1. Splash Screen

2. Lesson 1-1 Expressions and Formulas
Lesson 1-2 Properties of Real Numbers
Lesson 1-3 Solving Equations
Lesson 1-4 Solving Absolute Value Equations
Lesson 1-5 Solving Inequalities
Lesson 1-6 Solving Compound and Absolute Value Inequalities
Chapter Menu

3. Main Ideas and Vocabulary Key Concept: Order of Operations
Five-Minute Check
Main Ideas and Vocabulary
Key Concept: Order of Operations
Example 1: Evaluate Algebraic Expressions
Example 2: Use a Formula
Lesson 1 Menu

4. Use the order of operations to evaluate expressions.
Use formulas.
variable
power
polynomial
term
like terms
trinomial
binomial
formula
algebraic expression
order of operations
monomial
constant
coefficient
degree
Lesson 1 MI/Vocab

5. Lesson 1 KC1

6. Evaluate Algebraic Expressions
A. Evaluate (x – y)3 + 3 if x = 1 and y = 4.
(x – y)3 + 3 = (1 – 4)3 + 3 Replace x with 1 and y with 4.
= – Find (–3)3.
= –24 Add –27 and 3.
Answer: The value is –24.
Lesson 1 Ex1

7. Evaluate Algebraic Expressions
B. Evaluate s – t(s2 – t) if s = 2 and t = 3.4.
s – t(s2 – t) = 2 – 3.4(22 – 3.4) Replace s with 2 and t with 3.4.
= 2 – 3.4(4 – 3.4) Find 22.
= 2 – 3.4(0.6) Subtract 3.4 from 4.
= 2 – 2.04 Multiply 3.4 and 0.6.
= –0.04 Subtract 2.04 from 2.
Answer: The value is –0.04.
Lesson 1 Ex1

8. Evaluate Algebraic Expressions
x = 5, y = –2, and z = –1
Evaluate the numerator and the denominator separately.
Multiply 40 by –2.
Simplify the numerator and the denominator. Then divide.
Answer: The value is –9.
Lesson 1 Ex1

9. A. What is the value of (y – x)3 – 12 if x = –3 and y = –4?
B. –11
C. –13
D. 13 A B C D
Lesson 1 CYP1

10. B. What is the value of x – y2(x + 5) if x = 2 and y = 4?
C. –54
D. –25 A B C D
Lesson 1 CYP1

11. C.
A. –23
B. –19
C. 19
D. 23 A B C D
Lesson 1 CYP1

12. Answer: The area of the trapezoid is 152 square meters.
Use a Formula
Find the area of a trapezoid with base lengths of 13 meters and 25 meters and a height of 8 meters.
Answer: The area of the trapezoid is 152 square meters.
Lesson 1 Ex2

13. A. 450 cm3
B. 75 cm3
C. 50 cm3
D. 10 cm3 A B C D
Lesson 1 CYP2

14. End of Lesson 1

15. Five-Minute Check (over Lesson 1-1) Main Ideas and Vocabulary
Key Concept: Real Numbers
Example 1: Classify Numbers
Key Concept: Real Number Properties
Example 2: Identify Properties of Real Numbers
Example 3: Additive and Multiplicative Inverses
Example 4: Real-World Example
Example 5: Simplify an Expression
Lesson 2 Menu

16. Use the properties of real numbers to evaluate expressions.
Classify real numbers.
Use the properties of real numbers to evaluate expressions.
real numbers
rational numbers
irrational numbers
Lesson 2 MI/Vocab

17. Lesson 2 KC1

18. Answer: irrationals (I) and reals (R)
Classify Numbers
Answer: irrationals (I) and reals (R)
Lesson 2 Ex1

19. B. Name the sets of numbers to which 5 belongs.
Classify Numbers
B. Name the sets of numbers to which 5 belongs.
Answer: naturals (N), wholes (W), integers (Z), rationals (Q), reals (R)
Lesson 2 Ex1

20. Answer: rationals (Q) and reals (R)
Classify Numbers C.
Answer: rationals (Q) and reals (R)
Lesson 2 Ex1

21. D. Name the sets of numbers to which –43 belongs.
Classify Numbers
D. Name the sets of numbers to which –43 belongs.
Answer: integers (Z), rationals (Q), and reals (R)
Lesson 2 Ex1

22. E. Name the sets of numbers to which –23.3 belongs.
Classify Numbers
E. Name the sets of numbers to which –23.3 belongs.
Answer: rationals (Q) and reals (R)
Lesson 2 Ex1

23. A. The number belongs to which sets?
A. irrationals (I) and reals (R)
B. rationals (Q) and reals (R)
C. naturals (N), wholes (W), integers (Z), rationals (Q), and reals (R)
D. none of the above A B C D
Lesson 2 CYP1

