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: 65


$\displaystyle \frac{x}{x+3},\qquad x\neq-3,-2$

Work Step by Step

Complex rational expressions have rational expressions in the numerator or/and in the denominator. Here, the numerator contains $\displaystyle \frac{x}{x+3}$ (exclusion from the domain: $x\neq-3.\ )$ To get rid of $\displaystyle \frac{x}{x+3}$, we multiply both the numerator and denominator with $(x+3).$ $\displaystyle \frac{x+3}{x+3}\times\frac{x-\frac{x}{x+3}}{x+2}=\frac{x(x+3)-x}{(x+3)(x+2)},\qquad x\neq-3,-2$ $=\displaystyle \frac{x^{2}+3x-x}{(x+3)(x+2)}$ $=\displaystyle \frac{x^{2}+2x}{(x+3)(x+2)}$ $=\displaystyle \frac{x(x+2)}{(x+3)(x+2)}$ The expression has a common factor. Reduce. $=\displaystyle \frac{x}{x+3},\qquad x\neq-3,-2$
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.