College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 134: 13



Work Step by Step

$$\frac{4}{(x-2)(x+3)}-\frac{1}{(x+2)(x-2)}=\frac{2}{(x+3)(x+2)}$$ $$((x-2)(x+3)(x+2))(\frac{4}{(x-2)(x+3)}-\frac{1}{(x+2)(x-2)})=((x-2)(x+3)(x+2))(\frac{2}{(x+3)(x+2)})$$ $$4(x+2)-(x+3)=2(x-2)$$ $$4x+8-x-3=2x-4$$ $$3x+5=2x-4$$ $$x=-9$$ -9 is not an extraneous solution; $x=-9$.
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