Question 102654
In order to graph {{{x-3y<=0}}}, we need to graph the <b>equation</b> {{{x-3y=0}}} (just replace the inequality sign with an equal sign).
So lets graph the line {{{x-3y=0}}} (note: if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver<a>)

{{{ graph( 500, 500, -20, 20, -20, 20, (1/3)x) }}} graph of {{{x-3y=0}}} 

Now lets pick a test point, say (0,1). Any point will work, (just make sure the point doesn't lie on the line) but this point is the easiest to work with. Now evaluate the inequality {{{x-3y<=0}}} with the test point


Substitute (0,1) into the inequality

{{{(0)-3(1)<=0}}} Plug in {{{x=0}}} and {{{y=1}}}

{{{-3<=0}}} Simplify





Since this inequality is  true, we shade the entire region that contains (0,1). 


{{{drawing( 500, 500, -20, 20, -20, 20,
graph(  500, 500, -20, 20, -20, 20,(1/3)x,(1/3)x+1.5),
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+4.5),
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+7.5),
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+10.5),
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+13.5),
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+16.5),
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+19.5),
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+22.5),
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+25.5),
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+28.5))}}} Graph of {{{x-3y<=0}}} with the boundary (which is the line {{{x-3y=0}}} in red) and the shaded region (in green)