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: 68

Answer

$\dfrac{5x+2}{(x-3)(x+1)}$

Work Step by Step

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