Section 2.6 Absolute Value Equations 117 a. If the constant is positive, there are two solutions. b. If the constant is zero, there is one solution. c. If the constant is negative, there are no solutions. Step 3: Check the solutions by substituting them in the original equation. Objective 1 Examples Solve each equation. 1a. uxu = 5 1b. uau =-3 1c. uy - 4u = 7 1d. u2x - 1u - 3 = 2 1e. ` 4x + 6 5 ` + 1 = 1 Solutions 1a. We set the variable x equal to the numbers whose absolute value is 5, namely 5 and -5. uxu = 5 x = 5 or x = -5 So, the solution set is 5-5, 56 or 5±56. We can check by replacing x with 5 and -5 in the original equation. 1b. The equation asks us to find all real numbers a whose absolute value, or distance from zero, is -3. There is no real number whose absolute value is a negative number. Therefore, the solution set of this equation is the empty set, or . 1c. We set the expression y - 4 equal to the numbers whose absolute value is 7, namely 7 and -7. uy - 4u = 7 y - 4 = 7 or y - 4 = -7 Apply property 1. y - 4 + 4 = 7 + 4 y - 4 + 4=-7 + 4 Add 4 to each side. y = 11 y=-3 Simplify. Check: y = 11: y=-3: uy - 4u = 7 uy - 4u = 7 u11 - 4u = 7 u-3 - 4u = 7 u7u = 7 u-7u = 7 7 = 7 True 7 = 7 True So, the solution set is 5-3, 116. 1d. We must first isolate the absolute value expression. u2x - 1u - 3 = 2 u2x - 1u - 3 + 3 = 2 + 3 Add 3 to each side. u2x - 1u = 5 Simplify. 2x - 1 = 5 or 2x - 1 = -5 Apply property 1. 2x - 1 + 1 = 5 + 1 2x - 1 + 1=-5 + 1 Add 1 to each side. 2x =6 2x=-4 Simplify. 2x 6 2x = = -4 2 2 2 2 Divide each side by 2. x = 3 x=-2 Simplify.
hendricks_intermediate_algebra_1e_ch1_3
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