Solve the following system of equations by the method of cross-multiplication:
$mx-ny=m^2+n^2$
$x+y=2m$

Given:

The given system of equations is:

$mx-ny=m^2+n^2$

$x+y=2m$

 To do: 

Here, we have to solve the given system of equations by the method of cross-multiplication.

Solution:  

The given system of equations can be written as,

$mx-ny-(m^2+n^2)=0$

$x+y-2m=0$

The solution of a linear pair(standard form) of equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ is given by,

$\frac{x}{b_1c_2-b_2c_1}=\frac{-y}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1}$

Comparing the given equations with the standard form of the equations, we get,

$a_1=m, b_1=-n, c_1=-(m^2+n^2)$ and $a_2=1, b_2=1, c_2=-2m$

Therefore,

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

$\frac{x}{2mn+m^2+n^2}=\frac{-y}{-2m^2+m^2+n^2}=\frac{1}{m+n}$

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

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

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

$x=m+n$ and $y=m-n$

The solution of the given system of equations is $x=m+n$ and $y=m-n$. 

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Updated on: 10-Oct-2022

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