## Intermediate Algebra (12th Edition)

$\left( -\infty, -\dfrac{11}{2} \right] \cup \left[ 5,\infty \right)$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|4x+1|\ge21 ,$ 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} 4x+1\ge21 \\\\\text{OR}\\\\ 4x+1\le-21 .\end{array} Using the properties of inequality to isolate the variable results to \begin{array}{l}\require{cancel} 4x+1\ge21 \\\\ 4x\ge21-1 \\\\ 4x\ge20 \\\\ x\ge\dfrac{20}{4} \\\\ x\ge5 \\\\\text{OR}\\\\ 4x+1\le-21 \\\\ 4x\le-21-1 \\\\ 4x\le-22 \\\\ x\le-\dfrac{22}{4} \\\\ x\le-\dfrac{11}{2} .\end{array} In interval notation, the solution set is $\left( -\infty, -\dfrac{11}{2} \right] \cup \left[ 5,\infty \right) .$ The colored graph is the graph of the solution set.