Intermediate Algebra, Chapter 3, 3.4, Section 3.4, Problem 14

Illustrate the linear inequality $3x + 47 \geq 12$ in two variables.

To graph $3x + 4y \geq 12$ we must graph the boundary line $3x + 4y = 12$ first. To do this, we need to find the
intercepts of the line

$x$-intercept (set $y = 0$):

$
\begin{equation}
\begin{aligned}
3x + 4(0) &= 12 \\
\\
3x &= 12 \\
\\
x &= 4
\end{aligned}
\end{equation}
$


$y$-intercept (set $x = 0$):

$
\begin{equation}
\begin{aligned}
3(0) + 4y &= 12 \\
\\
4y &= 12 \\
\\
y &= 3
\end{aligned}
\end{equation}
$


Now, by using test point. Let's say point $(3,2)$ from the right of the boundary line.

$
\begin{equation}
\begin{aligned}
3x + 4y &\geq 12 \\
\\
3(3) + 4(2) &\geq 12 \\
\\
9 + 8 &\geq 12 \\
\\
17 &\geq 12
\end{aligned}
\end{equation}
$


Since the inequality symbol is $ \geq $, then the boundary line must be solid.
Moreover, since the test point satisfy the inequality, then we must shade the right
portion of the boundary line. So the graph is,