1. Warm Up Compare. Write <, >, or =. 1. −3 2 3. < 2. 6.5 6.3
1. −3 2 3.
<
>
> =
Tell whether the inequality x < 5 is true or false for the following values of x.
5. x = –10 T
6. x = 5 F
7. x = 4.99 T
8. x = T

2. An inequality is a statement that two quantities are not equal
An inequality is a statement that two quantities are not equal. The quantities are compared by using the following signs: ≤
A ≤ B
A is less
than or
equal to B.
<
A < B
than B.
>
A > B
A is greater ≥
A ≥ B ≠
A ≠ B
A is not
A solution of an inequality is any value that makes the inequality true.

3. An inequality like 3 + x < 9 has too many solutions to list
An inequality like 3 + x < 9 has too many solutions to list. You can use a graph on a number line to show all the solutions.
The solutions are shaded and an arrow shows that the solutions continue past those shown on the graph. To show that an endpoint is a solution, draw a solid circle at the number. To show an endpoint is not a solution, draw an empty circle.

5. Example 2: Graphing Inequalities
Graph each inequality.
Draw a solid circle at .
A. m ≥
Shade all the numbers greater than and draw an arrow pointing to the right. 1 – 2 3
B. t < 5(–1 + 3)
Simplify.
t < 5(–1 + 3)
t < 5(2)
t < 10
Draw an empty circle at 10.
Shade all the numbers less than 10 and draw an arrow pointing to the left. –4 –2 2 4 6 8 10 12 –6 –8

6. Check It Out! Example 2 Graph each inequality. a. c > 2.5
Draw an empty circle at 2.5.
Shade in all the numbers greater than 2.5 and draw an arrow pointing to the right.
a. c > 2.5 –4 –3 –2 –1 1 2 3 4 5 6
2.5
b. 22 – 4 ≥ w
Draw a solid circle at 0.
22 – 4 ≥ w
Shade in all numbers less than 0 and draw an arrow pointing to the left.
4 – 4 ≥ w
0 ≥ w –4 –3 –2 –1 1 2 3 4 5 6
c. m ≤ –3
Draw a solid circle at –3. –4 –2 2 4 6 8 10 12 –6 –8 −3
Shade in all numbers less than –3 and draw an arrow pointing to the left.

7. Example 3: Writing an Inequality from a Graph
Write the inequality shown by each graph.
x < 2
Use any variable. The arrow points to the left, so use either < or ≤. The empty circle at 2 means that 2 is not a solution, so use <.
x ≥ –0.5
Use any variable. The arrow points to the right, so use either > or ≥. The solid circle at –0.5 means that –0.5 is a solution, so use ≥.

8. Check It Out! Example 3
Write the inequality shown by the graph.
Use any variable. The arrow points to the left, so use either < or ≤. The empty circle at 2.5 means that 2.5 is not a solution, so use so use <.
x < 2.5

9. Reading Math
“No more than” means “less than or equal to.”
“At least” means “greater than or equal to”.

10. Turn on the AC when temperature is at least 85°F
Example 4: Application
Ray’s dad told him not to turn on the air conditioner unless the temperature is at least 85°F. Define a variable and write an inequality for the temperatures at which Ray can turn on the air conditioner. Graph the solutions.
Let t represent the temperatures at which Ray can turn on the air conditioner.
Turn on the AC when temperature
is at least
85°F t ≥ 85
Draw a solid circle at 85. Shade all numbers greater than 85 and draw an arrow pointing to the right.
t  85 75 80 85 90 70

11. Let w represent an employee’s wages.
Check It Out! Example 4
A store’s employees earn at least $8.50 per hour. Define a variable and write an inequality for the amount the employees may earn per hour. Graph the solutions.
Let w represent an employee’s wages.
An employee earns
at least
$8.50 w ≥
8.50
w ≥ 8.5 4 6 8 10 12 −2 2 14 16 18
8.5

12. 1. Describe the solutions of 7 < x + 4.
Lesson Quiz: Part I
1. Describe the solutions of 7 < x + 4.
all real numbers greater than 3
2. Graph h ≥ –4.75 –5
–4.75
–4.5
Write the inequality shown by each graph.
x ≥ 3 3. 4.
x < –5.5

13. Lesson Quiz: Part II
5. A cell phone plan offers free minutes for no more than 250 minutes per month. Define a variable and write an inequality for the possible number of free minutes. Graph the solution.
Let m = number of minutes
0 ≤ m ≤ 250
250

14. Solving one-step inequalities is much like solving one-step equations
Solving one-step inequalities is much like solving one-step equations. To solve an inequality, you need to isolate the variable using the properties of inequality and inverse operations.

