SOLUTION: Okay, I really need someone's help with this. I have asked everyone I know if they could help. And they CAN'T. Solve the linear inequalities by graphing: 3x+4y<=12 x+3y<=6

Algebra ->  Graphs -> SOLUTION: Okay, I really need someone's help with this. I have asked everyone I know if they could help. And they CAN'T. Solve the linear inequalities by graphing: 3x+4y<=12 x+3y<=6       Log On


   

Question 82635: Okay, I really need someone's help with this. I have asked everyone I know if they could help. And they CAN'T.
Solve the linear inequalities by graphing:
3x+4y<=12
x+3y<=6
x=>0
y=>0
PLEASE HELP!!!!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
3x+%2B+4y+%3C=+12
x+%2B+3y+%3C=+6
x+%3E=+0
y+%3E=+0
Look at the 3rd and 4th first. x+%3E=+0 simply says that x is
never negative. y+%3E=+0 says y is never negative. The area
on the graph where both are true is the upper right quadrant, so
all solutions must be there as a restriction.
Solve the 1st and 2nd for y.
3x+%2B+4y+%3C=+12
4y+%3C=+12+-+3x
y+%3C=+3+-+%283%2F4%29x
-------------
x+%2B+3y+%3C=+6
3y+%3C=+6+-+x
y+%3C=+2+-+x%2F3
now plot these
+graph%28+300%2C+300%2C+-2%2C+8%2C+-2%2C+8%2C+3+-+%283%2F4%29x%2C+2+-+x%2F3%29+
The solution is the area below the lines between (0,2) and (4,0)
Note that both these points are solutions and neither one
violates x+%3E=+0 or y+%3E=+0.