College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Section P.7 - Rational Expressions - P.7 Exercises: 63

Answer

$\frac{2}{(x-1)(x+3)}$

Work Step by Step

We multiply by $(x-1)(x+3)$, distribute and simplify: $\displaystyle \frac{\frac{1}{x-1}+\frac{1}{x+3}}{x+1}=\frac{(x-1)(x+3)(\frac{1}{x-1}+\frac{1}{x+3})}{(x-1)(x+3)(x+1)}=\frac{(x+3)+(x-1)}{(x-1)(x+1)(x+3)}=\frac{2(x+1)}{(x-1)(x+1)(x+3)}=\frac{2}{(x-1)(x+3)}$
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