8. Given the system of equations $\left\{\begin{array}{l}a_{1} x+y=c_{1} \\ a_{2} x+y=c_{2}\end{array}\right.$ has the solution $\left\{\begin{array}{l}x=5 \\ y=10\end{array}\right.$, then the solution to the system of equations $\left\{\begin{array}{l}a_{1} x+2 y=a_{1}-c_{1} \\ a_{2} x+2 y=a_{2}-c_{2}\end{array}\right.$ about $x, y$ is ( ).
A. $\left\{\begin{array}{l}x=4 \\ y=-5\end{array}\right.$
(B) $\left\{\begin{array}{l}x=-4 \\ y=5\end{array}\right.$
(C) $\left\{\begin{array}{l}x=-4 \\ y=-5\end{array}\right.$
D. $\left\{\begin{array}{l}x=4 \\ y=10\end{array}\right.$
Answer: (C)
Solution: The system of equations $\left\{\begin{array}{l}a_{1} x+2 y=a_{1}-c_{1} \\ a_{2} x+2 y=a_{2}-c_{2}\end{array}\right.$ can be transformed into $\left\{\begin{array}{l}a_{1}(1-x)+(-2 y)=c_{1} \\ a_{2}(1-x)+(-2 y)=c_{2}\end{array}\right.$,
Since the system of equations $\left\{\begin{array}{l}a_{1} x+y=c_{1} \\ a_{2} x+y=c_{2}\end{array}\right.$ has the solution
$\left\{\begin{array}{l}x=5 \\ y=10\end{array}\right.$,

Therefore,
$\left\{\begin{array}{l}1-x=5 \\ -2 y=10\end{array}\right.$,

Solving this, we get
$\left\{\begin{array}{l}x=-4 \\ y=-5\end{array}\right.$.

Hence, the answer is (C).