Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 3 - Equations and Problem Solving - Chapter 3 Test - Page 139: 13

Answer

$y=\dfrac{9x+47}{4}$

Work Step by Step

We first cross multiply. Then, using the properties of equality, in terms of $ y ,$ the given equation, $ \dfrac{x+3}{4}=\dfrac{y-5}{9} ,$ is equivalent to \begin{array}{l}\require{cancel} 9(x+3)=4(y-5) \\\\ 9x+27=4y-20 \\\\ 9x+27+20=4y \\\\ 9x+47=4y \\\\ \dfrac{9x+47}{4}=y \\\\ y=\dfrac{9x+47}{4} .\end{array}
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