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 141: 35



Work Step by Step

$\frac{2y}{3}$-$\frac{3}{4}$=$\frac{5}{12}$ 12($\frac{2y}{3}$-$\frac{3}{4}$)=12($\frac{5}{12}$) $\frac{24y}{3}$-$\frac{36}{4}$=5 8y-9=5 8y=5+9 8y=14 y=$\frac{14}{8}$ y=$\frac{7}{4}$ Check the answer. $\frac{2(\frac{7}{4})}{3}$-$\frac{3}{4}$=$\frac{5}{12}$ $\frac{\frac{14}{4}}{3}$-$\frac{3}{4}$=$\frac{5}{12}$ $\frac{14}{12}$-$\frac{3(3)}{4(3)}$=$\frac{5}{12}$ $\frac{14}{12}$-$\frac{9}{12}$=$\frac{5}{12}$ $\frac{5}{12}$=$\frac{5}{12}$
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