College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Review Exercises: 7


$\displaystyle x=\frac{11}{8}$

Work Step by Step

We solve: $\displaystyle \frac{1}{2}(x-\frac{1}{3})=\frac{3}{4}-\frac{x}{6} $ $\displaystyle \frac{12}{2}(x-\frac{1}{3})=(\frac{3}{4}-\frac{x}{6})(12)$ $\displaystyle 6(x-\frac{1}{3})=(\frac{3}{4}-\frac{x}{6})(12)$ $6x-2=9-2x$ $6x+2x=9+2$ $8x=11$ $x=\frac{11}{8}$
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