3/(2x-1)<-1,
If 2x-1>0, x>½:
3<-2x+1, 2x+2<0, x<-1, in contradiction to x>½.
Therefore 2x-1<0, so x<½.
When x=½, we have an asymptote.
Also, 3>-2x+1 because -2x+1 is negative.
And 2x+2>0, x>-1.
Therefore, -1<x<½ satisfies both conditions.
When x is very large and y=3/(2x-1)+1, the horizontal asymptote is y=1.