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

GIVEN: 

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

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

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

Here we have the pair of simultaneous equation

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

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

By cross multiplication method we get

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

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

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

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

So for x we have

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

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

$x=a$

And 

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

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

$y=b$

Hence we get the value of $x=a$ and $y=b$