## Intermediate Algebra (12th Edition)

$\left( -\infty, -20 \right) \cup \left( 40,\infty \right)$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|0.1x-1| \gt 3 ,$ use the definition of absolute value inequalities. Then use the properties of inequality to isolate the variable. For the interval notation, use a parenthesis for the symbols $\lt$ or $\gt.$ Use a bracket for the symbols $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} 0.1x-1 \gt 3 \\\\\text{OR}\\\\ 0.1x-1 \lt -3 .\end{array} Using the properties of inequality to isolate the variable results to \begin{array}{l}\require{cancel} 0.1x-1 \gt 3 \\\\ 0.1x \gt 3+1 \\\\ 0.1x \gt 4 \\\\ 10(0.1x) \gt 10(4) \\\\ x \gt 40 \\\\\text{OR}\\\\ 0.1x-1 \lt -3 \\\\ 0.1x \lt -3+1 \\\\ 0.1x \lt -2 \\\\ 10(0.1x) \lt 10(-2) \\\\ x \lt -20 .\end{array} In interval notation, the solution set is $\left( -\infty, -20 \right) \cup \left( 40,\infty \right) .$