Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.3 Rational Expressions - R.3 Exercises: 34

Answer

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

Work Step by Step

$\dfrac{y}{y^{2}+2y-3}-\dfrac{1}{y^{2}+4y+3}$ Factor the denominators of both fractions: $\dfrac{y}{y^{2}+2y-3}-\dfrac{1}{y^{2}+4y+3}=...$ $...=\dfrac{y}{(y+3)(y-1)}-\dfrac{1}{(y+3)(y+1)}=...$ Evaluate the subtraction of these two fractions using the LCD, which is $(y+3)(y-1)(y+1)$ in this case: $...=\dfrac{y(y+1)-(y-1)}{(y+3)(y-1)(y+1)}=...$ Simplify: $...=\dfrac{y^{2}+y-y+1}{(y+3)(y-1)(y+1)}=\dfrac{y^{2}+1}{(y+3)(y-1)(y+1)}$
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