题目内容
解下列方程(组)或不等式(组):
(1)2x+1=
(2)
(3)
<1-
(4)
.
(1)2x+1=
5x-1 |
2 |
|
(3)
3(y+1) |
8 |
y-1 |
4 |
|
(1)2x+1=
,
去分母,4x+2=5x-1,
移项合并,4x-5x=-1-2,
系数化1,x=3;
(2)
,
①+②×3得,5x+9x=13-27,
合并得,14x=-14,
得,x=-1;
(3)
<1-
,
去分母得,3(y+1)<8-2(y-1),
去括号得,3y+3<8-2y+2,
移项得,5y<7,
y<
;
(4)
,
由①得,5x>10,
x>2;
由②得,2x>13,
x>
;
∴不等式组的解为x>
;
5x-1 |
2 |
去分母,4x+2=5x-1,
移项合并,4x-5x=-1-2,
系数化1,x=3;
(2)
|
①+②×3得,5x+9x=13-27,
合并得,14x=-14,
得,x=-1;
(3)
3(y+1) |
8 |
y-1 |
4 |
去分母得,3(y+1)<8-2(y-1),
去括号得,3y+3<8-2y+2,
移项得,5y<7,
y<
7 |
5 |
(4)
|
由①得,5x>10,
x>2;
由②得,2x>13,
x>
13 |
2 |
∴不等式组的解为x>
13 |
2 |
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