How to solve this system of equation?

Question

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}

Answer 1

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 $$(x_0,y_0) =(0,0) $$

Answer 2

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 $x=-3$.