Question 23160
I'm not sure what you did in that.  It looks more complicated to me than it needs to be.  Let me start over.


Let x = amount of water to be added, and let's calculate the amount of alcohol in the mixture before and after the water is added.  Since you are only adding water, the amount of alcohol before equals the amount of alcohol after, and this is basis for the equation.


Amount of alcohol before = .80* 3500
Add x liters of water = No Alcohol
Amount of alcohol after = .50*(3500 + x)


Equation:
.80*3500 = .50(3500+x)
2800 = 1750+.50x
2800-1750 = .50x
1050 = .50x


Divide both sides by .50:
{{{1050/.50 = (.50x)/.50 }}}
{{{2100 = x}}}


Add 2100 ml of water.


Check:

Beginning alcohol= .80(3500) = 2800 ml.
After water added, alcohol = .50( 3500 + 2100)= .50(5600) = 2800 ml.


R^2 at SCC