Question 364777
Graph the lines {{{x=-3}}} and {{{x+2y=2}}}
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{{{x+2y=2}}}
{{{2y=-x+2}}}
{{{y=-(1/2)x+1}}}
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{{{drawing(300,300,-10,10,-10,10,blue(line(-3,10,-3,-10)),grid(1),graph(300,300,-10,10,-10,10,0,(2-x)/2))}}}

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The two lines break up the plane into 4 sections.
Since you know that {{{x>-3}}}, only the 2 right hand sections are the possible solution regions.
Choose a point in either of the two right hand sections, not on the line. 
({{{0}}},{{{0}}}) is a convenient choice.
Test the inequalities, if the inequalities are satisfied, shade the region containing ({{{0}}},{{{0}}}).
If not, shade the region that doesn't contain ({{{0}}},{{{0}}}).
{{{x>-3}}}
{{{0>-3}}}
True, but we already knew that one.
{{{x+2y>=2}}}
{{{0+2(0)>=2}}}
{{{0>=2}}}
False, so shade the region to the right of the blue line and above the green line.


*[invoke plot_any_inequality "x+2y>=2", -3, 10, -10, 10, 300, 300]