Intermediate Algebra (6th Edition)

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

Chapter 6 - Section 6.3 - Simplifying Complex Fractions - Exercise Set - Page 361: 29

Answer

$\dfrac{x-3}{9}$

Work Step by Step

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