SOLUTION: a chemist must mix 6 liters of a 30% acid solution with some of an 80% solution to get a 50% acid solution. How many liters of the 80% solution should be used. I think I have th

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Question 609196: a chemist must mix 6 liters of a 30% acid solution with some of an 80% solution to get a 50% acid solution. How many liters of the 80% solution should be used.
I think I have the equation set up correctly:
0.80x + 0.30*6=0.50(x+5)
but i honestly don't know how to go from there.

Found 2 solutions by ankor@dixie-net.com, Alan3354:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
0.80x + 0.30*6 = 0.50(x+5)
.8x + 1.8 = .5x + 2.5
.8x - .5x = 2.5 - 1.8
.3x = .7
x = .7%2F.3
x = 21%2F3 liters of the 80% stuff

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
a chemist must mix 6 liters of a 30% acid solution with some of an 80% solution to get a 50% acid solution. How many liters of the 80% solution should be used.
I think I have the equation set up correctly:
0.80x + 0.30*6=0.50(x+5)
--------------
0.8x + 0.30*6 = 0.50*(x + 6) not x+5, you got it right o/w
Solve for x
Multiply by 10 to eliminate fractions
8x + 18 = 5x + 30
x = 4 liters