4. Let real numbers $a, b, c$ satisfy $\left\{\begin{array}{l}a^{2}-b c-8 a+7=0 \\ b^{2}+c^{2}+b c-6 a+6=0,\end{array} \quad a\right.$, the range of values for $a$ is
4. $1 \leqslant a \leqslant 9$.
$b c=a^{2}-8 a+7$, and $(b+c)^{2}=a^{2}-2 a+1, b+c= \pm(a-1)$, then $b$ and $c$ are the two real roots of the equation $x^{2} \pm(a-1)x+a^{2}-8 a+7=0$, so $\Delta=(a-1)^{2}-4\left(a^{2}-8 a+7\right) \geqslant 0$, which gives $1 \leqslant a \leqslant 9$.