How do I get the inverse of this equation?

Question

The equation is the following: $$y = \frac{x}{x-2}$$

What I did is:

$$y = \frac{x}{x-2}$$

$$y \times (x-2) = x$$

$$\frac{x-2}{x} = \frac{1}{y}$$

$$\frac{x}{x}-\frac{2}{x} = \frac{1}{y}$$

$$1-\frac{2}{x} = \frac{1}{y}$$

$$1-\frac{x}{2} = y$$

I've got a lot of doubts about that last step. Besides that, the result according to the textbook is the following:

$$ f^{−1}(x)=-\frac{2x}{x−1} $$

Answer 1

Sorry, no, you can't take recirocals of the terms the way you have.

That is

$\frac {1}{a+b} \ne \frac 1a + \frac 1b$

Anyway

$y = \frac{x}{x-2}\\ y(x-2) = x\\ yx - 2y = x\\ yx - x = 2y\\ x(y-1) = 2y\\ x = \frac {2y}{y-1}$

$f^{-1}(y) = \frac {2y}{y-1}$

Answer 2

Assuming $x\neq2$, we can multiply both sides by $x-2$, giving $xy-2y=x\to x(y-1)=2y$. If $y=1$, there is no $x$ that satisfies the equation. If $y\neq1$, then $$x=\frac{2y}{y-1}$$

Answer 3

A different approach is to note that $$ y = \frac{x}{x-2} = \frac{(x-2)+2}{x-2} = 1+\frac{2}{x-2} $$ Hence, $(x-2)(y-1) = 2$ and $x = \frac{2}{y-1}+2 = \frac{2y}{y-1}$.

Answer 4

\begin{align} y &= \frac{x}{x-2} . \end{align}

\begin{align} y&=\frac{x+2-2}{x+2} ,\\ y&=1-\frac 2{x+2} ,\\ y-1&=-\frac 2{x+2} ,\\ -\frac{y-1}2&=\frac 1{x+2} ,\\ x+2 &=\frac 2{1-y} ,\\ x &=\frac 2{1-y}-2; ,\\ x &=\frac{2\,y}{1-y} , \end{align}

so the answer is indeed

\begin{align} f^{-1}(x)=\frac{2x}{1-x} . \end{align}