Exercise (6.4) - Absolute Value Inequality

1. Find the solution sets of the following inequalities. Illustrate each of the inequalities on the number line.
   
   (a) $|x-1|<3$
   
   
   
   

   $\begin{array}{l}\text{(a)}\ \ \ \ \ \ \ |x-1|\ <3\\\\\ \ \ \ \ \ \therefore \ \ -3<x-1<3\\\\\ \ \ \ \ \ \therefore \ \ -3+1<x-1+1<3+1\\\\\ \ \ \ \ \ \therefore \ \ -2<x<4\\\\\ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \{x\ |\ -2<x<4\}\end{array}$

   
   (b) $|x+5|>2$
   
   
   
   

   $\begin{array}{l}\text{(b)}\ \ \ \ \ \ \ \begin{array}{*{20}{l}} {|x+5|\ >2} \end{array}\\\\\ \ \ \ \ \ \therefore \ \ x+5<-2\ \ (\text{or})\ \ x+5>2\\\\\ \ \ \ \ \ \therefore \ \ x+5-5<-2-5\ \ (\text{or})\ \ x+5-5>2-5\\\\\ \ \ \ \ \ \therefore \ \ x<-7\ \ (\text{or})\ \ x>-3\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \{x\ |\ x<-7\ \ (\text{or})\ \ x>-3\}\end{array}$

   
   (c) $|x-2| \leq 4$
   
   
   
   

   $\begin{array}{l}\text{(c)}\ \ \ \ \ \ \ |x-2|\ \le \ 4\\\\\ \ \ \ \ \ \therefore \ \ -4\le x-2\le 4\\\\\ \ \ \ \ \ \therefore \ \ -4+2\le x-2+2\le 4+2\\\\\ \ \ \ \ \ \therefore \ \ -2\le x\le 6\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \{x\ |\ -2\le x\le 6\}\end{array}$

   
   (d) $|x+1| \geq 2$
   
   
   
   

   $\begin{array}{l}\text{(d)}\ \ \ \ \ \ \ |x+1|\ \ge 2\\\\\ \ \ \ \ \ \therefore \ \ x+1\le -2\ \ (\text{or})\ \ x+1\ge 2\\\\\ \ \ \ \ \ \therefore \ \ x+1-15\le -2-1\ \ (\text{or})\ \ x+1-1\ge 2-1\\\\\ \ \ \ \ \ \therefore \ \ x\le -3\ \ (\text{or})\ \ x\ge 1\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \{x\ |\ x\le -3\ \ (\text{or})\ \ x\ge 1\}\end{array}$

   
   (e) $|x-3|>0$
   
   
   
   

   $\begin{array}{l}\text{(e)}\ \ \ \ \ \ \ |x-3|\ >0\\\\\ \ \ \ \ \ \therefore \ \ x-3<0\ \ (\text{or})\ \ x-3>0\\\\\ \ \ \ \ \ \therefore \ \ x-3+3<0+3\ \ (\text{or})\ \ x-3+3>0+3\\\\\ \ \ \ \ \ \therefore \ \ x<3\ \ (\text{or})\ \ x>3\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \mathbb{R}\backslash \{3\}\end{array}$

   
   (f) $|x+3| \leq 0$
   
   
   
   

   $\begin{array}{l}\text{(f)}\ \ \ \ \ \ \ |x+3|\ \le 0\\\\\ \ \ \ \ \ \ \ \ \text{Since}\ |x+3|\ \text{cannot be negative,}\ \\\\\ \ \ \ \ \ \ \ \ x+3=0\\\\\ \ \ \ \ \ \therefore \ \ x=-3\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \{-3\}\end{array}$

   
   


2. Find the solution sets of the following inequalities.
   
   (a) $|2 x-1|<4$
   
   
   
   

   $\begin{array}{l}\text{(a)}\ \ \ \ \ \ |2x-1|<4\\\\\ \ \ \ \ \ \therefore \ \ -4<2x-1<4\\\\\ \ \ \ \ \ \therefore \ \ -4+1<2x-1+1<4+1\\\\\ \ \ \ \ \ \therefore \ \ -3<2x<5\\\\\ \ \ \ \ \ \therefore \ \ -\displaystyle\frac{3}{2}<\displaystyle\frac{{2x}}{2}<\displaystyle\frac{5}{2}\\\\\ \ \ \ \ \ \therefore \ \ -\displaystyle\frac{3}{2}<x<\displaystyle\frac{5}{2}\\\\\ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \left\{ {x\ |\ -\displaystyle\frac{3}{2}<x<\displaystyle\frac{5}{2}} \right\}\end{array}$

   
   (b) $|3 x+5|>6$
   
   
   
   

