# How would you graph inequalities 2y - 3x > 6?

Apr 1, 2015

There are a couple possible approaches. Here's one.

Start by graphing the equation: $2 y - 3 x = 6$
We will use a dotted (or dashed) line because the inequality we are working with is strict.

For this equation, it is straightforward to find the intercepts, so that's now I would graph this one.
(If you prefer to put it in slope-intercept form first, do that.)

$\left(0 , 3\right)$ and $\left(- 2 , 0\right)$ are the intercepts so draw the line through those two points. So you get this But a dotted or dashed line):

graph{2y-3x = 6 [-10, 10, -5, 5]}

The line $2 y - 3 x = 6$ cuts the plane into two regions. In one region, the value of $2 y - 3 x$ is $< 6$, in the other it is $> 6$. Our job now is to figure out which side is which so we can stay on the "greater than 6" side.

I see that the point $\left(0 , 0\right)$ (the origin) is not on the graph of the equation, so I'll just check to see if that side is the $< 6$ or $> 6$ side.
$2 \left(0\right) - 3 \left(0\right) = 0 - 0 = 0$ which is less than $6$. So the region below the line must be the $< 6$ side of the line.

The inequality we're looking at wants the $> 6$ side, so we shade that side. (If you wanted to double check, you could pick a point above the line. Say $\left(0 , 5\right)$ or (-10, 0$, \mathmr{and} \left(- 5. 5\right)$ and make sure that $2 y - 3 x > 6$

Your graph should look like this:

graph{2y-3x>6 [-14.24, 14.24, -7.12, 7.12]}