Solving a system of equations involving functions

Question

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 !.

Answer 1

When given such a system of equations, you are expected to find a function (here $f$) that satisfies all those equations for all values of $x$.

Let $\frac{x-1}x=y$ and $\frac1{1-x}=z$. Then $$f(x)+3f(y)=7x\tag1$$ $$f(y)+3f(z)=7y\tag2$$ $$f(z)+3f(x)=7z\tag3$$ Eliminate $f(z)$ by subtracting $(3)$ from $(2)$ thrice: $$f(y)-9f(x)=7y-21z\tag4$$ Eliminate $f(y)$ by subtracting $(4)$ from $(1)$ thrice: $$f(x)+27f(x)=28f(x)=7x-21y+63z$$ $$4f(x)=x-3y+9z$$ Hence we have the desired expression for $f(x)$: $$f(x)=\frac14\left(x-\frac{3(x-1)}x+\frac9{1-x}\right)$$ $$=\frac{x^2(1-x)-3(x-1)(1-x)+9x}{4x(1-x)}$$ $$=\frac{x^2-x^3+3-6x+3x^2+9x}{4x(1-x)}$$ $$=\frac{-x^3+4x^2+3x+3}{4x(1-x)}$$

Answer 2

From equation (C) you'll find f ($\frac {1}{1-x}$) next step put it in equation (B) and find f ($\frac{x-1}{x}$) and put it in (A) hence you have the value of f (x). So you have the answer(function) that applies in the system.

Answer 3

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 $f(x)$.