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

$-\dfrac{16}{3} \lt x \lt 4$
$\bf{\text{Solution Outline:}}$ To find all $x$ such that $f(x) \lt19 ,$ with $f(x)= 5+|3x+2| ,$ solve the inequality \begin{array}{l}\require{cancel} 5+|3x+2|\lt19 .\end{array} Then graph the solution set. In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Isolating the absolute value expression results to \begin{array}{l}\require{cancel} 5+|3x+2|\lt19 \\\\ |3x+2|\lt19-5 \\\\ |3x+2|\lt14 .\end{array} 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} -14\lt 3x+2 \lt14 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -14\lt 3x+2 \lt14 \\\\ -14-2 \lt 3x+2-2 \lt14-2 \\\\ -16 \lt 3x \lt 12 \\\\ -\dfrac{16}{3} \lt \dfrac{3x}{3} \lt \dfrac{12}{3} \\\\ -\dfrac{16}{3} \lt x \lt 4 .\end{array} Hence, the solution set $-\dfrac{16}{3} \lt x \lt 4 .$