Question

If $\mathsf{\;y = f(x) = \dfrac{ax - b}{bx - a}}$ then show that x = f(y) ​

Answer 1

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Answer 2

Given :-

• $\mathsf{\;y = f(x) = \dfrac{ax - b}{bx - a}}$

To Prove :-

• $\mathsf{x = f(y)}$

Solution :-

We need to replace x with y.

$\mathsf \red {{:\implies f(y) = \dfrac{ay - b}{by - a}}}$

$\mathsf{But,\;y = \dfrac{ax - b}{bx - a}}$

Substituting the value of y in f(y), We get :

$\mathsf{:\implies f(y) = \dfrac{a\bigg(\dfrac{ax - b}{bx - a}\bigg) - b}{b\bigg(\dfrac{ax - b}{bx - a}\bigg) - a}}$

$\mathsf{:\implies f(y) = \dfrac{\dfrac{a(ax - b)}{bx - a} - b}{\dfrac{b(ax - b)}{bx - a} - a}}$

$\mathsf{:\implies f(y) = \dfrac{\dfrac{a(ax - b) - b(bx - a)}{bx - a}}{\dfrac{b(ax - b) - a(bx - a)}{bx - a}}}$

$\mathsf{:\implies f(y) = \dfrac{a(ax - b) - b(bx - a)}{b(ax - b) - a(bx - a)}}$

$\mathsf{:\implies f(y) = \dfrac{a^2x - ab - b^2x + ab}{abx - b^2 - abx + a^2}}$

$\mathsf{:\implies f(y) = \dfrac{a^2x - b^2x}{a^2 - b^2}}$

$\mathsf{:\implies f(y) = \dfrac{x(a^2 - b^2)}{a^2 - b^2}}$

$\mathsf\red {{:\implies f(y) = x}}$

• Hence, Showed!