Question

If $y=f\left(x\right)=\frac{px+q}{px-p}$, then find $f\left(y\right)$ in terms of $x$.

Answer 1

$y=f\left(x\right)=\frac{px+q}{px-p}$

$\therefore f\left(y\right)=\frac{py+q}{py-p}$

$⟹f\left(y\right)=\frac{p\left(\frac{px+q}{px-p}\right)+q}{p\left(\frac{px+q}{px-p}\right)-p}$

$⟹f\left(y\right)=\frac{p(px+q)+q(px-p)}{p(px+q)-p(px-p)}$

$⟹f\left(y\right)=\frac{p^{2}x+pq+pqx-pq}{p^{2}x+pq-p^{2}x+p^2}$

$⟹f\left(y\right)=\frac{p^{2}x+pqx}{pq+p^2}$

$⟹f\left(y\right)=\frac{px(p+q)}{p(q+p)}=\frac{px}{p}$

$\therefore f\left(y\right)=x$