SOLUTION: Please show me how to solve. I'm having trouble understanding from when I asked a few days ago. Solve by graphing. 3x + 4y ≤ 12 x + 3y ≤ 6 x ≥ 0 y ≥

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Please show me how to solve. I'm having trouble understanding from when I asked a few days ago. Solve by graphing. 3x + 4y ≤ 12 x + 3y ≤ 6 x ≥ 0 y ≥      Log On


   

Question 173706: Please show me how to solve. I'm having trouble understanding from when I asked a few days ago.
Solve 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!
Solve by graphing.
3x + 4y ≤ 12
x + 3y ≤ 6
x ≥ 0
y ≥ 0
:
Plot the 1st two equations, put the equations in the slope/intercept form
3x + 4y =< 12
4y = 12 - 3x
y = 12%2F4 - 3%2F4x
y = -3%2F4x + 3
:
Calculate two points (substitute the values for x and find y):
x | y
-------
0 | 4
4 | 1
Graph should look like this:
+graph%28+300%2C+200%2C+-4%2C+5%2C+-4%2C+6%2C+-.75x%2B4%29+
:
Do the same with the 2nd equation
x + 3y => 6
3y = 6 - x
y = 6%2F3 - x%2F3
y = -1%2F3x + 2
Calculate two points (substitute the values for x and find y):
x | y
-------
0 | 2
3 | 1
Graph should look like this:
+graph%28+300%2C+200%2C+-4%2C+5%2C+-4%2C+6%2C+-.33x%2B2%29+
:
The last two equations tell you that only positive values for x and y are considered
:
Put both graphs on the same coordinate system:
+graph%28+300%2C+200%2C+-4%2C+5%2C+-4%2C+6%2C+-.75x%2B4%2C+-.33x%2B2%29+
Look at this graph
The 1st equation:3x + 4y =< ; 12, the purple line
Area of feasibility is at or below this line
:
The 2nd equation: x + 3y => 6, green line
Area of feasibility is at or above this line
:
x => 0
y => 0
As I said before, this means only positive values for x & y, inside the area of feasibility are considered.
:
The solution is the triangular area between the lines and to the right of the y axis.