<s>[INST] Question about functional equations 
Let $F(x)$ be the real-valued function defined for all real $x$ except for $x = 0$ and $x = 1$ and satisfying the functional equation $$F(x) + F\left(\frac{x-1}x\right) = 1+x$$ Find  $F(x)$ satisfying these conditions.

Write $F(x)$ as a rational function with expanded polynomials in the numerator and denominator.
I substituted $x = \frac{x-1}x, -\frac{1}{x-1}$. This gave me the system of equations
$F(x) + F\left(\frac{x-1}x\right) = 1+x, 
F( \frac{x-1}x ) + F(- \frac{1}{x-1} ) = 1+ \frac{x-1}x, 
F(- \frac{1}{x-1} ) + f(x) = 1- \frac{1}{x-1}.$
However, solving this system of equations gave me $F(x) = \frac{x^3-x^2-1}{2(x^2-2)}$, which is incorrect. Where did I go wrong?
 [/INST] It's just a simplification error. Let $g(x)=\frac{x-1}{x}$. Then, note
$$
g(x)=1-\frac{1}{x},\quad g(g(x))=\frac{1}{1-x},\quad g(g(g(x)))=x.
$$
And so you have the system:
\begin{align*}
1+x&=F(x)+F(g(x)),\\
1+g(x)&=F(g(x))+F(g(g(x))),\\
1+g(g(x))&=F(g(g(x)))+F(x)
\end{align*}
from which
\begin{align*}
F(x)&=[F(x)+F(g(x))+F(g(g(x)))]-[F(g(x))+F(g(g(x)))]\\
&=\frac{1}{2}[1+x+1+g(x)+1+g(g(x))]-(1+g(x))\\
&=\frac{x^3-x^2-1}{2(x^2-\color{red}{x})}.
\end{align*}
</s>