To get equal quantities of wine and water you need to get half of the wine into the jug with two pulls of the jug. So you need to get 5 gallons of wine:

First Jug + Second Jug = 5

The problem is the Second Jug will not be all wine it will be a ratio of wine and water that is contingent on the size of the jug. That ratio can be written as (10 - J) / 10. Plug that back into the original formula:

First Jug + Second Jug * Ratio

J + J ((10 - J) / 10) = 5, distribute the J

J + 10J/10 - J^2/10 = 5, combine terms

2J + 1/10 (J^2) = 5, rearrange into recognizable quadratic

(1/10)J^2 + 2J - 5 = 0

Plug that **** into a solver on the internet because ain't nobody remember the quadratic formula.

J = -5 (sqrt(2) - 2) and J = 5(sqrt(2) + 2)

Or J ~ {2.93, 17.07}

So the answer must be 2.93 because 17.07 is more than the keg holds.