Example 5 Let real numbers $a, b, c$ satisfy:
$$
\left\{\begin{array}{l}
a^{2}-b c-8 a+7=0, \\
b^{2}+c^{2}+b c-6 a+6=0 .
\end{array}\right.
$$

Find the range of values for $a$.
(1986, National High School Competition)
Proof: From (1) we get $b c=a^{2}-8 a+7$.
(2) - (1) gives $(b+c)^{2}=(a-1)^{2}$.

Thus $b+c= \pm(a-1)$.
By the inverse of Vieta's formulas, from (3) and (4) we get
$b, c$ are the roots of the equation $x^{2} \mp(a-1) x+\left(a^{2}-8 a+7\right)$ $=0$.
$$
\begin{array}{l}
\because b, c \in R, \\
\therefore \Delta=[\mp(a-1)]^{2}-4\left(a^{2}-8 a+7\right) \geqslant 0 .
\end{array}
$$

Solving this, we get $1 \leqslant a \leqslant 9$.