Solve the system $ \left\lbrace \begin{array}{ccc} \sqrt{1-x} + \sqrt{1-y} &=& \frac{y-x}{2} \\ x+y &=& 2xy \end{array}\right. $ I have to solve the following system:
$$
\left\lbrace \begin{array}{ccc}
\sqrt{1-x} + \sqrt{1-y} &=& \frac{y-x}{2} \\
x+y  &=& 2xy
\end{array}\right.
$$
I've showed that it is equivalent to
$$
\left\lbrace \begin{array}{ccc}
 \sqrt{1-x} - \sqrt{1-y} &=& 2 \\
 x+y  &=& 2xy
\end{array}\right. \text{ or } x=y= 1
$$
And i can't conclude nothing frome this system:
$$
\left\lbrace \begin{array}{ccc}
 \sqrt{1-x} - \sqrt{1-y} &=& 2 \\
 x+y  &=& 2xy
\end{array}\right.
$$

To solve the second equation system.
From $x+y=2xy$. We have $y=\frac{x}{2x-1}$.
Then use $\sqrt{1-x} = 2 + \sqrt{1-y}$ (I use the second equation, x=y=1 is excluded.)
$y-x-4 = 4 \sqrt{1-y}$
$\frac{x}{2x-1}-x-4 = 4 \sqrt{1-\frac{x}{2x-1}}$
$\frac{-2x^2-6x+4}{2x-1} = 4 \sqrt{\frac{x-1}{2x-1}}$
$(x^2+3x-2)^2 = 4(x-1)(2x-1)$
$x^4+6x^3-3x^2=0$
$x = -3 \pm 2 \sqrt{3}, 0$
rejected unsuitable roots, we have $x = -3 - 2 \sqrt{3}, y = -3 + 2 \sqrt{3}$
and $x=0, y=0$ but it does not satisfact first equation.
You need to square twice to get rid of all square roots and resulting a 4th order equation. Luckily it is solvable.
