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: 14

Answer

$$x=3$$

Work Step by Step

$$\frac{3}{(x+2)(x-1)}-\frac{1}{(x+1)(x-1)}=\frac{7}{2(x+2)(x+1)}$$ $$(2(x+2)(x+1)(x-1))(\frac{3}{(x+2)(x-1)}-\frac{1}{(x+1)(x-1)})=(2(x+2)(x+1)(x-1))(\frac{7}{2(x+2)(x+1)})$$ $$6(x+1)-2(x+2)=7(x-1)$$ $$6x+6-2x-4=7x-7$$ $$4x+2=7x-7$$ $$-3x=-9$$ $$x=3$$
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