有一道方程式,它的最后两步是这样的 3/5a1x+2/5b1y=c1 3/5a2x+2/5b2y=c2

3/5a1x+2/5b1y=c1 (1)
3/5a2x+2/5b2y=c2 (2)
3a1+4b1=c1 (3)
3a2+4b2=c2 (4)
(1)-(3)
(3/5 x-3)a1+(2/5 y-4)b1=0
(2)-(4)
(3/5 x-3)a2+(2/5 y-4)b2=0
由此可得3/5x-3=0 2/5y-4=0