1. 3-7 Warm Up Problem of the Day Lesson Presentation
Solving Inequalities with Rational Numbers
3-7
Warm Up
Problem of the Day
Lesson Presentation
Pre-Algebra

2. 3-7 Warm Up Solve. 1. t + 8.7 = –12.4 = 21.1 t r = 6 2. r + 4 = 11
Pre-Algebra
3-7
Solving Inequalities with Rational Numbers
Warm Up
Solve.
1. t = –12.4 = – t
4 9
1 3
r = 6
8 9
2. r + 4 = 11
x = 14.4 x
= 4.5
p 8.6
= 5.4 p
= 46.44

3. Problem of the Day
Arnie built a fence around a rectangular field that measures 120 feet by 90 feet. He put a post in each corner and every 6 feet along all four sides. How many fence posts did he use?
70 posts

4. Learn to solve inequalities with rational numbers.

5. The minimum size for a piece of first-class mail is 5 inches long, 3 inches wide, and inch thick. For a piece of mail, the combined length of the longest side and the distance around the thickest part may not exceed 108 inches. Many inequalities are used in determining postal rates.
1 2

6. Additional Examples 1: Solving Inequalities with Decimals
Solve.
A. 0.4x  0.8 x 
Divide both sides by 0.4.
x  2
B. y – 3.8 < 11
Add 3.8 to both sides of the equation.
y – <
y < 14.8

7. Try This: Examples 1 Solve. A. 0.6y  1.8 1.8 0.6 0.6 0.6 y 
Divide both sides by 0.6.
y  3
B. m – 4.2 < 15
Add 4.2 to both sides of the equation.
m – <
m < 19.2

8. Additional Example 2A: Solving Inequalities with Fractions
Solve.
x + > 2
2 3 A.
Subtract from both sides.
2 3
x – > 2 –
2 3
x > 1
1 3

9. Additional Example 2B: Solving Inequalities with Fractions Continued
Solve.
–2 n  9
1 4 B.
Rewrite –2 as the improper fraction – .
1 4
9 4
– n  9
9 4
– n  9
9 4
4 9 –
Multiply both sides by – .
4 9
n  –4
Change  to .
When multiplying or dividing an inequality by a negative number, reverse the inequality symbol.
Remember!

10. Subtract from both sides. v + – > 7 –
Try This: Example 2A
Solve.
1 5 A.
v > 7
Subtract from both sides.
1 5
v – > 7 –
1 5
v > 6
4 5

11. Try This: Example 2B Solve. –1 j ≤ 7 B. – j ≤ 7 – j ≤ 7 – j ≥ –5 2 5
Rewrite –1 as the improper fraction – .
2 5
7 5
– j ≤ 7
7 5
– j ≤ 7
7 5
5 7 –
Multiply both sides by – .
5 7
j ≥ –5
Change  to .

12. Additional Example 3: Problem Solving Application
The height of an envelope is 3.8 in. What are the minimum and maximum lengths to avoid an extra charge?
With first-class mail, there is an extra charge in any of these cases:
The length is greater than 11 inches
The height is greater than 6 inches
The thickness is greater than inch
The length divided by the height is less than 1.3 or greater than 2.5
1 2
1 8
1 4

13. Additional Example 3 Continued
Understand the Problem
The answer is the minimum and maximum lengths for an envelope to avoid an extra charge.
List the important information:
• The height of the piece of mail is 3.8 inches.
• If the length divided by the height is between 1.3 and 2.5, there will not be an extra charge.
Show the relationship of the information:
1.3
2.5 
length height

14. Additional Example 3 Continued
Make a Plan
You can use the model from the previous slide to write an inequality where l is the length and 3.8 is the height.
1.3
2.5  l
3.8

15. Additional Example 3 Continued
Solve
1.3 ≤ and ≤ 2.5
l 3.8
Multiply both sides of each inequality by 3.8.
3.8 • 1.3 ≤ l and l ≤ 2.5 • 3.8
l  4.94 and l ≤ 9.5
Simplify.
The length of the envelope must be between 4.94 in. and 9.5 in.

16. Additional Example 3 Continued
Look Back
4.94  3.8 = 1.3 and 9.5  3.8 = 2.5, so there will not be an extra charge.

17. Try This: Example 3
The height of an envelope is 4.9 in. What are the minimum and maximum lengths to avoid an extra charge?
With first-class mail, there is an extra charge in any of these cases:
The length is greater than 11 inches
The height is greater than 6 inches
The thickness is greater than inch
The length divided by the height is less than 1.3 or greater than 2.5
1 2
1 8
1 4

18. Try This: Example 3 Continued
Understand the Problem
The answer is the minimum and maximum lengths for an envelope to avoid an extra charge.
List the important information:
• The height of the piece of mail is 4.9 inches.
• If the length divided by the height is between 1.3 and 2.5, there will not be an extra charge.
Show the relationship of the information:
1.3
2.5 
length height

19. Try This: Example 3 Continued
Make a Plan
You can use the model from the previous slide to write an inequality where l is the length and 4.9 is the height.
1.3
2.5  l
4.9

20. Try This: Example 3 Continued
Solve
1.3 ≤ and ≤ 2.5
l 4.9
Multiply both sides of each inequality by 4.9.
4.9 • 1.3 ≤ l and l ≤ 2.5 • 4.9
l  6.37 and l ≤ 12.25
Simplify.
The length of the envelope must be between 6.37 in. and in.

21. Try This: Example 3 Continued
Look Back
6.37  4.9 = 1.3 and  4.9 = 2.5, so there will not be an extra charge.

22. Solve. Lesson Quiz: Part 1 1. w + 1.25  5.12 w ≤ 3.87 3 8 1 4 5 8
f ≥ 5
2. f – 1  4
x > 17.04
x > 14.2
h 0.7 4.
< 5.5
h < 3.85

23. Lesson Quiz: Part 2 5.
Rosa’s car gets between 20 and 21 mi/gal on the highway. She knows that her gas tank holds at least 18 gallons. Using a full tank of gas, what is the minimum distance Rosa could drive her car on the highway between fill-ups?
360 miles