Answer
$\frac{x+12}{x(x+3)} ; x \ne 0,-3$
Work Step by Step
$\frac{4}{x} - \frac{3}{(x+3)}$
$= \frac{4(x+3)-3x}{x(x+3)};x \ne 0,-3$
$= \frac{4x+12-3x}{x(x+3)};x \ne 0,-3$
$= \frac{x+12}{x(x+3)};x \ne 0,-3;$
College Algebra (6th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-32178-228-3
ISBN 13:
978-0-32178-228-1
Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6 - Page 86: 44
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