## College Algebra (11th Edition)

$\left( -\infty,\dfrac{6}{5} \right]\cup\left[ 2,\infty \right) .$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $|8-5x|\ge2 ,$ use the definition of absolute value inequalities. $\bf{\text{Solution Details:}}$ 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} 8-5x\ge2 \\\\\text{OR}\\\\ 8-5x\le-2 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 8-5x\ge2 \\\\ -5x\ge2-8 \\\\ -5x\ge-6 \\\\\text{OR}\\\\ 8-5x\le-2 \\\\ -5x\le-2-8 \\\\ -5x\le-10 .\end{array} Dividing both sides by a negative number and consequently reversing the inequality symbol results to \begin{array}{l}\require{cancel} x\le\dfrac{-6}{-5} \\\\ x\le\dfrac{6}{5} \\\\\text{OR}\\\\ x\ge\dfrac{-10}{-5} \\\\ x\ge2 .\end{array} Hence, the solution set is the interval $\left( -\infty,\dfrac{6}{5} \right]\cup\left[ 2,\infty \right) .$