Question 880266
<pre>
x + 3y &#8804; 6

First we draw the boundary line whose equation is

x + 3y = 6

We get the intercepts:

 x | y
 0 |
   | 0

If x = 0, x + 3y = 6
          0 + 3y = 6
               y = 2

 x | y
 0 | 2
   | 0
 
If y=0,   x + 3y = 6
        x + 3(0) = 6
           x + 0 = 6
               x = 6

 x | y
 0 | 2
 6 | 0

So the intercepts are (0,2) and (6,0)

So we plot those two points

{{{drawing(400,2800/11,-3,8,-3,4,

circle(0,2,0.15),circle(0,2,0.13),circle(0,2,0.11),circle(0,2,0.09),circle(0,2,0.07),circle(0,2,0.05),circle(0,2,0.03),circle(0,2,0.01),
circle(6,0,0.15),circle(6,0,0.13),circle(6,0,0.11),circle(6,0,0.09),circle(6,0,0.07),circle(6,0,0.05),circle(6,0,0.03),circle(6,0,0.01),


graph(400,2800/11,-3,8,-3,4) ) }}}

Then we notice that the inequality is &#8804; and not <.
If it were < we would draw the line dotted to show
that it was not part of the solution set, but since
it is &#8804; instead, the boundary line is part of the
solution so we draw the line solid through those
intercepts:

{{{drawing(400,2800/11,-3,8,-3,4,

circle(0,2,0.15),circle(0,2,0.13),circle(0,2,0.11),circle(0,2,0.09),circle(0,2,0.07),circle(0,2,0.05),circle(0,2,0.03),circle(0,2,0.01),
circle(6,0,0.15),circle(6,0,0.13),circle(6,0,0.11),circle(6,0,0.09),circle(6,0,0.07),circle(6,0,0.05),circle(6,0,0.03),circle(6,0,0.01),
line(18,-4,-12,6),


graph(400,2800/11,-3,8,-3,4) ) }}}

Next we must decide which side of the line to shade.

We pick a test point on either side of the line.  The test point
must NOT be ON the line.

Any point not on the line will do as a test point.  But the easiest
point to test, when it is not on the line is (0,0)

We substitute (x,y) = (0,0) in the original inequality:

  x + 3y &#8804; 6
0 + 3(0) &#8804; 6    
       0 &#8804; 6

That's true so since the origin is a solution and since it is
on the lower side of the line we shade the lower side, like
this:

{{{drawing(400,2800/11,-3,8,-3,4,

circle(0,2,0.15),circle(0,2,0.13),circle(0,2,0.11),circle(0,2,0.09),circle(0,2,0.07),circle(0,2,0.05),circle(0,2,0.03),circle(0,2,0.01),
circle(6,0,0.15),circle(6,0,0.13),circle(6,0,0.11),circle(6,0,0.09),circle(6,0,0.07),circle(6,0,0.05),circle(6,0,0.03),circle(6,0,0.01),
line(18,-4,-12,6),


graph(400,2800/11,-3,8,-3,4,15,y<(6-x)/3-.07),
graph(400,2800/11,-3,8,-3,4,15)

 ) }}} 
 
Edwin</pre>