Answer
$\left( -\dfrac{5}{3},\dfrac{4}{3} \right)$
Work Step by Step
Since for any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a,$ then the solution to the given equation, $
\left| 2x+\dfrac{1}{3} \right|+1\lt4
,$ is
\begin{array}{l}\require{cancel}
\left| 2x+\dfrac{1}{3} \right|\lt4-1
\\\\
\left| 2x+\dfrac{1}{3} \right|\lt3
\\\\
-3\lt 2x+\dfrac{1}{3} \lt3
\\\\
-3-\dfrac{1}{3}\lt 2x+\dfrac{1}{3}-\dfrac{1}{3} \lt3-\dfrac{1}{3}
\\\\
-\dfrac{10}{3}\lt 2x \lt \dfrac{8}{3}
\\\\
-\dfrac{\dfrac{10}{3}}{2}\lt \dfrac{2x}{2} \lt \dfrac{\dfrac{8}{3}}{2}
\\\\
-\dfrac{10}{6}\lt x \lt \dfrac{8}{6}
\\\\
-\dfrac{5}{3}\lt x \lt \dfrac{4}{3}
.\end{array}
Hence, the solution set is the interval $
\left( -\dfrac{5}{3},\dfrac{4}{3} \right)
.$
