SOLUTION: Could someone help me out? How much of a 40% antifreeze solution must a mechanical mix with an 80% antifreeze solution if 20 L of a 50% antifreeze solution are need? I tried th

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Could someone help me out? How much of a 40% antifreeze solution must a mechanical mix with an 80% antifreeze solution if 20 L of a 50% antifreeze solution are need? I tried th      Log On


   

Question 174976: Could someone help me out?
How much of a 40% antifreeze solution must a mechanical mix with an 80% antifreeze solution if 20 L of a 50% antifreeze solution are need?
I tried the following and was marked wrong.
Let c represent the 40% solution
___ y _____________ 80% ________
x + y= 20
40x+80y= 50
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40x+40y=8
40x+40y=.5
-4y=7.5
y=18.75
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x+18.75=20
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x=1.25 y= 18.75
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How much of a 40% antifreeze solution must a mechanic mix with an 80% antifreeze solution if 20 L of a 50% antifreeze solution are needed?
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active + active = active
0.40x + 0.80(20-x) = 0.50*20
40x + 1600 - 80x = 1000
-40x = -600
x = 15 liters (amount of 40% antifreeze needed in the mixture)
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20-x = 5 liters (amount of 80% antifreeze needed in the mixture)
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Cheers,
Stan H.