Intermediate Algebra (6th Edition)

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

Chapter 6 - Review - Page 404: 44

Answer

$\dfrac{x^2+9}{-6x}$

Work Step by Step

The given expression, $ \dfrac{\dfrac{x-3}{x+3}+\dfrac{x+3}{x-3}}{\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{\dfrac{(x-3)^2+(x+3)^2}{(x+3)(x-3)}}{\dfrac{(x-3)^2-(x+3)^2}{(x+3)(x-3)}} \\\\= \dfrac{\dfrac{(x-3)^2+(x+3)^2}{\cancel{(x+3)(x-3)}}}{\dfrac{(x-3)^2-(x+3)^2}{\cancel{(x+3)(x-3)}}} \\\\= \dfrac{(x-3)^2+(x+3)^2}{(x-3)^2-(x+3)^2} \\\\= \dfrac{(x^2-6x+9)+(x^2+6x+9)}{(x^2-6x+9)-(x^2+6x+9)} \\\\= \dfrac{x^2-6x+9+x^2+6x+9}{x^2-6x+9-x^2-6x-9} \\\\= \dfrac{(x^2+x^2)+(-6x+6x)+(9+9)}{(x^2-x^2)+(-6x-6x)+(9-9} \\\\= \dfrac{2x^2+18}{-12x} \\\\= \dfrac{2(x^2+9)}{-12x} \\\\= \dfrac{\cancel{2}(x^2+9)}{\cancel{2}\cdot(-6)x} \\\\= \dfrac{x^2+9}{-6x} .\end{array}
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