Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 2 - Section 2.3 - Solving Linear Equations - Exercise Set - Page 142: 75

Answer

x=$\frac{4}{3}$

Work Step by Step

$\frac{2}{3}$x=2-$\frac{5}{6}$x 6($\frac{2}{3}$x)=6(2-$\frac{5}{6}$x) $\frac{12}{3}$x=12-$\frac{6(5)}{6}$x 4x=12-5x 4x+5x=12 9x=12 x=$\frac{12}{9}$ x=$\frac{4}{3}$
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