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: 121

Answer

$\frac{3x+8}{3x+10}; x \ne -3, -\frac{10}{3};$

Work Step by Step

$\frac{3-\frac{1}{x+3}}{3+\frac{1}{x+3}}$ Take LCD in the numerator and denominator. $= \frac{\frac{3(x+3)-1}{x+3}}{\frac{3(x+3)+1}{x+3}}$ $= \frac{\frac{3x+9-1}{x+3}}{\frac{3x+9+1}{x+3}}$ $= \frac{\frac{3x+8}{x+3}}{\frac{3x+10}{x+3}}; x \ne -3, -\frac{10}{3};$ $= \frac{3x+8}{x+3} \times \frac{x+3}{3x+10}; x \ne -3, -\frac{10}{3};$ $=\frac{3x+8}{3x+10}; x \ne -3, -\frac{10}{3};$
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