## Intermediate Algebra (12th Edition)

$\left( -\dfrac{13}{3},3 \right)$ $\bf{\text{Solution Outline:}}$ To solve the given inequality, $|3x+2| \lt 11 ,$ use the definition of absolute value inequalities. Use the properties of inequalities 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 $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} -11 \lt 3x+2 \lt 11 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -11-2 \lt 3x+2-2 \lt 11-2 \\\\ -13 \lt 3x \lt 9 \\\\ -\dfrac{13}{3} \lt \dfrac{3x}{3} \lt \dfrac{9}{3} \\\\ -\dfrac{13}{3} \lt x \lt 3 .\end{array} In interval notation, the solution set is $\left( -\dfrac{13}{3},3 \right) .$ The colored graph is the graph of the solution set.