Graphing an involves two main steps.
- Plotting the .
- Shading one half-plane to show the solution set.

To graph the inequality, we have to draw the boundary line. The equation of a boundary line is obtained by replacing the inequality symbol with an equals sign.
$3x+2y=12$
To draw this line, we will first rewrite the equation in .
$3x+2y=12$

Write in slope-intercept form

$2y=-3x+12$

$y=2-3x+12 $

$y=2-3x +212 $

$y=-1.5x+6$

Now that the equation is in slope-intercept form, we can identify the $m$ and $b.$$m=-1.5b=6 $
We will plot the $y-$intercept and then use the slope to plot another point on the line. We will then connect these points with a straight line. This is our boundary line. Since the inequality has the symbol $≥$ the boundary should be **included** in the .
Now, to determine which region to shade we'll test a point to see if it's a solution to the inequality. Let's choose the origin as a test point, $(0,0).$
$y≥-1.5x+6$

$0≥? -1.5⋅0+6$

$0≥? 0+6$

$0≱6$

Since the substitution of the test point did **not** create a true statement, we will shade the region that does **not** contain $(0,0).$
### Determining the solutions

Finally, we will plot the given points to determine which ones belong to the solution set.

We can see that no of the points are a solution to the inequality.

To graph the inequality, we have to draw the boundary line. The equation of a boundary line is obtained by replacing the inequality symbol with an equals sign.
$8−4y=2x $
To draw this line, we will first rewrite the equation in .
$8−4y=2x$

Write in slope-intercept form

$-4y=2x−8$

$y=-42x+8 $

$y=-42x +-48 $

$y=-42x −48 $

$y=2x −2$

Now that the equation is in slope-intercept form, we can identify the $m$ and $b.$
$y=2x −2⇔y=21 x+(-2) $
We will plot the $y-$intercept $b=2,$ and then use the slope $m=21 $ to plot another point on the line. We will then connect these points with a straight line. This is our boundary line. Since the inequality has the symbol $≥,$ the boundary line will be included in the solution set.
To decide which region to shade, we will test a point, that's not on the boundary 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.
$8−4y≥2x$

$8−4⋅0≥? 2⋅0$

$8−0≥? 0$

$8≥0$

Since the substitution of the test point created a true statement, we will shade the region that contains $(0,0).$
Finally, we will plot the given points to determine which ones belong to the solution set.

Points that lie within the shaded region are solutions to the inequality.
$(-2,0),(0,0),and(1,4) $