Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Review: 17

Answer

$\dfrac{12}{5}$

Work Step by Step

Factoring the expressions and then cancelling the common factor/s between the numerator and the denominator, the given expression, $ \dfrac{4x+8y}{3}\div \dfrac{5x+10y}{9} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{4x+8y}{3}\cdot \dfrac{9}{5x+10y} \\\\= \dfrac{4(x+2y)}{3}\cdot \dfrac{3\cdot3}{5(x+2y)} \\\\= \dfrac{4(\cancel{x+2y})}{\cancel{3}}\cdot \dfrac{3\cdot\cancel{3}}{5(\cancel{x+2y})} \\\\= \dfrac{12}{5} .\end{array}
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