Example 6 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 $a$.
(1995, Jilin Province Junior High School Mathematics Competition)
$$
\begin{array}{l}
\text { Solution: Adding (1) and (2) gives } \\
b^{2}+c^{2}=-a^{2}+14 a-13 \text {. } \\
\text { From (1) we get } b^{2} c^{2}=\left(a^{2}-8 a+7\right)^{2} \text {. } \\
\text { (4)(3) Let } b^{2}=\frac{-a^{2}+14 a-13}{2}+t \text {, } \\
c^{2}=\frac{-a^{2}+14 a-13}{2}-t, \\
\end{array}
$$

Substituting into (4) and rearranging gives
$$
\begin{array}{l}
\left(\frac{-a^{2}+14 a-13}{2}\right)^{2}-\left(a^{2}-8 a+7\right)^{2} \\
=t^{2} \geqslant 0,
\end{array}
$$

which simplifies to $\left(a^{2}-10 a+9\right)(a-1)^{2} \leqslant 0$.
Solving this, we get $1 \leqslant a \leqslant 9$.
Therefore, the range of values for $a$ is $1 \leqslant a \leqslant 9$.