College Algebra (11th Edition)

Published by Pearson

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 134: 12

No solution

Work Step by Step

$$\frac{3}{x-2}+\frac{1}{x+2}=\frac{12}{(x+2)(x-2)}$$ $$((x+2)(x-2))(\frac{3}{x-2}+\frac{1}{x+2})=((x+2)(x-2))(\frac{12}{(x+2)(x-2)})$$ $$3(x+2)+(x-2)=12$$ $$3x+6+x-2=12$$ $$4x+4=12$$ $$4x=8$$ $$x=2$$ 2 is an extraneous solution; there is no solution.

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