College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Summary, Review, and Test - Review Exercises - Page 91: 120

Answer

$\frac{3x}{x-4}; x \ne 0,-4,4;$

Work Step by Step

$\frac{3+\frac{12}{x}}{1-\frac{16}{x^{2}}}$ $= \frac{\frac{3x+12}{x}}{\frac{x^{2}-16}{x^{2}}}; x \ne 0,-4,4;$ $= \frac{\frac{3(x+4)}{x}}{\frac{(x+4)(x-4)}{x^{2}}}; x \ne 0,-4,4;$ $= \frac{3(x+4)}{x} \times \frac{x^{2}}{(x+4)(x-4)}$ $=\frac{3x}{x-4}; x \ne 0,-4,4;$
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