24. A. irrationals (I) and reals (R) B. rationals (Q) and reals (R)
C. naturals (N), wholes (W), integers (Z), rationals (Q), and reals (R)
D. none of the above A B C D
Lesson 2 CYP1

25. C. The number belongs to which sets?
A. irrationals (I) and reals (R)
B. rationals (Q) and reals (R)
C. naturals (N), wholes (W), integers (Z), rationals (Q), and reals (R)
D. none of the above A B C D
Lesson 2 CYP1

26. D. The number belongs to which sets?
A. irrationals (I) and reals (R)
B. rationals (Q) and reals (R)
C. naturals (N), wholes (W), integers (Z), rationals (Q), and reals (R)
D. none of the above A B C D
Lesson 2 CYP1

27. E. The number 32.1 belongs to which sets?
A. irrationals (I) and reals (R)
B. rationals (Q) and reals (R)
C. naturals (N), wholes (W), integers (Z), rationals (Q), and reals (R)
D. none of the above A B C D
Lesson 2 CYP1

28. Lesson 2 KC2

29. Identify Properties of Real Numbers
Name the property illustrated by (–8 + 8) + 15 =
The Additive Inverse Property says that a number plus its opposite is 0.
Answer: Additive Inverse Property
Lesson 2 Ex2

30. What is the property illustrated by 3 + 0 = 3?
A. Distributive Property
B. Additive Inverse Property
C. Identity Property of Addition
D. Inverse Property of Multiplication A B C D
Lesson 2 CYP2

31. Additive and Multiplicative Inverses
Identify the additive inverse and multiplicative inverse for –7.
Since –7 + 7 = 0, the additive inverse of –7 is 7.
Answer:
Lesson 2 Ex3

32. What is the additive inverse and multiplicative inverse for the number 5?
A. B.
C. D. A B C D
Lesson 2 CYP3

33. There are two ways to find the total amount spent on stamps. Method 1
POSTAGE Audrey went to a post office and bought eight 39-cent stamps and eight 24-cent postcard stamps. What was the total amount of money Audrey spent on stamps?
There are two ways to find the total amount spent on stamps.
Method 1
Multiply the price of each type of stamp by 8 and then add.
S = 8(0.39) + 8(0.24) =
= 5.04
Lesson 2 Ex4

34. Answer: Audrey spent a total of $5.04 on stamps.
Method 2
Add the prices of both types of stamps and then multiply the total by 8.
S = 8( )
= 8(0.63)
= 5.04
Answer: Audrey spent a total of $5.04 on stamps.
Notice that both methods result in the same answer.
Lesson 2 Ex4

35. CHOCOLATE Joel went to the grocery store and bought 3 plain chocolate candy bars for $0.69 each and 3 chocolate-peanut butter candy bars for $0.79 each. How much did Joel spend altogether on candy bars?
A. $2.86
B. $4.44
C. $4.48
D. $7.48 A B C D
Lesson 2 CYP4

36. Simplify an Expression
Simplify 4(3a – b) + 2(b + 3a).
4(3a – b) + 2(b + 3a)
= 4(3a) – 4(b) + 2(b) + 2(3a) Distributive Property
= 12a – 4b + 2b + 6a Multiply.
= 12a + 6a – 4b + 2b Commutative Property (+)
= (12 + 6)a + (–4 + 2)b Distributive Property
= 18a – 2b Simplify.
Answer: 18a – 2b
Lesson 2 Ex5

37. Which expression is equivalent to 2(3x – y) + 4(2x + 3y)?
A. 14x + 10y
B. 14x + 2y
C. 14x + y
D. 11x + 2y A B C D
Lesson 2 CYP5

38. End of Lesson 2

39. Five-Minute Check (over Lesson 1-2) Main Ideas and Vocabulary
Example 1: Verbal to Algebraic Expression
Example 2: Algebraic to Verbal Sentence
Key Concept: Properties of Equality
Example 3: Identify Properties of Equality
Key Concept: Properties of Equality
Example 4: Solve One-Step Equations
Example 5: Solve a Multi-Step Equation
Example 6: Solve for a Variable
Example 7: Apply Properties of Equality
Example 8: Write an Equation
Lesson 3 Menu

40. Solve equations using the properties of equality.
Translate verbal expressions into algebraic expressions and equations, and vice versa.
Solve equations using the properties of equality.
open sentence
equation
solution
Lesson 3 MI/Vocab

41. Verbal to Algebraic Expression
A. Write an algebraic expression to represent the verbal expression 7 less than a number.
Answer: n – 7
Lesson 3 Ex1

