How can I find the positive solution of this system?

$\begingroup$

How can I find the positive solution for the system

$$x^{x+y}=y^n ;$$

$$y^{x+y}=y^n x^{2n}\quad ; n>0$$

I want help to find it's solutions.

$\endgroup$ 2

2 Answers

$\begingroup$

Go with just a observation : (x,y)=(1,1) satisfy!

Second: Multiply both eq. to get: $$(xy)^{x+y}=(xy)^{2n}$$

and done. $xy=1$ or $x+y=2n$ .

We substitute this in parent equations and get $y=x^2$ . so the solution is intersection of $y=x^2$ and $x+y=2$.

$\endgroup$ 9 $\begingroup$

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

$y$ = $x^{(x+y)/n}$. Substituting this in the $1st$ equation will give:

$x^{2n}$ = $x^{(x+y-n)(x+y)/n}$

When $x$ not equal to $1$ : $2n = (x+y-n)(x+y)/n $

Now let $x+y = p$ and you get a quadratic equation in $p$.By solving it you get the values for $p$ as : $-n$ and $2n$.

So, $p = 2n$ $\implies$ $x+y = 2n$ $\implies$ $y = 2n - x$

or $p = -n$ $\implies$ $x+y = -n$ $\implies$ $y = -n - x$

$\endgroup$ 0