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: 20

Answer

$ \displaystyle \frac{41}{24x} $

Work Step by Step

Step 1: Find the LCD. List of factors of the first denominator: $\qquad 2,3,x.$ List of factors of the second denominator: $\qquad 2,2,2,x$ Build the LCD: - write all factors of the 1st denominator:$\qquad $List$=2,3,x...\quad$ (for now) - add to the list factors of the second denominator that are not already on the list ( $x$ and one of the 2's are already in, we add the other two 2's to the list) List = $2,3,x,2,2$ $LCD=24x$ Step 2. Rewrite each expression with the LCD: $\displaystyle \frac{5}{6x}\cdot\frac{4}{4}+\frac{7}{8x}\cdot\frac{3}{3}=\frac{20}{24x}+\frac{21}{24x}=...$ Step 3. Combine numerators over the LCD $=\displaystyle \frac{20+21}{24x}$ Step 4. Simplify, if possible. =$ \displaystyle \frac{41}{24x} $
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