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: 33

Answer

$\displaystyle \frac{11y+15}{4y(y+5)}$

Work Step by Step

Step 1: Find the LCD. List of factors of the first denominator: $\qquad (y+5)$ List of factors of the second denominator: $\qquad 2,2,y$ Build the LCD: - write all factors of the 1st denominator:$\qquad $ List$= (y+5),...\quad$ (for now) - add to the list factors of the second denominator that are not already on the list ($2,2,$ and $y$ are added to the list) List = $(y+5),2,2,y$ $LCD=4y(y+5)$ Step 2. Rewrite each expression with the LCD: $=\displaystyle \frac{2}{y+5}\cdot\frac{4y}{4y}+\frac{3}{4y}\cdot\frac{(y+5)}{(y+5)}= \frac{2(4y)}{4y(y+5)} +\frac{3(y+5)}{4y(y+5)}=...$ Step 3. Combine numerators over the LCD $=$ $\displaystyle \frac{2(4y)+3(y+5)}{4y(y+5)}$ Step 4. Simplify, if possible. $= \displaystyle \frac{8y+3y+15}{4y(y+5)}$ = $\displaystyle \frac{11y+15}{4y(y+5)}$
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