42. Verbal to Algebraic Expression
B. Write an algebraic expression to represent the verbal expression the square of a number decreased by the product of 5 and the number.
Answer: x2 – 5x
Lesson 3 Ex1

43. A. Write an algebraic expression to represent the verbal expression 6 more than a number.
A. 6x
B. x + 6
C. x6
D. x – 6 A B C D
Lesson 3 CYP1

44. B. Write an algebraic expression to represent the verbal expression 2 less than the cube of a number.
A. x3 – 2
B. 2x3
C. x2 – 2
D. 2 + x3 A B C D
Lesson 3 CYP1

45. Algebraic to Verbal Sentence
A. Write a verbal sentence to represent 6 = –5 + x.
Answer: Six is equal to –5 plus a number.
Lesson 3 Ex2

46. Algebraic to Verbal Sentence
B. Write a verbal sentence to represent 7y – 2 = 19.
Answer: Seven times a number minus 2 is 19.
Lesson 3 Ex2

47. A. What is a verbal sentence that represents the equation n – 3 = 7?
A. The difference between a number and 3 is 7.
B. The sum of a number and 3 is 7.
C. The difference of 3 and a number is 7.
D. The difference of a number and 7 is 3. A B C D
Lesson 3 CYP2

48. B. What is a verbal sentence that represents the equation 5 = 2 + x?
A. Five is equal to the difference of 2 and a number.
B. Five is equal to twice a number.
C. Five is equal to the quotient of 2 and a number.
D. Five is equal to the sum of 2 and a number. A B C D
Lesson 3 CYP2

49. Lesson 3 KC1

50. Identify Properties of Equality
A. Name the property illustrated by the statement. a – 2.03 = a – 2.03
Answer: Reflexive Property of Equality
Lesson 3 Ex3

51. Identify Properties of Equality
B. Name the property illustrated by the statement. If 9 = x, then x = 9.
Answer: Symmetric Property of Equality
Lesson 3 Ex3

52. A. Reflexive Property of Equality B. Symmetric Property of Equality
A. What property is illustrated by the statement below? If x + 4 = 3, then 3 = x + 4.
A. Reflexive Property of Equality
B. Symmetric Property of Equality
C. Transitive Property of Equality
D. Substitution Property of Equality A B C D
Lesson 3 CYP3

53. A. Reflexive Property of Equality B. Symmetric Property of Equality
B. What property is illustrated by the statement below? If 3 = x and x = y, then 3 = y.
A. Reflexive Property of Equality
B. Symmetric Property of Equality
C. Transitive Property of Equality
D. Substitution Property of Equality A B C D
Lesson 3 CYP3

54. Lesson 3 KC2

55. Solve One-Step Equations
A. Solve s – 5.48 = Check your solution.
s – 5.48 = 0.02 Original equation
s – = Add 5.48 to each side.
s = 5.5 Simplify.
Check: s – 5.48 = 0.02 Original equation
5.5 – 5.48 = 0.02 Substitute 5.5 for s. ?
0.02 = 0.02 Simplify. 
Answer: The solution is 5.5.
Lesson 3 Ex4

56. Solve One-Step Equations
Original equation
Simplify.
Lesson 3 Ex4

57. Solve One-Step Equations
Check: Original equation ?
Substitute 36 for t. 
Simplify.
Answer: The solution is 36.
Lesson 3 Ex4

58. A. What is the solution to the equation x + 5 = 3?
B. –2
C. 2
D. 8 A B C D
Lesson 3 CYP4

59. B. What is the solution to the equation
C. 15
D. 30 A B C D
Lesson 3 CYP4

60. Solve a Multi-Step Equation
Solve 53 = 3(y – 2) – 2(3y –1).
53 = 3(y – 2) – 2(3y – 1) Original equation
53 = 3y – 6 – 6y + 2 Apply the Distributive Property.
53 = –3y – 4 Simplify the right side.
57 = –3y Add 4 to each side to isolate the variable.
–19 = y Divide each side by –3.
Answer: The solution is –19.
Lesson 3 Ex5

61. What is the solution to 25 = 3(2x + 2) – 5(2x + 1)?B. C.
D. 6 A B C D
Lesson 3 CYP5

62. Solve for a Variable
GEOMETRY The formula for the area of a trapezoid is where A is the area, b1 is the length of one base, b2 is the length of the other base, and h is the height of the trapezoid. Solve the formula for h.
Lesson 3 Ex6

63. Divide each side by (b1 + b2).
Solve for a Variable
Area of a trapezoid
Multiply each side by 2.
Simplify.
Divide each side by (b1 + b2).
Simplify.
Answer:
Lesson 3 Ex6

