$\forall x\in\mathbb R$, $|x|\neq 1$ 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.

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
