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 - Mid-Chapter Check Point - Page 155: 9

Answer

y=20

Work Step by Step

$\frac{3y}{5}$+$\frac{y}{2}$=$\frac{5y}{4}$-3 20($\frac{3y}{5}$+$\frac{y}{2}$)=20($\frac{5y}{4}$-3) $\frac{60y}{5}$+$\frac{20y}{2}$=$\frac{100y}{4}$-20(3) 12y+10y=25y-60 22y=25y-60 22y-25y=-60 -3y=-60 y=$\frac{-60}{-3}$ y=20
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