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，?$(a_{2}b_{1}?a_{1}b_{2})y=a_{2}c_{1}?a_{1}c_{2}⑤$?
當(dāng)?$a_{2}b_{1}?a_{1}b_{2}≠0$?時(shí)，整理得?$a_{1}a_{2}≠b_{1}b_{2}，$?方程⑤有唯一解，即此時(shí)方程組有唯一解
當(dāng)?$a_{2}b_{1}?a_{1}b_{2}=0，$??$a_{2}c_{1}?a_{1}c_{2}=0$?時(shí)，整理得，?$a_{1}a_{2}=b_{1}b_{2}=c_{1}c_{2}，$?方程⑤的解為任意解，
即此時(shí)方程組有無數(shù)個(gè)解
當(dāng)?$a_{2}b_{1}?a_{1}b_{2}=0，$??$a_{2}c_{1}?a_{1}c_{2}≠0$?時(shí)，整理得，?$a_{1}a_{2}=b_{1}b_{2}≠c_{1}c_{2}，$?方程⑤無解，即此時(shí)方程組無解
