Prealgebra (7th Edition)

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

Chapter 7 - Cumulative Review - Page 529: 21



Work Step by Step

$\frac{x}{2}$ =$\frac{x}{3}$+$\frac{1}{2}$ subtract $\frac{x}{3}$ from both sides $\frac{x}{2}$ -$\frac{x}{3}$ =$\frac{x}{3}$+$\frac{1}{2}$ -$\frac{x}{3}$ $\frac{x}{2}$ -$\frac{x}{3}$=$\frac{1}{2}$ LCD is 6 $\frac{x}{2}$ $\times$$\frac{3}{3}$ - $\frac{x}{3}$$\times$$\frac{2}{2}$=$\frac{1}{2}$$\times$$\frac{3}{3}$ $\frac{3x}{6}$ -$\frac{2x}{6}$ =$\frac{3}{6}$ $\frac{3x-2x}{6}$ =$\frac{3}{6}$ $\frac{x}{6}$=$\frac{3}{6}$ x=3 The solution is 3 check $\frac{x}{2}$ =$\frac{x}{3}$+$\frac{1}{2}$ $\frac{3}{2}$ =$\frac{3}{3}$+$\frac{1}{2}$ 1$\frac{1}{2}$=1+$\frac{1}{2}$ 1$\frac{1}{2}$=1$\frac{1}{2}$
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