Question: If we were to solve a system of equations numerically , we would be able to find the time it took for our calculations , but what If I wanted to solve it manually ? Is there some sort of a formula or a relation that illustrates how much time does it take ?

Answer: 
Factorising we get $xy(x-y)(x+y)(x^2 + y^2) \equiv 0 \mod 2^4\cdot 3\cdot 5$

Because $x, y$ have the same parity then, if $x,y$ even all factors are divisible by 2, else if $x,y$ odd all but $xy$ are divisible by 2 and $(x^2 - y^2)$ is divisible by 8. Therefore $xy(x-y)(x+y)(x^2 + y^2) \equiv 0 \mod 2^4$ 

Using Fermat's little theorem, we have $x^5 \equiv x \mod 5$ and $y^5 \equiv y \mod 5$ therefore $x^5y\equiv y^5x\pmod {5}$ also $x^3 \equiv x \mod 3$ and $y^3 \equiv y \mod 3$ therefore $x^5 \equiv x \mod 3$ and $y^5 \equiv y \mod 3$
Thanks to Chinese Reminder Theorem we need to check mod $3$, $5$ and $16$ only.
Using that $a^4 \equiv 1 (mod 5)$ for every invertible $a$ in $ \mathbb Z\_5$, and $a^4 \equiv a^2 \cdot a^2 \equiv 1 (mod 3)$ for every invertible $a$ in $ \mathbb Z\_3$ we are almost done. Now notice that if $x$, $y$ have the same parity, if they are both even we are done mod $16$. Otherwise $(2d+1)^4 = (4d^2 + 4d +1)^2\equiv 1 + 8(d^2+d) \equiv 1 (mod 16)$.