications?


all real numbers

at least 1.98 cm and no more than 2.02 cm

1-4

–<x 1

>–

>– >–


Solve each equation.

1. 5(x – 6) = 40 2. 5b = 2(3b – 8) 3. 2y + 6y = 15 – 2y + 8

4. 4x + 8 > 20

(For help, go to Lessons 1-3 and 1-4.)

Absolute Value Equations and InequalitiesAbsolute Value Equations and Inequalities

Solve each inequality.

6. 4(t – 1) < 3t + 5

1-5

5. 3a – 2 a + 6>–

1. 5(x – 6) = 40 2. 5b = 2(3b – 8)

= 5b = 6b – 16

x – 6 = 8 –b = –16x = 14 b = 16

Solutions



5(x – 6)5

405

5. 3a – 2 a + 6 6. 4(t – 1) < 3t + 53a – a 6 + 2 4t – 4 < 3t + 5

2a 8 4t – 3t < 5 + 4a 4 t < 9

3. 2y + 6y = 15 – 2y + 8 4. 4x + 8 > 202y + 6y + 2y = 15 + 8 4x > 12

10y = 23 x > 3y = 2.3

1-5

>–>–>–>–

Absolute Value Equations and InequalitiesAbsolute Value Equations and InequalitiesALGEBRA 2 LESSON 1-5ALGEBRA 2 LESSON 1-5

Solve |15 – 3x| = 6.

|15 – 3x| = 6

15 – 3x = 6 or15 – 3x = –6The value of 15 – 3x can

be 6 or –6 since |6| and |–6| both equal 6.

x = 3 or x = 7Divide each side of both

equations by –3.

–3x = –9 –3x = –21Subtract 15 from each

side of both equations.

Check: |15 – 3x| = 6 |15 – 3x| = 6|15 – 3(3)| 6 |15 – 3(7)| 6

|6| 6 |–6| 66 = 6 6 = 6

1-5


Solve 4 – 2|x + 9| = –5.

4 – 2|x + 9| = –5

–2|x + 9| = –9 Add –4 to each side.

x = –4.5 or x = –13.5Subtract 9 from

each side of both equations. Check: 4 – 2 |x + 9| = –5 4 – 2|x + 9| = –5

4 – 2 |–4.5 + 9| –5 4 – 2 |–13.5 + 9| –5

4 – 2 |4.5| –5 4 – 2 |–4.5| –5

4 – 2 (4.5) –5 4 – 2 (4.5) –5

–5 = –5 –5 = –5

1-5

|x + 9| = Divide each side by –2.92

x + 9 = or x + 9 = – Rewrite as two equations.92

92


Solve |3x – 4| = –4x – 1.

|3x – 4| = –4x – 1

1-5

3x – 4 = –4x – 1or 3x – 4= –(–4x –1)

Rewrite as two equations.

7x – 4 = – 13x – 4

= 4x + 1Solve each equation.

7x = 3–x = 5

x = orx = –5

37


(continued)

Check: |3x – 4| = –4x – 1 |3x – 4| = –4x – 1

|3( ) – 4| –4( ) – 1 |3(–5) – 4| (–4(–5) –1)37

37

|– | – |–19| 19197

197

– 19 = 19197

197

1-5

=/

The only solution is –5. is an extraneous solution.37


Solve |2x – 5| > 3. Graph the solution.

|2x – 5| > 3

2x – 5 < –3 or 2x – 5 > 3 Rewrite as a compound inequality.

x < 1 or x > 4

2x < 2 2x > 8 Solve for x.

1-5


Solve –2|x + 1| + 5 –3. Graph the solution.

1-5

>–

|x + 1| 4 Divide each side by –2 and reverse the inequality.

<–

–2|x + 1| + 5 –3>–

–2|x + 1| –8Isolate the absolute value expression. Subtract 5 from each side.

>–

–4 x + 1 4Rewrite as a compound inequality.

<– <–

–5 x 3Solve for x.

<– <–


The area A in square inches of a square photo is required to

satisfy 8.5 A 8.9. Write this requirement as an absolute value

inequality.


1-5

<– <–

Write an inequality.–0.2 A – 8.7 0.2<– <–

Rewrite as an absolute value inequality.|A – 8.7| 0.2<–

Find the tolerance.8.9 – 8.5

2 =0.42 = 0.2

Find the average of the maximum andminimum values.

8.9 + 8.52 =

17.42 = 8.7




1. –6, 6

2. –8, 8

3. –6, 12

4. 3, –

5. no solution

6. no solution

7. –18, 10

8. –7, 17

9. –7, 15

10. –

11.

