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

Solution:

GIVEN:

$2(a x-b y)+a+4 b=0$

 

$2(b x+a y)+b-4 a=0$

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

Here we have the pair of simultaneous equation

$2(a x-b y)+a+4 b=0$

 

$2(b x+a y)+b-4 a=0$

After rewriting equations

$2 a x-2 b y+(a+4 b)=0$

 

$2 b x+2 a y+(b-4 a)=0$

By cross multiplication method we get

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

$\frac{x}{\left(-2 b^{2}+8 a b\right)-\left(2 a^{2}+8 a b\right)}=\frac{-y}{\left(2 a b-8 a^{2}\right)-\left(2 b a+8 b^{2}\right)}=\frac{1}{4 a^{2}+4 b^{2}}$

$\frac{x}{-2\left(b^{2}+a^{2}\right)}=\frac{y}{8\left(a^{2}+b^{2}\right)}=\frac{1}{4 a^{2}+4 b^{2}}$

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

$\Rightarrow x=\frac{-1}{2}$

For y consider the following

$\frac{x}{-2\left(b^{2}+a^{2}\right)}=\frac{y}{8\left(a^{2}+b^{2}\right)}=\frac{1}{4 a^{2}+4 b^{2}}$

$\frac{y}{8\left(a^{2}+b^{2}\right)}=\frac{1}{4 a^{2}+4 b^{2}}$

$\Rightarrow y=2$

Hence we get the value of $x=-\frac{1}{2}$ and $y=2$