a/b+c/d+f*g=19+e. The RHS can be 21, 24, 25, 26, 28, 29.
Any products of f*g that exceed 29 can be eliminated, in other words fg<29.
The minimum value of a/b+c/d is 36/10+9/7 or about 5, while the maximum is 36/2+10/5 or about 20.
So fg<29-5 and fg<24 and fg>21-20 and fg>1.
Knowing that fg<24 we can work out possible values for f and g: (2,5), (2,6), (2,7), (2,9), (2,10). Since f and g are interchangeable we can put f=2 then g is in {5 6 7 9 10}. a and c are interchangeable as are b and d. If a/b+c/d are both to be integers then b is a factor of a and d is a factor of c. 36 has two factors in the list: 4 and 9; 10 has only one factor, 5. The other numbers have no listed factors. Let a=10 and b=5; c=36 and d can be 6 or 9.
Let's take stock. We have isolated 2, 5, 10, 36 so 6, 7, 9 remain and we know that d has to be 6 or 9. We have also reduced g to 6, 7 or 9. If g=6, d=9 and e=7; if g=7, d and e are 6 or 9; if g=9, d=6 and e=7.
So far we have 2*{6 7 9}={11 13}+{6 7 9} by substituting known values. The only way to get an even number on the RHS is to put e=7, so d and g are 6 and 9. The RHS then becomes 18 or 20. We eliminate 20 because we cannot make 20 on the LHS, so RHS=18 and g=9. That leaves d=6. Will it work?
10/5+36/6-7+2*9=2+6-7+18=19 YES!
So one solution is a=10, b=5, c=36, d=6, e=7, f=2, g=9. Interchangeability allows us to present other solutions:
a=36, b=6, c=10, d=5, e=7, f=9, g=2; a=36, b=6, c=10, d=5, e=7, f=2, g=9.
a/b and c/d are interchangeable