64. GEOMETRY The formula for the perimeter of a rectangle is where P is the perimeter, and w is the width of the rectangle. What is this formula solved for w? A. B. C. D. A B C D
Lesson 3 CYP6

65. Apply Properties of Equality
A B
C D
Read the Test Item
You are asked to find the value of the expression 4g – 2. Your first thought might be to find the value of g and then evaluate the expression using this value. Notice that you are not required to find the value of g. Instead, you can use the Subtraction Property of Equality.
Lesson 3 Ex7

66. Apply Properties of Equality
Solve the Test Item
Original equation
Subtract 7 from each side.
Answer: C
Lesson 3 Ex7

67. If 2x + 6 = –3, what is the value of 2x –3?
B. 6
C. –6
D. –12 A B C D
Lesson 3 CYP7

68. the cost for installation
Write an Equation
HOME IMPROVEMENT Carl wants to replace the five windows in the bedrooms of his two-story house. His neighbor has agreed to help install them for $250. If Carl has budgeted $1000 for the total cost of replacing the windows, what is the maximum amount he can spend on each window?
Explore Let c represent the cost of each window.
Plan
The number of windows
times
the cost per window
plus
the cost for installation
equals
the total cost.
5 ● c = 1000
Lesson 3 Ex8

69. Solve Original equation
Write an Equation
Solve Original equation
Subtract 250 from each side.
Simplify.
Divide each side by 5.
Simplify.
Answer: Carl can afford to spend $150 on each window.
Lesson 3 Ex8

70. Write an Equation
Check The total cost to replace five windows at $150 each is 5(150) or $750. Add the $250 cost of the installation to that, and the total bill to replace the windows is or $1000. Thus, the answer is correct.
Lesson 3 Ex8

71. HOME IMPROVEMENT Kelly wants to repair the siding on her house
HOME IMPROVEMENT Kelly wants to repair the siding on her house. Her contractor will charge her $300 plus $150 per square foot of siding. How much siding can she repair for $1500?
A. 100 ft2
B. 10 ft2
C. 8 ft2
D. 4.5 ft2 A B C D
Lesson 3 CYP8

72. End of Lesson 3

73. Five-Minute Check (over Lesson 1-3) Main Ideas and Vocabulary
Key Concept: Absolute Value
Example 1: Evaluate an Expression with Absolute Value
Example 2: Solve an Absolute Value Equation
Example 3: No Solution
Example 4: One Solution
Lesson 4 Menu

74. Evaluate expressions involving absolute values.
Solve absolute value equations.
absolute value
empty set
Lesson 4 MI/Vocab

75. Lesson 4 KC1

76. Evaluate an Expression with Absolute Value
Replace x with 4.
Simplify 2(4) first.
Subtract 8 from 6.
Add.
Answer: The value is 4.7.
Lesson 4 Ex1

77. A. 18.3
B. 1.7
C. –1.7
D. –13.7 A B C D
Lesson 4 CYP1

78. Solve an Absolute Value Equation
Case 1 a = b
y + 3 = 8
y + 3 – 3 = 8 – 3
y = 5
or Case 2 a = –b
y + 3 = –8
y + 3 – 3 = –8 – 3
y = –11
Check: |y + 3| = 8
|y + 3| = 8 ?
|5 + 3| = 8 ?
|–11 + 3| = 8 ?
|8| = 8 ?
|–8| = 8 
8 = 8 
8 = 8
Answer: The solutions are 5 or –11. Thus, the solution set is –11, 5.
Lesson 4 Ex2

79. What is the solution to |2x + 5| = 15?
B. –10, 5
C. –5, 10
D. –5 A B C D
Lesson 4 CYP2

80. |6 – 4t| + 5 = 0 Original equation
No Solution
Solve |6 – 4t| + 5 = 0.
|6 – 4t| + 5 = 0 Original equation
|6 – 4t| = –5 Subtract 5 from each side.
This sentence is never true.
Answer: The solution set is .
Lesson 4 Ex3

81. A. B. C. D. A B C D
Lesson 4 CYP3

82. One Solution
Case 1 a = b
8 + y = 2y – 3
8 = y – 3
11 = y
Lesson 4 Ex4

83. One Solution
Check: 
Answer:
Lesson 4 Ex4

84. A. B. C. D. A B C D
Lesson 4 CYP4

85. End of Lesson 4

86. Five-Minute Check (over Lesson 1-4) Main Ideas and Vocabulary
Key Concept: Properties of Inequality
Example 1: Solve an Inequality Using Addition or Subtraction
Example 2: Solve an Inequality Using Multiplication or Division
Example 3: Solve a Multi-Step Inequality
Example 4: Real-World Example: Write an Inequality
Lesson 5 Menu