16. Example 1A: Using Addition and Subtraction to Solve Inequalities
Solve the inequality and graph the solutions.
x + 12 < 20
x + 12 < 20
Since 12 is added to x, subtract 12 from both sides to undo the addition.
–12 –12
x + 0 < 8
x < 8
Draw an empty circle at 8.
–10 –8 –6 –4 –2 2 4 6 8 10
Shade all numbers less than 8 and draw an arrow pointing to the left.

17. Example 1B: Using Addition and Subtraction to Solve Inequalities
Solve the inequality and graph the solutions.
d – 5 > –7
d + 0 > –2
d > –2
d – 5 > –7
Since 5 is subtracted from d, add 5 to both sides to undo the subtraction.
Draw an empty circle at –2.
–10 –8 –6 –4 –2 2 4 6 8 10
Shade all numbers greater than –2 and draw an arrow pointing to the right.

18. Example 1C: Using Addition and Subtraction to Solve Inequalities
Solve the inequality and graph the solutions.
0.9 ≥ n – 0.3
0.9 ≥ n – 0.3
Since 0.3 is subtracted from n, add 0.3 to both sides to undo the subtraction.
1.2 ≥ n – 0
1.2 ≥ n 
1.2
Draw a solid circle at 1.2. 1 2
Shade all numbers less than 1.2 and draw an arrow pointing to the left.

19. Solve each inequality and graph the solutions.
Check It Out! Example 1
Solve each inequality and graph the solutions.
a. s + 1 ≤ 10
Since 1 is added to s, subtract 1 from both sides to undo the addition.
s + 1 ≤ 10
–1 –1 9
–10 –8 –6 –4 –2 2 4 6 8 10
s + 0 ≤ 9
s ≤ 9
b > –3 + t
Since –3 is added to t, add 3 to both sides to undo the addition.
> –3 + t +3
> 0 + t
–10 –8 –6 –4 –2 2 4 6 8 10
t <

20. Solve the inequality and graph the solutions.
Check It Out! Example 1c
Solve the inequality and graph the solutions.
q – 3.5 < 7.5
Since 3.5 is subtracted from q, add 3.5 to both sides to undo the subtraction.
q – 3.5 < 7.5
q – 0 < 11
q < 11 –7 –5 –3 –1 1 3 5 7 9 11 13

21. Since there can be an infinite number of solutions to an inequality, it is not possible to check all the solutions. You can check the endpoint and the direction of the inequality symbol.
The solutions of x + 9 < 15 are given by x < 6.

22. Lesson Quiz: Part I
Solve each inequality and graph the solutions.
1. 13 < x + 7
x > 6
2. –6 + h ≥ 15
h ≥ 21
y ≤ –2.1
y ≤ –8.8

23. Lesson Quiz: Part II
4. A certain restaurant has room for 120 customers. On one night, there are 72 customers dining. Write and solve an inequality to show how many more people can eat at the restaurant.
x + 72 ≤ 120; x ≤ 48, where x is a natural number

25. Example 1A: Multiplying or Dividing by a Positive Number
Solve the inequality and graph the solutions.
7x > –42
7x > –42
>
Since x is multiplied by 7, divide both sides by 7 to undo the multiplication.
1x > –6
x > –6
–10 –8 –6 –4 –2 2 4 6 8 10

26. Example 1B: Multiplying or Dividing by a Positive Number
Solve the inequality and graph the solutions.
Since m is divided by 3, multiply both sides by 3 to undo the division.
3(2.4) ≤ 3
7.2 ≤ m
(or m ≥ 7.2) 2 4 6 8 10 12 14 16 18 20

27. Example 1C: Multiplying or Dividing by a Positive Number
Solve the inequality and graph the solutions.
Since r is multiplied by , multiply both sides by the reciprocal of .
r < 16 2 4 6 8 10 12 14 16 18 20

28. Solve the inequality and graph the solutions.
Check It Out! Example 1a
Solve the inequality and graph the solutions.
4k > 24
Since k is multiplied by 4, divide both sides by 4.
k > 6 2 4 6 8 10 12 16 18 20 14

29. Solve the inequality and graph the solutions.
Check It Out! Example 1b
Solve the inequality and graph the solutions.
–50 ≥ 5q
Since q is multiplied by 5, divide both sides by 5.
–10 ≥ q 5 –5
–10
–15 15

30. Solve the inequality and graph the solutions.
Check It Out! Example 1c
Solve the inequality and graph the solutions.
Since g is multiplied by , multiply both sides by the reciprocal of .
g > 36 36 25 30 35 20 40 15

31. If you multiply or divide both sides of an inequality by a negative number, the resulting inequality is not a true statement! You need to reverse the inequality symbol to make the statement true.

32. This means there is another set of properties of inequality for multiplying or dividing by a negative number.

34. Caution!
Do not change the direction of the inequality symbol just because you see a negative sign. For example, you do not change the symbol when solving 4x < –24.

