x^x +y^y =31 x+y = 5 Find x and y


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Question Number 42906 by Cheyboy last updated on 04/Sep/18

$x^{x} + y^{y} = 31$ $x + y = 5$ $F i n d x a n d y$
	Answered by Joel578 last updated on 04/Sep/18

	$∴ (x, y) = (3, 2), (2, 3)$
	Commented by Cheyboy last updated on 04/Sep/18

	$T h a t z o b v i o u s s i r b t i n e e d w o r k i n g$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18

	$s o l v e b y r e a s o n i n g$ $x y x^{x} y^{y} x^{x} + y^{y} r e m a r k s$ $1 4 1 256 257 n o t s a t i s f y$ $2 3 4 27 31 s a t i s f y$ $3 2 27 4 31 s a t i s f y$ $4 1 256 1 257 n o t s a t i s f y$ $h e n c e t w o s e t o f s o l u t i o n$ $x = 2 y = 3$ $x = 3 y = 2$
	Commented by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18

	Answered by MJS last updated on 05/Sep/18

	$y = 5 - x$ $x^{x} + {(5 - x)}^{5 - x} - 31 = 0$ $put f (x) = x^{x} + {(5 - x)}^{5 - x} - 31$ $this is symmetric :$ $set x \to (5 - x) and f (x) stays the same$ $\Rightarrow minimum at (\begin{matrix} \frac{5}{2} \\ - 31 + \frac{25 \sqrt{10}}{4} \end{matrix})$ $we can only approximate zeros and find$ $x_{1} = 2; x_{2} = 3$ $f^{'} (x) = (1 + \ln x) x^{x} - (1 + \ln (5 - x)) {(5 - x)}^{5 - x}$ ${it}^{'} s easy to see that f^{'} (\frac{5}{2}) = 0 and$ $x < \frac{5}{2} \Rightarrow f^{'} (x) < 0; x > \frac{5}{2} \Rightarrow f^{'} (x) > 0$ $\Rightarrow no other solutions \in R$
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