87. Solve inequalities with one operation.
Solve multi-step inequalities.
set-builder notation
Lesson 5 MI/Vocab

88. Lesson 5 KC1

89. Solve an Inequality Using Addition or Subtraction
Solve 4y – 3 < 5y + 2. Graph the solution set on a number line.
4y – 3 < 5y + 2 Original inequality
4y – 3 – 4y < 5y + 2 – 4y Subtract 4y from each side.
– 3 < y + 2 Simplify.
–3 – 2 < y + 2 – 2 Subtract 2 from each side.
–5 < y Simplify.
y > –5 Rewrite with y first.
Lesson 5 Ex1

90. Solve an Inequality Using Addition or Subtraction
Answer: Any real number greater than –5 is a solution of this inequality.
A circle means that this point is not included in the solution set.
Lesson 5 Ex1

91. Which graph represents the solution to 6x – 2 < 5x + 7?B. C. D. A B C D
Lesson 5 CYP1

92. Lesson 5 KC2

93. Solve an Inequality Using Multiplication or Division
Solve 12  –0.3p. Graph the solution set on a number line.
Original inequality
Divide each side by –0.3, reversing the inequality symbol.
Simplify.
Rewrite with p first.
Lesson 5 Ex2

94. Solve an Inequality Using Multiplication or Division
Answer: The solution set is p | p  –40.
A dot means that this point is included in the solution set.
Lesson 5 Ex2

95. What is the solution to –3x  21?
A. x | x  –7
B. x | x  –7
C. x | x  7
D. x | x  7 A B C D
Lesson 5 CYP2

96. Solve a Multi-Step Inequality
Original inequality
Multiply each side by 2.
Add –x to each side.
Divide each side by –3, reversing the inequality symbol.
Lesson 5 Ex3

97. Solve a Multi-Step Inequality
Lesson 5 Ex3

98. A. B. C. D. A B C D
Lesson 5 CYP3

99. Explore Let g = the number of gallons of gasoline that Alida buys.
Write an Inequality
CONSUMER COSTS Alida has at most $15.00 to spend today. She buys a bag of potato chips and a can of soda for $1.59. If gasoline at this store costs $2.89 per gallon, how many gallons of gasoline, to the nearest tenth of a gallon, can Alida buy for her car?
Explore Let g = the number of gallons of gasoline that Alida buys.
Plan The total cost of the gasoline is 2.89g. The cost of the chips and soda plus the total cost of the gasoline must be less than or equal to $ Write an inequality.
Lesson 5 Ex4

100. Subtract 1.59 from each side.
Write an Inequality
The cost of chips & soda
plus
the cost of gasoline
is less than or equal to
$15.00.
g  15.00
Original inequality
Subtract 1.59 from each side.
Simplify.
Divide each side by 2.89.
Simplify.
Lesson 5 Ex4

101. Answer: Alida can buy up to 4.6 gallons of gasoline for her car.
Write an Inequality
Answer: Alida can buy up to 4.6 gallons of gasoline for her car.
Check
Lesson 5 Ex4

102. RENTAL COSTS Jeb wants to rent a car for his vacation
RENTAL COSTS Jeb wants to rent a car for his vacation. Value Cars rents cars for $25 per day plus $0.25 per mile. How far can he drive for one day if he wants to spend no more that $200 on car rental?
A. up to 700 miles
B. up to 800 miles
C. more than 700 miles
D. more than 800 miles A B C D
Lesson 5 CYP4

103. End of Lesson 5

104. Five-Minute Check (over Lesson 1-5) Main Ideas and Vocabulary
Key Concept: “And” Compound Inequalities
Example 1: Solve an “and” Compound Inequality
Key Concept: “Or” Compound Inequalities
Example 2: Solve an "or" Compound Inequality
Example 3: Solve an Absolute Value Inequality (<)
Example 4: Solve an Absolute Value Inequality (>)
Key Concept: Absolute Value Inequalities
Example 5: Solve a Multi-Step Absolute Value Inequality
Example 6: Real-World Example: Write an Absolute Value Inequality
Lesson 6 Menu

105. Solve compound inequalities.
Solve absolute value inequalities.
compound inequality
intersection
union
Lesson 6 MI/Vocab

106. Animation: Compound Inequalities
Lesson 6 KC1

107. Solve an “and” Compound Inequality
Solve 10  3y – 2 < 19. Graph the solution set on a number line.
Method 1 Write the compound inequality using the word and. Then solve each inequality.
10  3y – 2 and 3y – 2 < 19
12  3y 3y < 21
4  y y < 7
4  y < 7
Lesson 6 Ex1

