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

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$y = \frac{1 + 4 n - \sqrt{1 + 8 n}}{2}$

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

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

$y = \frac{1 + 4 n + \sqrt{1 + 8 n}}{2}$

answer is A, C.

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Detailed Solution

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

$y^{x + y} = x^{2 π} \cdot y^{n}$ ----(2)

$Multiplying (1) & (2)$

$(x y)^{x + γ} = (x y)^{2 π}$

$\Rightarrow x + y = 2 n (∵ x y \neq 1)$

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

${(\frac{x^{2}}{y})}^{n} = 1 \Rightarrow x^{2} = y (n > 0) h e n c e t h e q u a d r a t i c e q u a t i o n x^{2} + x - 2 n - 0$

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

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

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