35. Example 2A: Multiplying or Dividing by a Negative Number
Solve the inequality and graph the solutions.
–12x > 84
Since x is multiplied by –12, divide both sides by –12. Change > to <.
x < –7
–10 –8 –6 –4 –2 2 4 6
–12
–14 –7

36. Example 2B: Multiplying or Dividing by a Negative Number
Solve the inequality and graph the solutions.
Since x is divided by –3, multiply both sides by –3. Change to .
24  x
(or x  24) 16 18 20 22 24 10 14 26 28 30 12

37. Solve each inequality and graph the solutions.
Check It Out! Example 2
Solve each inequality and graph the solutions.
a. 10 ≥ –x
Multiply both sides by –1 to make x positive. Change  to .
–1(10) ≤ –1(–x)
–10 ≤ x
–10 –8 –6 –4 –2 2 4 6 8 10
b > –0.25h
Since h is multiplied by –0.25, divide both sides by –0.25. Change > to <.
–20
–16
–12 –8 –4 4 8 12 16 20
–17
–17 < h

38. Lesson Quiz
Solve each inequality and graph the solutions.
1. 8x < –24
x < –3
2. –5x ≥ 30
x ≤ –6 3.
x > 20 4.
x ≥ 6
5. A soccer coach plans to order more shirts for her team. Each shirt costs $9.85. She has $77 left in her uniform budget. What are the possible number of shirts she can buy?
0, 1, 2, 3, 4, 5, 6, or 7 shirts

39. Inequalities that contain more than one operation require more than one step to solve.

40. Example 1A: Solving Multi-Step Inequalities
Solve the inequality and graph the solutions.
45 + 2b > 61
Since 45 is added to 2b, subtract 45 from both sides to undo the addition.
45 + 2b > 61
– –45
2b > 16
Since b is multiplied by 2, divide both sides by 2 to undo the multiplication.
b > 8 2 4 6 8 10 12 14 16 18 20

41. Example 1B: Solving Multi-Step Inequalities
Solve the inequality and graph the solutions.
8 – 3y ≥ 29
Since 8 is added to –3y, subtract 8 from both sides to undo the addition.
8 – 3y ≥ 29
– –8
–3y ≥ 21
Since y is multiplied by –3, divide both sides by –3 to undo the multiplication. Change ≥ to ≤.
y ≤ –7
–10 –8 –6 –4 –2 2 4 6 8 10 –7

42. Solve the inequality and graph the solutions.
Check It Out! Example 1a
Solve the inequality and graph the solutions.
–12 ≥ 3x + 6
Since 6 is added to 3x, subtract 6 from both sides to undo the addition.
–12 ≥ 3x + 6
– – 6
–18 ≥ 3x
Since x is multiplied by 3, divide both sides by 3 to undo the multiplication.
–6 ≥ x
–10 –8 –6 –4 –2 2 4 6 8 10

43. Solve the inequality and graph the solutions.
Check It Out! Example 1b
Solve the inequality and graph the solutions.
Since x is divided by –2, multiply both sides by –2 to undo the division. Change > to <.
–5 –5
x + 5 < –6
Since 5 is added to x, subtract 5 from both sides to undo the addition.
x < –11
–20
–12 –8 –4
–16
–11

44. Solve the inequality and graph the solutions.
Check It Out! Example 1c
Solve the inequality and graph the solutions.
Since 1 – 2n is divided by 3, multiply both sides by 3 to undo the division.
1 – 2n ≥ 21
Since 1 is added to −2n, subtract 1 from both sides to undo the addition.
– –1
–2n ≥ 20
Since n is multiplied by −2, divide both sides by −2 to undo the multiplication. Change ≥ to ≤.
n ≤ –10
–10
–20
–12 –8 –4
–16

45. To solve more complicated inequalities, you may first need to simplify the expressions on one or both sides by using the order of operations, combining like terms, or using the Distributive Property.

46. Example 2A: Simplifying Before Solving Inequalities
Solve the inequality and graph the solutions.
2 – (–10) > –4t
12 > –4t
Combine like terms.
Since t is multiplied by –4, divide both sides by –4 to undo the multiplication. Change > to <.
–3 < t
(or t > –3) –3
–10 –8 –6 –4 –2 2 4 6 8 10

47. Example 2B: Simplifying Before Solving Inequalities
Solve the inequality and graph the solutions.
–4(2 – x) ≤ 8
−4(2 – x) ≤ 8
Distribute –4 on the left side.
−4(2) − 4(−x) ≤ 8
Since –8 is added to 4x, add 8 to both sides.
–8 + 4x ≤ 8
4x ≤ 16
Since x is multiplied by 4, divide both sides by 4 to undo the multiplication.
x ≤ 4
–10 –8 –6 –4 –2 2 4 6 8 10

48. Example 2C: Simplifying Before Solving Inequalities
Solve the inequality and graph the solutions.
Multiply both sides by 6, the LCD of the fractions.
Distribute 6 on the left side.
4f + 3 > 2
Since 3 is added to 4f, subtract 3 from both sides to undo the addition.
–3 –3
4f > –1