108. Solve an “and” Compound Inequality
Method 2 Solve both parts at the same time by adding 2 to each part. Then divide each part by 3.
10  3y – 2 < 19
12  3y < 21
4  y < 7
Lesson 6 Ex1

109. Solve an “and” Compound Inequality
Graph the solution set for each inequality and find their intersection.
4  y < 7
y < 7
y  4
Lesson 6 Ex1

110. What is the solution to 11  2x + 5 < 17?B. C. D. A B C D
Lesson 6 CYP1

111. Lesson 6 KC2

112. Solve an “or” Compound Inequality
Solve x + 3 < 2 or – x  –4. Graph the solution set on a number line.
Solve each inequality separately.
–x  –4 or
x + 3 < 2
x < –1
x  4
x < –1 or x  4
x < –1
x  4
Answer: The solution set is x | x < –1 or x  4.
Lesson 6 Ex2

113. What is the solution to x + 5 < 1 or –2x  –6
What is the solution to x + 5 < 1 or –2x  –6? Graph the solution set on a number line. A. B. C. D. A B C D
Lesson 6 CYP2

114. Solve an Absolute Value Inequality (<)
Solve 3 < |d|. Graph the solution set on a number line.
You can interpret 3 < |d| to mean that the distance between d and 0 on a number line is greater than 3 units. To make 3 < |d| true, you must substitute values for d that are greater than 3 units from 0.
Notice that the graph of 3 < |d| is the same as the graph of d < –3 or d > 3.
All of the numbers not between –3 and 3 are greater than 3 units from 0.
Answer: The solution set is d | d < –3 or d > 3.
Lesson 6 Ex3

115. What is the solution to |x| < 5?
A. {x|x > 5 or x < –5}
B. {x|–5 < x < 5}
C. {x|x < 5}
D. {x|x > –5} A B C D
Lesson 6 CYP3

116. Solve an Absolute Value Inequality (>)
Solve 3 > |d|. Graph the solution set on a number line.
You can interpret 3 > |d| to mean that the distance between d and 0 on a number line is less than 3 units. To make 3 > |d| true, you must substitute numbers for d that are fewer than 3 units from 0.
Notice that the graph of 3 > |d| is the same as the graph of d > –3 and d < 3.
All of the numbers between –3 and 3 are less than 3 units from 0.
Answer: The solution set is d | –3 < d < 3.
Lesson 6 Ex4

117. What is the solution to |x| > 5?B. C. D. A B C D
Lesson 6 CYP4

118. Lesson 6 KC3

119. Solve a Multi-Step Absolute Value Inequality
Solve |2x – 2|  4. Graph the solution set on a number line.
|2x – 2|  4 is equivalent to 2x – 2  4 or 2x – 2  –4.
Solve the inequality.
2x – 2  4 or 2x – 2  –4
2x  6 2x  –2
x  3 x  –1
Answer: The solution set is x | x  –1 or x  3.
Lesson 6 Ex5

120. What is the solution to |3x – 3| > 9
What is the solution to |3x – 3| > 9? Graph the solution set on a number line. A. B. C. D. A B C D
Lesson 6 CYP5

121. The rent for an apartment can differ from the average
Write an Absolute Value Inequality
A. HOUSING According to a recent survey, the average monthly rent for a one-bedroom apartment in one city is $750. However, the actual rent for any given one-bedroom apartment in the area might vary as much as $250 from the average. Write an absolute value inequality to describe this situation.
Let r = the actual monthly rent.
The rent for an apartment can differ from the average
by as much as
$250.
|750 – r|  250
Answer: |750 – r|  250
Lesson 6 Ex6

122. Write an Absolute Value Inequality
B. Solve the inequality |750 – r|  250 to find the range of monthly rent.
Rewrite the absolute value inequality as a compound inequality. Then solve for r.
–250  750 – r  250
–250 – 750  –r  250 – 750
–1000  –r  –500
1000  r  500
Answer: The solution set is r | 500  r  1000. The actual rent falls between $500 and $1000, inclusive.
Lesson 6 Ex6

123. A. HEALTH The average birth weight of a newborn baby is 7 pounds
A. HEALTH The average birth weight of a newborn baby is 7 pounds. However, this weight can vary by as much as 4.5 pounds. What is an absolute value inequality to describe this situation?
A. |4.5 – w|  7
B. |w – 4.5|  7
C. |w – 7|  4.5
D. |7 – w|  4.5 A B C D
Lesson 6 CYP6

