How to solve this system of equation? I need to solve the following system of $(x,y)$:
\begin{cases}
3y^3+3x\sqrt{1-x}=5\sqrt{1-x}-2y\\
x^2-y^2\sqrt{1-x}=\sqrt{2y+5}-\sqrt{1-x}
\end{cases}

The first equation can be written as
$x=1-y^{2}$
because it is manipulated in this way.
We bring to the first member the terms in x and to the second member the terms in y:
$(5-3x)\sqrt{1-x}=y(3y^{2}+2)$.
We see this equation as the equality of two products, that is:
$\sqrt{1-x}=y$,
$-3x+5=3y^{2}+2$.
The solution of this system is:
$x=1-y^{2}$.
We take the second equation and replace the value of $x$, obtaining an equation in y of the eighth degree:
$y^{8}-2y^{7}-4y^{6}+6y^{5}+4y^{4}-6y^{3}-3y^{2}-4=0$,
which splits into the product of two polynomies:
$(y-2)(y^{7}-3y^{5}+4y^{3}+2y^{2}+y+2)=0$.
From here we deduce that $y=2$, value that replaced in the equation $x=1-y^2$, as a result $x=-3$.
