Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.5 Rational Expressions - R.5 Exercises - Page 54: 84

Answer

$\frac{x^2-x-8}{x^2+x-2}$

Work Step by Step

Step 1. Perform operations to the numerator: $\frac{x+4}{x}-\frac{3}{x-2}=\frac{x^2-2x+4x-8-3x}{x(x-2)}=\frac{x^2-x-8}{x(x-2)}$ Step 2. Perform operations to the denominator: $\frac{x}{x-2}+\frac{1}{x}=\frac{x^2+x-2}{x(x-2)}$ Step 3. Use the above results in the original expression, we have: $\frac{\frac{x+4}{x}-\frac{3}{x-2}}{\frac{x}{x-2}+\frac{1}{x}}=\frac{\frac{x^2-x-8}{x(x-2)}}{\frac{x^2+x-2}{x(x-2)}}=\frac{x^2-x-8}{x(x-2)}\cdot\frac{x(x-2)}{x^2+x-2}=\frac{x^2-x-8}{x^2+x-2}$
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.