## Intermediate Algebra (12th Edition)

$\left( -\dfrac{9}{5},3 \right)$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|-5x+3| \lt 12 ,$ 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} -12 \lt -5x+3 \lt 12 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -12-3 \lt -5x+3-3 \lt 12-3 \\\\ -15 \lt -5x \lt 9 .\end{array} Dividing by a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-15}{-5} \gt \dfrac{-5x}{-5} \gt \dfrac{9}{-5} \\\\ 3 \gt x \gt -\dfrac{9}{5} \\\\ -\dfrac{9}{5} \lt x \lt 3 .\end{array} In interval notation, the solution set is $\left( -\dfrac{9}{5},3 \right) .$ The colored graph is the graph of the solution set.