49. Example 2C Continued
4f > –1
Since f is multiplied by 4, divide both sides by 4 to undo the multiplication.

50. Solve the inequality and graph the solutions.
Check It Out! Example 2a
Solve the inequality and graph the solutions.
2m + 5 > 52
Simplify 52.
2m + 5 > 25
– 5 > – 5
Since 5 is added to 2m, subtract 5 from both sides to undo the addition.
2m > 20
m > 10
Since m is multiplied by 2, divide both sides by 2 to undo the multiplication. 2 4 6 8 10 12 14 16 18 20

51. Solve the inequality and graph the solutions.
Check It Out! Example 2b
Solve the inequality and graph the solutions.
3 + 2(x + 4) > 3
Distribute 2 on the left side.
3 + 2(x + 4) > 3
3 + 2x + 8 > 3
Combine like terms.
2x + 11 > 3
Since 11 is added to 2x, subtract 11 from both sides to undo the addition.
– 11 – 11
2x > –8
Since x is multiplied by 2, divide both sides by 2 to undo the multiplication.
x > –4
–10 –8 –6 –4 –2 2 4 6 8 10

52. Check It Out! Example 2c
Solve the inequality and graph the solutions.
Multiply both sides by 8, the LCD of the fractions.
Distribute 8 on the right side.
5 < 3x – 2
Since 2 is subtracted from 3x, add 2 to both sides to undo the subtraction.
7 < 3x

53. Check It Out! Example 2c Continued
Solve the inequality and graph the solutions.
7 < 3x
Since x is multiplied by 3, divide both sides by 3 to undo the multiplication. 4 6 8 2 10

54. Lesson Quiz: Part I
Solve each inequality and graph the solutions.
– 2x ≥ 21
x ≤ –4
2. – < 3p
p > –3
3. 23 < –2(3 – t)
t > 7 4.

55. Lesson Quiz: Part II
5. A video store has two movie rental plans. Plan A includes a $25 membership fee plus $1.25 for each movie rental. Plan B costs $40 for unlimited movie rentals. For what number of movie rentals is plan B less than plan A?
more than 12 movies

56. Some inequalities have variable terms on both sides of the inequality symbol. You can solve these inequalities like you solved equations with variables on both sides.
Use the properties of inequality to “collect” all the variable terms on one side and all the constant terms on the other side.

57. Example 1A: Solving Inequalities with Variables on Both Sides
Solve the inequality and graph the solutions.
y ≤ 4y + 18
y ≤ 4y + 18
–y –y
0 ≤ 3y + 18
To collect the variable terms on one side, subtract y from both sides.
Since 18 is added to 3y, subtract 18 from both sides to undo the addition.
– – 18
–18 ≤ 3y
Since y is multiplied by 3, divide both sides by 3 to undo the multiplication.
–6 ≤ y (or y  –6)
–10 –8 –6 –4 –2 2 4 6 8 10

58. Example 1B: Solving Inequalities with Variables on Both Sides
Solve the inequality and graph the solutions.
4m – 3 < 2m + 6
To collect the variable terms on one side, subtract 2m from both sides.
–2m – 2m
2m – 3 <
Since 3 is subtracted from 2m, add 3 to both sides to undo the subtraction
2m <
Since m is multiplied by 2, divide both sides by 2 to undo the multiplication. 4 5 6

59. Solve the inequality and graph the solutions.
Check It Out! Example 1a
Solve the inequality and graph the solutions.
4x ≥ 7x + 6
4x ≥ 7x + 6
–7x –7x
To collect the variable terms on one side, subtract 7x from both sides.
–3x ≥ 6
x ≤ –2
Since x is multiplied by –3, divide both sides by –3 to undo the multiplication. Change ≥ to ≤.
–10 –8 –6 –4 –2 2 4 6 8 10

60. Solve the inequality and graph the solutions.
Check It Out! Example 1b
Solve the inequality and graph the solutions.
5t + 1 < –2t – 6
5t + 1 < –2t – 6
+2t t
7t + 1 < –6
To collect the variable terms on one side, add 2t to both sides.
Since 1 is added to 7t, subtract 1 from both sides to undo the addition.
– 1 < –1
7t < –7
Since t is multiplied by 7, divide both sides by 7 to undo the multiplication.
7t < –7
t < –1 –5 –4 –3 –2 –1 1 2 3 4 5

61. You may need to simplify one or both sides of an inequality before solving it. Look for like terms to combine and places to use the Distributive Property.

