118 Chapter 2 Linear Equations and Inequalities in One Variable So, the solution set is 5-2, 36. We can check by replacing x with -2 and 3 in the original equation. 1e. We must first isolate the absolute value expression. ` 4x + 6 5 ` + 1 = 1 ` 4x + 6 5 ` + 1 - 1 = 1 - 1 Subtract 1 from each side. ` 4x + 6 5 ` = 0 Simplify. 4x + 6 5 = 0 Apply property 1. 5a4x + 6 5 b = 5(0) Multiply each side by 5. 4x + 6 = 0 Simplify. 4x + 6 - 6 = 0 - 6 Subtract 6 from each side. 4x=-6 Simplify. 4x = -6 4 4 Divide each side by 4. x=- 3 2 Simplify. So, the solution set is e- 3 2 f . We can check by replacing x with 0 in the original equation. Student Check 1 Solve each equation. a. uxu = 7 b. uau =-10 c. uy + 1u = 6 d. u3x + 2u - 4 = 8 e. ` 5x - 9 4 ` + 3 = 3 Absolute Value Equations, |X| = |Y| To solve equations containing two absolute values equal to one another, we need to understand when the absolute value of two expressions is equal. This occurs when the expressions have the same distance from zero. So, the expressions must either be the same or opposite. The following are examples of absolute value expressions that are equal. u5u = u5u u-5u = u-5u The expressions inside the absolute values are the same. u5u = u-5u The expressions inside the absolute values are opposites. We will use the following property to solve equations containing two absolute values. Property: Property 2 for Absolute Value Equations If uXu = uYu, then X = Y or X=-Y. Objective 2 ▶ Solve absolute value equations of the form |X| = |Y|.
hendricks_intermediate_algebra_1e_ch1_3
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