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?

We know that
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
x> the volume in gallons of solution taken from Tank A.
y>the volume in gallons of solution taken from Tank B
Then
the volume taken from Tank B will be y=(8 -x)
remember that we are making a total of 8 gallons of solutions.

The alcohol content of the mixture is(1/3)x +(1/5)(8 -x) = (1/4)*85x + 3(8 -x) = 30 > 5x+24-3x=30> 5x-3x=30-24
2x = 6 > x=3
y=8-x> y=8-3> y=5

the answer is
3 gallons should be taken from Tank A;
5 gallons should be taken from Tank B