Section 2.8 Linear Inequalities 187 Solving a Linear Inequality Solve the inequality and graph the solution set. Express the solution set in setbuilder notation and in interval notation. Solution: Addition property of inequality (add 3p to both sides). Simplify. Subtraction property of inequality. 2p 5 6 3p 6 2p 3p 5 6 3p 3p 6 p 5 6 6 p 5 5 6 6 5 Graph: p 6 1 5p 0 p 6 16 Set-builder notation: Interval notation: 1, 12 2p 5 6 3p 6 Example 4 Skill Practice Solve the inequality. Graph the solution set and express in setbuilder notation and interval notation. 10. 2y 5 6 y 11 Classroom Example: p. 194, Exercise 52 Instructor Note: Students should be encouraged to check the solution set by replacing several values less than 1 into the original problem. Instructor Note: Ask students whether 1 is a solution to the inequality. Answer 10. 5y |y 6 66; 1, 62 ( 6 5 4 3 2 1 0 1 2 3 4 5 6 ( 6 TIP: The solution to an inequality gives a set of values that make the original inequality true. Therefore, you can test your final answer by using test points. That is, pick a value in the proposed solution set and verify that it makes the original inequality true. Furthermore, any test point picked outside the solution set should make the original inequality false. For example, p 4 p 3 Pick as an arbitrary test point Pick as an arbitrary test point outside within the proposed solution set. the proposed solution set. 2p 5 6 3p 6 2p 5 6 3p 6 2142 5 6 ? 3142 6 2132 5 6 ? 3132 6 ? 8 5 6 12 6 6 5 6 9 6 ? ? 13 6 18 ✔ True 1 6 3 False 4. Multiplication and Division Properties of Inequality Multiplying both sides of an equation by the same quantity results in an equivalent equation. However, the same is not always true for an inequality. If you multiply or divide an inequality by a negative quantity, the direction of the inequality symbol must be reversed. ( 6 5 4 3 2 1 0 1 2 3 4 5 6
miller_introductory_algebra_3e_ch1_3
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