How do you graph the inequality #x + 2y < 3#, #2x – 3y<6#?

1 Answer
Jul 3, 2016

The lines meet at (3, 0) and the y-intercepts are #3/2 and -2#.
Shade the region on the left of (3, 0), in-between the lines. For any point (x, y) here, both the inequalities are satisfied.


Rearrange to the forms #y<3/2-x/2 and y>2/3x-2#.

Now, the line #y=3/2-x/2# cuts the axes at (3, 0) and (0, 3/2) and the

line #y+2/3x-2 cuts the axes at (3, 0) and (0, -2). (3, 9) is the common

point. For any point (x, y) in the region enclosed by these lines, on

the left of the common point (3,

#2/3x-2 < y < 3/2-x/2#.

Separately, this is the given pair of inequalities.

The opposite region, on the right of (3, 0), is for the reversed inequalities.