12.

13. –4, 8

14. –1,

15. 1

53

32

23

32

32

16. x < –12 or x > 6

17. x –3 or x 13

18. y –9 or y 15


1-5

>–<–

>–<–


20. x –3 or x 4

21. z < –4 or z > 4

22. 0 < y < 18

23. –2 < y < 3

24. –2 < x < 6

25. no solution

26. –3 w

27. – t

28. |h – 1.4| 0.1

29. |k – 50.5| 0.5

30. |C – 27.5| 0.25

23

13

12

12

1115

1715

1-5

>–<–

<– <–

<– <–

<–

<–

<–


31. |b – 52.2| 2.5

32. |m – 1250| 5033. |d – 0.11885|

0.00015

34. no solutions

35. – , –1

36. – ,

37. –

38. no solutions

39.

40.

41. –1, –3

42. –

43. 2

44. –6 x 8

45. x –8 or x 5

32

143

163

13

52

118

7136

23


47. t < – or t > 2


49. x < – or x >

32

32

12

1-5

>–

<–

<–

<–

<– <–

<–


50. x –8.4 or x 9.6

51. –3.5 x 7.5

52. –5 < x < 11

53. x < –32 or x > 22

54. Region 10 s 0.1, |s – 0.05| 0.05;

Region 20.1 s 1, |s – 0.55| 0.45;

Region 31 s 3, |s – 2| 1;

Region 43 s 6, |s – 4.5| 1.5;

55. |C – 28.75| 0.25; 28.5 C 29.0

56. The graph of | x | < a (where a > 0) is the set of all points on the number line that lie between the points for a and –a. The graph of | x | > a has two parts: the left part consists of the points to the left of the point for –a, and the right part consists of the points to the right of the point for a.

1-5

>–<–

<– <–

<– <– <–

<– <– <–

<– <– <–

<– <– <–

<– <– <–


57. Answers may vary. Sample: |x – 1| 0; | x | < –5

58. |x – 36.8| 0.05, 36.75 x 36.85

59. |x – 9.55| 0.02, 9.53 x 9.57

60. x is in inches: |x – 3600| 4,3596 x 3604.

61. – ,

62. ,

63. ,

64. (–6 x –5) or (5 x 6)

65. (x –6) or (–5 < x < 5) or (x 6)

66. x

b + ca

b + ca

ab + dc

–ab + dc

ac + dab

ac – dab

52

67. D

68. F

69. C

70. G

71. [2] |x – 3| 5

–5 x – 3 5

–2 x 8

–1, 0, 1, 2, 3, 4, 5, 6, 7, 8

[1] only includes –2 x 8

and does not show work1-5

>–

<–<– <–

<–<– <–

<–<– <–

<– <–<– <–

<–>–

>–

<–<–<–<–

<–

<– <–


72. [4] 3 |2x – 4| + 5< 41 3 |2x – 4| < 36 Subtr. Prop. of

Ineq. |2x – 4| < 12 Div. Prop. of Ineq. –12 < 2x – 4 < 12 Def. of absolute

value –8 < 2x < 16 Add. Prop. of Ineq. –4 < x < 8 Div. Prop. of Ineq.

[3] appropriate methods, but with one computational error

[2] does not include steps

[1] only includes final answer of –4 < x < 8, with no steps or justification of steps

73. y < 6

74. s > –

75. a 4

76. x 12

75

1-5

>–

<–


77. x

78. t 7

79. 36

80. –

81. Inverse Prop. of Add.

82. Closure Prop. Of Mult.


1114

415

1-5

>–

<–


1. Solve |3x + 1| = 4.

2. Solve |–2x + 3| + 7 > 9. Graph the solution.

3. Solve |4x – 12| 8. Graph the solution.

4. A machinist is to drill a hole with diameter D inches, that satisfies 0.16 D 0.19. Express the specification using an absolute value inequality.

1-5

<–

<– <–

1 x 5–< –<

|D – 0.175| 0.015–<

– , 153

x < or x >12

52

1-6


Write each number as a percent.

1. 2. 1

3. 0.0043 4.

