设有方程组2X+ay=1, X+2y=2,(1)求a为何值时,y小于0?(2)a为何值时x小于0,y小于0?

2x+ay=1..............1
x+2y=2...............2
1-2*2
(a-4)y=-3
y=-3/(a-4)
x=(2a-2)/(a-4)

y<0
y=-3/(a-4)<0
y=3/(a-4)>0
a-4>0
a>4
当a>4时，y<0

x<0
(2a-2)/(a-4)<0
[(2a-8)+6]/(a-4)<0
2+6/(a-4)<0
6/(a-4)<-2
3<-(a-4)
a-4<-3
a<1

a>4,a<1,a的解集为空集，即不论为何值，x,y都不可能同时小于0