SOLUTION: Mixing water and antifreeze: How much water should be added to i gallon of pure antifreeze to obtain a solution that is 60% antifreeze? I need help setting this problem up.

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Question 157631This question is from textbook College Algebra
: Mixing water and antifreeze: How much water should be added to i gallon of pure antifreeze to obtain a solution that is 60% antifreeze?
I need help setting this problem up.This question is from textbook College Algebra

Answer by ptaylor(1894) About Me  (Show Source):
You can put this solution on YOUR website!

I AM ASSUMING THAT THE "i" SHOULD BE A "1"
Let x=amount of water needed
Now we know that the amount of pure antifreeze that we have before the water is added (1 gal) has to equal the amount of pure antifreeze that we have after the water is added(0.60(1+x)). So our equation to solve is
1=0.60(1+x) get rid of parens
1=0.60+0.60x subtract 0.60 from each side
1-0.60=0.60-0.60+0.60x collect like terms
0.40=0.60x divide each side by 0.60
x=2/3 gal-------------------------amount of pure water needed
CK
1=0.60(1 2/3)
1=0.60(5/3)
1=0.20*5
1=1

Hope this helps---ptaylor