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: 20

Answer

$$x=-\frac{7}{2},\frac{1}{3}$$

Work Step by Step

$$\frac{7}{x^2}-\frac{19}{x}=6$$ $$(x^2)(\frac{7}{x^2}-\frac{19}{x})=(x^2)(6)$$ $$7-19x=6x^2$$ $$0=6x^2+19x-7$$ $$(2x+7)(3x-1)=0$$ $$x=-\frac{7}{2},\frac{1}{3}$$
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