Example 12 Let real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{2}-y z-8 x+7=0, \\
y^{2}+z^{2}+y z-6 x+6=0 .
\end{array}\right.
$$

Find the range of values for $x$.
(1980, National High School Mathematics Competition)
Solution: From (1), we get $y z=x^{2}-8 x+7$.
From (2), we get
$$
\begin{array}{l}
(y+z)^{2}=y z+6 x-6=(x-1)^{2} . \\
\therefore y+z= \pm(x-1) . \\
\because y 、 \pm \frac{x-1}{2}, z \text { form an arithmetic sequence, so let } \\
y= \pm \frac{x-1}{2}+d, z= \pm \frac{x-1}{2}-d . \text { Substituting into }
\end{array}
$$
(3), we get
$$
\begin{array}{l}
3 x^{2}-30 x+27=-d^{2} . \\
\therefore x^{2}-10 x+9 \leqslant 0 .
\end{array}
$$

Solving, we get $1 \leqslant x \leqslant 9$.