College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.2 - Page 118: 31

Answer

$\frac{4}{x} = \frac{5}{2x} + 3 ; x \ne 0;$ $x = \frac{1}{2} $

Work Step by Step

$\frac{4}{x} = \frac{5}{2x} + 3 ; x \ne 0;$ Least Common Denominator is $2x$ Multiply both sides by $2x$ $2x(\frac{4}{x}) = 2x(\frac{5}{2x} + 3 ) ; x \ne 0;$ $8 = 2x(\frac{5}{2x}) + 2x(3)$ $8 = 5+ 6x$ $6x = 8-5$ $6x = 3$ $x = \frac{3}{6}$ $x = \frac{1}{2}$ Our only restriction is $x \ne 0$ $x = \frac{1}{2}$ is the solution of the equation.
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