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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.

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$x=y=1$ is a solution. What class has assigned this as homework? –  Gerry Myerson Mar 21 '13 at 12:17
hint: multiply them together. –  Raymond Manzoni Mar 21 '13 at 12:27

2 Answers 2

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 $x+y=2$.

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there are another set of solutions in: $x + y = -n$ –  lsp Mar 21 '13 at 12:45
But we don't want those solutions..... they give -ve solutions –  Mr.ØØ7 Mar 21 '13 at 12:53
You do not want negative solution in y, but i guess 'x' can be negative as it is not mentioned in the question. –  lsp Mar 21 '13 at 12:56
@lsp I don't see what you mean. It's mentioned we want +ve solutions for system and "system" constitute (x,y) –  Mr.ØØ7 Mar 21 '13 at 12:58
Even i thought the same before, but i am not sure. In that case only $x+y=2n$ is the solution. –  lsp Mar 21 '13 at 13:03

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

$y$ = $x^{(x+y)/n}$. 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 $p = -n$ $\implies$ $x+y = -n$ $\implies$ $y = -n - x$

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