1. Chapter 1: Equations and inequalities
BIG IDEAS:
Use properties to evaluate and simplify expressions
Use problem solving strategies and verbal models
Solve linear and absolute value equations and inequalities

2. What is the difference between a daily low temperature of -5 and a daily high temperature of 18?

3. Lesson 1: Apply Properties of Real

4. Essential question
How are addition and subtraction related and how are multiplication and division related?

5. Opposite: additive inverse: the opposite of b is –b
VOCABULARY
Opposite: additive inverse: the opposite of b is –b
Reciprocal: multiplicative inverse: the reciprocal of a = 1/a.

7. Properties

8. 5 4 EXAMPLE 1 Graph real numbers on a number line 5
Graph the real numbers – and 3 on a number line. 5 4
SOLUTION
Note that – = –1.25. Use a calculator to approximate
3 to the nearest tenth: 5 4
(The symbol means is approximately equal to.)
So, graph – between –2 and –1, and graph 3 between
1 and 2, as shown on the number line below. 5 4

9. EXAMPLE 3
Identify properties of real numbers
Identify the property that the statement illustrates.
7 + 4 = 4 + 7
SOLUTION
Commutative property of addition
= 1 1 13
SOLUTION
Inverse property of multiplication

10. Definition of subtraction
EXAMPLE 4
Use properties and definitions of operations
Use properties and definitions of operations to show that a + (2 – a) = 2. Justify each step.
SOLUTION
a + (2 – a)
= a + [2 + (– a)]
Definition of subtraction
= a + [(– a) + 2]
Commutative property of addition
= [a + (– a)] + 2
Associative property of addition
= 0 + 2
Inverse property of addition
= 2
Identity property of addition

11. GUIDED PRACTICE
for Examples 3 and 4
Identify the property that the statement illustrates.
(2 3) 9 = 2 (3 9)
SOLUTION
Associative property of multiplication.
= 15
SOLUTION
Identity property of addition.

12. GUIDED PRACTICE
for Examples 3 and 4
Identify the property that the statement illustrates.
4(5 + 25) = 4(5) + 4(25)
SOLUTION
Distributive property.
= 500
SOLUTION
Identity property of multiplication.

13. Comm. prop. of multiplication
GUIDED PRACTICE
for Examples 3 and 4
Use properties and definitions of operations to show that the statement is true. Justify each step.
b (4 b) = 4 when b = 0
SOLUTION 1 b
= b ( )
b (4 b)
Def. of division 1 b
= b ( )
Comm. prop. of multiplication 1 b
= (b )
Assoc. prop. of multiplication
= 1 4
Inverse prop. of multiplication
= 4
Identity prop. of multiplication

14. Comm. prop. of addition Assoc. prop. of addition Combine like terms.
GUIDED PRACTICE
for Examples 3 and 4
Use properties and definitions of operations to show that the statement is true. Justify each step.
3x + (6 + 4x) = 7x + 6
SOLUTION
3x + (6 + 4x)
= 3x + (4x + 6)
Comm. prop. of addition
= (3x + 4x) + 6
Assoc. prop. of addition
= 7x + 6
Combine like terms.

15. Essential question
How are addition and subtraction related and how are multiplication and division related?
They are inverses of one another. Subtraction is defined as adding the opposite of the number being subtracted. Division is defined as multiplying by the reciprocal of the divisor.

16. Jill has enough money for a total of 32 table decorations and wall decorations. If n is the number of table decorations, write an expression for the number of wall decorations she can buy.

17. Lesson 2: Evaluate and simplify algebraic expression

18. Essential question
When an expression involves more than one operation, in what order do you do the operations?

19. Variable: a letter that is used to represent one or more numbers
VOCABULARY
Power: an expression formed by repeated multiplication of the same factor
Variable: a letter that is used to represent one or more numbers
Term: each part of an expression separated by + and – signs
Coefficient: the number that leads a variable
Identity: a statement that equates to two equivalent expressions

21. EXAMPLE 1
Evaluate powers
(–5)4
= (–5) (–5) (–5) (–5)
= 625
–54
= –( )
= –625

22. Substitute –3 for x. Evaluate power. Multiply. Add. EXAMPLE 2
Evaluate an algebraic expression
Evaluate –4x2 – 6x + 11 when x = –3.
–4x2 – 6x + 11
= –4(–3)2 – 6(–3) + 11
Substitute –3 for x.
= –4(9) – 6(–3) + 11
Evaluate power.
= –
Multiply.
= –7
Add.

23. EXAMPLE 3
Use a verbal model to solve a problem
Craft Fair
You are selling homemade candles at a craft fair for $3 each. You spend $120 to rent the booth and buy materials for the candles.
Write an expression that shows your profit from
selling c candles.
Find your profit if you sell 75 candles.

24. EXAMPLE 3
Use a verbal model to solve a problem
SOLUTION
STEP 1
Write: a verbal model. Then write an algebraic expression. Use the fact that profit is the difference between income and expenses. – 3 c –
120
An expression that shows your profit is 3c – 120.

