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

1 Answer
Apr 1, 2015

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

Start by graphing the equation: #2y-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)# and #(-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 = 6# cuts the plane into two regions. In one region, the value of #2y-3x# 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 #(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# or #>6# side.
#2(0)-3(0)=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 #(0, 5)# or (-10, 0#, or (-5. 5)# and make sure that #2y-3x > 6#

Your graph should look like this:

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