$\forall x\in\mathbb R$, $|x|\neq 1$ it is known that $f\left(\frac{x-3}{x+1}\right)+f\left(\frac{3+x}{1-x}\right)=x$. Find $f(x)$. 
$\forall x\in\mathbb R$, $|x|\neq 1$ $$f\left(\frac{x-3}{x+1}\right)+f\left(\frac{3+x}{1-x}\right)=x$$Find $f(x)$.

Now what I'm actually looking for is an explanation of a solution to this problem. I haven't really ever had any experience with such equations.

The solution:
Let $t=\frac{x-3}{x+1}$. Then $$f(t)+f\left(\frac{t-3}{t+1}\right)=\frac{3+t}{1-t}$$
Now let $t=\frac{3+x}{1-x}$. Then $$f\left(\frac{3+t}{1-t}\right)+f(t)=\frac{t-3}{t+1}$$
Add both equalities: $$\frac{8t}{1-t^2}=2f(t)+f\left(\frac{t-3}{t+1}\right)+f\left(\frac{3+t}{1-t}\right)=2f(t)+t$$
Hence the answer is $$f(x)=\frac{4x}{1-x^2}-\frac{x}{2}$$

This is unclear to me. For instance, how come we can assign a different value to the same variable? Does anyone understand this? I'd appreciate any help.
 A: I don't know how they came up with this solution; but it is correct. 
Look at the first line: For all but finitely many $t$ the number $x:={3+t\over 1-t}$ is admissible, and one has ${3+x\over 1-x}={t-3\over t+1}$ and  ${x-3\over x+1}=t$.  Since your functional equation is true for all but finitely many $x$ it follows that
$$f(t)+f\left(\frac{t-3}{t+1}\right)=\frac{3+t}{1-t}\tag{1}$$
is true for all but finitely many $t$. Similarly, the equation
$$f\left(\frac{3+t}{1-t}\right)+f(t)=\frac{t-3}{t+1}\tag{2}$$
is true for all but finitely many $t$.
In $(1)$ and $(2)$ we can write as well $x$ in place of $t$. Adding these equations (with $t$ replaced by $x$) and comparing with the original functional equation we now can say that
$$2f(x)+x={8x\over 1-x^2}$$
for all but finitely many $x$. It follows that
$$f(x)=\frac{4x}{1-x^2}-\frac{x}{2}\tag{3}$$
for all but finally many $x$.
But it's not over yet: We only have proven that any solution to the original functional equation is of the form $(3)$ for most $x$. We now have to test whether $(3)$ is actually a solution. This is a simple verification which I can leave to you.
A: This is one way to think about it:
For any $t ≠ \pm 1$ you can write $t$ as $t = \tfrac{x-3}{x+1}$ or as $t = \tfrac{3+x}{1-x}$ if you set $x = \tfrac{3+t}{1-t}$ or $x = \tfrac{t-3}{t+1}$ respectively (which means that $z ↦ \tfrac{z-3}{z+1}$ and $z ↦ \tfrac{3+z}{1-z}$ are really involutons inverse functions on $ℝ \setminus \{\pm 1\}$). In your solution, the $x$’s of the two fractions are actually different ones, the $t$ stays the same.
So you can translate your solution to

Let $t ≠ \pm 1$.
Write $t = \tfrac{x_1-3}{x_1+1}$ (such $x_1$ exists), then $f(t) + …$
Write $t = \tfrac{3+x_2}{x_2-1}$ (such $x_2$ exists), then $f(\tfrac{3+t}{1-t}) + …$

The conclusions are still true because the functional equations still hold for the $x_i$’s you used to write $t$ (you have to check they are not $\pm 1$, though). Then you can safely add both equalities without contradiction.

Also, it is worth mentioning that the Mobius transformation
$$g \colon ℝ \setminus \{\pm 1\} → ℝ \setminus \{\pm 1\},\, x ↦ \tfrac{x - 3}{x + 1}$$
actually has order $3$, that is $g^3 = \operatorname{id}$, or $g^2 = g^{-1}$, where $g^{-1}$ is actually given by $g^{-1} (x) = \tfrac{3 + x}{1 - x}$. So $g$ and $g^{-1}$ correspond to the fractions you are examining. Then you can think of it that way:
It is given that:
$$f∘g + f∘g^{-1} = \operatorname{id}$$
But then, since $g$ has order $3$, you have.
\begin{align*}
(f + f∘g)∘g &= f∘g + f∘g^2 &=&  \operatorname{id}&, \quad \text{and}\\
(f∘g^{-1} + f)∘g^{-1} &= f∘g^{-2} + f∘g^{-1} &=&  \operatorname{id}&
\end{align*}
And so, by multiplying with $g$ or $g^{-1}$ from the right you have:
$$f + f∘g = g^{-1}, \quad \text{and} \quad f∘g^{-1} + f = g$$
And so, by adding those and using the functional equation again:
$$g + g^{-1} = (f + f∘g) + (f∘g^{-1} + f) = 2f + \operatorname{id},$$
from which you can derive your result $f = \frac{g + g^{-1} - \operatorname{id}}{2}$.
A: It's a bit unclear, but if you look for what $t$'s each of the equalities hold, it starts to make sense.
Note that if $x\ne-1$ and $t_0\ne1$, then $t_0=\frac{x-3}{x+1}\Longleftrightarrow(x+1)t_0=x-3\Longleftrightarrow t_0+3=x(1-t_0)\Longleftrightarrow x=\frac{3+t_0}{1-t_0}$
So the first equality holds for all $t_0\in\mathbb R\setminus\{1\}$.
Similarly if $x\ne1$ and $t_1\ne-1$, then $t_1=\frac{3+x}{1-x}\Longleftrightarrow(1-x)t_1=3+x\Longleftrightarrow t_1-3=x(1+t_1)\Longleftrightarrow x=\frac{t_1-3}{t_1+1}$
So the second equality holds for all $t_1\in\mathbb R\setminus\{-1\}$.
Now for any $x\in\mathbb R$, $|x|\ne1$ we know that $$f(x)+f\left(\frac{x-3}{x+1}\right)=\frac{3+x}{1-x}$$ by the first equation, substituting $x$ for $t_0$ and $$f\left(\frac{3+x}{1-x}\right)+f(x)=\frac{x-3}{x+1}$$ by the second equation, substituting $x$ for $t_1$.
Now add these to get the solution.
A: t is a symbolic name for the variable, you can rename $x-3\over x+1$ to z and $3+x \over 1-x$ to y and then rename the z and y in the equations to t...
A: Since it works for all $x$ it means that $t$ has to take all the values of the domain,since t will be equal to all of those values,both $$\frac{x-3}{x+1},\frac{x+3}{1-x}$$
Will take all those values.so basically $f(t)$ in first will be equal to the $f(t)$ in other,or you can take that for example $$a=\frac{x-3}{x+1},b=\frac{x+3}{1-x}$$
and for example $f(a)=7a$ then $f(b)=7b$ so just changing the letter b into a it will become the same equation