25. Substitute 75 for c. Multiply. Subtract. EXAMPLE 3
Use a verbal model to solve a problem
STEP 2
Evaluate: the expression in Step 1 when c = 75.
3c – 120
= 3(75) – 120
Substitute 75 for c.
= 225 – 120
Multiply.
= 105
Subtract.
ANSWER
Your profit is $105.

26. GUIDED PRACTICE
for Examples 1, 2, and 3
Evaluate the expression. 63
SOLUTION
216
–26
SOLUTION
–64

27. GUIDED PRACTICE
for Examples 1, 2, and 3
(–2)6
SOLUTION 64
5x(x – 2) when x = 6
SOLUTION
120

28. GUIDED PRACTICE
for Examples 1, 2, and 3
3y2 – 4y when y = – 2
SOLUTION 20
(z + 3)3 when z = 1
SOLUTION 64

29. GUIDED PRACTICE
for Examples 1, 2, and 3
What If? In Example 3, find your profit if you sell 135 candles.
ANSWER
Your profit is $285.

30. Distributive property
EXAMPLE 4
Simplify by combining like terms
8x + 3x
= (8 + 3)x
Distributive property
= 11x
Add coefficients.
5p2 + p – 2p2
= (5p2 – 2p2) + p
Group like terms.
= 3p2 + p
Combine like terms.
3(y + 2) – 4(y – 7)
= 3y + 6 – 4y + 28
Distributive property
= (3y – 4y) + (6 + 28)
Group like terms.
= –y + 34
Combine like terms.

31. Group like terms. Combine like terms. EXAMPLE 4
Simplify by combining like terms
2x – 3y – 9x + y
= (2x – 9x) + (– 3y + y)
Group like terms.
= –7x – 2y
Combine like terms.

32. GUIDED PRACTICE
for Example 5
8. Identify the terms, coefficients, like terms, and constant terms in the expression 2 + 5x – 6x2 + 7x – 3. Then simplify the expression.
SOLUTION
Terms:
2, 5x, –6x2 , 7x, –3
Coefficients:
5 from 5x, –6 from –6x2 , 7 from 7x
Like terms:
5x and 7x, 2 and –3
Constants:
2 and –3
Simplify:
–6x2 +12x – 1

33. Simplify the expression.
GUIDED PRACTICE
for Example 5
Simplify the expression.
15m – 9m
SOLUTION 6m
2n – 1 + 6n + 5
SOLUTION
8n + 4

34. GUIDED PRACTICE
for Example 5
3p3 + 5p2 – p3
SOLUTION
2p3 + 5p2
2q2 + q – 7q – 5q2
SOLUTION
–3q2 – 6q

35. GUIDED PRACTICE
for Example 5
8(x – 3) – 2(x + 6)
SOLUTION
6x – 36
–4y – x + 10x + y
SOLUTION
9x –3y

36. Essential question
When an expression involves more than one operation, in what order do you do the operations?
Order of operations: Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction

37. On a blank sheet of paper, complete #1-13 ODD on P16 in the blue Quiz section. Please turn into the homework bin when finished.

38. Lesson 3: solve linear equations

39. What are the steps for solving a linear equation?
Essential question
What are the steps for solving a linear equation?

40. Equation: a statement that two expressions are equal
VOCABULARY
Equation: a statement that two expressions are equal
Linear equation: may be written in the form ax + b = 0; no exponents
Solution: a number that makes a true statement when substituted into an equation
Equivalent equations: two equations that have the same solutions

41. Write original equation.
EXAMPLE 1
Solve an equation with a variable on one side
Solve 4 5
x + 8 = 20. 4 5
x + 8 = 20
Write original equation. 4 5
x = 12
Subtract 8 from each side.
x = (12) 5 4
Multiply each side by , the reciprocal of 5 4
x = 15
Simplify.
ANSWER
The solution is 15.
CHECK
x = 15 in the original equation. 4 5
x + 8 =
(15) + 8 = = 20

42. EXAMPLE 2
Write and use a linear equation
During one shift, a waiter earns wages of $30 and gets an additional 15% in tips on customers’ food bills. The waiter earns $105. What is the total of the customers’ food bills?
Restaurant
SOLUTION
Write a verbal model. Then write an equation. Write 15% as a decimal.

43. Subtract 30 from each side.
EXAMPLE 2
Write and use a linear equation
105 = x
Write equation.
75 = 0.15x
Subtract 30 from each side.
500 = x
Divide each side by 0.15.
The total of the customers’ food bills is $500.
ANSWER

44. GUIDED PRACTICE
for Examples 1 and 2
Solve the equation. Check your solution.
x + 9 = 21
ANSWER
The solution is x = 3.
x – 41 = – 13
ANSWER
The solution is x = 4. 3. 3 5 –
x + 1 = 4
ANSWER
The solution is 5.

45. GUIDED PRACTICE
for Examples 1 and 2 4.
REAL ESTATE
A real estate agent’s base salary is $22,000 per year. The agent earns a 4% commission on total sales. How much must the agent sell to earn $60,000 in one year?
The agent must sell $950,000 in a year to each $ 60000
ANSWER

46. Write original equation.
EXAMPLE 3
Standardized Test Practice
SOLUTION
7p + 13 = 9p – 5
Write original equation.
13 = 2p – 5
Subtract 7p from each side.
18 = 2p
Add 5 to each side.
9 = p
Divide each side by 2.
ANSWER
The correct answer is D

47. Write original equation.
EXAMPLE 3
Standardized Test Practice
CHECK
7p + 13 = 9p – 5
Write original equation.
7(9) (9) – 5 = ?
Substitute 9 for p.
– 5 = ?
Multiply.
76 = 76
Solution checks.

