点(r, s)在直线y=2x+3上?fficeffice" />

1. (2r-s+3)(4r+2s-6)=0

2. (3r+2s-5) (2r-s+3)=0

E

解释中说可推出三个方程组, 为什么是这样算呢, 为什么不是求上面两个方程的交集呢, 为什么求下面三个的交呢?

2r-s+3=0, 4r+2s-6=0

3r+2s-5=0 , 2r-s+3=0 4r+2s-6=0, 3r+2s-5=0[em39]

第一行两个方程式与第二行的两个方程式的交集结果包括：

2r-s+3=0 两组方程共有的方程式fficeffice" /> 及 4r+2s-6=0与3r+2s-5=0的交集

I think that in #1condition, You absolutely can get solution about r and s by the equation 2r-s+3=0 and 4r+2s-6=0. similarly, You can get other solution about r and s by the equation 3r+2s-5=0 and 2r-3+3 in the #2condition.

Therefore, I personally think that the answer maybe D

What 's niuniu" opinion?