试题

题目:
解方程
(1)4(x-2)+10=2(x-2)+5           
(2)
y+1
2
-1=2-
2y-1
3

答案
(1)解:去括号得:4x-8+10=2x-4+5                            
                 移项得:4x-2x=5-4+8-10                              
                     合并同类项得:2x=-1                                      
                     系数化为1得:x=-
1
2
                                       
(2)解:去分母得:3(y+1)-6=12-2(2y-1)(1分)
             去括号得:3y+3-6=12-4y+2                        (2分)
                移项得:3y+4y=12+2-3+6                        (3分)
        合并同类项得:7y=17                                 (4分)
          系数化为1得:y=
17
7
                               (5分)
(1)解:去括号得:4x-8+10=2x-4+5                            
                 移项得:4x-2x=5-4+8-10                              
                     合并同类项得:2x=-1                                      
                     系数化为1得:x=-
1
2
                                       
(2)解:去分母得:3(y+1)-6=12-2(2y-1)(1分)
             去括号得:3y+3-6=12-4y+2                        (2分)
                移项得:3y+4y=12+2-3+6                        (3分)
        合并同类项得:7y=17                                 (4分)
          系数化为1得:y=
17
7
                               (5分)
考点梳理
解一元一次方程.
(1)通过去分母、去括号、移项、合并同类项及系数化为1解方程即可得到答案;
(2)通过去括号、去分母、去括号、移项、合并同类项及系数化为1解方程即可得到答案;
本题考查了解一元一次方程的知识,容易在去括号和移项上出错,要注意:移项、去括号时要变号.
计算题.
找相似题