Intermediate Algebra (6th Edition)

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

Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set: 49

Answer

$\dfrac{5a+1}{(a+1)(a+1)(a-1)}$

Work Step by Step

Factoring the given expression, $ \dfrac{2}{a^2+2a+1}+\dfrac{3}{a^2-1} ,$ results to \begin{array}{l}\require{cancel} \dfrac{2}{(a+1)(a+1)}+\dfrac{3}{(a+1)(a-1)} .\end{array} Using the $LCD= (a+1)(a+1)(a-1) ,$ the expression, $ \dfrac{2}{(a+1)(a+1)}+\dfrac{3}{(a+1)(a-1)} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{(a-1)(2)+(a+1)(3)}{(a+1)(a+1)(a-1)} \\\\= \dfrac{2a-2+3a+3}{(a+1)(a+1)(a-1)} \\\\= \dfrac{5a+1}{(a+1)(a+1)(a-1)} .\end{array}
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