How can I solve this system of equations?

Question

Here is a system of equations:

$$\begin{cases} x^2 + 10y = 41\\ y^2-2z = 23\\ z^2-6x = 17 \end{cases} $$

What's the value of $x$ and $y$ and $z$?

Answer 1

You can do this is with substitution.

$$y = \frac{41-x^2}{10}$$

$$z = \frac{20-y^2}{-2}$$

Now plug this $z$ into the third equation. This gives you an equation in terms of $x$ and $y$. Substitute the above equation into $y$, and you have a polynomial in $x$. Solve the polynomial equation and find $x$.

Answer 2

It is straightforward to determine the solutions over $\mathbb{C}$. We obtain indeed a polynomial of degree $8$, hence $8$ complex roots. More precisely, we obtain $$ x=\frac{z^2 - 17}{6},\; y=\frac{ - z^4 + 34z^2 + 1187}{360}, $$ where $z$ is a root of the polynomial $$ t^8 - 68t^6 - 1218t^4 + 37516t^2 - 837431. $$ This may not look nice, but it is a valid answer.

Is it possible that the system has a typo ? If you, say, consider the equations \begin{align*} x^2+10y & = 41, \\ y^2-2z & = 32, \\ z^2-6x & = 10, \end{align*} then $(x,y,z)=(9,-4,-8)$ is a nice solution.