Question 70071
solve the following system of linear inequalities by graphing. 
x+2y<=3
2x-3y<=6
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Solve each for y as follows:
y<=(-1/2)x+(3/2)
y>=(2/3)x-2
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Graph the EQUALITIES involved as they are the boundary of the solution sets
y=(-1/2)x+(3/2) passes thru (0,3/2) and (2,1/2); graph it and shade half-plane
below the equality line.
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y=(2/3)x-2 passes thru (0,-2) and (3,0); graph it and shade half-plane above
the equality line.
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The solution set is all the points that satisfy both conditions for "y".
The equality lines look like this:
{{{graph(600,400,-10,10,-10,10,(-1/2)x+(3/2),(2/3)x-2)}}}
Cheers,
Stan H.