Solving a system of equations involving functions Solve this system of equations:
$$
\left\{\begin{array}{cccccr}
\displaystyle{\mathrm{f}\left(x\right)} & \displaystyle{+} &
\displaystyle{3\mathrm{f}\left(x - 1 \over x\right)} & \displaystyle{=} & 
\displaystyle{7x} & \displaystyle{\qquad\qquad\qquad\qquad\left(\mathrm{A}\right)}
\\
\displaystyle{\mathrm{f}\left(x - 1 \over x\right)} & \displaystyle{+} &
\displaystyle{3\mathrm{f}\left(1 \over 1 - x\right)} & \displaystyle{=} & 
\displaystyle{7x - 7 \over x} & \displaystyle{\left(\mathrm{B}\right)}
\\
\displaystyle{\mathrm{f}\left(1 \over 1 - x\right)} & \displaystyle{+} &
\displaystyle{3\mathrm{f}\left(x\right)} & \displaystyle{=} & 
\displaystyle{7 \over 1 - x} & \displaystyle{\left(\mathrm{C}\right)}
\end{array}\right.
$$
I've never solved a system of equations with functions, so I'm brand new to this concept. Could someone clue me out or provide a solution ?. Thanks !.

Considering $$f(x)+3f(\tfrac{x-1}x)=7x\tag 1$$
$$f(\tfrac{x-1}x)+3f(\tfrac1{1-x})=\tfrac{7x-7}x\tag 2$$
$$f(\tfrac1{1-x})+3f(x)=\tfrac7{1-x}\tag 3$$ just rename $A=f(x)$, $B=f(\tfrac{x-1}x)$, $C=f(\tfrac1{1-x})$.
This gives $$A+3B=7x \tag 4$$ $$B+3C=\tfrac{7x-7}x\tag 5$$ $$C+3A=\tfrac7{1-x}\tag 6$$ that is to say three linear equations. Eliminate $C$ from $(6)$, replace in $(5)$ to get $B$, replace in $(4)$ to get $A$ which is $f(x)$.
