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

GIVEN: 

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

$x+y=2 a b$

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

Here we have the pair of simultaneous equation

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

$x+y-2 a b=0$

By cross multiplication method we get 

$\frac{x}{(-2 a b)\left(\frac{a}{b}\right)-\left(-\left(a^{2}+b^{2}\right)\right)}=\frac{-y}{(-2 a b)\left(\frac{b}{a}\right)-\left(-\left(a^{2}+b^{2}\right)\right)}=\frac{1}{\frac{b}{a}-\frac{a}{b}}$

$\frac{x}{\left(-2 a^{2}\right)+\left(a^{2}+b^{2}\right)}=\frac{-y}{\left(-2 b^{2}\right)+\left(a^{2}+b^{2}\right)}=\frac{1}{\frac{\left(b^{2}-a^{2}\right)}{a b}}$

$\frac{x}{\left(b^{2}-a^{2}\right)}=\frac{y}{\left(b^{2}-a^{2}\right)}=\frac{1}{\frac{\left(b^{2}-a^{2}\right)}{a b}}$

$\frac{x}{\left(b^{2}-a^{2}\right)}=\frac{y}{\left(b^{2}-a^{2}\right)}=\frac{a b}{\left(b^{2}-a^{2}\right)}$

$x=y=a b$

Hence we get the value of $x=\mathrm{y}=\mathrm{ab}$