If $f(x)=y=\frac{a x-b}{c x-a}$, then prove that $f(y)=x$.
We have, $\quad f(x)=y=\frac{a x-b}{c x-a}$
$\begin{aligned} \therefore \quad f(y) & =\frac{a y-b}{c y-a}=\frac{a\left(\frac{a x-b}{c x-a}\right)-b}{c\left(\frac{a x-b}{c x-a}\right)-a} \\ & =\frac{a(a x-b)-b(c x-a)}{c(a x-b)-a(c x-a)}=\frac{a^2 x-a b-b c x+a b}{a c x-b c-a c x+a^2} \\ & =\frac{a^2 x-b c x}{a^2-b c}=\frac{x\left(a^2-b c\right)}{\left(a^2-b c\right)}=x \\ \therefore \quad f(y) & =x\quad \text{Hence proved.}\end{aligned}$
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