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.
$-2x+y=-3 $
To draw this line, we will first rewrite the equation in .
Now that the equation is in slope-intercept form, we can identify the $m$ and the $b.$
$y=2x−3⇔y=2x+(-3) $
We will plot the $y-$intercept $b=-3,$ 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.
### 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.
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, $y+3x=21 .$
### Boundary line

First, we need to rewrite the equation in by isolating $y.$
$y+3x=21 ⇔y=-3x+21 $
Thus, the slope is $3$ and the $y$-intercept is $21 .$ 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≤-3x+21 $

$0≤? -3⋅0+21 $

$0≤? 0+21 $

$0≤21 $

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