Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Cumulative Review Exercises - Page 880: 14


The solution of the given equation is $\underline{\left( 1,7 \right)}$

Work Step by Step

We simplify the inequality further to obtain the value of x: $\begin{align} & \frac{x+5}{x-1}\gt2 \\ & x+5\gt2x-2 \\ & -x\gt-7 \\ & x\lt7 \end{align}$ For $ x=1$, $\frac{x+5}{x-1}$ is undefined as the denominator becomes zero. To make the left hand side positive, the value of x should be greater than 1. Hence, $x\gt1$ The solution of the given equation is $\underline{\left( 1,7 \right)}$
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