1. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Warm Up
Problem of the Day
Lesson Presentation
Course 3

2. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Warm Up
Solve.
1. x + 6 = 13
2. 8n = 48
3. t  2 = 56
4. 6 =
x = 7
n = 6
t = 58 z 6
z = 36

3. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Problem of the Day
Bill and Brad are taking drivers education. Bill drives with his instructor for one and a half hours three times a week. He needs a total of 27 hours. Brad drives two times a week, two hours each time. He needs 26 hours. Who will finish his hours first?
Bill

4. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Learn to solve and graph inequalities.

5. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Vocabulary
inequality
algebraic inequality
solution of an inequality
solution set

6. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
An inequality compares two quantities and typically uses one of these symbols: 
<
is less than
is greater than  
is less than or equal to
is greater than or equal to

7. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Additional Example 1: Completing an Inequality
Compare. Write < or >.
A. 23 –
>
B. 5(12)
<

8. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Try This: Example 1
Compare. Write < or >.
A. 19 –
<
B. 4(15)
>

9. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
An inequality that contains a variable is an algebraic inequality.
A number that makes an inequality true is a solution of the inequality.
The set of all solutions is called the solution set. The solution set can be shown by graphing it on a number line.

10. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Word Phrase
Inequality
Sample Solutions
Solution Set
x is less than 5
x < 5
x = 4
4 < 5
x = 2.1
2.1 < 5

11. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Word Phrase
Inequality
Sample Solutions
Solution Set
a is greater than 0
a is more than 0
a > 0
a = 7
7 > 0
a = 25
25 > 0
–3 –2 –

12. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Word Phrase
Inequality
Sample Solutions
Solution Set
y is less than or equal to 2
y is at most 2
y  2
y = 0
0  2
y = 1.5
1.5  2
–3 –2 –

13. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Word Phrase
Inequality
Sample Solutions
Solution Set
m is greater than or equal to 3
m is at least 3
m  3
m = 17
17  3
m = 3
3  3 –

14. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Most inequalities can be solved the same way equations are solved.
Use inverse operations on both sides of the inequality to isolate the variable.
There are special rules when multiplying or dividing by a negative number, which you will learn in the next chapter.

15. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Additional Example 2A: Solving and Graphing Inequalities
Solve and graph the inequality.
A. x  8
–2.5
–2.5
Subtract 2.5 from both sides.
x  5.5
According to the graph, 5.4 is a solution, since 5.4 < 5.5, and 6 should not be solution because 6 > 5.5.

16. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Additional Example 2B: Solving and Graphing Inequalities
Solve and graph the inequality.
B. 5t > 15
5t > 15
Divide both sides by 5. 5 5
t > 3

17. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Additional Example 2C: Solving and Graphing Inequalities
Solve and graph the inequality.
C. w – 1 < 8
Add 1 to both sides.
w < 9 –

18. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Additional Example 2D: Solving and Graphing Inequalities
Solve and graph the inequality. p 4
D. 3 
3  p 4
4 •
4 •
Multiply both sides by 4.
12  p

19. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Try This: Examples 2A and 2B
Solve and graph each inequality.
A. x + 2  3.5 –2 –2
Subtract 2 from both sides.
x  1.5
B. 6u > 72
6u > 72
Divide both sides by 6. 6 6
u > 12

20. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Try This: Examples 2C and 2D
Solve and graph each inequality.
C. z – 6 < 15
Add 6 to both sides.
z < 21
–21 –14 –
D. 2  b 9
2  b 9
9 •
9 •
Multiply both sides by 9.
18  b

21. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Additional Example 3: Problem Solving Application
An interior designer is planning to place a wallpaper border along the edges of all four walls of a room. The total distance around the room is 88 feet. The border comes in packages of 16 feet. What is the least number of packages that must be purchased to be sure that there is enough border to complete the room?

22. Understand the Problem
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Solving Simple Inequalities
Course 3
Additional Example 3 Continued 1
Understand the Problem
The answer will be the least number of packages of border needed to wallpaper a room.
List the important information:
The total distance around the room is 88 feet.
The border comes in packages of 16 feet.
Show the relationship of the information:
the number of packages of border
the length of one package of border
Distance around the room • 

23. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Additional Example 3 Continued 2
Make a Plan
Use the relationship to write an inequality. Let x represent the number of packages of border. x •
16 ft 
88 ft

24. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Additional Example 3 Continued
Solve 3
16x  88
16x  88
Divide both sides by 16. 16 16
x  5.5
At least 5.5 packages of border must be used to complete the room.

25. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Additional Example 3 Continued 4
Look Back
Because whole packages of border must be purchased, at least 6 packages of border must be purchased to ensure that there is enough to complete the room.

26. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Try This: Example 3
Ron will provide 130 cookies for the school fundraiser. He has to buy the cookies in packages of 20. What is the least number of packages Ron must buy to be sure to have enough cookies?

27. Understand the Problem
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Solving Simple Inequalities
Course 3
Try This: Example 3 Continued 1
Understand the Problem
The answer will be the number of packages of cookies a customer needs to purchase.
List the important information:
Cookies are sold in packages of 20 cookies.
A customer needs to purchase 130 cookies.
Show the relationship of the information:
the number of packages of cookies to be purchased
the number of cookies in one package
130 cookies • 

28. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Try This: Example 3 Continued 2
Make a Plan
Use the relationship to write an inequality. Let x represent the number of packages of cookies.  x •
20 cookies
130 cookies

29. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Try This: Example 3 Continued
Solve 3
20x  130
20x  130
Divide both sides by 20. 20 20
x  6.5
At least 6.5 packages of cookies need to be purchased.

30. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Try This: Example 3 Continued 4
Look Back
Because whole packages of cookies must be purchased, at least 7 packages of cookies must be purchased for the party.

31. Solving Simple Inequalities
1-5
Solving Simple Inequalities
Course 3
Lesson Quiz
Use < or > to compare each inequality.
(2) –
Solve and graph each inequality.
3. k + 9 < 12
4. 3 
5. A school bus can hold 64 passengers. Three classes would like to use the bus for a field trip. Each class has 21 students. Write and solve an inequality to determine whether all three classes will fit on the bus.
>
>
k< 3
–5 –4–3–2– m 2
6  m
–4 –3–2–
3(21)  64; 63  64; yes ?