7. Given $\alpha, \beta \in \mathbf{R}$, the intersection point of the lines $\frac{x}{\sin \alpha+\sin \beta}+\frac{y}{\sin \alpha+\cos \beta}=1$ and $\frac{x}{\cos \alpha+\sin \beta}+\frac{y}{\cos \alpha+\cos \beta}=1$ lies on the line $y=-x$, then $\sin \alpha+\cos \alpha+\sin \beta+\cos \beta=$ $\qquad$ .
7.0 Detailed Explanation: From the given information, we can assume the intersection point of the two lines to be $\left(x_{0},-x_{0}\right)$, and $\sin \alpha, \cos \alpha$ are the roots of the equation $\frac{x_{0}}{t+\sin \beta}+$ $\frac{-x_{0}}{t+\cos \beta}=1$, which means they are the roots of the equation $t^{2}+(\cos \beta+\sin \beta) t+\sin \beta \cos \beta-x_{0}(\cos \beta-\sin \beta)=0$. Therefore, $\sin \alpha+\cos \alpha=-(\sin \beta+\cos \beta)$, which implies $\sin \alpha+\cos \alpha+\sin \beta+\cos \beta=0$.