62. Example 3A: Simplify Each Side Before Solving
Solve the inequality and graph the solutions.
2(k – 3) > 6 + 3k – 3
Distribute 2 on the left side of the inequality.
2(k – 3) > 3 + 3k
2k + 2(–3) > 3 + 3k
2k – 6 > 3 + 3k
To collect the variable terms, subtract 2k from both sides.
–2k – 2k
–6 > 3 + k
Since 3 is added to k, subtract 3 from both sides to undo the addition.
–3 –3
–9 > k

63. Example 3A Continued
–9 > k
–12 –9 –6 –3 3

64. Example 3B: Simplify Each Side Before Solving
Solve the inequality and graph the solution.
0.9y ≥ 0.4y – 0.5
0.9y ≥ 0.4y – 0.5
To collect the variable terms, subtract 0.4y from both sides.
–0.4y –0.4y
0.5y ≥ – 0.5
0.5y ≥ –0.5
Since y is multiplied by 0.5, divide both sides by 0.5 to undo the multiplication.
y ≥ –1 –5 –4 –3 –2 –1 1 2 3 4 5

65. Check It Out! Example 3a
Solve the inequality and graph the solutions.
5(2 – r) ≥ 3(r – 2)
Distribute 5 on the left side of the inequality and distribute 3 on the right side of the inequality.
5(2 – r) ≥ 3(r – 2)
5(2) – 5(r) ≥ 3(r) + 3(–2)
10 – 5r ≥ 3r – 6
Since 6 is subtracted from 3r, add 6 to both sides to undo the subtraction.
16 − 5r ≥ 3r
Since 5r is subtracted from 16 add 5r to both sides to undo the subtraction.
+ 5r +5r
≥ 8r

66. Check It Out! Example 3a Continued
16 ≥ 8r
Since r is multiplied by 8, divide both sides by 8 to undo the multiplication.
2 ≥ r –6 –2 2 –4 4

67. Check It Out! Example 3b
Solve the inequality and graph the solutions.
0.5x – x < 0.3x + 6
2.4x – 0.3 < 0.3x + 6
Simplify.
2.4x – 0.3 < 0.3x + 6
Since 0.3 is subtracted from 2.4x, add 0.3 to both sides.
2.4x < 0.3x + 6.3
Since 0.3x is added to 6.3, subtract 0.3x from both sides.
–0.3x –0.3x
2.1x <
Since x is multiplied by 2.1, divide both sides by 2.1.
x < 3

68. Check It Out! Example 3b Continued–5 –4 –3 –2 –1 1 2 3 4 5

69. There are special cases of inequalities called identities and contradictions.

71. Example 4A: Identities and Contradictions
Solve the inequality.
2x – 7 ≤ 5 + 2x
2x – 7 ≤ 5 + 2x
–2x –2x
Subtract 2x from both sides.
–7 ≤ 5 
True statement.
The inequality 2x − 7 ≤ 5 + 2x is an identity. All values of x make the inequality true. Therefore, all real numbers are solutions.

72. Example 4B: Identities and Contradictions
Solve the inequality.
2(3y – 2) – 4 ≥ 3(2y + 7)
Distribute 2 on the left side and 3 on the right side.
2(3y – 2) – 4 ≥ 3(2y + 7)
2(3y) – 2(2) – 4 ≥ 3(2y) + 3(7)
6y – 4 – 4 ≥ 6y + 21
6y – 8 ≥ 6y + 21
–6y –6y
Subtract 6y from both sides. 
–8 ≥ 21
False statement.
No values of y make the inequality true. There are no solutions.

73.  Check It Out! Example 4a Solve the inequality. 4(y – 1) ≥ 4y + 2
Distribute 4 on the left side.
4(y) + 4(–1) ≥ 4y + 2
4y – 4 ≥ 4y + 2
–4y –4y
Subtract 4y from both sides. 
–4 ≥ 2
False statement.
No values of y make the inequality true. There are no solutions.

74.  Check It Out! Example 4b Solve the inequality. x – 2 < x + 1
Subtract x from both sides.
–2 < 1 
True statement.
All values of x make the inequality true.
All real numbers are solutions.

75. Lesson Quiz: Part I
Solve each inequality and graph the solutions.
1. t < 5t + 24
t > –6
2. 5x – 9 ≤ 4.1x – 81
x ≤ –80
3. 4b + 4(1 – b) > b – 9
b < 13

76. Lesson Quiz: Part II
4. Rick bought a photo printer and supplies for $186.90, which will allow him to print photos for $0.29 each. A photo store charges $0.55 to print each photo. How many photos must Rick print before his total cost is less than getting prints made at the photo store?
Rick must print more than 718 photos.

77. Lesson Quiz: Part III
Solve each inequality.
5. 2y – 2 ≥ 2(y + 7)
contradiction, no solution
6. 2(–6r – 5) < –3(4r + 2)
identity, all real numbers

78. The inequalities you have seen so far are simple inequalities
The inequalities you have seen so far are simple inequalities. When two simple inequalities are combined into one statement by the words AND or OR, the result is called a compound inequality.

