解$：①×a_{2}$得，$a_{1}a_{2}x+a_{2}b_{1}y=a_{2}c_{1}③$
$②×a_{1}$得，$a_{1}a_{2}x+a_{1}b_{2}y=a_{1}c_{2}④$
③−④得，$(a_{2}b_{1}−a_{1}b_{2})y=a_{2}c_{1}−a_{1}c_{2}⑤$
当$a_{2}b_{1}−a_{1}b_{2}≠0$时，整理得$a_{1}a_{2}≠b_{1}b_{2}，$方程⑤有唯一解，即此时方程组有唯一解
当$a_{2}b_{1}−a_{1}b_{2}=0，$$a_{2}c_{1}−a_{1}c_{2}=0$时，整理得，$a_{1}a_{2}=b_{1}b_{2}=c_{1}c_{2}，$方程⑤的解为任意解，
即此时方程组有无数个解
当$a_{2}b_{1}−a_{1}b_{2}=0，$$a_{2}c_{1}−a_{1}c_{2}≠0$时，整理得，$a_{1}a_{2}=b_{1}b_{2}≠c_{1}c_{2}，$方程⑤无解，即此时方程组无解
