Find all functions $f:\mathbb R \to \mathbb R$ that satisfy $$ f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x $$ for all nonzero $x$.
My thoughts so far are to plug in $(x - 1)/x$ for $x,$ but it always comes out so complicated, so I don't really know where to go from there.
$$f(x) + 3 f\left( 1-\frac {1}{x} \right) = 7x$$ $x\to\frac{1}{1-x}$: $$ f\left(\frac{1}{1-x}\right)+3f(x)=7\left(\frac{1}{1-x}\right) $$ $x\to1-\frac{1}{x}$: $$ f\left(1-\frac{1}{x}\right)+3f\left(\frac{1}{1-x}\right)=7\left(1-\frac{1}{x}\right) $$ Hence, we have: $$ \begin{cases} \begin{align} f(x)+3 f\left( 1-\frac {1}{x} \right)+0 &= 7x\\ 3f(x)+0+f\left(\frac{1}{1-x}\right)&=7\left(\frac{1}{1-x}\right)\\ 0+f\left(1-\frac{1}{x}\right)+3f\left(\frac{1}{1-x}\right)&=7\left(1-\frac{1}{x}\right) \end{align} \end{cases} $$ Solve for $f(x)$: $$ f(x)=-\frac{3}{4}\left(1-\frac{1}{x}\right)+\frac{x}{4}+\frac{9}{4}\left(\frac{1}{1-x}\right) $$