Algebra: A Combined Approach (4th Edition)

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

Chapter 7 - Test - Page 561: 17

Answer

$\frac{4y^2 + 13y - 15}{(y + 5)(y + 1)(y + 4)}$

Work Step by Step

$\frac{4y}{y^2 + 6y + 5}$ - $\frac{3}{y^2 +5y + 4}$ = $\frac{4y}{(y + 5)(y + 1)}$ - $\frac{3}{(y + 4)(y + 1)}$ = $\frac{4y(y + 4)}{(y + 5)(y + 1)(y + 4)}$ - $\frac{3(y + 5)}{(y + 4)(y + 1)(y + 5)}$ = $\frac{4y(y + 4) - 3(y + 5)}{(y + 5)(y + 1)(y + 4)}$ = $\frac{4y^2 + 16y - 3y - 15}{(y + 5)(y + 1)(y + 4)}$ = $\frac{4y^2 + 13y - 15}{(y + 5)(y + 1)(y + 4)}$
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