   $\begin{array}{l}\text{(b)}\ \ \ \ \ \ \ |3x+5|>6\\\\\ \ \ \ \ \ \therefore \ \ 3x+5<-6\ \ (\text{or})\ \ 3x+5>6\\\\\ \ \ \ \ \ \therefore \ \ 3x+5-5<-6-5\ \ (\text{or})\ \ 3x+5-5>6-5\\\\\ \ \ \ \ \ \therefore \ \ 3x<-11\ \ (\text{or})\ \ 3x>1\\\\\ \ \ \ \ \ \therefore \ \ \displaystyle\frac{{3x}}{3}<-\displaystyle\frac{{11}}{3}\ \ (\text{or})\ \ \displaystyle\frac{{3x}}{3}>\displaystyle\frac{1}{3}\\\\\ \ \ \ \ \ \therefore \ \ x<-\displaystyle\frac{{11}}{3}\ \ (\text{or})\ \ x>\displaystyle\frac{1}{3}\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \left\{ {x\ |\ x<-\displaystyle\frac{{11}}{3}\ \ (\text{or})\ \ x>\displaystyle\frac{1}{3}} \right\}\end{array}$

   
   (c) $|4 x-2| \leq 4$
   
   
   
   

   $\begin{array}{l}\text{(c)}\ \ \ \ \ \ |4x-2|\ \le 4\\\\\ \ \ \ \ \ \therefore \ \ -4\le 4x-2\le 4\\\\\ \ \ \ \ \ \therefore \ \ -4+2\le 4x-2+2\le 4+2\\\\\ \ \ \ \ \ \therefore \ \ -2\le 4x\le 6\\\\\ \ \ \ \ \ \therefore \ \ -\displaystyle\frac{2}{4}\le \displaystyle\frac{{4x}}{4}\le \displaystyle\frac{6}{4}\\\\\ \ \ \ \ \ \therefore \ \ -\displaystyle\frac{1}{2}\le x\le \displaystyle\frac{3}{2}\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \left\{ {x\ |\ -\displaystyle\frac{1}{2}\le x\le \displaystyle\frac{3}{2}} \right\}\end{array}$

   
   (d) $|2 x+1| \geq 2$
   
   
   
   

   $\begin{array}{l}\text{(d)}\ \ \ \ \ \ |2x+1|\ \ge 2\\\\\ \ \ \ \ \ \therefore \ \ 2x+1\le -2\ \ (\text{or})\ \ 2x+1\ge 2\\\\\ \ \ \ \ \ \therefore \ \ 2x+1-1\le -2-1\ \ (\text{or})\ \ 2x+1-1\ge 2-1\\\\\ \ \ \ \ \ \therefore \ \ 2x\le -3\ \ (\text{or})\ \ 2x\ge 1\\\\\ \ \ \ \ \ \therefore \ \ \displaystyle\frac{{2x}}{2}\le -\displaystyle\frac{3}{2}\ \ (\text{or})\ \ \displaystyle\frac{{2x}}{2}\ge \displaystyle\frac{1}{2}\\\\\ \ \ \ \ \ \therefore \ \ x\le -\displaystyle\frac{3}{2}\ \ (\text{or})\ \ x\ge \displaystyle\frac{1}{2}\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \left\{ {x\ |\ x\le -\displaystyle\frac{3}{2}\ \ (\text{or})\ \ x\ge \displaystyle\frac{1}{2}} \right\}\end{array}$

   
   (e) $|2 x-3|>0$
   
   
   
   

   $ \begin{array}{l}\text{(e)}\ \ \ \ \ \ \ |2x-3|\ >0\\\\\ \ \ \ \ \ \therefore \ \ 2x-3<0\ \ (\text{or})\ \ 2x-3>0\\\\\ \ \ \ \ \ \therefore \ \ 2x-3+3<0+3\ \ (\text{or})\ \ 2x-3+3>0+3\\\\\ \ \ \ \ \ \therefore \ \ 2x<3\ \ (\text{or})\ \ 2x>3\\\\\ \ \ \ \ \ \therefore \ \ \displaystyle\frac{{2x}}{2}<\displaystyle\frac{3}{2}\ \ (\text{or})\ \ \displaystyle\frac{{2x}}{2}>\displaystyle\frac{3}{2}\\\\\ \ \ \ \ \ \therefore \ \ x<\displaystyle\frac{3}{2}\ \ (\text{or})\ \ x>\displaystyle\frac{3}{2}\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \mathbb{R}\backslash \left\{ {\displaystyle\frac{3}{2}} \right\}\end{array}$

   
   (f) $|5 x+3| \leq 0$
   
   
   
   

   $\begin{array}{l}\text{(f)}\ \ \ \ \ \ \ |5x+3|\ \le 0\\\\\ \ \ \ \ \ \ \ \ \text{Since}\ |5x+3|\ \text{cannot be negative,}\ \\\\\ \ \ \ \ \ \ \ \ 5x+3=0\\\\\ \ \ \ \ \ \therefore \ \ x=-\displaystyle\frac{3}{5}\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \left\{ {-\displaystyle\frac{3}{5}} \right\}\end{array}$