Beginning Algebra With Applications, Chapter 5, Review Exercises, Section Review Exercises, Problem 24

Illustrate the solution set of $3x + 2y \leq 12$

$
\begin{equation}
\begin{aligned}
3x + 2y &\leq 12
&& \text{Solve the inequality for } y \\
\\
2y &\leq -3x + 12 \\
\\
y &\leq -\frac{3}{2}x + 6
\end{aligned}
\end{equation}
$


To graph the inequality, we first find the intercepts of the line $\displaystyle y = -\frac{3}{2}x + 6$.
In this case, the $x$-intercept (set $y = 0$) is $(4,0)$

$
\begin{equation}
\begin{aligned}
0 &= -\frac{3}{2}x + 6\\
\\
\frac{3}{2}x &= 6\\
\\
x &= 6 \left( \frac{2}{3} \right)\\
\\
x &= 4
\end{aligned}
\end{equation}
$

And the $y$-intercept (set $x = 0$) is $(0,6)$

$
\begin{equation}
\begin{aligned}
y &= -\frac{1}{2} (0) + 6\\
\\
y &= 6
\end{aligned}
\end{equation}
$

So the graph is



Graph $\displaystyle y = -\frac{3}{2}x + 6 $ as a solid line. Shade
the lower half of the plane.