SOLUTION: I need to translate this problem into a pair of linear equations in two variables. How many liters of a 40%-alcohol solution must be mixed with 10 liters of a solution that is 80%

Question 335003: I need to translate this problem into a pair of linear equations in two variables. How many liters of a 40%-alcohol solution must be mixed with 10 liters of a solution that is 80% alcohol to get a solution that is 60% alcohol? Can anyone please help me?

Answer by nyc_function(2741) (Show Source): You can put this solution on YOUR website!
You get two equations, one based on total volume, and one based on the mix:
x + 10 = y
x(0.4) + 10(0.8) = y(0.6)
Substitute the first equation for "y" into the second equation and solve for "x":
x(0.4) + 10(0.8) = (x + 10)(0.6)
x(0.4) + 8 = x(0.6) + 6
x(0.4) - x(0.6) = 6 - 8
x(-0.2) = -2
x = -2/-0.2 = 10
Therefore, y = 20.
10 liters of a 40% alcohol solution must be mixed with 10 liters of a 80% alcohol solution to get a 20 liters of a 60% alcohol solution.

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