System of quadratic equations with three variables The problem is as follows:
For $x,y,z \in R$,
$$
\left\{ 
\begin{array}{l}
x^{2} -yz-8x+7=0 \\ 
y^{2}+z^{2}+yz-6x+6=0
\end{array}
\right. 
$$
What is the domain of $x$?
One way to solve this is to use another variable. This is shown in this answer. What other ways are there to solve this?
Addition:
The answer to this question is $1 \leq x \leq 9$.

We have that $yz=x^2-8x+7$ and, substituting into the 2nd equation, $y^2+z^2=6x-6-yz=6x-6-(x^2-8x+7)=-x^2+14x-13$.
Since the system of equations $y^2+z^2=a$ and $yz=b$ has a solution iff $a\ge2b$ and $a\ge-2b$,
as shown below,
the given system has a solution iff 
1) $-x^2+14x-13\ge2(x^2-8x+7)$ and 2) $-x^2+14x-13\ge -2(x^2-8x+7)$.
Since 1)  $-x^2+14x-13\ge2(x^2-8x+7) \iff -3x^2+30x-27\ge0 \iff x^2-10x+9\le0 \iff
(x-1)(x-9)\le0 \iff 1\le x\le9$
and
2) $-x^2+14x-13\ge -2(x^2-8x+7) \iff x^2-2x+1\ge0 \iff (x-1)^2\ge0$, which is true $\;\;\;\;\;\;\;\;\;$for all $x\in\mathbb{R}$,
the values of $x$ for which this system has a solution are the values of $x$ in $[1,9]$.
$----------------------------------------$
$\Longrightarrow$ If $y^2+z^2=a$ and $yz=b$ has a solution, then $y^2-2yz+z^2=(y-z)^2\ge0\implies a\ge2b$, and $y^2+2yz+z^2=(y+z)^2\ge0\implies a\ge-2b$.
$\Longleftarrow$ If $a\ge2b$ and $a\ge-2b$, then $a\ge2|b|\ge0$ and $a^2\ge4b^2$.
If $a=0$, then $b=0$ and the system has the solution $y=0, z=0$. 
If $a>0$, then $y=\big(\frac{a+\sqrt{a^2-4b^2}}{2}\big)^{1/2}$ satisfies $y^4-ay^2+b^2=0$, so $y^2-a+\frac{b^2}{y^2}=0$ and therefore letting $z=\frac{b}{y}$ gives a solution of $y^2+z^2=a$ and $yz=b$.
