(I)     AB  +   C  =  AC
         +      +      +
(II)    AD  +   E  =  FB
       ------------------
(III)   FD  +   G  =  EC
       (IV)    (V)   (VI)


Solution:
(I.1)   B = 0
(II.1)  D + E = 10
(II.2)  A + 1 = F
(III.2) F + 1 = E
(VI.2)  A + F = E
The last two equations determine A.
 A = 1
 F = 2
 E = 3
 D = 7
We are left with equation (V) C + 3 = G.
There are two possible solutions
C=5, G=8 and C=6, G=9.


        10  +   5  =  15
         +      +      +
        17  +   3  =  20
       ------------------
        27  +   8  =  35 


        10  +   6  =  16
         +      +      +
        17  +   3  =  20
       ------------------
        27  +   9  =  36 


References:

W. N. Bolchowitinow, B. I. Koltowoi, I. K. Lagowski;
  Spass für freie Stunden - Rätsel Spiele Denkaufgaben,
  Verlag für die Frau, Leipzig, 1980, 2. Auflage
  Problem 194 A: (with E and G already given)

Johannes Lehmann;
  Mathe mit Pfiff,
  Manz Verlag 1977, ISBN 3-7863-0383-5
  Chap: Kryptarithmetik p71-72
  Problem 23: (with A, E, and G already given)

Johannes Lehmann;
  Mathe mit Herz,
  Urania Verlag 1991, ISBN 3-332-00479-4, new edition of "Mathe mit Pfiff"
  Chap: Kryptarithmetik p81-83
  Problem 23: (with A, E, and G already given)