Question 675208
<pre>
y < {{{7/3}}}x - 5

First we draw the graph of the boundary line y = {{{7/3}}}x - 5.
Two points on that line are (0,-5) and (3,2).  We must draw the line
dotted because the inequality is < and therefore does not include 
the boundary line:

{{{drawing(400,400,-7,13,-15,5, graph(400,400,-7,13,-15,5,((7/3)x-5)*sqrt(sin(10x))/sqrt(sin(10x))),circle(0,-5,.1),circle(3,2,.1), locate(3,2,"(3,2)"),
locate(0,-5,"(0,-5)")

 )}}}

Since the line does not go through the origin we can use (0,0) as
a test point to see whether the origin is a solution.  If it is a
solution then all points are on the same side of the boundary line 
as the origin and if the origin is not a solution, all the solutions
are on the side the origin is not on.  So we substitute (x,y) = (0,0)
in the original inequality:

y < {{{7/3}}}x - 5
0 < {{{7/3}}}·0 - 5
0 < -5

That is false so we must shade the side of the line that the origin
is not on:

{{{drawing(400,400,-7,13,-15,5, graph(400,400,-7,13,-15,5,((7/3)x-5)*sqrt(sin(10x))/sqrt(sin(10x))),
graph(400,400,-7,13,-15,5,25,y<(7/3)x-5.3), graph(400,400,-7,13,-15,5)

 )}}}

Edwin</pre>