SOLUTION: please help me set up this equation : a chemistry student mixes a 10% solution with a 20% solution to get 20L of a 13% solution. how much of each is used? i tried to do x + y =2

Question 405455: please help me set up this equation : a chemistry student mixes a 10% solution with a 20% solution to get 20L of a 13% solution. how much of each is used?
i tried to do x + y =20(.13), .1x + .2y = .13
and y= -2.47, so i don't think i did it right...
also, i have several more problems like this, so it would be useful to have a generic equation to plug the numbers into...

thanks
Found 2 solutions by stanbon, mananth:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A chemistry student mixes a 10% solution with a 20% solution to get 20L of a 13% solution. how much of each is used?
---------
Equations
active + active = active
0.10x + 0.20(20-x) = 0.13*20
Multiply thru by 100 to get:
10x + 20*20 - 20x = 13*20
-10x = -7*20
x = (1/2)7
x = 3.5L (amt of 10% solution needed)
---
20-x = 16.5L (amt of 20% solution)
======================================
Cheers,
Stan H.
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Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
percent ---------------- quantity
10.00% ---------------- x
20.00% ---------------- 20-x
13.00% ---------------- 20
...
10x+20(20-x)=13*20

10x+400 -20x=260
10x-20x =260-400
-10x-=-140
/-10
x=14 liters=====> 10% solution
balance = 6 liters 20%

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