We have had solutions to this problem from Alan Hemphill, Atharv Tillu from Bedford Modern School in Bedford, England, Andrei Lazanu from School 205 in Bucharest, Romania, Julia Collins from Langley Park School for Girls in Bromley, England, Clement Goh from River Valley High School in Singapore, Christopher Tynan from St. Bees School in Cumbria, England, Rosamund Feeney from Riccarton High School in New Zealand and from Aditya Sahu. Well done to you all.

Everyone approached the question in a similar way. Aditya Sahu's approach follows:

We can simplify most of these equations first, before we start giving values to all of the letters.

As A + C = A we know that C = 0.

As F * D = F we know that D = 1.

B - G = G, therefore B = 2G (add G to both sides of the equation).
B / H = G therefore B = HG (multiply both sides by H).

Since B = HG, and B also equals 2G,
H must be 2.

Since B is twice G, the options are:

So B = 6 and G = 3

E - G = F, therefore E - 3 = F.
The smallest possible value for F is 4, and this means that E would be 7.
F cannot be any greater than 4 because this would mean that E > 7 and this is not allowed.
Therefore F = 4 and E = 7

Now we can solve A + H = E.
This is A + 2 = 7, so A = 5.

Here are the solved equations:

And all the values of A-H: