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}
 A: Let's resurrect this old question. Two solutions to
\begin{cases}
3y^3+3xz=5z-2y\\
x^2-y^2z=\sqrt{2y+5}-z\end{cases}
are $(x,\,y)= (-3,\,2)$ for $z=\sqrt{1-x}$ and,
$$x = \text{Real root}(7 - x - 3 x^2 + 7 x^3 - 8 x^4 + 2 x^5 - x^6 + x^7=0)$$
$$y = \text{Real root}(2 + y + 2 y^2 + 4 y^3 - 3 y^5 + y^7=0)$$
for $z=-\sqrt{1-x}$.
Unless your system has special symmetries, I'm afraid the only way to find solutions that algebraic numbers of high degree is to use resultants.
A: 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 $x=-3$.
A: you can try the solve it numerically first with newton algorithm:
$$
f(x,y) = (3y^3-2x\sqrt{1-x}-2y, x^2+(1-y^2)\sqrt{1-x}-\sqrt{2y+5})
$$
$$
(x_{n+1}, y_{n+1}) = (x_n,y_n) +Df(x_n,y_n)^{-1}.f(x_n,y_n)
$$
with $$(x_0,y_0) =(0,0) $$
