Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set - Page 426: 13

Answer

$\dfrac{(4y+3)}{(2y-1)}$

Work Step by Step

Since,in the given rational expressions denominator are same, thus take it as common and subtract the rational terms by changing the sign of each term of the second rational. Given: $\dfrac{20y^2+5y+1}{6y^2+y-2}-\dfrac{8y^2-12y-5}{6y^2+y-2}$ or, $=\dfrac{20y^2+5y+1-8y^2+12y+5}{6y^2+y-2}$ or, $=\dfrac{12y^2+9y+8y+6}{(3y+2)(2y-1)}$ or, $=\dfrac{(3y+2)(4y+3)}{(3y+2)(2y-1)}$ or, $=\dfrac{(4y+3)}{(2y-1)}$
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