Question

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

Answer 1

The given system of equations is:

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

On simplification$,$

$mx-ny-(m^2+n^2)=0....\left(1\right)x+y-2m=0....\left(2\right)$

By cross multiplication method, we know that, for a system of linear equations in $x$ and $y$, of the form $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we have:

$\frac{x}{b_1{c}_2-b_2{c}_1}=\frac{y}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

Taking the values of $a_1,b_1,c_1,a_2,b_{2}and{c}_2$ from $\left(1\right)$ and $\left(2\right)$, we get:

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

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

Using the algebraic identities, $(a+b{)}^2=a^2+b^2+2ab$ and $a^2-b^2=(a-b)(a+b)$, we get:

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

Now, on comparing $\frac{x}{(m+n{)}^2}$ and $\frac{1}{m+n}$, we get:

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

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

$\therefore x=m+n$

Similarly, on comparing $\frac{y}{(m+n)(m-n)}$ and $\frac{1}{m+n}$, we get:

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

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

$\therefore y=m-n$

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