Simple solution to system of equations

Question

I have a following system of equations:

$ x^2 + y + z = 0 \\ y^2 + x + z = 0 \\ 2z + 3y + 3x = 0 $

I am able to find solutions, but nothing really simple and straightforward. Do you have any ideas for a simple, clever solution?

Answer 1

Almost identical to what @marty cohen wrote in a now deleted answer.

$$x^2 + y + z = 0 \tag 1$$ $$y^2 + x + z = 0 \tag 2$$ $$2z + 3y + 3x = 0\tag 3$$

From $(3)$, we have $z=-\frac{3}{2} (x+y)$.

Replace in $(1)$ to obtain $y=2 x^2-3 x$. This gives $z=3 (x-1) x$. Replace now in $(2)$ to obtain $$4 x^4-12 x^3+6 x^2+4 x=2 (x-2) x \left(2 x^2-2 x-1\right)$$

So the roots $$x_1=0 \qquad x_2=2 \qquad x_{3,4}=\frac{1}{2} \left(1\pm\sqrt{3}\right)$$

Answer 2

On my phone.

$x^2 + y + z = 0 \\ y^2 + x + z = 0 \\ 2z + 3y + 3x = 0 $

$z=-3y/2-3x/2$

$0=x^2+y-(3/2)y-(3/2)x =x^2-3x/2-y/2$

$y=2x^2-3x$

$y^2=4x^4-12x^3+9x^2$

$0=y^2+x-(3/2)y-(3/2)x =4x^4-12x^3+9x^2-x/2-(3/2)(2x^2-3x) =4x^4-12x^3+6x^2+4x $

$x=0$ or $0=4x^3-12x^2+6x+4 =2x^3-6x^2+3x+2 $

$x=2, (1\pm\sqrt{3})/2$

(from Wolfram)

Substitute these to get y and z.

Nothing elegant, but couldn't think of a better way.

Answer 3

HINT

I do not know if it is a clever way to solve it, but it is an alternative way indeed.

To begin with, add the first and second equations together.

Then subtract the third equation from them.

Hence we get that

\begin{align*} x^{2} + y^{2} - 2x - 2y = 0 \Longleftrightarrow (x-1)^{2} + (y-1)^{2} = 2 \end{align*}

Then we can make the change of variable $x = 1 + \sqrt{2}\cos(\theta)$ and $y = 1 + \sqrt{2}\sin(\theta)$.

Consequently, \begin{align*} z = - 3 - \frac{3\sqrt{2}\cos(\theta) + 3\sqrt{2}\sin(\theta)}{2} \end{align*}

Now solve for $\theta$.

Can you take it from here?