System of equations with Lambert $W$ function I have a system of equations with two equations containing the Lambert $W$ function as follows,
$$\begin{cases} x = 1 - W_0(\frac{C_1 e^{y + 1}}{y + 1}) \\ y = 1 - W_0(\frac{C_2 e^{x + 1}}{x + 1}) \end{cases}$$
I have already solved the system numerically using a simple iterative method. A gentleman told me that the answer to the system is $y = \frac{k_1 - k_2 x}{k_1 x - k_2}$ where $k_1 = C_2 - C_1$ and $k_2 = C_2 - C_1$ but I have no clue how can one compute the answer.


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*How can I get the answer to the system (i.e the steps)?

*While we have two curves in the system of equation, the intersection (i.e the answer of the system) must be a point not a curve. But $y = \frac{k_1 - k_2 x}{k_1 x - k_2}$ is a hyperbola. How can I describe it?

 A: $$\begin{cases} 1-x =  W_0(\frac{C_1 e^{y + 1}}{y + 1}) \\ 1-y = W_0(\frac{C_2 e^{x + 1}}{x + 1}) \end{cases}$$
Using the poperty of the Lambert function  $Y=W(X)$ invertion : $Ye^Y=X$
$$\begin{cases} (1-x)e^{1-x}=\frac{C_1 e^{y + 1}}{y + 1}   \\ (1-y)e^{1-y}=\frac{C_2 e^{x + 1}}{x + 1} \end{cases}$$
$$C_1C_2e^{x+y}=C_2(1-x)(y+1)=C_1(1-y)(x+1)$$
$$C_2(1-x)y+C_1(x+1)y= C_1(x+1)-C_2(1-x)$$
$$y=\frac{C_1(x+1)-C_2(1-x)}{C_2(1-x)+C_1(x+1)}$$
This leads to the expected relationship :
$$y=\frac{(C_1+C_2)x+C_1-C2}{(C_1-C_2)x+C_1+C_2}$$
Then we have to separate the equation with unknown $x$ from the equation with unknown $y$.
$$y+1= \frac{2C_1(x+1)x}{ (C_1-C_2)x+C_1+C_2}$$
$$(1-x)e^{1-x}= \frac{C_1 e^{y + 1}}{y + 1} =\frac{C_1 e^{\frac{2C_1(x+1)x}{ (C_1-C_2)x+C_1+C_2}}}{\frac{2C_1(x+1)x}{ (C_1-C_2)x+C_1+C_2}}$$
$$2x(x+1)(1-x)e^{1-x}=((C_1-C_2)x+C_1+C_2) e^{\frac{2C_1(x+1)x}{ (C_1-C_2)x+C_1+C_2}}$$
This equation contains the unknown $x$ only. A similar one can be found containing $y$ only. But, the more likely, those equations cannot be solved for the unknown expressed on closed form. 
