8. Given real numbers $a, b, c$ satisfy
$$
a-b+c=7, a b+b c+b+c^{2}+16=0 \text {. }
$$

Then the value of $\frac{b}{a}$ is $\qquad$ .
8. $-\frac{4}{3}$.

From $a-b+c=7$, we get $a=b-c+7$. Substituting into $a b+b c+b+c^{2}+16=0$, we get $(b-c+7) b+b c+b+c^{2}+16=0$. Simplifying, we get $(b+4)^{2}+c^{2}=0$. Therefore, $b=-4, c=0, a=3$. Thus, $\frac{b}{a}=-\frac{4}{3}$.