(I)     ABC  :     C  =    DE
         -        *        +
(II)     AE  +    AF  =    DD
       ----------------------
(III)   BGA  -   FHC  =    IJ
       (IV)      (V)     (VI)


Solution:
First of all I show that there is no standard solution.
That means no leading zero and different letters represent
different digits. Assume there were a solution.
Then F>=1 from (III) and further B>=2 from (III).
Now A>=3 from (IV). With (II) we see D>=2A>=6.
But now (VI) shows I >= 2D >= 12 which is too big for a 
digit in a decimal number system.

So we are looking for a non standard solution.
When I say A==B, that means A=B (mod 10).
(IV.1)  E + A == C
(III.1) J + C == A
(IV.1) + (III.1)  E + J == 0
(VI.1)  E + D == J
(IV.1) + (III.1) + (VI.1)  2E + D == 0  =>  D is even
(II.1)  E + F == D
(IV.1) + (III.1) + (VI.1) + (II.1)  3E + F == 0
(III)  B >= F
(IV)   A >= B
(II)   D >= A
  D >= A >= B >= F
(II.2)  2A = D  no carry as D>=F
(VI)   D <= 4  and therefore A <= 2.
There are three cases for A.
Case A=0, D=0: Then E=F=0 from (II) and I=J=0 from (VI).
  Then all others letters get zero from (III).
Case A=1, D=2: Then E is 4 or 9 from 2E + D == 0.
  But this contradicts (II).
Case A=2, D=4: Then E is 3 or 8 from 2E + D == 0.
  But this E=8 contradicts (II).
  The case E=3 gives the only solution.

       215  :     5  =    43
        -        *        +
        23  +    21  =    44
      ----------------------
       192  -   105  =    87


References:

J. Petigk, 
  Mathematik in der Freizeit, 
  Aulis Verlag Deubner & Co, Köln, 1998, 
  p151