Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Problems - Page 76: 3



Work Step by Step

$$\left | 2x-1 \right \vert - \left | x+5 \right \vert=3$$ $$\left | 2x-1 \right \vert=\begin{cases} 2x-1, & \mbox{if }\; 2x-1 \geq0 \;\rightarrow\; x\geq1/2\\ 1-2x, & \mbox{if }\; 2x-1\lt0 \;\rightarrow\; x\lt1/2 \end{cases}$$ $$\left | x+5 \right \vert=\begin{cases} x+5, & \mbox{if }\; x+5\geq0\; \rightarrow\; x\geq-5\\ -x-5, & \mbox{if }\; x+5\lt0 \;\rightarrow\; x\lt-5 \end{cases}$$ $$x\lt5 \qquad -5\leq x\lt1/2 \qquad x\geq1/2$$ $$\mbox{If} \; x\lt-5 \\ 1-2x-(-x-5)=3\\ 1-2x+x+5=3\\ \color{Red}{x=3} $$ $$\mbox{If} \; -5\leq x\lt1/2 \\ 1-2x-(x+5)=3\\ 1-2x-x-5=3\\ \color{limegreen}{x=-7/3}$$ $$\mbox{If} \; x\geq1/2 \\ 2x-1-(x+5)=3\\ 2x-1-x-5=3\\ \color{limegreen}{x=9}$$ $$\mbox{Therefore, the solutions are} \; x=\frac{-7}{3}, 9$$
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