48. Write original equation.
EXAMPLE 4
Solve an equation using the distributive property
Solve 3(5x – 8) = –2(–x + 7) – 12x.
3(5x – 8) = –2(–x + 7) – 12x
Write original equation.
15x – 24 = 2x – 14 – 12x
Distributive property
15x – 24 = – 10x – 14
Combine like terms.
25x – 24 = –14
Add 10x to each side.
25x = 10
Add 24 to each side.
x = 2 5
Divide each side by 25 and simplify.
ANSWER
The solution 2 5

49. Substitute for x. Simplify. Solution checks. EXAMPLE 4
Solve an equation using the distributive property
CHECK 2 5
3( – 8) –2(– ) – 12 = ? 2 5
Substitute for x.
3(–6) –14 – 4 5 = ? 24
Simplify.
– 18 = – 18
Solution checks.

50. EXAMPLE 5
Solve a work problem
Car Wash
It takes you 8 minutes to wash a car and it takes a friend 6 minutes to wash a car. How long does it take the two of you to wash 7 cars if you work together?
SOLUTION
STEP 1
Write a verbal model. Then write an equation.

51. Multiply each side by the LCD, 24.
EXAMPLE 5
Solve a work problem
Solve the equation for t.
STEP 2 1 8
t t = 7 6
Write equation.
24( t t) = 24 (7) 1 8 6
Multiply each side by the LCD, 24.
3t + 4t = 168
Distributive property
7t = 168
Combine like terms.
t = 24
Divide each side by 7.
ANSWER
It will take 24 minutes to wash 7 cars if you work together.

52. EXAMPLE 5
Solve a work problem
CHECK
You wash = 3 cars and your friend washes = 4 cars in 24 minutes. Together, you wash 7 cars. 1 6 8

53. GUIDED PRACTICE
for Examples 3, 4, and 5
Solve the equation. Check your solution.
5. –2x + 9 = 2x – 7
ANSWER
The correct answer is 4.
– x = –6x + 15
ANSWER
The correct answer is 1.
(x + 2) = 5(x + 4)
ANSWER
The solution is –7.

54. GUIDED PRACTICE
for Examples 3, 4, and 5
Solve the equation. Check your solution.
8. –4(2x + 5) = 2(–x – 9) – 4x
ANSWER
The solution x = – 1
x x = 39 1 4 2 5 9.
ANSWER
The correct answer is 60

55. GUIDED PRACTICE
for Examples 3, 4, and 5
Solve the equation. Check your solution.
x = x – 1 2 3 5 6
ANSWER
The correct answer is 4
What If? In Example 5, suppose it takes you 9 minutes to wash a car and it takes your friend 12 minutes to wash a car. How long does it take the two of you to wash 7 cars if you work together?
11.
ANSWER
It will take 36 minutes to wash 7 cars if you work together.

56. What are the steps for solving a linear equation?
Essential question
What are the steps for solving a linear equation?
If the equation involves an expression in parenthesis, remove the parentheses by using the distributive property. Then use the properties of equality to obtain equivalent equations in a series of steps until you obtain an equation of the form x = a.

57. Use the distributive property to rewrite xy-5y as a product.

58. Lesson 4: Rewrite formulas and equations

59. What are formulas, and how are formulas used?
Essential question
What are formulas, and how are formulas used?

60. VOCABULARY
Formula: an equation that relates two or more quantities, usually represented by variables
Solve for a variable: to rewrite an equation as an equivalent equation in which the variable is on one side and does not appear on the other side; isolate the variable

61. Write circumference formula.
EXAMPLE 1
Rewrite a formula with two variables
Solve the formula C = 2πr for r. Then find the radius of a circle with a circumference of 44 inches.
SOLUTION
STEP 1
Solve the formula for r.
C = 2πr
Write circumference formula. C 2π
= r
Divide each side by 2π.
STEP 2
Substitute the given value into the rewritten formula.
r = C 2π = 44 7
Substitute 44 for C and simplify.
The radius of the circle is about 7 inches.
ANSWER

62. GUIDED PRACTICE
for Example 1
Find the radius of a circle with a circumference of 25 feet. 1.
The radius of the circle is about 4 feet.
ANSWER

63. GUIDED PRACTICE
for Example 1
The formula for the distance d between opposite vertices of a regular hexagon is d = where a is the distance between opposite sides. Solve the formula for a. Then find a when d = 10 centimeters. 2. 2a 3
SOLUTION d 3
a = 2 3 5
When d = 10cm, a = or cm

64. Write perimeter formula.
EXAMPLE 2
Rewrite a formula with three variables
Solve the formula P = 2l + w for w. Then find the width of a rectangle with a length of 12 meters and a perimeter of 41 meters.
SOLUTION
Solve the formula for w.
STEP 1
P = 2l + 2w
Write perimeter formula.
P – 2l = 2w
Subtract 2l from each side.
P – 2l 2
= w
Divide each side by 2.

