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

$a\le-\dfrac{10}{3} \text{ or } a\ge\dfrac{2}{3}$
$\bf{\text{Solution Outline:}}$ Solve the given inequality, $|3a+4|+2\ge8 ,$ by isolating first the absolute value expression. Then use the definition of a greater than absolute value inequality. Finally, graph the solution set. In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Using the properties of inequality, the given expression is equivalent to \begin{array}{l}\require{cancel} |3a+4|+2\ge8 \\\\ |3a+4|\ge8-2 \\\\ |3a+4|\ge6 .\end{array} 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 given inequality is equivalent to \begin{array}{l}\require{cancel} 3a+4\ge6 \\\\\text{OR}\\\\ 3a+4\le-6 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 3a+4\ge6 \\\\ 3a\ge6-4 \\\\ 3a\ge2 \\\\ a\ge\dfrac{2}{3} \\\\\text{OR}\\\\ 3a+4\le-6 \\\\ 3a\le-6-4 \\\\ 3a\le-10 \\\\ a\le-\dfrac{10}{3} .\end{array} Hence, the solution set is $a\le-\dfrac{10}{3} \text{ or } a\ge\dfrac{2}{3} .$