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

Solution:

GIVEN:

$(a+2 b) x+(2 a-b) y=2$

$(a-2 b) x+(2 a+b) y=3$

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

Here we have the pair of simultaneous equation

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

$(a-2 b) x+(2 a+b) y-3=0$

By cross multiplication method we get

$\frac{x}{((2 a-b) \times-3)-((2 a+b) \times(-2))}=\frac{-y}{(-3) \times(a+2 b)-((-2) \times(a-2 b))}$

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

$\frac{x}{(-6 a+3 b)-(-4 a-2 b)}=\frac{-y}{(-3 a-6 b)-(-2 a+4 b)}$

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

$\frac{x}{(-2 a+5 b)}=\frac{-y}{(-a-10 b)}=\frac{1}{10 a b}$

$\frac{x}{(-2 a+5 b)}=\frac{y}{(a+10 b)}=\frac{1}{10 a b}$

$\Rightarrow \frac{x}{(-2 a+5 b)}=\frac{1}{10 a b}$

$\Rightarrow x=\frac{(5 b-2 a)}{10 a b}$

And

$\frac{-y}{(-a-10 b)}=\frac{1}{10 a b}$

$\Rightarrow \frac{y}{(a+10 b)}=\frac{1}{10 a b}$

$\Rightarrow y=\frac{(a+10 b)}{10 a b}$

Hence we get the value of $x=\frac{5 b-2 a}{10 a b}$ and $y=\frac{a+10 b}{10 a b}$