How much pure antifreeze must be added to 12 gallons of 10% antifreeze to make a 50% antifreeze solution?

How much pure antifreeze must be added to 12 gallons of 10% antifreeze to make a 50% antifreeze solution?

I got 14.4 gallons.

x + .1(12 gallons) = .5y

x + 1.2 = .5y

I then used substitution,

x+1.2=.5(x+12)

.5x=7.2, which then gives me 14.4.

That's not the answer according to the answer key, and it say's it should be 9.6 gallons. What did I do wrong? There's a sample problem in the textbook that I used to help me work through this problem, but I have no idea where I've gone wrong where I'm not getting the correct answer.