#### Answer

The solution is $\Big(-\dfrac{5}{3},\dfrac{4}{3}\Big)$

#### Work Step by Step

$\Big|2x+\dfrac{1}{3}\Big|+1\lt4$
Take $1$ to the right side of the inequality:
$\Big|2x+\dfrac{1}{3}\Big|\lt4-1$
$\Big|2x+\dfrac{1}{3}\Big|\lt3$
Solving this absolute value inequality is equivalent to solving the following inequality:
$-3\lt2x+\dfrac{1}{3}\lt3$
Multiply the whole inequality by $3$:
$3\Big(-3\lt2x+\dfrac{1}{3}\lt3\Big)$
$-9\lt6x+1\lt9$
Subtract $1$ from all three parts of the inequality:
$-9-1\lt6x+1-1\lt9-1$
$-10\lt6x\lt8$
Divide all three parts of the inequality by $6$:
$-\dfrac{10}{6}\lt\dfrac{6x}{6}\lt\dfrac{8}{6}$
$-\dfrac{5}{3}\lt x\lt\dfrac{4}{3}$
Expressing the solution in interval notation:
$\Big(-\dfrac{5}{3},\dfrac{4}{3}\Big)$