5. 1.04 6. 3


ProbabilityProbability

38

56

1 400

1. = 3 ÷ 8 = 0.375 = 0.375(100%) = 37.5%

2. 1 = = 11 ÷ 6 = 1.83 = 1.83(100%) = 183 %

3. 0.0043 = 0.0043(100%) = 0.43%

4. = 1 ÷ 400 = 0.0025 = 0.0025(100%) = 0.25%

5. 1.04 = 1.04(100%) = 104%

6. 3 = 3(100%) = 300%

Solutions



1 400

56

116

13

38

1-6

ProbabilityProbabilityALGEBRA 2 LESSON 1-6ALGEBRA 2 LESSON 1-6

1-6

A player hit the bull’s eye on a circular dartboard 8 times out of

50. Find the experimental probability that the player hits the bull’s eye.

P(bull’s eye) = = 0.16, or 16% 8 50


Describe a simulation you could use that involves flipping a coin

to find the experimental probability of guessing exactly 2 answers out of

6 correctly on a true-false quiz.

Getting heads with a flip of a coin has the same probability as guessing the correct answer to a question on a true-false test. So, let heads represent a correct answer and tails represent an incorrect answer. To simulate guessing the answers for a six-question true-false test, flip a coin six times. Record the number of heads. Repeat 100 times. Count to see how many times heads came up exactly twice. Divide this number by 100. The result is the experimental probability that the simulation gives for guessing 2 correct answers out of 6.

1-6


Find the theoretical probability of rolling a multiple of 3 with a

number cube.

To roll a multiple of 3 with a number cube, you must roll 3 or 6.

1-6

26 6 equally likely outcomes are in

the sample space.

2 outcomes result in a multiple of 3.

13=


Brown is a dominant eye color for human beings. If a father and mother each carry a gene for brown eyes and a gene for blue eyes, what is the probability of their having a child with blue eyes?

B b

B BB Bb

b Bb bb

Gene fromFather

Gene fromMother

Let B represent the dominant gene for brown eyes. Let b represent the recessive gene for blue eyes.

1-6

The sample space contains four equally likely outcomes {BB, Bb, Bb, bb}.

14

The outcome bb is the only one for which a child will have blue eyes. So,

P(blue eyes) = .14The theoretical probability that the child will have blue eyes is , or 25%.


For the dartboard in Example 5, find the probability that a dart

that lands at random on the dartboard hits the outer ring.

P(outer ring) = area of outer ring area of circle with radius 4r

= (area of circle with radius 4r) – (area of circle with radius 3r)area of circle with radius 4r

=16 r 2 – 9 r 2

16 r 2

1-6

=16 r 2

7 r 2

=(4r)2 – (3r)2

(4r2)

= 7 16

The theoretical probability of hitting the outer ring is , or about 44%. 7 16



1. or about 47%; or about 53%

2. the number 1: or about 15.7%;

the number 2: or about 16.4%;




the number 6: or about 17.2%


3. Answers may vary. Sample: Generate random numbers between 0 and 1 using a graphing calculator. In each random number, examine the first five digits. Let even digits represent correct answers and odd digits incorrect answers. If there are two or more even digits, make a tally mark for that number. Do this 100 times. Find the total number of tally marks. This, as a percent, gives the experimental probability. The simulated probability should be about 70%.

161340

179340

21134116745

268116745

26823

134

1-6


4. Answers may vary. Sample: Toss 5 coins. Keep a tally of the times 3 or more heads are tossed. (A head represents a correct answer.) Do this 100 times. The total number of tally marks, as a percent, gives the experimental probability. The simulated probability should be about 40%.

5. Answers may vary. Sample: Generate 100 random numbers with a calculator. Record the first five digits of each number. Let 0 and 1 represent correct answers and the other digits incorrect answers. Tally the recorded numbers with exactly one digit that represents a correct answer. Tally the recorded numbers with exactly two digits that represent correct answers. Tally the recorded numbers with exactly three digits that represent correct answers. The tally totals, as a percent, give the experimental probabilities. They should be about 40%, 20%, and 5%, respectively.

1-6


6. , or 30%

7. , or 50%

8. , or 80%

9. , or 80%

10. , or 38.4%11. , or 15.2%

12. , or 82.4%13. , or 56%

14. , or 61.6%

310

12

45

4548

125

19125

103125

142577

125

12

15. {Gg, Gg, gg, gg}; , or 50%

16. {Gg, Gg, Gg, Gg}; 1, or 100%

17. , or 6.25%18. , or 37.5%

19. , or 25%

20. , or 75%

21. a. 1

b. 0

116

38

14

34

22. Answers may vary. Sample: Let the digits 1–6 correspond to good eggs, and 7–9 correspond to bad eggs. Ignoring the digit 0, start in the first row of the table. Circle groups of three digits, 20 groups in all. Tally the circled groups that do not have a 7, 8, or 9. The experimental probability of getting 3 cartons with only unbroken eggs is

, or .5

2014

1-6


23. Answers may vary. Sample: Let odd digits represent heads and even digits represent tails. Use the first 50 digits of the table. The experimental probability of heads

is .

