Solve the following system of equations by using the method of cross-multiplication:
$\frac{a}{x} - \frac{b}{y} = 0 \frac{a b^{2}}{x} + \frac{a^{2} b}{y} = a^{2} + b^{2}, where x \neq 0 and y \neq 0$

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Solution

Substituting $\frac{1}{x} = u and \frac{1}{y} = v$ in the given equations, we get
$a u - b v + 0 = 0 . . . . . (i) a b^{2} u + a^{2} b v - (a^{2} + b^{2}) = 0 . . . . . (i i)$
Here, $a_{1} = a, b_{1} = - b, c_{1} = 0 and a_{2} = a b^{2}, b_{2} = a^{2} b, c_{2} = - (a^{2} + b^{2})$ .
So, by cross-multiplication, we have
$\frac{u}{b_{1} c_{2} - b_{2} c_{1}} = \frac{v}{c_{1} a_{2} - c_{2} a_{1}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}} \Rightarrow \frac{u}{(- b) [- (a^{2} + b^{2})] - (a^{2} b) (0)} = \frac{v}{(0) (a b^{2}) - (- a^{2} - b^{2}) (a)} = \frac{1}{(a) (a^{2} b) - (a b^{2}) (- b)} \Rightarrow \frac{u}{b (a^{2} + b^{2})} = \frac{v}{a (a^{2} + b^{2})} = \frac{1}{a b (a^{2} + b^{2})}$
$\Rightarrow u = \frac{b (a^{2} + b^{2})}{a b (a^{2} + b^{2})}, v = \frac{a (a^{2} + b^{2})}{a b (a^{2} + b^{2})} \Rightarrow u = \frac{1}{a}, v = \frac{1}{b} \Rightarrow \frac{1}{x} = \frac{1}{a}, \frac{1}{y} = \frac{1}{b} \Rightarrow x = a, y = b$
Hence, x = a and y = b.

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