Solve each of the following systems of equations by the method of cross-multiplication :

GIVEN: 

$\frac{a^{2}}{x}-\frac{b^{2}}{y}=0$

$\frac{a^{2} b}{x}+\frac{a b^{2}}{y}=a+b$

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

$\frac{a^{2}}{x}-\frac{b^{2}}{y}=0$

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

Let $\frac{1}{x}=u$ and $v=\frac{1}{y}$

Rewriting equations 

$a^{2} u-b^{2} v=0 \ldots \ldots(1)$

$a^{2} b u+a b^{2} v-(a+b)=0 \ldots \ldots(2)$

Now, by cross multiplication method we get

$\frac{u}{\left(-(a+b)\left(-b^{2}\right)\right)-(0)}=\frac{-v}{\left(-(a+b)\left(a^{2}\right)\right)-(0)}=\frac{1}{\left(a^{3} b^{2}\right)+\left(a^{2} b^{3}\right)}$

$\frac{u}{\left(a b^{2}+b^{3}\right)}=\frac{v}{\left(a^{3}+b a^{2}\right)}=\frac{1}{\left(a^{3} b^{2}+a^{2} b^{3}\right)}$

For u consider the following

$\frac{u}{\left(a b^{2}+b^{3}\right)}=\frac{1}{\left(a^{3} b^{2}+a^{2} b^{3}\right)}$

$\frac{l l}{(a+b)}=\frac{1}{a^{2}(a+b)}$

$u=\frac{1}{a^{2}}$

For y consider

$\frac{v}{\left(a^{3}+b a^{2}\right)}=\frac{1}{\left(a^{3} b^{2}+a^{2} b^{3}\right)}$

$\frac{v}{(a+b)}=\frac{1}{b^{2}(a+b)}$

$v=\frac{1}{b^{2}}$

We know that

$\frac{1}{x}=u$ and $v=\frac{1}{y}$

Now 

$\frac{1}{x}=\frac{1}{a^{2}}$ and $\frac{1}{b^{2}}=\frac{1}{y}$

$x=a^{2}$ and $y=b^{2}$

Hence we get the value of $x=\mathrm{a}^{2}$ and $\mathrm{y}=b^{2}$