Find $f(5)$ where $f$ satisfies $f(x)+f(1/(1-x))=x $ Question:
How do you Find $f(5)$ in which the function satisfies 
$$f(x)+f\left(\frac{1}{1-x}\right)=x $$
where $x\in\Bbb{R}$ and  $x\neq 0,1$?

My steps:
Step 1)
Substitute $5$ into the equation to get:
$$f(5)+f\left(\frac{1}{-4}\right)=5$$
But then I had gotten stuck there and I could not find $f(5)$
Please write detailed steps.
 A: You can use the fact that$$\left( \frac{1}{1-x} \right)^{-1}=1-\frac{1}{x},$$where the exponent $-1$ stands for the reverse. If you substitute $1/(1-x)$ and $1-1/x$ in the functional equation and solve three simultaneous equations, you can find general form of $f(x)$.
A: $$f\left(5\right)+f\left(-\frac14\right)=5$$
If we know $f\left(-\frac14\right)$, we can solve the problem.
$$f\left(-\frac14\right)+f\left(\frac45\right)=-\frac14$$
If we know $f\left(\frac45 \right)$, we can solve the problem.
$$f\left(\frac45\right)+f\left(5\right)=\frac45$$
Why don't we just solve the linear system? Are you able to solve it?
A: by using $x=5,4/5,-1/4$ we have:
$$f\left(-\frac 1 4\right)+f\left(\frac 4 5\right)=-\frac 1 4$$
$$ f(5)+f\left(\frac 4 5\right)=\frac 4 5$$
$$f(5)+f\left(-\frac 1 4\right)=5$$
Then sum the last two expressions and subtract the first to get:
$$ 2f(5)+f\left(\frac 4 5\right)+f\left(-\frac 1 4\right)-f\left(-\frac 1 4\right)-f\left(\frac 4 5\right)=5+\frac 45+\frac 1 4$$
hence $2f(5)=6.05$ and then $f(5)=3.025$.
