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

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

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

Solution:

On substituting $\frac{1}{x}=u$ and $\frac{1}{y}=v$ in the equations given, we obtain $\mathrm{au}-\mathrm{bv}+0=0$
$a b^{2} u+a^{2} b v-\left(a^{2}+b^{2}\right)=0$
Here, $a_{1}=a, b_{1}=-b, c_{1}=0, a_{2}=a b^{2}, b_{2}=a^{2} b$ and $c_{2}=-\left(a^{2}+b^{2}\right)$
On cross-multiplication, we get:

$\frac{u}{b_{1} c_{2}-b_{2} c_{1}}=\frac{v}{c_{1} a_{2}-c_{2} a_{1}}=\frac{1}{a_{1} b_{2}-a_{2} b_{1}}$
$\Rightarrow \frac{u}{(-b)\left[-\left(a^{2}+b^{2}\right)\right]-\left(a^{2} b\right)(0)}=\frac{v}{(0)\left(a b^{2}\right)-\left(-a^{2}-b^{2}\right)(a)}=\frac{1}{(a)\left(a^{2} b\right)-\left(a b^{2}\right)(-b)}$
$\begin{array}{l} \Rightarrow \frac{u}{b\left(a^{2}+b^{2}\right)}=\frac{v}{a\left(a^{2}+b^{2}\right)}=\frac{1}{a b\left(a^{2}+b^{2}\right)} \\ \Rightarrow \mathrm{u}=\frac{b\left(a^{2}+b^{2}\right)}{a b\left(a^{2}+b^{2}\right)}, \mathrm{V}=\frac{a\left(a^{2}+b^{2}\right)}{a b\left(a^{2}+b^{2}\right)} \\ \Rightarrow \mathrm{u}=\frac{1}{a}, \mathrm{v}=\frac{1}{b} \\ \Rightarrow \frac{1}{x}=\frac{1}{a}, \frac{1}{y}=\frac{1}{b} \\ \Rightarrow \mathrm{x}=\mathrm{a}, \mathrm{y}=\mathrm{b} \end{array}$
As a result, $x=a$ and $y=b$