Solve the following simultaneous equations:

$\dfrac{a}{x} - \dfrac{b}{y} = 0, \dfrac{ab^2}{x} + \dfrac{a^2 b}{y} = (a^2 + b^2)$

Substituting $\dfrac{1}{x} = m \text{ and } \dfrac{1}{y} = n$ in $\dfrac{a}{x} - \dfrac{b}{y} = 0$,

⇒ am - bn = 0

⇒ am = bn

⇒ m = $\dfrac{bn}{a}$     ……..(1)

Substituting $\dfrac{1}{x} = m \text{ and } \dfrac{1}{y} = n$ in $\dfrac{ab^2}{x} + \dfrac{a^2b}{y} = (a^2 + b^2)$

⇒ ab²m + a²bn = a² + b²     ………(2)

Substituting value of m from equation (1) in (2), we get :

⇒ ab² $\Big(\dfrac{bn}{a}\Big)$ + a²bn = a² + b²

⇒ b³n + a²bn = a² + b²

⇒ bn(b² + a²) = a² + b²

⇒ bn = $\dfrac{(a^2 + b^2)}{(a^2 + b^2)}$

⇒ bn = 1

⇒ n = $\dfrac{1}{b}$

Substituting value of n in equation (1), we get:

⇒ m = $\dfrac{b}{a} \times \dfrac{1}{b}$

⇒ m = $\dfrac{1}{a}$

$\Rightarrow \dfrac{1}{x} = m \\[1em] \ \Rightarrow \dfrac{1}{x} = \dfrac{1}{a} \\[1em] \Rightarrow x = a \\[1em] \Rightarrow \dfrac{1}{y} = n \\[1em] \Rightarrow \dfrac{1}{y} = \dfrac{1}{b} \\[1em] \Rightarrow y = b.$

Hence, x = a, y = b.