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borislav_mirchev

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borislav_mirchev

#1 h May 20, 2016, 9:01 AM • 2 Y

Y by Adventure10, Mango247

Solve the system:
$x\sqrt{x-y}+y\sqrt{x+y}=xy$
$x\sqrt{y-1}+y\sqrt{x-1}=x+y$

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#2 h May 27, 2016, 8:24 AM • 2 Y

Y by borislav_mirchev, Adventure10

From the under square values to be non-negative we have $x \ge 1, y \ge 1, x \ge y$ . Rearranging second equation of the system we have
$x(\sqrt{y-1}-1) + y(\sqrt{x-1}-1) = 0(*)$ .From here we can have 3 possible cases:
Case 1: $2 \ge x \ge y$
Case 2: $x \ge y \ge 2$
Case 3: $x \ge 2 \ge y$
Cases 1 and 2 and condition $(*)$ easily lead to $x=y=2$ (because then we have at the same time: $\sqrt{x-1}-1 \ge 0, \sqrt{y-1}-1 \ge 0$ or $\sqrt{x-1}-1 \le 0, \sqrt{y-1}-1 \le 0$ )
So it left to deal with the Case 3. From the first equation of the system we have consequently:
$y\sqrt{x+y}=x(y-\sqrt{x-y}) \rightarrow y \ge \sqrt{x-y} \rightarrow f(y) = y^2 + y - x \ge 0$ . The last one inequality should be executed for every y ( $y \in [1,2]$ ). This last is equivalent with :
$\begin{cases} \frac{-1}{2} < 1 \\\ 1+4x > 0 \\\ 1(1^2 + 1 -x) \ge 0 \end{cases}$ from where we have $x \le 2$ and from the second equation of the system we have $2\sqrt{y-1} + y\sqrt{2-1} = 2 + y$ , from where we conclude $y=2$ .
So finally we receive only one solution (x,y)=(2,2)

This post has been edited 1 time. Last edited by georgi111, May 27, 2016, 8:25 AM
Reason: technical mistake

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borislav_mirchev

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borislav_mirchev

#3 h May 27, 2016, 8:35 AM • 1 Y

Y by Adventure10

More solutions can be seen here http://dxdy.ru/post1126434.html#p1126434.

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