分类：教育/科学-学习帮助
问题：初二已知y满足不等式[(y+1)/2]
已知y满足不等式[(y+1)/2]-y>2+[(y+2)/3],试话件式|y+1|+|2y-1|.
回答：[(y+1)/2]-y>2+[(y+2)/3]
[(y+1)/2]>(y+2)+[(y+2)/3]   [移项]
[(y+1)/2]>(y+2)*4/3         [合并同类项（y+2)]
两边同时乘以6，解得y<-13/5
因为y<-13/5
所以y+1<0,2y-1<0
所以|y+1|+|2y-1|
=-(y+1)-(2y-1)
=-y-1-2y+1
=-3y