Answer
$\text{ the interval }
\left[ -3,\dfrac{3}{2} \right]
$
Work Step by Step
Using properties of inequality, then,
\begin{array}{l}
1\le \dfrac{2}{3}x+3\le4\\\\
1-3\le \dfrac{2}{3}x+3-3\le4-3\\\\
-2\le \dfrac{2}{3}x\le1\\\\
-2\cdot\dfrac{3}{2}\le \dfrac{3}{2}\cdot\dfrac{2}{3}x\le1\cdot\dfrac{3}{2}\\\\
-3\le x\le\dfrac{3}{2}
.\end{array}
Hence, the solution set is $
\text{ the interval }
\left[ -3,\dfrac{3}{2} \right]
$.
