Answer
$\text{the interval }
\left[ -\dfrac{4}{3},\dfrac{7}{6} \right]
$
Work Step by Step
Using the properties of inequality, the solution to the given inequality, $
0 \le \dfrac{2(3x+4)}{5} \le 3
,$ is
\begin{array}{l}\require{cancel}
0\cdot5 \le \dfrac{2(3x+4)}{5}\cdot5 \le 3\cdot5
\\\\
0 \le 2(3x+4) \le 15
\\\\
0 \le 6x+8 \le 15
\\\\
0-8 \le 6x+8-8 \le 15-8
\\\\
-8 \le 6x \le 7
\\\\
-\dfrac{8}{6} \le \dfrac{6}{6}x \le \dfrac{7}{6}
\\\\
-\dfrac{4}{3} \le x \le \dfrac{7}{6}
.\end{array}
In interval notation, the solution is $
\text{the interval }
\left[ -\dfrac{4}{3},\dfrac{7}{6} \right]
.$
