Section 2.4 Linear Inequalities and Applications 97 3. Applications are solved by translating the given statement into an appropriate inequality. Key phrases are “is at least” and “is at most.” A good way to remember this is to think of money. If we have at least $10, we would have $10 or more. If we have at most $10, we would have $10 or less. GRAPHING CALCULATOR SKILLS The graphing calculator can be used as a reference to check the work we do by hand. We can do this using the Test menu and the Store feature. Example: x - 5 > 2 Solution: The solution of this inequality is x > 7. To check the work on the calculator, we should determine if the inequality is true for a number larger than 7, is false for a number less than 7, and determine what happens at 7. SECTION 2.4 EXERCISE SET Write About It! Use complete sentences in your answer to each exercise. 1. If a boundary number of an inequality is not included in the solution set of the inequality, how is this represented when graphing the solution set on a number line? 2. If a boundary number of an inequality is included in the solution set of the inequality, how is this represented when graphing the solution set on a number line? 3. Explain how to express an inequality in interval notation. 4. What similarities exist between the solution sets of inequalities on a number line and the solution sets of inequalities represented in interval notation? 5. Should parentheses or brackets be used with ∞ in interval notation? Why? 6. When solving an inequality, what operations cause the direction of the inequality to change? Practice Makes Perfect! Graph the solution set of each inequality on a number line and express the solution set in interval notation and set-builder notation. (See Objective 1.) 7. x > 5 8. x > 7 9. x<-4 10. x<-8 11. x ≥ 10 12. x≥-12 13. x ≤ 1 2 14. x≤- 3 4 Let x = 8 (a value larger than 7). 8 T X,T,u,n ENTER X,T,u,n 2 5 2nd ATH 3 2 ENTER The result of 1 confirms that 8 is a solution of the inequality. The shaded portion of the graph should contain 8 (to the right of 7). Let x = 6 (a value less than 7). 6 T X,T,u,n ENTER X,T,u,n 2 5 2nd ATH 3 ENTER 2 The result of 0 means that 6 is not a solution of the inequality. The shaded portion of the graph should not include the side with 6. Let x = 7. T X,T,u,n ENTER X,T,u,n 2 5 2nd ATH 3 ENTER 2 The result is 0 which means that 7 is not a solution of the inequality. Therefore, we should use a parenthesis on the value 7 on the graph.
hendricks_intermediate_algebra_1e_ch1_3
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