#### Answer

$\left[ -\dfrac{5}{2},1 \right]$

#### Work Step by Step

Since for any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a,$ then the given inequality, $
|4x+3|\le7
,$ is equivalent to
\begin{array}{l}\require{cancel}
-7\le 4x+3\le7
.\end{array}
Using the properties of inequality, then
\begin{array}{l}\require{cancel}
-7-3\le 4x+3-3\le7-3
\\\\
-10\le 4x\le4
\\\\
-\dfrac{10}{4}\le \dfrac{4x}{4}\le\dfrac{4}{4}
\\\\
-\dfrac{5}{2}\le x\le1
.\end{array}
Hence, the solution is the interval $
\left[ -\dfrac{5}{2},1 \right]
.$