65. Substitute 41 for P and 12 for l.
EXAMPLE 2
Rewrite a formula with three variables
Substitute the given values into the rewritten formula.
STEP 2
41 – 2(12) 2
w =
Substitute 41 for P and 12 for l.
w = 8.5
Simplify.
The width of the rectangle is 8.5 meters.
ANSWER

66. GUIDED PRACTICE
for Example 2
Solve the formula P = 2l + 2w for l. Then find the length of a rectangle with a width of 7 inches and a perimeter of 30 inches. 3.
Length of rectangle is 8 in.
ANSWER
Solve the formula A = lw for w. Then find the width of a rectangle with a length of 16 meters and an area of 40 square meters. 4.
Write of rectangle is 2.5 m
w = A l
ANSWER

67. GUIDED PRACTICE
for Example 2
Solve the formula for the variable in red. Then use the given information to find the value of the variable.
A = 1 2 bh 5.
Find h if b = 12 m
and A = 84 m2.
ANSWER = h 2A b

68. GUIDED PRACTICE
for Example 2
Find the value of h if b = 12m and A = 84m2. = h 2A b
Find h if b = 12 m
and A = 84 m2.
ANSWER
h = 14m

69. GUIDED PRACTICE
for Example 2
Solve the formula for the variable in red. Then use the given information to find the value of the variable.
A = 1 2 bh 6.
and A = 9 cm2.
Find b if h = 3 cm
ANSWER = b 2A h

70. GUIDED PRACTICE
for Example 2
Solve the formula for the variable in red. Then use the given information to find the value of the variable.
A = 1 2 bh 6.
and A = 9 cm2.
Find b if h = 3 cm
ANSWER
b = 6cm

71. GUIDED PRACTICE
for Example 2
Solve the formula for the variable in red. Then use the given information to find the value of the variable.
A = 1 2 7.
(b1 + b2)h
Find h if b1 = 6 in.,
b2 = 8 in., and A = 70 in.2
ANSWER
h = 2A
(b1 + b2)

72. GUIDED PRACTICE
for Example 2
Solve the formula for the variable in red. Then use the given information to find the value of the variable.
A = 1 2 7.
(b1 + b2)h
Find h if b1 = 6 in.,
b2 = 8 in., and A = 70 in.2
ANSWER
h = 10 in.

73. Write original equation.
EXAMPLE 3
Rewrite a linear equation
Solve 9x – 4y = 7 for y. Then find the value of y when x = –5.
SOLUTION
Solve the equation for y.
STEP 1
9x – 4y = 7
Write original equation.
–4y = 7 – 9x
Subtract 9x from each side.
y = 9 4 7 – + x
Divide each side by –4.

74. Write original equation.
EXAMPLE 3
Rewrite a linear equation
STEP 2
Substitute the given value into the rewritten equation.
y = 9 4 7 – +
(–5)
Substitute –5 for x.
y = 45 4 7 –
Multiply.
y = –13
Simplify.
CHECK
9x – 4y = 7
Write original equation.
9(–5) – 4(–13) 7 = ?
Substitute –5 for x and –13 for y.
7 = 7
Solution checks.

75. Write original equation.
EXAMPLE 4
Rewrite a nonlinear equation
Solve 2y + xy = 6 for y. Then find the value of y when x = –3.
SOLUTION
Solve the equation for y.
STEP 1
2y + x y = 6
Write original equation.
(2+ x) y = 6
Distributive property
y = 6
2 + x
Divide each side by (2 + x).

76. Substitute –3 for x. Simplify. EXAMPLE 4 Rewrite a nonlinear equation
Substitute the given value into the rewritten equation.
STEP 2
y = 6
2 + (–3)
Substitute –3 for x.
y = – 6
Simplify.

77. GUIDED PRACTICE
for Examples 3 and 4
Solve the equation for y. Then find the value of y when x = 2.
8. y – 6x = 7
9. 5y – x = 13
10. 3x + 2y = 12
y = x
y = 19
ANSWER
y = x 5 13 +
y = 5
ANSWER
y = – 3x 2
+ 6
ANSWER
y = 3

78. GUIDED PRACTICE
for Examples 3 and 4
Solve the equation for y. Then find the value of y when x = 2.
11. 2x + 5y = –1
12. 3 = 2xy – x
13. 4y – xy = 28 2x 5 –1 –
y = –1
y =
ANSWER
y = 1 4
3 +x 2x
ANSWER
y = 14 28
4 – x
y =
ANSWER

79. What are formulas, and how are formulas used?
Essential question
What are formulas, and how are formulas used?
Formulas are equations that relate two or more quantities, usually represented by variables. Formulas can be used to solve many real-world problems, such as problems about investment, temperature, perimeter, area and volume.

80. A balloon is released from a height of 5 feet above the ground
A balloon is released from a height of 5 feet above the ground. Its altitude (in feet) after t minutes is given by the expression 5+82t. What is the altitude of the balloon after 6 minutes.

