Question 325750: A 40% antifreeze solution is to be mixed with a 70% solution to get 90 L of a 50% solution. How many liters of the 40% and 70% solutions will be needed?
I must solve this problem using a system of equations.
my first equation is x + y = 90, am I on the right track? what would the second equation be?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A 40% antifreeze solution is to be mixed with a 70% solution to get 90 L of a 50% solution. How many liters of the 40% and 70% solutions will be needed?
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Quantity Equation: x + y = 90 L
Alcohol Equation:0.4x+0.7y = 0.5*90
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Multiply thru 1st by 4
Multiply thru 2nd by 10
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4x + 4y = 4*90
4x + 7y = 5*90
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Subtract 1st from 2nd and solve for "y":
3y = 90
y = 30 L (amt of 70% antifreeze needed in the mixture)
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Since x + y = 90, x = 60 L (amt of 40% antifreeze needed in the mixture)
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Cheers,
Stan H.
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