MCQ
If $y = f(x) = \frac{{ax + b}}{{cx - a}}$, then $x$ is equal to
- A$1/f(x)$
- B$1/f(y)$
- C$yf(x)$
- ✓$f(y)$
$⇒ x(cy - a) = b + ay$
$⇒ x = \frac{{ay + b}}{{cy - a}} = f(y)$.
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$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5} \text { and } \frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is :