81. Lesson 5: use problem solving strategies and models

82. Essential question
How can problem solving strategies be used to find verbal and algebraic models?

83. VOCABULARY
Verbal model: a word equation that may be written before an equation is written in mathematical symbols

84. EXAMPLE 1
Use a formula
High-speed Train
The Acela train travels between Boston and Washington, a distance of 457 miles. The trip takes 6.5 hours. What is the average speed?
SOLUTION
You can use the formula for distance traveled as a verbal model.
Distance (miles) =
Rate (miles/hour)
Time (hours)
457 = r
6.5

85. Write equation. Divide each side by 6.5. EXAMPLE 1 Use a formula
An equation for this situation is 457 = 6.5r. Solve for r.
457 = 6.5r
Write equation.
70.3 r
Divide each side by 6.5.
The average speed of the train is about 70.3 miles per hour.
ANSWER
CHECK
You can use unit analysis to check your answer.
457 miles
6.5 hours
70.3 miles
1 hour

86. GUIDED PRACTICE
for Example 1 1.
AVIATION: A jet flies at an average speed of 540 miles per hour. How long will it take to fly from New York to Tokyo, a distance of 6760 miles?
Jet takes about 12.5 hours to fly from New York to Tokyo.
ANSWER

87. EXAMPLE 2
Look for a pattern
Paramotoring
A paramotor is a parachute propelled by a fan-like motor. The table shows the height h of a paramotorist t minutes after beginning a descent. Find the height of the paramotorist after 7 minutes.

88. EXAMPLE 2
Look for a pattern
SOLUTION
The height decreases by 250 feet per minute.
You can use this pattern to write a verbal model for the height.
An equation for the height is h = 2000 – 250t.

89. EXAMPLE 2
Look for a pattern
So, the height after 7 minutes is h = 2000 – 250(7) = 250 feet.
ANSWER

90. EXAMPLE 3
Draw a diagram
Banners
You are hanging four championship banners on a wall in your school’s gym. The banners are 8 feet wide. The wall is 62 feet long. There should be an equal amount of space between the ends of the wall and the banners, and between each pair of banners. How far apart should the banners be placed?
SOLUTION
Begin by drawing and labeling a diagram, as shown below.

91. Subtract 32 from each side.
EXAMPLE 3
Draw a diagram
From the diagram, you can write and solve an equation to find x.
x x x x x = 62
Write equation.
5x + 32 = 62
Combine like terms. 5x = 30
Subtract 32 from each side. x = 6
Divide each side by 5.
The banners should be placed 6 feet apart.
ANSWER

92. EXAMPLE 4
Standardized Test Practice
SOLUTION
STEP 1
Write a verbal model. Then write an equation.
An equation for the situation is 460 = 30g + 25(16 – g).

93. Distributive property
EXAMPLE 4
Standardized Test Practice
Solve for g to find the number of gallons used on the highway.
STEP 2
460 = 30g + 25(16 – g)
Write equation.
460 = 30g – 25g
Distributive property
460 = 5g + 400
Combine like terms.
60 = 5g
Subtract 400 from each side.
12 = g
Divide each side by 5.
The car used 12 gallons on the highway.
ANSWER
The correct answer is B.
CHECK:
(16 – 12) =
= 460

94. GUIDED PRACTICE
for Examples 2, 3 and 4 2.
PARAMOTORING: The table shows the height h of a paramotorist after t minutes. Find the height of the paramotorist after 8 minutes.
So, the height after 8 minutes is h = 2400 – 210(8) = 720 ft
ANSWER

95. GUIDED PRACTICE
for Examples 2, 3 and 4 3.
WHAT IF? In Example 3, how would your answer change if there were only three championship banners?
The space between the banner and walls and between each pair of banners would increase to 9.5 feet.
ANSWER

96. GUIDED PRACTICE
for Examples 2, 3 and 4 4.
FUEL EFFICIENCY A truck used 28 gallons of gasoline and traveled a total distance of 428 miles. The truck’s fuel efficiency is 16 miles per gallon on the highway and 12 miles per gallon in the city. How many gallons of gasoline were used in the city?
Five gallons of gas were used.
ANSWER

97. Essential question
How can problem solving strategies be used to find verbal and algebraic models?
The problem solving strategies use a formula and look for a pattern that can be used to write verbal models which can then be used to write algebraic models. The strategy “draw a diagram” can be used to write an algebraic model directly.

98. On a blank sheet of paper complete #2-12 Even on P40 in the blue quiz section. Turn into the homework bin when finished.

99. Lesson 6: solve linear inequalities

100. Essential question
How are the rules for solving linear inequalities similar to those for solving linear equations, and how are they different?

101. Linear inequality: an inequality using <, >, ≥, ≤
VOCABULARY
Linear inequality: an inequality using <, >, ≥, ≤
Compound inequality: consists of two simple inequalities joined by “and” or “or”
Equivalent inequalities: inequalities that have the same solutions as the original inequality

102. When multiplying or dividing by a negative, flip the inequality sign.
Note:
Solve inequalities just the same as equalities using the Order of Operations
When multiplying or dividing by a negative, flip the inequality sign.

