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 - Page 354: 67



Work Step by Step

Factoring the given expression, $ \dfrac{3}{x+3}+\dfrac{5}{x^2+6x+9}-\dfrac{x}{x^2-9} ,$ results to \begin{array}{l}\require{cancel} \dfrac{3}{x+3}+\dfrac{5}{(x+3)(x+3)}-\dfrac{x}{(x+3)(x-3)} .\end{array} Using the $LCD= (x+3)(x+3)(x-3) $, the expression above simplifies to \begin{array}{l} \dfrac{(x+3)(x-3)(3)+(x-3)(5)-(x+3)(x)}{(x+3)(x+3)(x-3)} \\\\= \dfrac{(x^2-9)(3)+5x-15-x^2-3x}{(x+3)(x+3)(x-3)} \\\\= \dfrac{3x^2-27+5x-15-x^2-3x}{(x+3)(x+3)(x-3)} \\\\= \dfrac{(3x^2-x^2)+(5x-3x)+(-27-15)}{(x+3)(x+3)(x-3)} \\\\= \dfrac{2x^2+2x-42}{(x+3)(x+3)(x-3)} \\\\= \dfrac{2(x^2+x-21)}{(x+3)^2(x-3)} .\end{array}
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