Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 7 - Section 7.4 - Adding and Subtracting Rational Expressions with Different Denominators - Exercise Set - Page 516: 47


Answer: $\frac{2y^{2}-5y+3}{(2y+3)^{2}}$ = $\frac{2y^{2}-5y+3}{4y^{2}+12y+ 9}$ (Both answers are the same, only the second one has its denominator expanded).

Work Step by Step

Given: $\frac{y-1}{2y+3}$ + $\frac{3}{(2y+3)^{2}}$ Evaluating the sum and simplifying: $\frac{y-1}{2y+3}$ + $\frac{3}{(2y+3)^{2}}$ = $\frac{(y-1)(2y-3) + 3}{(2y+3)^{2}}$ = $\frac{y(2y)-3(y)-1(2y)-1(-3)}{(2y+3)^{2}}$ = $\frac{2y^{2}-3y-2y+3}{(2y+3)^{2}}$ = $\frac{2y^{2}-5y+3}{(2y+3)^{2}}$ = $\frac{2y^{2}-5y+3}{(2y)^{2}+2(2y)(3)+ 3^{2}}$ = $\frac{2y^{2}-5y+3}{4y^{2}+12y+ 9}$
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