## Intermediate Algebra (12th Edition)

$\left( -\infty, -\dfrac{12}{5} \right) \cup \left( \dfrac{8}{5},\infty \right)$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|5x+2|\gt10 ,$ 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 $\ge.$ For graphing inequalities, use a hollowed dot for the symbols $\lt$ or $\gt.$ Use a solid dot 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} 5x+2\gt10 \\\\\text{OR}\\\\ 5x+2\lt-10 .\end{array} Using the properties of inequality to isolate the variable results to \begin{array}{l}\require{cancel} 5x+2\gt10 \\\\ 5x\gt10-2 \\\\ 5x\gt8 \\\\ x\gt\dfrac{8}{5} \\\\\text{OR}\\\\ 5x+2\lt-10 \\\\ 5x\lt-10-2 \\\\ 5x\lt-12 \\\\ x\lt-\dfrac{12}{5} .\end{array} In interval notation, the solution set is $\left( -\infty, -\dfrac{12}{5} \right) \cup \left( \dfrac{8}{5},\infty \right) .$ The colored graph is the graph of the solution set.