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

Solution:

GIVEN:

$m x-n y=m^{2}+n^{2}$

$x+y=2 m$

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

Here we have the pair of simultaneous equation

$m x-n y-\left(m^{2}+n^{2}\right)=0$

$x+y-2 m=0$

By cross multiplication method we get

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

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

$\frac{x}{(m+n)^{2}}=\frac{1}{m+n}$

$x=m+n$

Now for y

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

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

$\frac{y}{(m-n)(m+n)}=\frac{1}{m+n}$

$y=m-n$

Hence we get the value of $x=m+n$ and $y=m-n$