Proving that the inverse of $\frac{(1-x)y}{x(1-y)}=n$ is $\frac{(1-x)y}{x(1-y)}=\frac{1}{n}$

Question

I wish to prove that the inverse of $$\frac{(1-x)y}{x(1-y)}=n$$ is $$\frac{(1-x)y}{x(1-y)}=\frac{1}{n}$$

for some positive real constant $n,$ with $x,y \in \Bbb R \cap(0,1).$

I started off by rewriting the first implicit function in explicit form:

$$ y=\frac{n(x-1)}{n(x-1)-x}. $$

Next I swapped $x$ and $y.$

$$x=\frac{n(y-1)}{n(y-1)-y}.$$

Then I solved for $y.$

$$ y=\frac{n(x-1)}{(n-1)x-n}. $$

Is there an easier way to do this? I found the inverse but not in the form that I wanted it in.

Answer 1

Just swap $x$ and $y$: $$\frac{(1 - y)x}{y(1 - x)} = n$$ This is already the inverse. Take the reciprocal of both sides and you'll have it in the form given in the statement.

Of course, we are ignoring any divide-by-zero issue.

Answer 2

Assume $x,y\ne1$.

Then $\dfrac y{1-y}=\dfrac{nx}{1-x}\iff y-yx=nx-nxy$

$\iff x=\dfrac y{(1-y)n+y} \iff \dfrac x{1-x}=\dfrac y{(1-y)n}$.

Now switch the roles of $x$ and $y$ to get the desired result.