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

GIVEN:

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

$(a+b)(x+y)=4 a b$

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

Here we have the pair of simultaneous equation

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

$(a+b) x+(a+b) y-4 a b=0$

By cross multiplication method we get

$\frac{x}{(-4 a b)(a+b)-(a+b)\left(-2 a^{2}+2 b^{2}\right)}=\frac{-y}{(-4 a b)(a-b)-(a+b)\left(-2 a^{2}+2 b^{2}\right)}$

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

$\frac{x}{(a+b)\left((-4 a b)-\left(-2 a^{2}+2 b^{2}\right)\right)}=\frac{-y}{(-4 a b)(a-b)-(a+b)\left(-2 a^{2}+2 b^{2}\right)}$

$=\frac{1}{(a+b)((a-b)-(a+b))}$

Consider the following for x

$\frac{x}{(a+b)\left((-4 a b)-\left(-2 a^{2}+2 b^{2}\right)\right)}=\frac{1}{(a+b)((a-b)-(a+b))}$

$\frac{x}{(-4 a b)-\left(-2 a^{2}+2 b^{2}\right)}=\frac{1}{(a-b)-(a+b)}$

$x=\frac{4 a b+2 a^{2}-2 b^{2}}{2 b}$

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

Now consider the following for y 

$\frac{-y}{(-4 a b)(a-b)-(a+b)\left(-2 a^{2}+2 b^{2}\right)}=\frac{1}{(a+b)((a-b)-(a+b))}$

$\frac{-y}{(-4 a b)(a-b)-(a+b)\left(-2 a^{2}+2 b^{2}\right)}=\frac{1}{(a+b)((a-b)-(a+b))}$

$\frac{y}{(4 a b)(a-b)+(a+b)\left(-2 a^{2}+2 b^{2}\right)}=\frac{1}{(a+b)(-2 b)}$

$\frac{y}{(4 a b)(a-b)+(a+b)\left(-2 a^{2}+2 b^{2}\right)}=\frac{1}{(a+b)(-2 b)}$

$\frac{y}{(4 a b)(a-b)+(a+b)(-2)\left(a^{2}-b^{2}\right)}=\frac{1}{(a+b)(-2 b)}$

$\frac{y}{(4 a b)(a-b)+(a+b)(-2)(a-b)(a+b)}=\frac{1}{(a+b)(-2 b)}$

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

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

Hence we get the value of $x=\frac{2 a b+b^{2}-a^{2}}{b}$ and $y=\frac{(a-b)\left(a^{2}+b^{2}\right)}{(a+b) b}$