Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 7 - Section 7.2 - Multiplying and Dividing Rational Expressions - Exercise Set - Page 499: 27

Answer

$\displaystyle \frac{10-y}{y-7}$

Work Step by Step

Step by step multiplication of rational expressions: 1. Factor completely what you can 2. Reduce (divide) numerators and denominators by common factors. 3. Multiply the remaining factors in the numerators and multiply the remaining factors in the denominators. $(\displaystyle \frac{P}{Q}\cdot\frac{R}{S}=\frac{PR}{QS})$ --- Factor what we can: ... difference of squares, factor out $(-1)$ first $25-y^{2}=-(y^{2}-5^{5})=-(y+5)(y-5)$ ... factor $y^{2}+by+c$ by searching for two factors of $c$ whose sum is $b$. $y^{2}-8y-20=(y-10)(y+2)$ $y^{2}-2y-35=(y-7)(y+5)$ $y^{2}-3y-10=(y-5)(y+2)$ The problem becomes $...=\displaystyle \frac{-(y+5)(y-5)\cdot(y-10)(y+2)}{(y-7)(y+5)\cdot(y-5)(y+2)}\qquad$ ... divide out the common factors $=\displaystyle \frac{-\fbox{$(y+5)$}\fbox{$(y-5)$}\cdot(y-10)\fbox{$(y+2)$}}{(y-7)\fbox{$(y+5)$}\cdot\fbox{$(y-5)$}\fbox{$(y+2)$}}\qquad$ $-\displaystyle \frac{-(y-10)}{y-7}$ = $\displaystyle \frac{10-y}{y-7}$
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