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: 70

Answer

$\dfrac{-6x}{(x+3)(x-3)(x-3)}$

Work Step by Step

The given expression, $ \dfrac{3}{x^2-9}-\dfrac{x}{x^2-6x+9}+\dfrac{1}{x+3} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{3}{(x+3)(x-3)}-\dfrac{x}{(x-3)(x-3)}+\dfrac{1}{x+3} \\\\ \dfrac{(x-3)(3)-(x+3)(x)+(x-3)(x-3)(1)}{(x+3)(x-3)(x-3)} \\\\ \dfrac{3x-9-x^2-3x+x^2-6x+9}{(x+3)(x-3)(x-3)} \\\\ \dfrac{(-x^2+x^2)+(3x-3x-6x)+(-9+9)}{(x+3)(x-3)(x-3)} \\\\ \dfrac{-6x}{(x+3)(x-3)(x-3)} .\end{array}
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