24. , or 78.9%25. , or 35.4%26. , or 29.3%

12

116147

52147

43147

27. , or 21.1%28.

29.

30.

31. 0

32.

33. 1

34.

35.

31147

16

12

23

16

13

49

36.

37.

38. a. (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

b. 36 outcomes

c.

d.

49

49

13616

1-6


39. 6.4%

40. a. ;

b. Answers may vary. Sample: Variables

such as injuries make probability a poor predictor.

41. if there are any restrictions on the last digit of a ZIP code

42. or 6.7%

14

34

115

43. a.

b. a to b – a or

c. A game where the probability of

winning is ;

when the odds of

winning are , the

probability of winning is

only .

44. Check students’ work.

45. a. about

b. about

aa + b

ab – a

12

12

13

1323

46. B

47. H

48. [2] (11, 12, 13, 21, 22, 23, 31, 32, 33); all the

pairs are equally likely because the three regions have the same area, and the outcome of the first spin does not affect the

outcome of the second spin.

[1] answers one of the two parts

1-6


49. [2] ; there is one favorable outcome (11) out of 9 possible outcomes.

[1] no explanation included

50. [4] a. The favorable outcomes for an even sum are 11, 13, 22, 31, and 33.

b. The probability of an even sum is

c. The favorable outcomes for a sum that is not prime are 13, 22, 31, and

33.

d. The probability of a sum that is not prime is .

e. > , so an even sum is more likely than a sum that is not prime.

[3] omits one of the five parts of the answer

[2] omits two OR three of the five parts of the answer

[1] omits four of the five parts of the answer

19

no. of favorable outcomestotal no. of outcomes

= .59

49

59

49

1-6


51. No; not all odds are 50–50. In this case they are, which means 1 out every 250 people will have defective chromosomes.

52. –12, 6

53. – , 5

54. –13, 6

55. –5, 9

56. –13, 10

57. – ,

58. x –4 or x 4

59. –2 x 6

60. x < 1 or x > 2

61. –5 < x < 1

62. x –8 or x 263. –15 x –3

53

125

245

13

1-6

>–

<– <–

<–

<– >–

<– <–


1. A bowler rolled the ball 35 times and got 5 strikes. What is the experimental probability that the bowler gets a strike?

2. Describe a simulation you could use that involves flipping a coin to find the experimental probability of guessing at least four correct answers on a five-question true or false quiz.

3. What is the theoretical probability of rolling a sum less than 5 using two number cubes?

4. Segments parallel to the sides are used to divide a square board 3 ft on each side into 9 equal-size smaller squares. If the board is in a level position and a grain of rice lands on the board at a random point, what is the probability that it lands on a corner section?

Answers may vary. Sample: Let heads represent a correct answer and tails represent an incorrect answer. Flip the coin five times. Record the number of heads. Repeat 100 times. Divide the number of times you got 4 or 5 heads by 100.

1-6

49

16

17

Tools of AlgebraTools of AlgebraALGEBRA 2 CHAPTER 1ALGEBRA 2 CHAPTER 1

1-A

1. Check students’ answers.

2. Dist. Prop., Assoc. Prop. of Add., Identity Prop. of Mult., Dist. Prop., arithmetic

3. 29

4. 84

5. 15

6. 2a2 + a

7. –3x + 5y

8. 2a – 4b

9. 11x – 27

10. 7

11.

12.

13. 5

14. 45

163

83

15. 4

16. x = , a b17. x = 3c2, c 0

18. x = a(b – 1) + 5, a 0

19. x = , a b

20. r =

21. $138

22. 9

2aa – b

c – 2a – b

s2 h

=/

=/

=/

=/

Tools of AlgebraTools of AlgebraALGEBRA 2 CHAPTER 1ALGEBRA 2 CHAPTER 1

23. x –4

24. x > 7

25. t >

26. x < –16 or x >

12

25

27. 1 < d < 2

28. –1,

29. –

30. 3 p 7

32

53

31. w < – or w > 0

32.

33. 0

34.

35. 1

36. about 40%

43

25

35

1-A

>–

<– <–

Documents

1-1 ALGEBRA 2 LESSON 1-1 Simplify. 1.–(–7.2)2.1 – (–3) 3.–9 + (–4.5) 4.(–3.4)(–2) (For help, go to Skills Handbook page 845.) Properties of Real Numbers