Review/Finish Chapter 10-4

Review/Finish Chapter 10-4 Page 517 #14-26 Solve and Graph- Due today!

Student Learning Goal Chart Lesson Reflections 3-7

Pre-Algebra Learning GoalStudents will understand rational and real numbers.

Students will understand rational and real numbers by being able to do the following: • Learn to write rational numbers in equivalent forms (3.1) • Learn to add and subtract decimals and rational numbers with like denominators (3.2) • Learn to add and subtract fractions with unlike denominators (3.5) • Learn to multiply fractions, decimals, and mixed numbers (3.3) • Learn to divide fractions and decimals (3.4) • Learn to solve equations with rational numbers (3.6) • Learn to solve inequalities with rational numbers (3-7)

Today’s Learning Goal Assignment Learn to solve inequalities with rational numbers.

Pre-Algebra HW Page 142 #14-26 and Page 143 #47-56 Spiral Review

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

3-7 Solving Inequalities with Rational Numbers t = 21.1 – 8 9 r = 6 x = 4.5 p = 46.44 Pre-Algebra Warm Up Solve. 1. t+ 8.7 = –12.4 4 9 1 3 2.r+ 11 4 = 3. 3.2x = 14.4 p 8.6 4. =5.4

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

1 2 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.

0.4 0.4 0.8 0.4 x  Additional Examples 1: Solving Inequalities with Decimals Solve. A. 0.4x 0.8 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

0.6 0.6 1.8 0.6 y  Try This: Examples 1 Solve. A. 0.6y  1.8 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

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

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

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

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

1 2 1 8 1 4 • 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 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:

Understand the Problem length width   2.5 1.3 Additional Example 3 Continued 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:

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

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

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

1 2 1 8 1 4 • 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 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:

Understand the Problem length width   2.5 1.3 Try This: Example 3 Continued 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:

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

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

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

5 8 f ≥ 5 Lesson Quiz: Part 1 Solve. w ≤ 3.87 1.w + 1.25  5.12 1 4 3 8 2.f – 1 4 x > 14.2 3. 1.2x > 17.04 h 0.7 < 5.5 h < 3.85 4.

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

Review/Finish Chapter 10-4