124. B. HEALTH The average birth weight of a newborn baby is 7 pounds
B. HEALTH The average birth weight of a newborn baby is 7 pounds. However, this weight can vary by as much as 4.5 pounds. What is the range of birth weights for newborn babies?
A. w | w  11.5
B. w | w  2.5
C. w | 2.5  w  11.5
D. w | 4.5  w  7 A B C D
Lesson 6 CYP6

125. End of Lesson 6

126. Five-Minute Checks Image Bank Math Tools Algebra Tiles
Compound Inequalities
CR Menu

127. Lesson 1-2 (over Lesson 1-1) Lesson 1-3 (over Lesson 1-2)
5Min Menu

128. 1. Exit this presentation.
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IB 1

129. IB 2

130. IB 3

131. IB 4

132. Animation 1

133. Animation 2

134. Evaluate (12 – 9)3.
A. 9
B. 27
C. 63
D. 909 A B C D
5Min 1-1

135. Simplify 5(4 + n) + 6n. A. 7n + 9 B. 7n + 20 C. 11n + 20 D. 35n + 20 A
5Min 1-2

136. A. B. C. D. A B C D
5Min 1-3

137. What is the area of a square with sides 15 centimeters?
A cm2
B. 225 cm2
C. 60 cm2
D. 30 cm2 A B C D
5Min 1-4

138. Write an expression for the area of a rectangle with length 2x and width x.
A. 2x2
B. 3x
C. 4x2
D. 6x A B C D
5Min 1-5

139. Three consecutive integers have a sum of 114
Three consecutive integers have a sum of 114. What is the second integer?
A. 40
B. 39
C. 38
D. 37 A B C D
5Min 1-6

140. Find the value of the expression (13 + 5) ● 3 – 42.
(over Lesson 1-1)
Find the value of the expression (13 + 5) ● 3 – 42.
A. 12
B. 18
C. 37
D. 38 A B C D
5Min 2-1

141. (over Lesson 1-1) A.
B. –5
C. 2 D. A B C D
5Min 2-2

142. Evaluate 2(a + b) – 5w if a = –3, b = 2, and w = –1.
(over Lesson 1-1)
Evaluate 2(a + b) – 5w if a = –3, b = 2, and w = –1.
A. –8
B. –5
C. 3
D. 6 A B C D
5Min 2-3

143. (over Lesson 1-1)
A. 24 B. C.
D. –24 A B C D
5Min 2-4

144. (over Lesson 1-1)
One way to write the formula for the surface area of a rectangular prism is A = 2h(l + w) + 2lw. Find the surface area of a rectangular prism with a length of 6 inches, a width of 4 inches, and a height of 2 inches.
A. 48 in2
B. 60 in2
C. 68 in2
D. 88 in2 A B C D
5Min 2-5

145. (over Lesson 1-1)
Suppose x = 8 ÷ (9 + 2) – 3 and y = 8 ÷ [9 + 2(–4)]. Which statement is true?
A. x > y
B. x < y
C. x = y
D. x = –4y A B C D
5Min 2-6

146. A. I B. Z, N C. N, W, Z, Q, R D. R, I, W, Q, Z, N (over Lesson 1-2) A
5Min 3-1

147. A. W, Z, Q, R, N B. N, W, Z, I C. W, Z, Q D. Q, R (over Lesson 1-2) A
5Min 3-2

148. (over Lesson 1-2)
Name the property illustrated by the equation a + (7 + c) = (a + 7) + c.
A. Associative (+)
B. Commutative (+)
C. Distributive
D. Inverse (×) A B C D
5Min 3-3

149. (over Lesson 1-2)
Name the property illustrated by the equation 3( ) = 3(4) + 3(0.2).
A. Associative (+)
B. Commutative (+)
C. Distributive
D. Inverse (×) A B C D
5Min 3-4

150. Simplify (2c)(3d) + c + 5cd + 3c2.
(over Lesson 1-2)
Simplify (2c)(3d) + c + 5cd + 3c2.
A. 3c2 + 6cd + c + 6
B. 3c2 + 5cd + 3c + 3d
C. 3c2 + 10cd + c
D. 3c2 + 11cd + c A B C D
5Min 3-5

151. (over Lesson 1-2)
Which statements are true? I. All whole numbers are rational numbers. II. All natural numbers are integers. III. All real numbers are irrational.
A. I only
B. I and II only
C. II and III only
D. I, II, and III A B C D
5Min 3-6

152. (over Lesson 1-3)
Write an algebraic expression to represent the verbal expression, three times the sum of a number and its square.
A. 3x + x2
B. 3(x + x2)
C. 3x2 + x
D. x2 + x + 3 A B C D
5Min 4-1

