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If $f(x)=y=\frac{a x-b}{c x-a}$, then prove that $f(y)=x$.
Solution:
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Verified Answer
$\operatorname{Given} f(x)=y=\frac{a x-b}{c x-a}$
$$
\begin{aligned}
\operatorname{Now} f(y) &=\frac{a y-b}{c y-a}=\frac{a\left(\frac{a x-b}{c x-a}\right)-b}{c\left(\frac{a x-b}{c x-a}\right)-a} \\
&=\frac{a(a x-b)-b(c x-a)}{c(a x-b)-a(c x-a)} \\
&=\frac{a^2 x-b c x}{a^2-b c}=\frac{x\left(a^2-b c\right)}{\left(a^2-b c\right)}=x
\end{aligned}
$$
$$
\therefore \quad f(y)=x \quad \text { Hence proved. }
$$
$$
\begin{aligned}
\operatorname{Now} f(y) &=\frac{a y-b}{c y-a}=\frac{a\left(\frac{a x-b}{c x-a}\right)-b}{c\left(\frac{a x-b}{c x-a}\right)-a} \\
&=\frac{a(a x-b)-b(c x-a)}{c(a x-b)-a(c x-a)} \\
&=\frac{a^2 x-b c x}{a^2-b c}=\frac{x\left(a^2-b c\right)}{\left(a^2-b c\right)}=x
\end{aligned}
$$
$$
\therefore \quad f(y)=x \quad \text { Hence proved. }
$$
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