how to inverse my formula in bitwise xor to get the value of 53?

Question

This is my formula in hex.
(53+3C+7-1)xor (45+9B)
And the result of this formula is : 75 <-- hex
My problem is I need to get 53 back using another formula.

Answer 1

some pseudo-code:

int startValue = 0x53
int resultValue = (startValue + 3C + 7 - 1) xor (45 + 9B)
int initialValue = (resultValue xor (45 + 9B)) - 3C - 7 + 1

so, you just xor the same 45+9B and reverse other operations (+ and -)