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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}

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  • $\begingroup$ Substitute $z=\sqrt{1-x}$, $u=\sqrt{2y+5}$. This will give you a polynomial in $z$. $\endgroup$ Oct 3, 2013 at 12:27
  • $\begingroup$ How about $u$? I have already done but there is no more results. Can you explain more clearly? Thanks. $\endgroup$ Oct 3, 2013 at 13:21

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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.

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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 $x=-3$.

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

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    $\begingroup$ Oh, my! This looks (looks ...) as discouraging hopeless as the original equations...or maybe more! $\endgroup$
    – DonAntonio
    Oct 3, 2013 at 12:12
  • $\begingroup$ It's really quite complicated. Actually, there exists a method which can solve this system more simply but I can't find now. $\endgroup$ Oct 3, 2013 at 12:29

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