Question 75653
{{{3x-y<2}}} solve for y
{{{-y<2-3x}}} Subtract 3x from both sides
{{{y>3x-2}}} Divide both sides by -1. When dividing by a negative, flip the inequality
Solve the 2nd equation for y:
{{{x+y>2}}}
{{{y>2-x}}} Subtract x from both sides
So we can now graph {{{y>2-x}}} and {{{y>3x-2}}}. We just graph the equations as we normally would if they were {{{y=3x-2}}} and {{{y=2-x}}}. Here's the graph:

{{{ graph( 300, 200, -6, 5, -10, 10, 3x-2,2-x) }}}

Now lets pick the test point (0,0) to see which region we should shade:
{{{0>2-0}}}
{{{0>2}}} false. so we don't shade the region containing (0,0) for that inequality. This means we shade above the line {{{y=2-x}}} (green line). Now lets test it on the 2nd inequality:
{{{0>3(0)-2}}}
{{{0>-2}}} true. So we shade the region containing (0,0) for the inequality {{{y>3x-2}}} (the upper side of the red line). These regions intersect in the top most region above both of the lines. This is where the solution set lies.