But you are combining two different samples beginning in different states to form a third. You can tap this general idea, but not without accounting for the difference in amount of matter in the two samples as jon_mx's solution does above.

I think that since the two individual tanks are the same volume and temp, you could simplify it by assuming all the initial gas and it's corresponding pressure was in one of the two tanks.

V1 = 100 gal (tank 1)

P1 = 100 psi + 14.7 psi (1 atm from tank 1) + 14.7 psi (actually from tank 2 but assumed to be in tank 1) = 129.4

V2 = 200 gal (both tanks together)

P2 = ?

P1V1 = P2V2

(129.4)(100) = (P2)(200)

P2 = 64.7 psi ----- pressure gauge reads 64.7-14.7 = 50

I think this also makes it clearer why the initial hidden atmospheric pressures shouldn't change anything in the final result.

If PG = pressure gauge reading and A= atmospheric pressure

then P1 = PG readings of the two tanks plus atmospheric pressure for each of the tanks assumed all in one tank

and P2 = the final PG reading plus the atmospheric pressure for the combined tank

P1V1 = P2V2

(PG1 + 0 + A + A)(100) = (PG2 + A)(200)

(PG1 + 2A) = (PG2 + A)(2)

PG1 + 2A = 2(PG2) +2A

PG1 = 2(PG2)