There are a couple possible approaches. Here's one.
Start by graphing the equation: 2y-3x = 62y−3x=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.)
(0, 3)(0,3) and (-2, 0)(−2,0) 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 2y-3x = 62y−3x=6 cuts the plane into two regions. In one region, the value of 2y-3x2y−3x is <6<6, in the other it is >6>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 (0,0)(0,0) (the origin) is not on the graph of the equation, so I'll just check to see if that side is the <6<6 or >6>6 side.
2(0)-3(0)=0-0=02(0)−3(0)=0−0=0 which is less than 66. So the region below the line must be the <6<6 side of the line.
The inequality we're looking at wants the >6>6 side, so we shade that side. (If you wanted to double check, you could pick a point above the line. Say (0, 5)(0,5) or (-10, 0, or (-5. 5),or(−5.5) and make sure that 2y-3x > 62y−3x>6
Your graph should look like this:
graph{2y-3x>6 [-14.24, 14.24, -7.12, 7.12]}
