7. Let $a, b, c$ be three distinct real numbers such that the equations $x^{2}+a x+1=0$ and $x^{2}+b x+c=0$ have a common real root, and the equations $x^{2}+x+a=0$ and $x^{2}+c x+b=0$ also have a common real root, then the value of $a+b+c$ is $\qquad$.
7. -3 Explanation: From $\left\{\begin{array}{l}x_{1}^{2}+a x_{1}+1=0 \\ x_{1}^{2}+b x_{1}+c=0\end{array} \Rightarrow(a-b) x_{1}+(1-c)=0 \Rightarrow x_{1}=\frac{c-1}{a-b}\right.$.

Similarly, from $\left\{\begin{array}{l}x_{2}^{2}+x_{2}+a=0 \\ x_{2}^{2}+c x_{2}+b=0\end{array} \Rightarrow(1-c) x_{2}+(a-b)=0 \Rightarrow x_{2}=\frac{a-b}{c-1}\right.$.
Thus, $x_{1}=\frac{1}{x_{2}}$. On the other hand, by Vieta's formulas, $\frac{1}{x_{1}}$ is also a root of the first equation, which indicates that $x_{2}$ is a common root of the equations $x^{2}+a x+1=0$ and $x^{2}+x+a=0$. Therefore, $(a-1)\left(x_{2}-1\right)=0$. However, when $a=1$, the two equations have no real roots, so $x_{2}=x_{1}=1$,
Thus, $a=-2, b+c=-1$, hence $a+b+c=-3$.