Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
$x=y=1$ 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

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 $x+y=2$.

share|cite|improve this answer
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 – ABC 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) – ABC 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 $p = -n$ $\implies$ $x+y = -n$ $\implies$ $y = -n - x$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.