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 427: 39



Work Step by Step

$y^2-5y+6=y^2-2y-3y+6=y(y-2)-3(y-3)=(y-3)(y-2)$, so LCD$=(y-3)(y-2)$, thus $\frac{3y+7}{y^2-5y+6}-\frac{3}{y-3}=\frac{3y+7}{(y-3)(y-2)}-\frac{3(y-2)}{(y-3)(y-2)}=\frac{3y+7-3y+6}{(y-3)(y-2)}=\frac{13}{(y-3)(y-2)}$
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