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.
 A: 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
A: A much more direct way is to consider that this kind of system is composed of an eigensystem condition + (the last equation which is plainly a normalizing condition.
Let us take your example written under the form of a slightly simplified system
$$\begin{cases}2C_1+4C_2+8C_3&=&C_1\\
3 C_1+9C_2+27C_3&=&C_2\\
4C_1+16C_2+64C_3&=&C_3\end{cases}$$
giving the matrix-vector eigen-equation (of the form $MV=XV$)
$$\underbrace{\begin{pmatrix}2&4&8\\
3&9&27\\
4&16&64\end{pmatrix}}_M\underbrace{\begin{pmatrix}C_1\\
C_2\\
C_3\end{pmatrix}}_V=X \underbrace{\begin{pmatrix}C_1\\
C_2\\
C_3\end{pmatrix}}_V$$
where there are only 3 possibilities for $X$ and $(C_1,C_2,C_3)$ to be chosen among the eigenvalues and associated eigenvectors with normalizing condition $C_1+C_2+C_3=1$ giving (for example using the matlab program below):
$$X = 71.5723 \ or \ X=3.2194 \ or \ X=0.2083$$
associated resp. to normalized vectors:
$$\begin{pmatrix}C_1\\
C_2\\
C_3\end{pmatrix}= \begin{pmatrix}0.0888\\
0.2776\\
0.6335\end{pmatrix} \ \ or \  \ \begin{pmatrix}0.6198\\
0.5713\\
-0.1912\end{pmatrix}  \ \ or \  \ \begin{pmatrix}2.2101\\
-1.4302\\
0.2201\end{pmatrix}$$
Matlab program:
 M=[2, 4,  8
    3, 9, 27
    4,16, 64];
 eig(M), % list of eigenvalues
 [P,~]=eig(M); % P is a matrix whose columns are eigenvectors
 for k=1:3
     V=P(:,k);V=V/sum(V), % normalization of eigenvectors
 end;

