Is there a certain formula that shows how much time does it take to solve a system of more than $200$ linear equation manually? 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 ?
 A: The number of arithmetical operations required to solve a system of $n$ linear equations in $n$ unknowns by Gaussian elimination is on the order of $n^3$, so the question is, how long does it take you to do a multiplication by hand? But the problem is that the numbers you have to work with may grow very large, so you'll either be doing your arithmetic with 200-digit numbers, or you'd be using only a few significant digits and get into problems with rounding errors. 
A: Unless the system has a special structure, Gaussian elimination takes $\frac{n^3-n}3$ additions, $\frac{n(n-1)(2n+5)}6$ multiplications and $\frac{n(n+1)}2$ divisions.
When done by hand the usual way and keeping $d$ significant digits, additions take an effort grossly proportional to $d$, multiplications and division proportional to $d^2$.
As the proportionality constant will vary from person to person, and possibly slighty with $d$, it is better to measure an average value by timing for the desired $d$.
A very crude approximation is $$Cn^3d^2.$$

Assuming you can process a $4\times4$ system to the desired accuracy in an hour (professional calculator ?), it would take like $125000$ hours for $200\times200$, i.e. $52$ years, counting $8$ hours a day and working $300$ days a year.

From the answer by @GerryMyerson, we should indeed take into account that larger systems require more digits of accuracy. By backward error analysis, the error is on the order of $8n^3g(A)u$ where $u$ is the precision of the individual computations and $g(A)$ is the so-called growth factor of the matrix. It can reasonably be taken to equal $n$.
So the formula could be written
$$C'n^3\log^2n.$$
There is certainly a residual dependency on $d=\log n$ due to the fact that carries involve larger and larger numbers and other nonlinear effects such as tendinitis and aging of the calculator. But this is probably not measurable for "small" sizes like $n=200$.

Note that for large $d$, multiplications and divisions are better done using specialized tables of logarithms, so that the complexity lowers from $O(d^2)$ to $O(d)$ or so. (Though the cost of the interpolation process is hard to evaluate.)
