$\text{ For }x>0,y>0\text{ and }xy\ne 1\text{ , the solution of the system of equation }{x}^{x+y}=y^x\text{ and }$$y^{x+y}=x^{2x}{y}^n\text{ is, where }n>0$

$x^{x+y}=y^n$------(1)

$y^{x+y}=x^{2\pi }·y^n$----(2)

$\text{ Multiplying }\left(1\right)\&\left(2\right)$

$(xy{)}^{x+\gamma }=(xy{)}^{2\pi }$

$\Rightarrow x+y=2n(\because xy\ne 1)$

$\text{substitue the above in (1) we get }{x}^{2n}=y^n$

$$\begin{gathered} {\left(\frac{x^2}{y}\right)}^n=1 \\ \Rightarrow x^2=y\left(n>0\right) \\ hence the quadratic equation x^2+x-2n-0 \end{gathered}$$

$\Rightarrow x=\frac{-1+\sqrt{1+8n}}{2}$

$\begin{array}{rcl} & \Rightarrow & y=2n-x\\ & =& \frac{1+4n-\sqrt{1+8n}}{2}\end{array}$