153. (over Lesson 1-3)
Write an algebraic expression to represent the verbal expression, five less than the product of the cube of a number and –4.
A. –4n3 – 5
B. (–4n)3 – 5
C. (–43)n – 5
D. –4(n – 5)3 A B C D
5Min 4-2

154. Write an equation to represent the sum of 23 and twice a number is 65.
(over Lesson 1-3)
Write an equation to represent the sum of 23 and twice a number is 65.
A. 2(n + 23) = 65
B. n + 2(23) = 65
C. 2n + 23 = 65
D. (n + 2) + 23 = 65 A B C D
5Min 4-3

155. Solve 12f – 4 = 7 + f. A. –1 B. C. D. 1 (over Lesson 1-3) A B C D
5Min 4-4

156. Solve 10y + 1 = 3(–2y – 5). A. –4 B. –1 C. 1 D. 4 (over Lesson 1-3) A
5Min 4-5

157. If 3x – 2 = 16, what is the value of 15x –10?
(over Lesson 1-3)
If 3x – 2 = 16, what is the value of 15x –10?
A. 6
B. 48
C. 80
D. 120 A B C D
5Min 4-6

158. Evaluate |4w + 3| if w = –2. A. 4 B. 5 C. 11 D. 20 (over Lesson 1-4) A
5Min 5-1

159. Evaluate |2x + y| if x = 1.5 and y = 4.
(over Lesson 1-4)
Evaluate |2x + y| if x = 1.5 and y = 4.
A. 7
B. 7.5
C. 11
D. 14 A B C D
5Min 5-2

160. Evaluate 5|xy – w| if w = –2, x = 1.5, and y = 4.
(over Lesson 1-4)
Evaluate 5|xy – w| if w = –2, x = 1.5, and y = 4.
A. 27.5
B. 20
C. 37.5
D. 40 A B C D
5Min 5-3

161. Solve |b + 20| = 21. A. {–41, –1} B. {–41, 1} C. {41, –1} D. {41, 1}
(over Lesson 1-4)
Solve |b + 20| = 21.
A. {–41, –1}
B. {–41, 1}
C. {41, –1}
D. {41, 1} A B C D
5Min 5-4

162. Solve –4|a + 5| = –8. A. {–7, –3} B. {–7, 3} C. {7, –3} D. {7, 3}
(over Lesson 1-4)
Solve –4|a + 5| = –8.
A. {–7, –3}
B. {–7, 3}
C. {7, –3}
D. {7, 3} A B C D
5Min 5-5

163. A. No value of n satisfies the equation.
(over Lesson 1-4)
Consider the equation |n – 8| = |8 – n|. Which of the following is true about the value(s) of n?
A. No value of n satisfies the equation.
B. Only positive values of n satisfy the equation.
C. Only negative values of n satisfy the equation.
D. All values of n satisfy the equation. A B C D
5Min 5-6

164. Which choice shows the solution set and the graph of 3x + 7 > 22?
(over Lesson 1-5)
Which choice shows the solution set and the graph of 3x + 7 > 22?
A. {x | x > 5}
B. {x | x ≥ 5}
C. {x | x ≤ 5}
D. {x | x ≥ 6} A B C D
5Min 6-1

165. Which choice shows the solution set and the graph of 3(3w + 1) ≥ 4.8?
(over Lesson 1-5)
Which choice shows the solution set and the graph of 3(3w + 1) ≥ 4.8?
A. {w | w < 0.2}
B. {w | w > 0.2}
C. {w | w ≥ 0.2}
D. {w | w ≥ 0.4} A B C D
5Min 6-2

166. (over Lesson 1-5)
Which choice shows the solution set and the graph of 7 + 3y > 4(y + 2)?
A. {y | y > –1}
B. {y | y ≤ –1}
C. {y | y ≤ –2}
D. {y | y < –1} A B C D
5Min 6-3

167. A. {w | w < –3} B. {w | w ≤ –3} C. {w | w ≥ –3} D. {w | w > –3}
(over Lesson 1-5)
A. {w | w < –3}
B. {w | w ≤ –3}
C. {w | w ≥ –3}
D. {w | w > –3} A B C D
5Min 6-4

168. (over Lesson 1-5)
Tom makes $4.50 an hour. He worked 12 hours in one week. If at least one-third of this pay is taken out for taxes and other deductions, what is the greatest amount of money he can take home?
A. $18
B. $27
C. $36
D. $54 A B C D
5Min 6-5

169. Which inequality is graphed on the number line in the figure?
(over Lesson 1-5)
Which inequality is graphed on the number line in the figure?
A. 4x < –8
B. x > –2
C. 6 > –3x
D. –2x < 4 A B C D
5Min 6-6

170. This slide is intentionally blank.
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