# 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}

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

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.

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 $$x=-3$$.

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

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