Question 82664
Solve the following ststem of linear inequalities by graphing
x+2y<3
2x-3y<6
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solve each equation for y:
y<(-1/2)x+(3/2)
y>(2/3)x-2
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Graph the EQUALTIY which is the boundary of each of the INEQUALITIES:
Each graph is drawn as a dashed line because the EQUALITY Is not part
of the final answer:
{{{graph(400,300,-10,10,-10,10,(-1/2)x+(3/2),(2/3)x-2)}}}
Pick a test point in-between the lines, like (0,0)
Substitute the test point values into each INEQUALITY to 
see where the solution half-plane is:
0<(-1/2)*0+(3/2); 0<(3/2); that is true so the test point satisfies 
0>(2/3)*0-2; 0>-2; that is true so the test point satisfies.
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The half-plane below the 1st equation and above the 2nd equation
is the solution set; that is a wedge-shaped area; see it?
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Cheers,
Stan H.