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

Try to find the range of values for $a$.
From the given conditions, we have
$$
\left\{\begin{array}{l}
b c=a^{2}-8 a+7, \\
(b+c)^{2}=(a-1)^{2} .
\end{array}\right.
$$

Therefore, from equation (1), we get $b+c= \pm(a-1)$.
Assume $b=\frac{ \pm(a-1)}{2}+t, c=\frac{ \pm(a-1)}{2}-t$.
Multiplying the two equations, we get
$$
b c=a^{2}-8 a+7=\frac{(a-1)^{2}}{4}-t^{2} .
$$

Simplifying and rearranging, we get
$$
3 a^{2}-30 a+27=-4 t^{2} \text {. }
$$

Therefore, $3 a^{2}-30 a+27 \leqslant 0$.
Solving this, we get $1 \leqslant a \leqslant 9$.