SOLUTION: write the equatio of the line with slope -2 and y-intercept (0,0). Then the graph the line. Solve the following system of linear inequalities by graphing. 3x+4y<12 X+3y<6

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: write the equatio of the line with slope -2 and y-intercept (0,0). Then the graph the line. Solve the following system of linear inequalities by graphing. 3x+4y<12 X+3y<6       Log On


   

Question 83543: write the equatio of the line with slope -2 and y-intercept (0,0). Then the graph the line.


Solve the following system of linear inequalities by graphing.
3x+4y<12
X+3y<6
x>0
y>0
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
write the equation of the line with slope -2 and y-intercept (0,0).
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It's just y = -2x
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If you use the point/slope equation: y - 0 = -2(x - 0), y = -2x
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Then the graph the line.
+graph%28+300%2C+200%2C+-6%2C+5%2C+-10%2C+10%2C+-2x%29+

Solve the following system of linear inequalities by graphing.
3x+4y<12
4y < -3x + 12
y < (-3/4)x + 12/4
y < (-3/4)x + 3; plot this graph from x = 0 to x = +6
:
x + 3y < 6
3y < -x + 6
y < (-1/3)x + 6/3
y < (-1/3)x + 2; plot this graph from x = 0 to x = +6
x>0
y>0
It should look like this
+graph%28+300%2C+200%2C+-2%2C+6%2C+-2%2C+6%2C+%28-3%2F4%29x%2B3%2C+%28-1%2F3%29x%2B2%29+
:
The area of feasibility is below the lowest line, but positive x & y
The corners of this area are: 0,0; 4,0; 0,2; and at the intersection 2.4, 1.2
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