This is a shaded region graph.

Start out by drawing the graph of the boundary line, which is
just like the inequality y > 3x except that you replace the 
inequality symbol > by an equal sign =.

That is we draw the line whose equation is 

    y = 3x  Which contains the points (-1,3), (0,0), and (1,3) 

Since the inequality symbol is > and not ≥, the graph will 
not include its boundary line so we draw the graph of the boundary 
line as a dotted line like this:



Since the graph of the line y = 3x is the boundary of the graph 
of y > 3x, all the solutions are points on one side of the line.

So we arbitrarily choose a test point, which can be on either side 
of the line, but not on the line. Let's arbitrarily choose the test
point (6,5).

Now if (6,5) is a solution to y > 3x, then all the other points on the
same side of the line as (6,5) will also be solutions.  So we will
shade that side of the line.  On the other hand, if it (6,5) is not
a solution to y > 3x, then none of the other points on that side of
the line will be solutions, either, and so we will then know that
all the solutions are on the other side.  So we would shade the other
side from the side the test point is on.

So let's test the arbitrarily chosen point (6,5), marked on the graph below:



We test it by substituting x=6 and y=5 into the inequality:

y > 3x
5 > 3(6)
5 > 18

That is false so we know that none of the points on that
side of the dotted line are solutions, so we know that all
the solutions are on the other side than the side (6,5) is
on.

Therefore we shade the side of the line that the test point 
(6,5) was NOT on, which is the upper left region, so we
shade it like this green region:



Notice that if we had arbitrarily picked a test point on the other
side of the line say (-4,3),

We would have tested it by substituting x=-4 and y=3 into the inequality:

y > 3x
3 > 3(-4)
3 > -12

and that is true, so we would have known that all the solutions
are on the same side as that test point.  So we would have ended
up with the same graph, and shaded the same side of the line.

Edwin