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: 39

Answer

a. Restriction values $: 1 $ b. Solution set $ :\{3\}$

Work Step by Step

$\frac{1}{x-1} +5 = \frac{11}{x-1}$ a. $x = 1 $ makes the denominator zero. So, $x \ne 1$ b. $\frac{1}{x-1} +5 = \frac{11}{x-1}; x \ne 1;$ $\frac{1+5(x-1)}{x-1} = \frac{11}{x-1}; x \ne 1;$ $\frac{1+5x-5}{x-1} = \frac{11}{x-1}; x \ne 1;$ $\frac{5x-4}{x-1} = \frac{11}{x-1}; x \ne 1;$ Multiply both sides by $(x-1)$ to clear fraction. $(x-1)(\frac{5x-4}{x-1} )= (x-1)(\frac{11}{x-1});$ $5x-4 = 11$ $5x = 11 +4$ $5x = 15$ Divide both sides by $5$ $x = 3$
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