SOLUTION: How much of a 40% antifreeze solution must a mechanic mix with an 80% antifreeze solution if 20 gallons of a 50% antifreeze solution are needed? Here's what I tried: Let x= a

Question 607388: How much of a 40% antifreeze solution must a mechanic mix with an 80% antifreeze solution if 20 gallons of a 50% antifreeze solution are needed?
Here's what I tried:
Let x= amount of 40%, y= amount of 80%
0.40x + 0.80y = 0.5(20)
x + y = 20

Multiplied by 10 to get rid of decimals in first equation:
4x + 8y = 10
x + y = 20
Put into matrices and got:
(37.5, -17.5)
The answer is supposed to equal to 15 gal of 40%, 5 gal of 80% according to the lab worksheet I'm working on. I don't know what I am doing wrong.
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

 

Hi

Let x= amount of 40%, y= amount of 80% 

0.40x + 0.80y = 0.5(20)

 x + y = 20 OR x = 20-y

good Work! substituting (20-y) for x

0.40(20-y) + 0.80y = 0.5(20)

  .40y = .5(20)-.40(20)

     y = .10(20)/.40

     y = 5 gal of the 80% solution and 15gal of the 40% solution (5 + 15 = 20) 

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