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 1.3 - Page 133: 17

Answer

$x=5$

Work Step by Step

Given, $y_{1}=\frac{1}{x}, y_{2}=\frac{1}{2x} , y_{3}=\frac{1}{x-1}$ The equation is $3y_{1}+4y_{2}=4y_{3}$ Substitute $y_{1}, y_{2}$ and $y_{3}$ values in the equation. $3y_{1}+4y_{2}=4y_{3}$ $3(\frac{1}{x})+4(\frac{1}{2x})=4(\frac{1}{x-1}) $ $x=0$ and $x=1$ makes the denominator zero. So,$0$ and $1$ are restricted values. $\frac{3}{x}+\frac{4}{2x}=\frac{4}{x-1} ; x\ne 0, x\ne 1;$ $\frac{3}{x}+\frac{2}{x}=\frac{4}{x-1} ; x\ne 0, x\ne 1;$ $\frac{5}{x}=\frac{4}{x-1} ; x\ne 0, x\ne 1;$ $5(x-1) = 4x$ $5x-5=4x$ $5x-4x=5$ $x=5$ $x=5$ is not the restricted value. The solution is $x=5$
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