Đáp án:
$\begin{array}{l}
\left( {4 - x} \right).\left( {2x + 3y} \right) \le 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
4 - x \le 0\\
2x + 3y \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
4 - x \ge 0\\
2x + 3y \le 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 4\\
x \ge - \frac{{3y}}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 4\\
x \le - \frac{{3y}}{2}
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 4\\
4 \ge - \frac{{3y}}{2} \Rightarrow y \ge - \frac{8}{3}
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 4\\
4 \le - \frac{{3y}}{2} \Rightarrow y \ge - \frac{8}{3}
\end{array} \right.
\end{array} \right.\\
Vậy\left\{ \begin{array}{l}
x \ge 4\\
y \ge - \frac{8}{3}
\end{array} \right.\,hoac\,\,\left\{ \begin{array}{l}
x \le 4\\
y \le - \frac{8}{3}
\end{array} \right.
\end{array}$
