## College Algebra (11th Edition)

Published by Pearson

# Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 134: 15

#### Answer

$$x=-2$$

#### Work Step by Step

$$\frac{2x+1}{x-2}+\frac{3}{x}=-\frac{6}{x(x+2)}$$ $$(x(x-2))(\frac{2x+1}{x-2}+\frac{3}{x})=(x(x-2))(-\frac{6}{x(x+2)})$$ $$x(2x+1)+3(x-2)=-6$$ $$2x^2+x+3x-6=-6$$ $$2x^2+4x=0$$ $$x^2+2x=0$$ $$x(x+2)=0$$ $$x=0,-2$$ 0 is an extraneous solution. $$x=-2$$

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