Question 965308
First graph 
{{{2x-y=-2}}}
{{{y=2x+2}}}
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{{{graph(300,300,-10,10,-10,10,2x+2)}}}
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Then choose a point not on the line and test the inequality.
(0,0) is convenient.
{{{2(0)-0>-2}}}
{{{0>-2}}}
True, shade the region containing (0,0).
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{{{graph(300,300,-10,10,-10,10,2x-y>-2)}}}
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Now do the same for the other inequality,
{{{x-y=-1}}}
{{{y=x+1}}}
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{{{graph(300,300,-10,10,-10,10,x+1)}}}
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Then choose a point not on the line and test the inequality.
(0,0) is convenient.
{{{0-0>-1}}}
{{{0>-1}}}
True, shade the region containing (0,0).
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{{{graph(300,300,-10,10,-10,10,x-y>-1)}}}
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So now putting it all together,
{{{graph(300,300,-10,10,-10,10,2x+2,x+1)}}}
Find the point of intersection,
{{{2x+2=x+1}}}
{{{x=-1}}}
Then,
{{{y=-1+1=0}}}
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{{{drawing(300,300,-10,10,-10,10,blue(line(-10,-18,-1,0)),
blue(line(-1,0,20,21)),graph(300,300,-10,10,-10,10,0))}}}
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The solution region would still be the region containing (0,0) but now its boundaries have been modified.
I can't show it here like the graphs above due to software limitations.
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*[illustration eqa.JPG].