SOLUTION: Solve the following system of linear inequalities by showing work and graphing. x – 3y > 6 3x + 2y > 12 Can anyone offer any help on this problem?

Algebra ->  Linear-equations -> SOLUTION: Solve the following system of linear inequalities by showing work and graphing. x – 3y > 6 3x + 2y > 12 Can anyone offer any help on this problem?       Log On


   



Question 83464: Solve the following system of linear inequalities by showing work and graphing.
x – 3y > 6
3x + 2y > 12
Can anyone offer any help on this problem?

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the following system of linear inequalities by showing work and graphing.
x – 3y > 6
3x + 2y > 12
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solve eqch inequality for y, as follows:
y < (1/3)x-2
y > (-3/2)x+6
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Now, graph the EQUALITY associated with each inequality:
graph%28-10%2C10%2C-10%2C10%2C%281%2F3%29x-2%2C%28-3%2F2%29x%2B6%29
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Each of these is a boundary of the corresponding INEQUALITY.
They are not part of the graph so should be drawn as dashed lines.
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Now, determine where the half-plane INEQUALITY lies relative to
each boundary. How???
1st Pic a test point, like (0,0)
Substitute those values into the INEQUALTIY to determine if the
point is in the half-plane solution.
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Substituting in the 1st InEQUALITY you get 0<(1/3)x-2; 0<-2
That is false so the half-plane on the other side of y=(1/3)x-2
should be shaded.
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Substituting in the 2nd INEQUALITY you get 0>(-3/2)*0+6
That is true so the half-plane on the side of y>(-3/2)x+6
which contains the test point is the half-plane graph of
y>(-3/2)x+6. That half-plane should be darkened.
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the intersection of the two darkened half-plnes is the
graph of the solution set of the system of inequalities.
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Cheers,
Stan H.