Question 82996
Lets graph the equations first

{{{x-y=3}}} Simply replace the inequality sign with an equal sign
{{{x+2y=6}}}


So lets graph {{{x-y=3}}} first

*[invoke graphing_linear_equations "standard", 1, -1, 3, 2, 1]


Now lets graph {{{x+2y=6}}} 

*[invoke graphing_linear_equations "standard", 1, 2, 6, 2, 1]


So lets graph {{{y=x-3}}} and {{{y=(-1/2)x+3}}} together


{{{drawing( 300, 300, -5, 5, -5, 5,
  grid( 1 ),
  graph( 300, 300, -5, 5, -5, 5, x-3, (-1/2)x+3)
)}}}


Now lets pick the test point (0,0) and evaluate {{{x-y<=3}}}

{{{(0)-(0)<=3}}}

{{{0<=3}}} true. Shade the region that does contain (0,0) for {{{x-y=3}}}. So this means you shade above the line {{{y=x-3}}}



Now lets use the same test point (0,0) and evaluate {{{x+2y<=6}}}

{{{(0)+2(0)<=6}}}

{{{0<=6}}} true. Shade the region that does contain (0,0) for {{{x+2y=6}}}. So this means you shade below the line {{{y=(-1/2)x+3}}}


So the overlapping region of these two regions would represent the solution set.