Question 79316
4x - 3y <= 3 or
2x + y >= 2 
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You first have to graph the EQUALITIES:
y=(1/3)(4x-3=(4/3)x-1
y=-2x+2
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{{{graph(300.200,-10,10,-10,10,(4/3)x-1,-2x+2)}}}
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Now you need to find the half-plane solution set for each INEQUALITY:
Pick a test point that is not on the boundary line y=(4/3)x-1
Pick (0,0)
Substitute that into the INEQUALITY, y>(4/3)x-1 to see if the side of the line
that contains (0,0) satisfies the INEQUALITY:
0>(3/4)*0-1
This is true so the half-plane containing (0,0) is the solution set for 
4x - 3y <= 3; shade that half-plane that is above the boundary line.
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Pick a test point that is not on the boundary line for y=2x+2.
Pick (0,0)
Substitute that into the INEQUALITY, 2x + y >= 2 :
2*0+0>2; 0>2
That is false so the half-plane that does NOT contain (0,0) is the solution
set for 2x+y>2; shade that half-plane.
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Finally, find the intersection of the two shaded half planes to get
the solution for the system of inequalities.
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Cheers,
Stan H.