Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.8 Absolute Value Equations and Inequalities - 1.8 Exercises - Page 167: 17

Answer

The solutions are $x=\dfrac{2}{9}$ and $x=-\dfrac{4}{3}$

Work Step by Step

$\Big|\dfrac{6x+1}{x-1}\Big|=3$ Solving this absolute value equation is equivalent to solving two equations, which are: $\dfrac{6x+1}{x-1}=3$ and $\dfrac{6x+1}{x-1}=-3$ Solve the first equation: $\dfrac{6x+1}{x-1}=3$ Take $x-1$ to multiply the right side: $6x+1=3(x-1)$ $6x+1=3x-3$ Take $3x$ to the left side and $1$ to the right side: $6x-3x=-3-1$ $3x=-4$ Solve for $x$: $x=-\dfrac{4}{3}$ Solve the second equation: $\dfrac{6x+1}{x-1}=-3$ Take $x-1$ to multiply the right side: $6x+1=-3(x-1)$ $6x+1=-3x+3$ Take $3x$ to the left side and $1$ to the right side: $6x+3x=3-1$ $9x=2$ Solve for $x$: $x=\dfrac{2}{9}$ The solutions are $x=\dfrac{2}{9}$ and $x=-\dfrac{4}{3}$
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