(I)     HJ  +  AD  =  DF
         +      +      +
(II)    BG  +  EC  =  DF
       ------------------
(III)   AC  +  GE  = HEK 
       (IV)    (V)   (VI)


Solution:
From (III) or (VI) follows H=1.
The only variable which can be zero is K.
As there are ten variables one must be zero. So K = 0. 
Looking at the last digits of (VI) we get F=5.
Now collect the other equations of the last digit
(I)    J + D == 5
(II)   G + C == 5 
(III)  C + E == 0
(IV)   J + G == C
(V)    D + C == E
where a == b means a = b (modulo 10).
This system is underdetermined as
(I) + (II) - (III) = (IV) + (V).
But (VI) gives a further equation 
20 D + 10 = 100 + 10 E that is
(VII)  2 D + 1 == E
Now combine (VII) - 2(V) + 2(III) 
 3 E + 1 == 0
which gives E=3. Substitute in the congruence
(III) gives C=7, then (II) gives G=8,
then (V) gives D=6 and (I) gives J=9.
Note that (VII) is not usefull to determine D.
The last variables are easily found.

        19  +  46  =  65
         +      +      +
        28  +  37  =  65
       ------------------
        47  +  83  = 130 



References:

Hg. Richard Zehl;
  Denken mit Spass, Die besten trend-Nüsse
  Orac Pietsch, Wien 1981, ISBN 3-85368-893-4
  - Addition von Gleichungen, p52