Question 83028
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}}}


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

{{{(0)+2(0)>6}}}

{{{0>6}}} false. Shade the region that doesn't contain (0,0) for {{{x+2y=6}}}. So this means you shade above the line {{{y=(-1/2)x+3}}}


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Now the overlapping regions is the complete shading. In other words, every point that satisfies
 
{{{x-y<3}}} 
{{{x+2y>6}}}

is in this region. So this is what the overlapped shaded region looks like:


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