|3x-6y| ≦ 9

Factor out 3

3|x-2y| ≦ 9

Divde both sides by 3

|x-2y| ≦ 3

First we get the boundary graph:

|x-2y| = 3

x-2y = 3  and x-2y = -3

We graph those two lines



Now we use the origin (0,0) as a test point:

|x-2y| ≦ 3
|0-0y| ≦ 3
     0 ≦ 3
That's true so the region that should be shaded is
the strip between the two lines, and including
the two lines as well.

Now we'll draw the box with the corners:

(1,1), (2,1), (2,2,), and (1,2)



The strip between the two parallel lines contains all of the
box, and a whole lot more too.  So the answer is "both in and out".
[The upper left corner of the box (1,2) is on the boundary, but the
inequality includes its bondary.]


Edwin