(2y-1)/3-1=(5y+1)/2-(3y+1)/2-(3y+1)/6
1个回答
展开全部
(2y-1)/3-1=(5y+1)/2-(3y+1)/2-(3y+1)/6
(2y-1)/3-1=(5y+1)/2-[(3y+1)/2+(3y+1)/6]
(2y-1)/3-1=(5y+1)/2-2(3y+1)/3
(2y-1)/3+2(3y+1)/3-1=(5y+1)/2
[(2y-1)+2(3y+1)]/3-1=(5y+1)/2
(2y-1+6y+2)/3-1=(5y+1)/2
(8y+1)/3-1=(5y+1)/2
6*(8y+1)/3-6*1=6*(5y+1)/2
2(8y+1)/3-6=3(5y+1)
16y+2-6=15y+3
16y-4=15y+3
16y-15y=3+4
y=7
(2y-1)/3-1=(5y+1)/2-[(3y+1)/2+(3y+1)/6]
(2y-1)/3-1=(5y+1)/2-2(3y+1)/3
(2y-1)/3+2(3y+1)/3-1=(5y+1)/2
[(2y-1)+2(3y+1)]/3-1=(5y+1)/2
(2y-1+6y+2)/3-1=(5y+1)/2
(8y+1)/3-1=(5y+1)/2
6*(8y+1)/3-6*1=6*(5y+1)/2
2(8y+1)/3-6=3(5y+1)
16y+2-6=15y+3
16y-4=15y+3
16y-15y=3+4
y=7
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
