Question 151858
First, graph the line {{{y=x/3}}}. Note: the equation is the same as {{{y=(1/3)x}}}



{{{ drawing(500, 500, -20, 20, -20, 20,
 graph( 500, 500, -20, 20, -20, 20,(1/3)x)

)}}} Graph of {{{y=x/3}}}



Now let's pick a test point (that's not on the line), say (0,1) and plug it into the inequality {{{y>x/3}}}



{{{y>x/3}}} Start with the given inequality.



{{{1>0/3}}} Plug in {{{x=0}}} and {{{y=1}}}



{{{1>0}}} Simplify



Since the inequality is true, this means that we shade the <font size=4><b>entire</b></font> region that the point (0,1) is in. So we simply shade everything above the line {{{y=x/3}}} like this:


{{{drawing( 500, 500, -20, 20, -20, 20,
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+3),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+6),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+9),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+12),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+15),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+18),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+21),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+24),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+27),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+30),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+33),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+36),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+39),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+42),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+45),
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+48))}}} Graph of {{{y>x/3}}} with the shaded region in green