College Algebra (11th Edition)

$\left( -\infty, \dfrac{2}{3}\right] \cup\left[ 2,\infty \right)$
Since for any $a\gt0$, $|x|\gt a$ implies $x \ge a$ or $x\le -a$ then the solution to the given inequality, $\left| 3x-4 \right|\ge 2 ,$ is \begin{array}{l}\require{cancel} 3x-4\ge 2 \\\\ 3x\ge 2+4 \\\\ 3x\ge 6 \\\\ x\ge \dfrac{6}{3} \\\\ x\ge 2 ,\\\\\text{OR}\\\\ 3x-4\le -2 \\\\ 3x\le -2+4 \\\\ 3x\le 2 \\\\ x\le \dfrac{2}{3} .\end{array} Hence, the solution set is the interval $\left( -\infty, \dfrac{2}{3}\right] \cup\left[ 2,\infty \right) .$