80. Example 1: Chemistry Application
The pH level of a popular shampoo is between 6.0 and 6.5 inclusive. Write a compound inequality to show the pH levels of this shampoo. Graph the solutions.
Let p be the pH level of the shampoo.
6.0
is less than or equal to
pH level
6.5
≤ p ≤
6.0 ≤ p ≤ 6.5
5.9
6.1
6.2
6.3
6.0
6.4
6.5

81. Let c be the chlorine level of the pool.
Check It Out! Example 1
The free chlorine in a pool should be between 1.0 and 3.0 parts per million inclusive. Write a compound inequality to show the levels that are within this range. Graph the solutions.
Let c be the chlorine level of the pool.
1.0
is less than or equal to
chlorine
3.0
≤ c ≤
1.0 ≤ c ≤ 3.0 2 3 4 1 5 6

82. In this diagram, oval A represents some integer solutions of x < 10 and oval B represents some integer solutions of x > 0. The overlapping region represents numbers that belong in both ovals. Those numbers are solutions of both x < 10 and x > 0.

83. You can graph the solutions of a compound inequality involving AND by using the idea of an overlapping region. The overlapping region is called the intersection and shows the numbers that are solutions of both inequalities.

84. Example 2A: Solving Compound Inequalities Involving AND
Solve the compound inequality and graph the solutions.
–5 < x + 1 < 2
Since 1 is added to x, subtract 1 from each part of the inequality.
–5 < x + 1 < 2
– – 1 – 1
–6 < x < 1
Graph –6 < x.
Graph x < 1.
Graph the intersection by finding where the two graphs overlap.
–10 –8 –6 –4 –2 2 4 6 8 10

85. Example 2B: Solving Compound Inequalities Involving AND
Solve the compound inequality and graph the solutions.
8 < 3x – 1 ≤ 11
8 < 3x – 1 ≤ 11
9 < 3x ≤ 12
Since 1 is subtracted from 3x, add 1 to each part of the inequality.
Since x is multiplied by 3, divide each part of the inequality by 3 to undo the multiplication.
3 < x ≤ 4

86. Graph the intersection by finding where the two graphs overlap.
Example 2B Continued
Graph 3 < x.
Graph x ≤ 4.
Graph the intersection by finding where the two graphs overlap. –5 –4 –3 –2 –1 1 2 3 4 5

87. Solve the compound inequality and graph the solutions.
Check It Out! Example 2a
Solve the compound inequality and graph the solutions.
–9 < x – 10 < –5
Since 10 is subtracted from x, add 10 to each part of the inequality.
–9 < x – 10 < –5
1 < x < 5
Graph 1 < x.
Graph x < 5.
Graph the intersection by finding where the two graphs overlap. –5 –4 –3 –2 –1 1 2 3 4 5

88. Graph the intersection by finding where the two graphs overlap.
Check It Out! Example 2b
Solve the compound inequality and graph the solutions.
–4 ≤ 3n + 5 < 11
–4 ≤ 3n + 5 < 11
– – 5 – 5
–9 ≤ 3n <
Since 5 is added to 3n, subtract 5 from each part of the inequality.
Since n is multiplied by 3, divide each part of the inequality by 3 to undo the multiplication.
–3 ≤ n < 2
Graph –3 ≤ n.
Graph n < 2.
Graph the intersection by finding where the two graphs overlap. –5 –4 –3 –2 –1 1 2 3 4 5

89. In this diagram, circle A represents some integer solutions of x < 0, and circle B represents some integer solutions of x > 10. The combined shaded regions represent numbers that are solutions of either x < 0 or x >10.

90. You can graph the solutions of a compound inequality involving OR by using the idea of combining regions. The combine regions are called the union and show the numbers that are solutions of either inequality.
>

91. Example 3A: Solving Compound Inequalities Involving OR
Solve the inequality and graph the solutions.
8 + t ≥ 7 OR 8 + t < 2
8 + t ≥ 7 OR 8 + t < 2
Solve each simple inequality.
– –8 – −8
t ≥ –1 OR t < –6
Graph t ≥ –1.
Graph t < –6.
Graph the union by combining the regions.
–10 –8 –6 –4 –2 2 4 6 8 10

92. Example 3B: Solving Compound Inequalities Involving OR
Solve the inequality and graph the solutions.
4x ≤ 20 OR 3x > 21
4x ≤ 20 OR 3x > 21
x ≤ 5 OR x > 7
Solve each simple inequality.
Graph x ≤ 5.
Graph x > 7.
Graph the union by combining the regions.
–10 –8 –6 –4 –2 2 4 6 8 10

93. Solve the compound inequality and graph the solutions.
Check It Out! Example 3a
Solve the compound inequality and graph the solutions.
2 +r < 12 OR r + 5 > 19
2 +r < 12 OR r + 5 > 19
Solve each simple inequality.
– – –5 –5
r < 10 OR r > 14
Graph r < 10.
Graph r > 14.
Graph the union by combining the regions. –4 –2 2 4 6 8 10 12 14 16

