## Intermediate Algebra (12th Edition)

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