Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 4 - Test - Page 325: 32

Answer

x = 1

Work Step by Step

$\frac{2}{3}$ + $\frac{x}{4}$ = $\frac{5}{12}$ + $\frac{x}{2}$ Multiply both sides by the LCD (least common denominator), which is 12: 12($\frac{2}{3}$ + $\frac{x}{4}$) = 12($\frac{5}{12}$ + $\frac{x}{2}$) 12($\frac{2}{3}$) + 12($\frac{x}{4}$) = 12($\frac{5}{12}$) + 12($\frac{x}{2}$) Now, cancel out the denominators: 12($\frac{2}{3}$) + 12($\frac{x}{4}$) = 12($\frac{5}{12}$) + 12($\frac{x}{2}$) --> 4(2) + 3x = 5 + 6x --> 8 + 3x = 5 + 6x Now, combine like terms: 8 + 3x - 8 = 5 + 6x - 8 --> 3x = -3 + 6x 3x - 6x = -3 + 6x -6x -3x = -3 Divide both sides by -3 to get x by itself: -3x $\div$ -3 = -3 $\div$ -3 x = 1
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