94. Solve the compound inequality and graph the solutions.
Check It Out! Example 3b
Solve the compound inequality and graph the solutions.
7x ≥ 21 OR 2x < –2
7x ≥ 21 OR 2x < –2
x ≥ 3 OR x < –1
Solve each simple inequality.
Graph x ≥ 3.
Graph x < −1.
Graph the union by combining the regions. –5 –4 –3 –2 –1 1 2 3 4 5

95. Every solution of a compound inequality involving AND must be a solution of both parts of the compound inequality. If no numbers are solutions of both simple inequalities, then the compound inequality has no solutions.
The solutions of a compound inequality involving OR are not always two separate sets of numbers. There may be numbers that are solutions of both parts of the compound inequality.

96. Example 4A: Writing a Compound Inequality from a Graph
Write the compound inequality shown by the graph.
The shaded portion of the graph is not between two values, so the compound inequality involves OR.
On the left, the graph shows an arrow pointing left, so use either < or ≤. The solid circle at –8 means –8 is a solution so use ≤.
x ≤ –8
On the right, the graph shows an arrow pointing right, so use either > or ≥. The empty circle at 0 means that 0 is not a solution, so use >.
x > 0
The compound inequality is x ≤ –8 OR x > 0.

97. Example 4B: Writing a Compound Inequality from a Graph
Write the compound inequality shown by the graph.
The shaded portion of the graph is between the values –2 and 5, so the compound inequality involves AND.
The shaded values are on the right of –2, so use > or ≥. The empty circle at –2 means –2 is not a solution, so use >.
m > –2
The shaded values are to the left of 5, so use < or ≤. The empty circle at 5 means that 5 is not a solution so use <.
m < 5
The compound inequality is m > –2 AND m < 5 (or -2 < m < 5).

98. Check It Out! Example 4a
Write the compound inequality shown by the graph.
The shaded portion of the graph is between the values –9 and –2, so the compound inequality involves AND.
The shaded values are on the right of –9, so use > or . The empty circle at –9 means –9 is not a solution, so use >.
x > –9
The shaded values are to the left of –2, so use < or ≤. The empty circle at –2 means that –2 is not a solution so use <.
x < –2
The compound inequality is –9 < x AND x < –2
(or –9 < x < –2).

99. Check It Out! Example 4b
Write the compound inequality shown by the graph.
The shaded portion of the graph is not between two values, so the compound inequality involves OR.
On the left, the graph shows an arrow pointing left, so use either < or ≤. The solid circle at –3 means –3 is a solution, so use ≤.
x ≤ –3
On the right, the graph shows an arrow pointing right, so use either > or ≥. The solid circle at 2 means that 2 is a solution, so use ≥.
x ≥ 2
The compound inequality is x ≤ –3 OR x ≥ 2.

100. Lesson Quiz: Part I
1. The target heart rate during exercise for a 15 year-old is between 154 and 174 beats per minute inclusive. Write a compound inequality to show the heart rates that are within the target range. Graph the solutions.
154 ≤ h ≤ 174

101. Lesson Quiz: Part II
Solve each compound inequality and graph the solutions.
2. 2 ≤ 2w + 4 ≤ 12
–1 ≤ w ≤ 4
r > −2 OR 3 + r < −7
r > –5 OR r < –10

102. Lesson Quiz: Part III
Write the compound inequality shown by each graph. 4.
x < −7 OR x ≥ 0 5.
−2 ≤ a < 4

103. Now You Try… Solve and Graph the Compound Inequality
-3 < x + 2 < 7
x – 1 < -1 OR x – 5 > -1
2 < x + 2 < 5
11 < 2x + 3 < 21
n + 2 < 3 OR n + 3 > 7 -5 5
-5 < x < 5
x < 0 OR x > 4 2 4 2 4
0 < x < 3
4 < x < 9 4 6 8
x < 1 OR x > 4 1 3 5

104. The compound inequality is x ≤ –8 OR x > 0.
Write the compound inequality shown by the graph.
The compound inequality is x ≤ –8 OR x > 0.

105. The compound inequality is –9 < x AND x < –2
Write the compound inequality shown by the graph.
The compound inequality is –9 < x AND x < –2
(or –9 < x < –2).

