Presentation is loading. Please wait.

Presentation is loading. Please wait.

Section 1.6 Solving Linear Inequalities Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.

Similar presentations


Presentation on theme: "Section 1.6 Solving Linear Inequalities Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc."— Presentation transcript:

1 Section 1.6 Solving Linear Inequalities Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.

2 Objectives  Solve linear inequalities.  Solve compound inequalities.  Solve applied problems using inequalities.

3 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.

4 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.

5 Example Solve each of the following. Then graph the solution set. a. 3x – 5 < 6 – 2x b. 13 – 7x ≥ 10x – 4

6 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

7 Example Solve – 3 < 2x + 5 ≤ 7. Then graph the solution set.

8 Example Solve: 2x – 5 ≤ –7 or 2x – 5 > 1. Then graph the solution set.

9 Application - Example 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?

10 Example (continued) 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.

11 Example (continued) 2. Translate. Translate to an inequality. Income plan B is greater than Income plan A 20n > 250 + 10n 3. Carry out. Solve the inequality.

12 Application continued 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.


Download ppt "Section 1.6 Solving Linear Inequalities Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc."

Similar presentations


Ads by Google