Beginning Algebra With Applications, Chapter 5, 5.6, Section 5.6, Problem 28

Illustrate the solution set $2x -3 (y + 1) \geq y - (7 - x)$

$
\begin{equation}
\begin{aligned}
2x - 3y - 3 &\geq y - 7 + x
&& \text{Solve the inequality for } y \\
\\
-3y - y &\geq -2x + x + 3 - 7
&& \text{Group terms}\\
\\
-4y &\geq -x - 4
&& \text{Evaluate}\\
\\
\frac{-4y}{-4} &\leq \frac{-x}{-4} - \frac{4}{-4}
&& \text{Remember that if you divide or multiply negative numbers ,the inequality symbol reverses}\\
\\
y &\leq \frac{x}{4} + 1
&& \text{Simplify}
\end{aligned}
\end{equation}
$


To graph the inequality, we first find the intercepts of the line $\displaystyle y = \frac{x}{4} + 1$.
In this case, the $x$-intercept (set $y = 0$) is $(-4,0)$

$
\begin{equation}
\begin{aligned}
0 &= \frac{x}{4} + 1 \\
\\
\frac{x}{4} &=- 1\\
\\
x &= -4
\end{aligned}
\end{equation}
$


And, the $y$-intercept (set $x = 0$) is $(0,1)$

$
\begin{equation}
\begin{aligned}
y &= \frac{0}{4} + 1 \\
\\
y &= 1
\end{aligned}
\end{equation}
$


So, the graph is



Graph $\displaystyle y = \frac{x}{4} + 1 $ as a solid line. Shade the lower half-plane.