Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 11 - Rational Expressions and Functions - Mid-Chapter Quiz - Page 678: 20


$\frac{2x^2 + 17x - 45}{(x-5)(x+4)}$

Work Step by Step

To solve: $\frac{2x}{x-5} + \frac{9}{x+4}$, you will first need to make sure both fractions have common denominators. To do this, we must find the LCD, or the Least Common Multiple. Examining both denominators, you can see that they have no common factors between them, therefore the LCD will be $(x-5)(x+4)$. Now we can rewrite the fractions using the new LCD we found. $\frac{2x}{x-5} + \frac{9}{x+4}$ = $\frac{2x \times (x+4)}{(x-5) \times (x+4)} + \frac{9 \times (x-5)}{(x+4) \times (x-5)}$ =$\frac{2x^2 - 8x}{(x-5)(x+4)} + \frac{9x - 45}{(x-5)(x+4)}$ Now that we have common denominators, we can add the numerators: $\frac{2x^2 + 8x}{(x-5)(x+4)} + \frac{9x - 45}{(x-5)(x+4)}$ = $\frac{(2x^2 + 8x) + (9x - 45)}{(x-5)(x+4)}$ = $\frac{2x^2 + 8x + 9x - 45}{(x-5)(x+4)}$ = $\frac{2x^2 + 17x - 45}{(x-5)(x+4)}$ The numerator cannot be factored, therefore, this is our final answer.
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.