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

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 + 8.7 = –12.4 = 21.1 – t 4 9 1 3 r = 6 8 9 2. r + 4 = 11 3. 3.2x = 14.4 x = 4.5 p 8.6 4. = 5.4 p = 46.44

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

Learn to solve inequalities with rational numbers.

The minimum size for a piece of first-class mail is 5 inches long, 3 inches wide, and 0.007 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

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

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 – 4.2 + 4.2 < 15 + 4.2 m < 19.2

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

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!

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

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 .

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

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

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

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.

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.

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

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

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

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 12.25 in.

Try This: Example 3 Continued Look Back 6.37  4.9 = 1.3 and 12.25  4.9 = 2.5, so there will not be an extra charge.

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 3. 1.2x > 17.04 x > 14.2 h 0.7 4. < 5.5 h < 3.85

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