Check: a 6 - 1 8 = a 12 Original equation 3 2 6 - 1 8 = 3 2 12 Replace a with 3 2 . 3 12 - 1 8 = 3 24 Simplify. Recall: 3 2 ÷ 6 = 3 2 · 1 6 = 3 12 . 3 2 ÷ 12 = 3 2 · 1 12 = 3 24 6 24 - 3 24 = 3 24 Write equivalent fractions with LCD, 24. 3 24 = 3 24 Simplify. Since a = 3 2 makes the equation true, the solution set is e 3 2 f . 1c. - 2 3 (x - 6) = 4 5 (x + 10) 15 c- 2 3 (x - 6)d = 15 c 4 5 (x + 10)d Multiply each side by the LCD, 15. c15 a- 2 3 b d (x - 6) = c15 a 4 5 b d (x + 10) Apply the associative property. -10(x - 6) = 12(x + 10) Multiply the coefficients. -10x + 60 = 12x + 120 Apply the distributive property. -10x + 60 - 12x = 12x + 120 - 12x Subtract 12x from each side. -22x + 60 = 120 Simplify. -22x + 60 - 60 = 120 - 60 Subtract 60 from each side. -22x = 60 Simplify. -22x 60 = -22 -22 Divide each side by -22. x=- 30 11 Simplify. Check: - 2 3 (x - 6) = 4 5 (x + 10) Original equation - 2 3 a- 30 11 - 6b = 4 5 a- 30 11 + 10b Replace x with - 30 11 . - 2 3 a- 30 11 b - a- 2 3 b(6) = 4 5 a- 30 11 b + 4 5 (10) Distribute. 20 11 + 4=- 24 11 + 8 Simplify. 20 11 + 44 11 =- 24 11 + 88 11 Add the fractions. 64 11 = 64 11 Simplify. Since x=- 30 11 makes the equation true, the solution set is e- 30 11 f . 126 Chapter 2 Linear Equations and Inequalities in One Variable
hendricks_beginning_algebra_1e_ch1_3
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