Answer
$\color{blue}{[-12, 6]}$
Work Step by Step
Multiply $3$ to each part:
\begin{array}{ccccc}
&3(-5)&\le &3 \cdot \dfrac{x-3}{3} &\le &3(1)
\\&-15 &\le &x-3 &\le &3
\end{array}
Add $3$ to each part:
\begin{array}{ccccc}
\\&-15+3 &\le &x-3+3 &\le &3+3
\\&-12 &\le &x &\le &6
\end{array}
Thus, the solution to the given inequality is:
$\color{blue}{[-12, 6]}$
