To solve this system of equations, we can start by adding the first two equations together to eliminate B:

A + B + C = 110
A - B - C = 2
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2A = 112
A = 56

Now we can substitute the value of A back into the first equation to solve for B:

56 + B + C = 110
B + C = 54
B = 54 - C

Next, we can substitute the values of A and B into the third equation:

B - 15 = C + 15
54 - C - 15 = C + 15
39 = 2C
C = 19.5

Finally, we can substitute the values of A and C back into the equation B = 54 - C to solve for B:

B = 54 - 19.5
B = 34.5

Therefore, the solution to the system of equations is A = 56, B = 34.5, and C = 19.5.