106. In Summary:
Two inequalities that are combined into one statement by the word AND or OR is called a compound inequality.
If it contains the word AND it is split into two equations and the graph is in between two points.
If it contains the word OR the graphs go in opposite directions from each point.
If you have a you are working with a conjunction, an ‘and’ statement.
Remember: “Less thand”
If you have a you are working with a disjunction, an ‘or’ statement.
Remember: “Greator”

107. Graphing Inequalities
If the variable is on the left, the arrow points the same direction as the inequality.
Parentheses/bracket method :
Parentheses: endpoint is not included <, >
Bracket: endpoint is included ≤, ≥
x < 2
x ≥ 2
Open Circle/closed circle method:
Open Circle: endpoint is not included <, >
Closed Circle: endpoint is included ≤, ≥
x < 2
x ≥ 2

108. Inequalities – Interval Notation
[( smallest, largest )]
Parentheses: endpoint is not included <, >
Bracket: endpoint is included ≤, ≥
Infinity: always uses a parenthesis
x < 2
( –∞, 2)
x ≥ 2
[2, ∞)
4 < x < 9
3-part inequality
(4, 9)

109. The set of all x such that x is greater than or equal to 5.
Inequalities – Set-builder Notation
{variable | condition }
pipe
{ x | x  5}
The set of all x such that x is greater than or equal to 5.
x < 2
( –∞, 2)
{ x | }
x < 2
x ≥ 2
[2, ∞)
{ x | x ≥ 2}
4 < x < 9
(4, 9)
{ x | 4 < x < 9}

110. Inequalities x ≥ 5 x < –3
Graph, then write interval notation and set-builder notation.
x ≥ 5 [
Interval Notation:
[ 5, ∞)
Set-builder Notation:
{ x | x ≥ 5}
x < –3 )
Interval Notation:
(– ∞, –3)
Set-builder Notation:
{ x | x < –3 }

111. Inequalities 1 < a < 6 –7 < x ≤ 3
Graph, then write interval notation and set-builder notation.
1 < a < 6 ( )
Interval Notation:
( 1, 6 )
Set-builder Notation:
{ a | 1 < a < 6 }
–7 < x ≤ 3 ( ]
Interval Notation:
(– 7, –3]
Set-builder Notation:
{ x | –7 < x ≤ 3 }

112. Solving Inequalities
Solve then graph the solution and write it in interval notation and set-builder notation.
Don’t write = ! (
Interval Notation:
( 1, ∞ )
Set-builder Notation:
{ x | x > 1 }

113. Solving Inequalities
Solve then graph the solution and write it in interval notation and set-builder notation. ]
(– ∞, –3 ]
Interval Notation:
Set-builder Notation:
{ k | k ≤ –3 }

114. Solving Inequalities
Solve then graph the solution and write it in interval notation and set-builder notation. )
Interval Notation:
(– ∞, 6 )
Set-builder Notation:
{ p | p < 6 }

115. Solving Inequalities [ Interval Notation: [– 3, ∞ )
Solve then graph the solution and write it in interval notation and set-builder notation. [
Interval Notation:
[– 3, ∞ )
Set-builder Notation:
{ m | m ≥ – 3 }

116. Solving Absolute Value Equations & Inequalities

117. Absolute Value (of x) Symbol lxl
The distance x is from 0 on the number line.
Always positive
Ex: l-3l=3

118. Ex: x = 5
What are the possible values of x?
x = or x = -5

119. To solve an absolute value equation:
ax+b = c, where c>0
To solve, set up 2 new equations, then solve each equation.
ax+b = c or ax+b = -c
** make sure the absolute value is by itself before you split to solve.

120. * Plug in answers to check your solutions!
Ex: Solve 6x-3 = 15
6x-3 = or 6x-3 = -15
6x = or 6x = -12
x = 3 or x = -2
* Plug in answers to check your solutions!

121. Get the abs. value part by itself first!
Ex: Solve 2x = 8
Get the abs. value part by itself first!
2x+7 = 11
Now split into 2 parts.
2x+7 = 11 or 2x+7 = -11
2x = 4 or 2x = -18
x = 2 or x = -9
Check the solutions.

122. Solving Absolute Value Inequalities
ax+b < c, where c>0
Becomes an “and” problem
Changes to: –c<ax+b<c
ax+b > c, where c>0
Becomes an “or” problem
Changes to: ax+b>c or ax+b<-c

123. Ex: Solve & graph.
Becomes an “and” problem

124. Solve & graph. Get absolute value by itself first.
Becomes an “or” problem

125. Competition Problems

126. How many integral values of x satisfy the inequality:
x – 4 < 5x + 3 < x + 30

127. Answer:
{-1,0,1,2,3,4,5,6} 8

128. Solve the inequality: 5(k – 4) – 2(k + 6) ≥ 4(k + 1) – 1

129. Answer: ≥ k

130. Solve for x: 8 – 3x > x + 20

131. Answer: -3 > x

132. Solve: |3x + 4| – 13 < 2x – 7

133. Answer: -2 < x < 2

134. Solve for x: –7x +18 ≤ 9 + 5x

135. Answer: x ≥ 3/4

136. Solve the compound inequality: 5x – 7 ≤ 4x – 5 and –3x +15 ≤ 9

137. Answer: x = 2

138. Find the sum of the solution(s): |2x – 1| = 3x + 2

139. Answer: -1/5 (Check your answers! The solution x=-3 does not work!)

140. Solve:

141. Answer:
Means union (or)