103. EXAMPLE 1
Graph simple inequalities
a. Graph x < 2.
The solutions are all real numbers
less than 2.
An open dot is used in the graph to indicate 2 is not a solution.

104. EXAMPLE 1
Graph simple inequalities
b. Graph x ≥ –1.
The solutions are all real numbers greater than or equal to –1.
A solid dot is used in the graph to indicate –1 is a solution.

105. EXAMPLE 2
Graph compound inequalities
a. Graph –1 < x < 2.
The solutions are all real numbers that are greater than –1 and less than 2.

106. EXAMPLE 2
Graph compound inequalities
b. Graph x ≤ –2 or x > 1.
The solutions are all real numbers that are less than or equal to –2 or greater than 1.

107. GUIDED PRACTICE
for Examples 1 and 2
Graph the inequality.
1. x > –5
The solutions are all real numbers greater than 5.
An open dot is used in the graph to indicate –5 is not a solution.

108. GUIDED PRACTICE
for Examples 1 and 2
Graph the inequality.
2. x ≤ 3
The solutions are all real numbers less than or equal to 3.
A closed dot is used in the graph to indicate 3 is a solution.

109. GUIDED PRACTICE
for Examples 1 and 2
Graph the inequality.
3. –3 ≤ x < 1
The solutions are all real numbers that are greater than or equalt to –3 and less than 1.

110. GUIDED PRACTICE
for Examples 1 and 2
Graph the inequality.
4. x < 1 or x ≥ 2
The solutions are all real numbers that are less than 1 or greater than or equal to 2.

111. EXAMPLE 3
Solve an inequality with a variable on one side
Fair
You have $50 to spend at a county fair. You spend $20 for admission. You want to play a game that costs $1.50. Describe the possible numbers of times you can play the game.
SOLUTION
STEP 1
Write a verbal model. Then write an inequality.

112. Subtract 20 from each side.
EXAMPLE 3
Solve an inequality with a variable on one side
An inequality is g ≤ 50.
STEP 2
Solve the inequality.
g ≤ 50
Write inequality.
1.5g ≤ 30
Subtract 20 from each side.
g ≤ 20
Divide each side by 1.5.
ANSWER
You can play the game 20 times or fewer.

113. Write original inequality.
EXAMPLE 4
Solve an inequality with a variable on both sides
Solve 5x + 2 > 7x – 4. Then graph the solution.
5x + 2 > 7x – 4
Write original inequality.
– 2x + 2 > – 4
Subtract 7x from each side.
– 2x > – 6
Subtract 2 from each side.
Divide each side by –2 and reverse the inequality.
x < 3
ANSWER
The solutions are all real numbers less than 3. The graph is shown below.

114. GUIDED PRACTICE
for Examples 3 and 4
Solve the inequality. Then graph the solution.
5. 4x + 9 < 25
7. 5x – 7 ≤ 6x
x < 4
ANSWER
x > – 7
ANSWER
6. 1 – 3x ≥ –14
8. 3 – x > x – 9
x ≤ 5
ANSWER
x < 6
ANSWER

115. Write original inequality.
EXAMPLE 5
Solve an “and” compound inequality
Solve – 4 < 6x – 10 ≤ 14.
Then graph the solution.
– 4 < 6x – 10 ≤ 14
Write original inequality.
– < 6x – ≤
Add 10 to each expression.
6 < 6x ≤ 24
Simplify.
1 < x ≤ 4
Divide each expression by 6.
ANSWER
The solutions are all real numbers greater than 1 and less than or equal to 4. The graph is shown below.

116. Write first inequality. Write second inequality.
EXAMPLE 6
Solve an “or” compound inequality
Solve 3x + 5 ≤
11 or 5x – 7 ≥ 23
. Then graph the solution.
SOLUTION
A solution of this compound inequality is a solution of either of its parts.
First Inequality
Second Inequality
3x + 5 ≤ 11
Write first inequality.
5x – 7 ≥ 23
Write second inequality.
3x ≤ 6
5x ≥ 30
Subtract 5 from each side.
Add 7 to each side.
x ≥ 6
x ≤ 2
Divide each side by 3.
Divide each side by 5.

117. EXAMPLE 6
Solve an “or” compound inequality
ANSWER
The graph is shown below. The solutions are all real numbers less than or equal to 2 or greater than or equal to 6.

118. EXAMPLE 7
Write and use a compound inequality
Biology
A monitor lizard has a temperature that ranges from 18°C to 34°C. Write the range of temperatures as a compound inequality. Then write an inequality giving the temperature range in degrees Fahrenheit.

119. Multiply each expression by , the reciprocal of . 9 5
EXAMPLE 7
Write and use a compound inequality
SOLUTION
The range of temperatures C can be represented by the inequality 18 ≤ C ≤ 34. Let F represent the temperature in degrees Fahrenheit.
18 ≤ C ≤ 34
Write inequality.
18 ≤ ≤ 34 5 9
(F – 32)
Substitute for C. 9 5
(F – 32)
Multiply each expression by ,
the reciprocal of 9 5
32.4 ≤ F – 32 ≤ 61.2
64.4 ≤ F ≤ 93.2
Add 32 to each expression.

