Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 7 - Section 7.4 - Adding and Subtracting Rational Expressions with Different Denominators - Exercise Set - Page 517: 29

Answer

$\displaystyle \frac{7x-20}{x(x-5)}$

Work Step by Step

Step 1: Find the LCD. List of factors of the first denominator: $\qquad x$ List of factors of the second denominator: $\qquad (x-5)$ Build the LCD: - write all factors of the 1st denominator:$\qquad $ List$=x,...\quad$ (for now) - add to the list factors of the second denominator that are not already on the list ($(x-5)$ is added to the list) List = $x,(x-5), $ $LCD=x(x-5)$ Step 2. Rewrite each expression with the LCD: $=\displaystyle \frac{4}{x}\cdot\frac{(x-5)}{(x-5)}+\frac{3}{(x-5)}\cdot\frac{x}{x}= \displaystyle \frac{4(x-5)}{x(x-5)}+\frac{3x}{x(x-5)}=...$ Step 3. Combine numerators over the LCD $= \displaystyle \frac{4(x-5)+3x}{x(x-5)} $ Step 4. Simplify, if possible. $= \displaystyle \frac{4x-20+3x}{x(x-5)} $ = $\displaystyle \frac{7x-20}{x(x-5)}$
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