Question 823297
<pre>

3x-y > -6

First you graph, just the bountary line, whose equation is
just like the inequality with an = sign instead of >

The equation of the boundary line is

3x-y = -6 

We getting a couple or three points, maybe the intercepts 
(-2,0) and (0,6), maybe one other (-3,-3).  Since the ineuality
symbol is > and not &#8806;, we draw the line dotted, not solid:

{{{drawing(400,400,-5,3,-3,8, graph(400,400,-5,3,-3,8,(3x+6)*sqrt(sin(20x))/sqrt(sin(20x))),
circle(0,6,0.15),circle(0,6,0.13),circle(0,6,0.11),circle(0,6,0.09),circle(0,6,0.07),circle(0,6,0.05),circle(0,6,0.03),circle(0,6,0.01),
circle(-2,0,0.15),circle(-2,0,0.13),circle(-2,0,0.11),circle(-2,0,0.09),circle(-2,0,0.07),circle(-2,0,0.05),circle(-2,0,0.03),circle(-2,0,0.01),

circle(-3,-3,0.15),circle(-3,-3,0.13),circle(-3,-3,0.11),circle(-3,-3,0.09),circle(-3,-3,0.07),circle(-3,-3,0.05),circle(-3,-3,0.03),circle(-3,-3,0.01)






 )}}} 

Now we select a test point to find out which side of the line we
are to shade.  We can test any point that isn't on the line; however
since the origin (0,0) is not on the line, we test to see if it is a
solutio to the original inequality.  If it is then all solutions will
be on the same side of the dotted line that the origin is on. If the
origin is not a solution, we will shade the other side.

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

    3x-y > -6
3(0)-(0) > -6
       0 > -6

That is true because 0 is greater than any negative number.  So we
shade the side of the line that the origin is on:

{{{drawing(400,400,-5,3,-3,8, graph(400,400,-5,3,-3,8,(3x+6)*sqrt(sin(20x))/sqrt(sin(20x)),10,10,10,10,y<3x+5.8),

graph(400,400,-5,3,-3,8)


 )}}} 

Edwin</pre>