Tank a contains a mixture of 10 gallons water and 5 gallons pure alcohol. tank b has 12 gallons water and 3 gallons alcohol. how many gallons should be taken from each tank and combined in order to obtain an 8 gallon solution containing 25% alcohol by volume?

The mixture in Tank A is 5/(5+10) = 1/3 pure alcohol. The mixture in Tank B is 3/(3+12) = 1/5 pure alcohol. We want to combine these to make a mixture that is 25% = 1/4 pure alcohol.

Let a represent the volume in gallons of solution taken from Tank A. Then the volume taken from Tank B will be (8 -a) because we are making a total of 8 gallons of solutions. The alcohol content of the mixture is
.. (1/3)a +(1/5)(8 -a) = (1/4)*8
.. 5a + 3(8 -a) = 30 . . . . . . . . multiply by 15
.. 2a = 6 . . . . . . . . . . . . . . . . . subtract 24, collect terms
.. a = 3

3 gallons should be taken from Tank A; 5 gallons should be taken from Tank B.