Solving non linear equations where equations are equals I wonder and tried to google it, but I am not sure what to google it, how to solve non linear equations where equations are equal between each other. I am able to write a specific algorithm for two equations but not dynamically for N equations. I will show the example of three (how my equations approximately looks like):

C1, C2, C3, X - are unknows, but in the end I do not need to know result of X.
It can be interpreted like this (last equation C1 + C2 + C3 = 1 is not included here):

Please, don't try to solve this, I am not sure if these equations have results. I just randomly typed coefficients. But this is how my equations can looks like. Only with different coefficients. I tried to calculated it with only two unknows and I have got quadratic equation in the end so with three unknowns there will be cubic in the end. With N unknows there will be polynomial equation with degree N. Also, I have to say, result do not have to be with 100% accuracy. I am not sure if its help somehow or not.
I found on google that maybe using iterative method could help. I look at few iterative methods but I am still not sure how to use it on this kind of problem. I also found, that non linear equation can by linearize. Maybe that would be a option but I am not sure how to do it here.

From $C_1 + C_2 + C_3 = 1 $, you get $C_3 = 1 - C_1 - C_2 $
Substitute that in your equations, you get
$\dfrac{ 4 C_1 + 8 C_2 + 16 (1 - C_1 - C_2) }{2 C_1} = \dfrac{ -12 C_1 - 8 C_2 + 16} {2 C_1} = \dfrac{ - 6 C_1 - 4 C_2 + 8} {C_1} $
and
$\dfrac{ 9 C_1 + 27 C_2 + 81(1 - C_1 - C_2) }{3 C_2} = \dfrac{ - 24 C_1 - 18 C_2 + 27 }{C_2} $
and
$\dfrac{16 C_1 + 64 C_2 + 256 (1 -C_1 - C_2) }{4 C_3} = \dfrac{ -60 C_1 - 48 C_2 + 64 }{C_3} $
Since these expressions are equal as you have in your question, we can cross multiply to get
$( - 6C_1 - 4C_2 +8 ) C_2 = (-24 C_1 - 18 C_2 + 27) C_1 \hspace{25pt}(1)$
and
$(- 6 C_1 - 4 C_2+8) (1 -C_1 - C_2) = (-60 C_1 - 48 C_2 + 64 ) C_1 \hspace{25pt}(2)$
Equations (1) and (2) are two quadratic equations in $C_1 $ and $C_2$ and can be solved using the method outlined in the solution of this problem
