College Algebra (6th Edition)

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: 46

Answer

$\frac{2x^{2}+5x+12}{(x-3)(x+2)} ; x \ne 3,-2;$

Work Step by Step

$\frac{3x}{(x-3)} - \frac{x+4}{(x+2)}$ $ = \frac{3x(x+2)-(x-3)(x+4)}{(x-3)(x+2)}; x \ne 3,-2;$ $ = \frac{3x^{2}+6x-(x^{2}+4x-3x-12)}{(x-3)(x+2)}; x \ne 3,-2;$ $ = \frac{3x^{2}+6x-(x^{2}+x-12)}{(x-3)(x+2)}; x \ne 3,-2;$ $ = \frac{3x^{2}+6x-x^{2}-x+12}{(x-3)(x+2)}; x \ne 3,-2;$ $=\frac{2x^{2}+5x+12}{(x-3)(x+2)} ; x \ne 3,-2;$
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