2. Copyright © 2009 Pearson Education, Inc. CHAPTER 1: Graphs, Functions, and Models 1.1 Introduction to Graphing 1.2 Functions and Graphs 1.3 Linear Functions, Slope, and Applications 1.4 Equations of Lines and Modeling 1.5 Linear Equations, Functions, Zeros and Applications 1.6 Solving Linear Inequalities

3. Copyright © 2009 Pearson Education, Inc. 1.6 Solving Linear Inequalities  Solve linear inequalities.  Solve compound inequalities.  Solve applied problems using inequalities.

4. Slide 1.6 - 4 Copyright © 2009 Pearson Education, Inc. Inequalities An inequality is a sentence with, , or  as its verb. Example: 3x  5 < 6 – 2x To solve an inequality is to find all values of the variable that make the inequality true. Each of these numbers is a solution of the inequality, and the set of all such solutions is its solution set. Inequalities that have the same solution set are called equivalent inequalities.

5. Slide 1.6 - 5 Copyright © 2009 Pearson Education, Inc. Principles for Solving Inequalities For any real numbers a, b, and c: The Addition Principle for Inequalities: If a < b is true, then a + c < b + c is true. The Multiplication Principle for Inequalities: If a 0 are true, then ac < bc is true. If a bc is true. Similar statements hold for a  b. When both sides of an inequality are multiplied or divided by a negative number, we must reverse the inequality sign.

6. Slide 1.6 - 6 Copyright © 2009 Pearson Education, Inc. Example Solve each of the following. Then graph the solution set. a. 3x – 5 < 6 – 2xb. 13 – 7x ≥ 10x – 4 Solution:

7. Slide 1.6 - 7 Copyright © 2009 Pearson Education, Inc. Compound Inequalities When two inequalities are joined by the word and or the word or, a compound inequality is formed. Conjunction contains the word and. Example:  3 < 2x + 5 and 2x + 5  7 The sentence –3 < 2x + 5 ≤ 7 is an abbreviation. Disjunction contains the word or. Example: 2x – 5  –7 or 2x – 5 > 1

8. Slide 1.6 - 8 Copyright © 2009 Pearson Education, Inc. Example Solve each of the following. Then graph the solution set. a. – 3 1 Solution a:

9. Slide 1.6 - 9 Copyright © 2009 Pearson Education, Inc. Example (continued) Solve each of the following. Then graph the solution set. a. – 3 1 Solution b:

10. Slide 1.6 - 10 Copyright © 2009 Pearson Education, Inc. Application – Example (optional) For her house painting job, Erica can be paid in one of two ways: Plan A: $250 plus $10 per hour Plan B: $20 per hour Suppose that a job takes n hours. For what values of n is plan B better for Erica?

11. Slide 1.6 - 11 Copyright © 2009 Pearson Education, Inc. Example (continued) (optional) Solution: 1. Familiarize. Read the problem. For a 30 hour job, n = 30: Plan A: $250 + $10 30 = $550 Plan B: $20 30 = $600 Plan B is better for a 30 hour job. For a 20 hour job, n = 20: Plan A: $250 + $10 20 = $450 Plan B: $20 20 = $400 Plan A is better for a 20 hour job.

12. Slide 1.6 - 12 Copyright © 2009 Pearson Education, Inc. Example (continued) (optional) 2. Translate. Translate to an inequality. Income plan B is greater than Income plan A 20n > 250 + 10n 3. Carry out. Solve the inequality.

13. Slide 1.6 - 13 Copyright © 2009 Pearson Education, Inc. Application continued (optional) 4. Check. For n = 25, the income from plan A is $250 + $10 25, or $500, and the income from plan B is $20 25, or $500, the same under either plan. We have seen that plan B pays more for a 30-hr job. Since 30 > 25, this provides a partial check. We cannot check all values of n. 5. State. For values of n greater than 25 hr, plan B is better for Erica.