Intermediate Algebra (6th Edition)

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

Chapter 6 - Review - Page 404: 40

Answer

$\dfrac{y}{x-y}$

Work Step by Step

The given expression, $ \dfrac{1+\dfrac{x}{y}}{\dfrac{x^2}{y^2}-1} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{\dfrac{y+x}{y}}{\dfrac{x^2-y^2}{y^2}} \\\\= \dfrac{y+x}{y}\div\dfrac{x^2-y^2}{y^2} \\\\= \dfrac{y+x}{y}\cdot\dfrac{y^2}{x^2-y^2} \\\\= \dfrac{x+y}{y}\cdot\dfrac{y^2}{(x+y)(x-y)} \\\\= \dfrac{\cancel{x+y}}{\cancel{y}}\cdot\dfrac{\cancel{y}\cdot y}{(\cancel{x+y})(x-y)} \\\\= \dfrac{y}{x-y} .\end{array}
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