How do you graph #-y+3x<=6#?

1 Answer
Apr 1, 2015

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

I finding leading negative signs hard to read, so I would begin by rewriting the inequality as: #3x-y <= 6#

Start by graphing the equation: #3x-y = 6#

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, -6)# and #(2, 0)# are the intercepts so draw the line through those two points. So you get this:

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

The line #3x-y = 6# cuts the plane into two regions. In one region, the value of #3x-y# 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 "less 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.
#3(0)-(0)=0-0=0# which is less than #6#. So the region above 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 #(5, 0)# or (10, 0)# and make sure that #3x - y < 6#

Your graph should look like this:

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