## Intermediate Algebra (12th Edition)

$\left( -\dfrac{11}{2},5 \right)$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|4x+1| \lt 21 ,$ use the definition of absolute value inequalities. Use the properties of inequalities 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 $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} -21 \lt 4x+1 \lt 21 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -21-1 \lt 4x+1-1 \lt 21-1 \\\\ -22 \lt 4x \lt 20 \\\\ -\dfrac{22}{4} \lt \dfrac{4x}{4} \lt \dfrac{20}{4} \\\\ -\dfrac{11}{2} \lt x \lt 5 .\end{array} In interval notation, the solution set is $\left( -\dfrac{11}{2},5 \right) .$ The colored graph is the graph of the solution set.