120. EXAMPLE 7
Write and use a compound inequality
ANSWER
The temperature of the monitor lizard ranges from 64.4°F to 93.2°F.

121. GUIDED PRACTICE
for Examples 5,6, and 7
Solve the inequality. Then graph the solution.
9. –1 < 2x + 7 < 19
ANSWER
The solutions are all real numbers greater than – 4 and less than 6.
–4 < x < 6

122. GUIDED PRACTICE
for Examples 5,6 and 7
Solve the inequality. Then graph the solution.
10. –8 ≤
–x – 5
≤ 6
The solutions are all real numbers greater than and equal to – 11 and less than and equal to 3.
ANSWER
–11 ≤ x ≤ 3

123. GUIDED PRACTICE
for Examples 5,6 and 7
Solve the inequality. Then graph the solution.
11. x + 4 ≤ 9
or x – 3 ≥ 7
ANSWER
The graph is shown below. The solutions are all real numbers.
less than or equal to 5 or greater than or equal to 10.
x ≤ 5 or x ≥ 10

124. GUIDED PRACTICE
for Examples 5,6 and 7
Solve the inequality. Then graph the solution.
12. 3x – 1< –1 or 2x + 5 ≥ 11
x < 0 or x ≥ 3
less than 0 or greater than or equal to 3.
ANSWER
The graph is shown below. The solutions are all real numbers.

125. GUIDED PRACTICE
for Examples 5,6 and 7
13.
WHAT IF? In Example 7, write a compound inequality for a lizard whose temperature ranges from 15°C to 30°C. Then write an inequality giving the temperature range in degrees Fahrenheit.
ANSWER
15 ≤ C ≤ 30 or 59 ≤ F ≤ 86

126. Essential question
How are the rules for solving linear inequalities similar to those for solving linear equations, and how are they different?
The addition and subtraction properties are the same, but if you multiply or divide both sides of an inequality by a negative number, the inequality symbol must be reversed.

127. What is unique about absolute value numbers that is not true of integers?

128. Lesson 7: Solve absolute value equations and inequalities

129. Essential question
How are absolute value equations and inequalities like linear equations and inequalities?

130. VOCABULARY
Absolute value: the distance an umber is from 0 on a number line; always positive
Extraneous solution: an apparent solution that must be rejected because it does not satisfy the original equation

131. Write original equation.
EXAMPLE 1
Solve a simple absolute value equation
Solve |x – 5| = 7. Graph the solution.
SOLUTION
| x – 5 | = 7
Write original equation.
x – 5 = – 7 or x – 5 = 7
Write equivalent equations.
x = 5 – 7 or x = 5 + 7
Solve for x.
x = –2 or x = 12
Simplify.

132. EXAMPLE 1
Solve a simple absolute value equation
ANSWER
The solutions are –2 and 12. These are the values of x that are 7 units away from 5 on a number line. The graph is shown below.

133. Write original equation.
EXAMPLE 2
Solve an absolute value equation
Solve |5x – 10 | = 45.
SOLUTION
| 5x – 10 | = 45
Write original equation.
5x – 10 = 45 or 5x – 10 = –45
Expression can equal 45 or –45 .
5x = 55 or x = –35
Add 10 to each side.
x = 11 or x = –7
Divide each side by 5.

134. Solve an absolute value equation
EXAMPLE 2
Solve an absolute value equation
ANSWER
The solutions are 11 and –7. Check these in the original equation.
Check:
| 5x – 10 | = 45
| 5x – 10 | = 45
| 5(11) – 10 | = 45 ?
| 5(–7) – 10 | = 45 ?
|45| = 45 ?
| – 45| = 45 ?
45 = 45
45 = 45

135. Write original equation.
EXAMPLE 3
Check for extraneous solutions
Solve |2x + 12 | = 4x. Check for extraneous solutions.
SOLUTION
| 2x + 12 | = 4x
Write original equation.
2x = 4x or 2x = – 4x
Expression can equal 4x or – 4 x
12 = 2x or 12 = –6x
Add –2x to each side.
6 = x or –2 = x
Solve for x.

136. Check for extraneous solutions
EXAMPLE 3
Check for extraneous solutions
Check the apparent solutions to see if either is extraneous.
CHECK
| 2x + 12 | = 4x
| 2x + 12 | = 4x
| 2(6) +12 | = 4(6) ?
| 2(–2) +12 | = 4(–2) ?
|24| = 24 ?
|8| = – 8 ?
24 = 24
8 = –8
The solution is 6. Reject –2 because it is an extraneous solution.
ANSWER

137. – 3 – 4 – 2 – 1 1 2 3 4 5 6 7 – 5 – 6 – 7 GUIDED PRACTICE
for Examples 1, 2 and 3
Solve the equation. Check for extraneous solutions.
1. | x | = 5
The solutions are –5 and 5. These are the values of x that are 5 units away from 0 on a number line. The graph is shown below.
ANSWER
– 3
– 4
– 2
– 1 1 2 3 4 5 6 7
– 5
– 6
– 7

