College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.5 - Rational Expressions - R.5 Exercises - Page 47: 34

Answer

$\dfrac{y+3}{y+4}$

Work Step by Step

The given expression, $ \dfrac{y^2+y-2}{y^2+3y-4}\div\dfrac{y^2+3y+2}{y^2+4y+3} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{y^2+y-2}{y^2+3y-4}\cdot\dfrac{y^2+4y+3}{y^2+3y+2} \\\\= \dfrac{(y+2)(y-1)}{(y+4)(y-1)}\cdot\dfrac{(y+3)(y+1)}{(y+2)(y+1)} \\\\= \dfrac{(\cancel{y+2})(\cancel{y-1})}{(y+4)(\cancel{y-1})}\cdot\dfrac{(y+3)(\cancel{y+1})}{(\cancel{y+2})(\cancel{y+1})} \\\\= \dfrac{y+3}{y+4} .\end{array}
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