Answer 1

The given system of equations may be written as

`mx - ny - (m^2 + n^2) = 0`

`x + y - 2m = 0

Here

`a_1 = m, b_1 = -n, c_1 = -(n^2 + n^2)`

`a_2 = 1, b_2 = 1, c_2 = -2m`

By cross multiplication, we have

`x/(2mn + (m^2 + n^2)) = (-y)/(-2m^2 + (m^2 + n^2)) = 1/(m + n)`

`=> x/(2mn + m^2 + n^2) = (-y)/(-m^2 + n^2) = 1/(m + n)`

`=> x/(m + n)^2 = (-y)/(-m^2 + n^2) = 1/(m + n)`

Now

`x/(m + n)^2 = 1/(m + n)`

`=> x = (m + n^2)/(m + n)`

=> x = m + n

And

`(-y)/(-m^2 + n^2)  = 1/(m + n)`

`=> -y = (-m^2 + n^2)/(n + n)`

`=> y = (m^2 - n^2)/(m + n)`

`=> y = ((m - n)(m + n))/(m + n)`

=> y = m - n

Hence, x = m + n, y = m - n is the solution of the given system of equation.