Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.6 - Rational Expressions - Exercise Set - Page 85: 44


$=\displaystyle \frac{x+12}{x(x+3)}, \quad x\neq-3,0$

Work Step by Step

The denominators are different. Find a common denominator. LCD=$x(x+3)$ $\displaystyle \frac{4}{x}-\frac{3}{x+3}=\frac{4(x+3)}{x(x+3)}-\frac{3x}{x(x+3)}$ $=\displaystyle \frac{4x+12-3x}{x(x+3)}$ $=\displaystyle \frac{x+12}{x(x+3)}$ ... exclude values that yield 0 in the denominator: $x\neq-3,0$ ... and, cancel common factors (here, there are none) $=\displaystyle \frac{x+12}{x(x+3)}, \quad x\neq-3,0$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.