Question 52998
Let x=amt of pure antifreeze that must be added

Solution (1)
Amt of pure antifreeze in the 8 gals (8)(.40) plus amt of pure antifreeze 
added (x)(1) equals the amt of pure antifreeze in the final mixture (x+8)(.60).
Thus, our equation to solve is:
(8)(.4)+(x)(1)=(x+8)(.60) simplifying, we get:
3.2+x=.6x+4.8 or
.4x=1.6
x=4 gal

Solution (2)
If we assume the mixture is antifreeze and water, notice that the amt of water does not change throughout the process.
Amt of water in the 8 gal (8)(.60) plus the amt of water added (0)equals the amt of water in the final mixture (x+8)(.40).
Thus, our equation to solve is:
(8)(.60)=(x+8)(.40) simplifying, we get:
4.8=.4x+3.2 or
.4x=1.6
x=4 gal

Hope this helps---ptaylor