Answer
$-\dfrac{16}{3} \lt x \lt 4$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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
.$
