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: 19


$$x=\frac{1}{9}, -\frac{5}{2}$$

Work Step by Step

$$\frac{5}{x^2}-\frac{43}{x}=18$$ $$\frac{5}{x^2}-\frac{43x}{x^2}=18$$ $$\frac{5-43x}{x^2}=18$$ $$5-43x=18x^2$$ $$0=18x^2+43x-5$$ $$0=18x^2+(45x-2x)-5$$ $$0=9x(2x+5)-1(2x+5)$$ $$(9x-1)(2x+5)=0$$ $$x=\frac{1}{9}, -\frac{5}{2}$$
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