Graphing an involves two main steps.
- Plotting the boundary line.
- Shading the solution.

### Boundary Line

To write the equation of the , replace the inequality sign with an equals sign.
$\begin{aligned} \text{-}2x+y = \text{-}3
\end{aligned}$
To draw this line, we will first rewrite the equation in .
$\text{-}2x+y=\text{-}3$

$y=2x-3$

Now that the equation is in slope-intercept form, we can identify the $m$ and the $b.$
$\begin{gathered}
y=2x-3\quad \Leftrightarrow \quad y=2x+(\text{-}3)
\end{gathered}$
We will plot the $y\text{-}$intercept $b=\text{-}3,$ and then use the slope $m=2$ to plot another point on the line. We will then connect these points with a straight line. This is our boundary line. Since we have a strict inequality, the boundary line will be dashed.
### Test a point

To decide which side of the boundary line to shade, we will substitute a test point that is not on this line. If the substitution creates a true statement, we shade the region that includes the test point. Otherwise, we shade the opposite region. Let's use $(0,0)$ as our test point.
$y\lt 2x - 3$

${\color{#009600}{0}} \lt 2\cdot{\color{#0000FF}{0}} - 3$

$0\nless\text{-}3$

### Shade the correct region

Since the substitution of the test point did **not** create a true statement, we will shade the region that does **not** contain the point.

To graph the given inequality, we will begin by graphing the boundary line, $y=x-1.$
### Boundary line

Note that, since the inequality is less than,

the boundary will not be included and the line should be dashed.

### Test a point

Let's now use the origin, $(0,0),$ as a test point to determine which region to shade.
$y<x-1$

${\color{#009600}{0}} <{\color{#0000FF}{0}} -1$

$0\nless\text{-}1$

Thus, the point is not a solution to the inequality. This means that we should shade the region **below** the graph.
To graph the given inequality, we will begin by graphing the boundary line, $y+3x=\frac{1}{2}.$
### Boundary line

First, we need to rewrite the equation in by isolating $y.$
$y+3x=\frac{1}{2} \quad \Leftrightarrow \quad y=\text{-}3x+\frac{1}{2}$
Thus, the slope is $3$ and the $y$-intercept is $\frac{1}{2}.$ We'll use this information to graph the boundary.

### Test a point

Let's us the origin as a test point to determine what region to should shade.
$y\leq \text{-}3x +\dfrac{1}{2}$

${\color{#009600}{0}} \stackrel{?}{\leq} \text{-}3\cdot{\color{#0000FF}{0}} +\dfrac{1}{2}$

$0 \stackrel{?}{\leq} 0 +\dfrac{1}{2}$

$0 \leq \dfrac{1}{2}$

Since the test point did produce a true statement, we will shade the region contains $(0,0).$