138. GUIDED PRACTICE
for Examples 1, 2 and 3
Solve the equation. Check for extraneous solutions.
2. |x – 3| = 10
The solutions are –7 and 13. These are the values of x that are 10 units away from 3 on a number line. The graph is shown below.
ANSWER
– 3
– 4
– 2
– 1 1 2 3 4 5 6 7
– 5
– 6
– 7 8 9 10 11 12 13

139. GUIDED PRACTICE
for Examples 1, 2 and 3
Solve the equation. Check for extraneous solutions.
3. |x + 2| = 7
The solutions are –9 and 5. These are the values of x that are 7 units away from – 2 on a number line.
ANSWER

140. GUIDED PRACTICE
for Examples 1, 2 and 3
Solve the equation. Check for extraneous solutions.
4. |3x – 2| = 13
ANSWER
The solutions are 5 and

141. GUIDED PRACTICE
for Examples 1, 2 and 3
Solve the equation. Check for extraneous solutions.
5. |2x + 5| = 3x
The solution of is 5. Reject 1 because it is an extraneous solution.
ANSWER

142. GUIDED PRACTICE
for Examples 1, 2 and 3
Solve the equation. Check for extraneous solutions.
6. |4x – 1| = 2x + 9
ANSWER
The solutions are – and 5. 3 1

143. Subtract 5 from each side.
EXAMPLE 4
Solve an inequality of the form |ax + b| > c
Solve |4x + 5| > 13. Then graph the solution.
SOLUTION
The absolute value inequality is equivalent to
4x +5 < –13 or 4x + 5 > 13.
First Inequality
Second Inequality
4x + 5 < –13
4x + 5 > 13
Write inequalities.
4x < –18
4x > 8
Subtract 5 from each side.
x < – 9 2
x > 2
Divide each side by 4.

144. EXAMPLE 4
Solve an inequality of the form |ax + b| > c
ANSWER
The solutions are all real numbers less than or greater than 2. The graph is shown below.
– 9 2

145. GUIDED PRACTICE
for Example 4
Solve the inequality. Then graph the solution.
7. |x + 4| ≥ 6
x < –10 or x > 2
The graph is shown below.
ANSWER

146. GUIDED PRACTICE
for Example 4
Solve the inequality. Then graph the solution.
8. |2x –7|>1
ANSWER
x < 3 or x > 4
The graph is shown below.

147. GUIDED PRACTICE
for Example 4
Solve the inequality. Then graph the solution.
9. |3x + 5| ≥ 10
ANSWER
x < –5 or x > 1 2 3
The graph is shown below.

148. EXAMPLE 5
Solve an inequality of the form |ax + b| ≤ c
A professional baseball should weigh ounces, with a tolerance of ounce. Write and solve an absolute value inequality that describes the acceptable weights for a baseball.
Baseball
SOLUTION
Write a verbal model. Then write an inequality.
STEP 1

149. Write equivalent compound inequality.
EXAMPLE 5
Solve an inequality of the form |ax + b| ≤ c
STEP 2
Solve the inequality.
|w – 5.125| ≤
Write inequality.
Write equivalent compound inequality.
– ≤ w – ≤
5 ≤ w ≤ 5.25
Add to each expression.
So, a baseball should weigh between 5 ounces and 5.25 ounces, inclusive. The graph is shown below.
ANSWER

150. EXAMPLE 6
Write a range as an absolute value inequality
The thickness of the mats used in the rings, parallel bars, and vault events must be between 7.5 inches and 8.25 inches, inclusive. Write an absolute value inequality describing the acceptable mat thicknesses.
Gymnastics
SOLUTION
STEP 1
Calculate the mean of the extreme mat thicknesses.

151. EXAMPLE 6
Write a range as an absolute value inequality
Mean of extremes = = 7.875 2
Find the tolerance by subtracting the mean from the upper extreme.
STEP 2
Tolerance = 8.25 – 7.875
= 0.375

152. EXAMPLE 6
Write a range as an absolute value inequality
STEP 3
Write a verbal model. Then write an inequality.
A mat is acceptable if its thickness t satisfies |t – 7.875| ≤
ANSWER

153. GUIDED PRACTICE
for Examples 5 and 6
Solve the inequality. Then graph the solution.
|x + 2| < 6
ANSWER
–8 < x < 4
The solutions are all real numbers less than – 8 or greater than 4. The graph is shown below.

154. GUIDED PRACTICE
for Examples 5 and 6
Solve the inequality. Then graph the solution.
11. |2x + 1| ≤ 9
ANSWER
–5 ≤ x ≤ 4
The solutions are all real numbers less than –5 or greater than 4. The graph is shown below.

155. GUIDED PRACTICE
for Examples 5 and 6
Solve the inequality. Then graph the solution.
|7 – x| ≤ 4
3 ≤ x ≤ 11
ANSWER
The solutions are all real numbers less than 3 or greater than 11. The graph is shown below.

156. GUIDED PRACTICE
for Examples 5 and 6
Gymnastics: For Example 6, write an absolute value inequality describing the unacceptable mat thicknesses.
A mat is unacceptable if its thickness t satisfies |t – 7.875| >
ANSWER

157. Essential question
How are absolute value equations and inequalities like linear equations and inequalities?
An absolute value equation can be rewritten as two linear equations, and an absolute value inequality can be rewritten as two linear inequalities.