## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x\le-\dfrac{11}{3} \text{ or } x\ge\dfrac{19}{3}$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|3x-4|\ge15 ,$ use the definition of a greater than (greater than or equal to) absolute value inequality. Then use the properties of inequality to isolate the variable in each resulting equation. $\bf{\text{Solution Details:}}$ 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} 3x-4\ge15 \\\\\text{OR}\\\\ 3x-4\le-15 .\end{array} Using the properties of inequality to isolate the variable in each equation results to \begin{array}{l}\require{cancel} 3x-4\ge15 \\\\ 3x\ge15+4 \\\\ 3x\ge19 \\\\ x\ge\dfrac{19}{3} \\\\\text{OR}\\\\ 3x-4\le-15 \\\\ 3x\le-15+4 \\\\ 3x\le-11 \\\\ x\le-\dfrac{11}{3} .\end{array} Hence, the solution set is $x\le-\dfrac{11}{3} \text{ or } x\ge\dfrac{19}{3} .$