\(a + b = c\)
\(b - a = d\)
\(a < b\)
\(a \neq 0\)
(If a was 0, then b and d would be the same)
If a was 1, b could be any number from 3 - 8 (6 numbers)
If a was 2, b could be 3, 5, 6, 7 (4 numbers)
If a was 3, b could be 4, 5 (2 numbers)
If a was 4, b could only be 5 (1 number)
There are no other possible values for a
Therefore there are 6 + 4 + 2 + 1 = 13 possibilities.