SOLUTION: Could you please help me with this problem. I am supposed to use a system of equations to solve this question. A chemist needs 10 liters of a 60% alcohol solution. She has 30% and

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A chemist needs 10 liters of a 60% alcohol solution. He has 30% and 80% alcohol solutions available to mix. How many liters of each need to be mixed to obtain 10 liters of a 60% alcohol solution.
:
Let x = amt of 30% solution
Let y = amt of 80% solution
:
Total amt equation:
x + y = 10
x = (10-y); we can use this for substitution
:
Alcohol equation
30x + 80y = 60(10)
30x + 80y = 600
Simplify divide by 10
3x + 8y = 60
:
Substitute (10-y) for x in the above equation, find y
3(10-y) + 8y = 60
30 - 3y + 8y = 60
5y = 60 - 30
y =
y = 6 liters of 80% solution
then
x = 10 - 6
x = 4 liters of 30% solution
:
:
check solution by using the actual amt of alcohol
.3x + .8y = .6(10)
.3(4) + .8(6) = .6