Question 463978
<pre>
|3x-6y| &#8806; 9

Factor out 3

3|x-2y| &#8806; 9

Divde both sides by 3

|x-2y| &#8806; 3

First we get the boundary graph:

|x-2y| = 3

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

We graph those two lines

{{{drawing(400,400,-4,4,-4,4,

graph(400,400,-4,4,-4,4),line(-11,-7,11,4), line(11,7,-11,-4)

)}}}

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

|x-2y| &#8806; 3
|0-0y| &#8806; 3
     0 &#8806; 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)

{{{drawing(400,400,-4,4,-4,4,

graph(400,400,-4,4,-4,4),line(-11,-7,11,4), line(11,7,-11,-4),
green(line(1,1,2,1),line(2,1,2,2),line(2,2,1,2),line(1,2,1,1))



)}}}

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</pre>