How do you graph #9x-9y> -36# on the coordinate plane?

1 Answer
Jo S.
Oct 30, 2017

The half plane above and to the left of #y=x+4#.

Explanation:

Simplify first, to give #x-y> -4# then temporarily convert it to an equation so we can draw the critical line: #x-y=-4#.
If you like to use the #y=mx+c# format, you can rearrange your equation to #y=x+4#.
Depending on the requirement of the curriculum you're following, you might draw the line dotted to indicate strict inequality (the editor here does not allow this):
graph{x+4 [-9.125, 10.875, -2.24, 7.76]}
Next you need to decide which side of the line agrees with the inequality. I like to use #(0, 0)# as a test (as long as it doesn't sit on the line).
Is it true that #9xx0-9xx0< -36#?
It is not true, therefore the half of the plane split by the critical line which contains the point #(0, 0)# does not satisfy the inequality.
The correct half of the plane is the part above and to the left of the critical line.

Related questions
See all questions in Linear Inequalities in Two Variables
Impact of this question
1582 views around the world
You can reuse this answer
Creative Commons License