Prealgebra (7th Edition)

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

Chapter 4 - Review - Page 320: 67

Answer

$\frac{7y}{36}$

Work Step by Step

$9=\underline3\times\underline3$ $12=\underline2\times\underline2\times3$ $lcd=2\times2\times3\times3=36$ $\frac{5y}{12}-\frac{2y}{9}=\frac{5y\times3}{12\times3}-\frac{2y\times4}{9\times4}=\frac{15y}{36}-\frac{8y}{36}=\frac{15y-8y}{36}=\frac{7y}{36}$
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