∵a1b2-a2b1≠0∴(1)×b2-(2)×b1,得x=(b2c1-b1c2)/(a1b2-a2b1) (1)×a2-(2)×a1,得y=(a1c2-a2c1)/(a1b2-a2b1)故原方程组的解是：x=(b2c1-b1c2)/(a1b2-a2b1) y=(a1c2-a2c1)/(a1b2-a2b1)
