A math (absolute value)

Case 1: x < -1/3
Then |3x-1|=-(3x-1), and |3x+1|=-(3x+1).
-(3x-1)-(3x+1)=2
-6x=2
x=-1/3 (rejected since x must be < -1/3)

Case 2: -1/3 <= x < 1/3
Then |3x-1|=-(3x-1), and |3x+1|=3x+1.
-(3x-1)+3x+1=2
2=2
therefore, there is infinite number of solutions in this case.

Case 3: x>=-1/3
Then |3x-1|=3x-1, and |3x+1|=3x+1.
3x-1+3x+1=2
6x=2
x=1/3

From the above cases, -1/3 <= x <= 1/3.

2006-12-07 21:43:14 補充：
To 樓上既 Amy，1-3x 3X 1=2, not 0 !所以2=2，i.e. 無限個solution (when -1/3

2006-12-07 21:44:31 補充：
仲有，when x

要分case 答ge lee 題
｜3x-1｜ + ｜3x+1｜ =2
when x>1/3
3x-1 + 3x+1 =2
x=1/3
when -1/3< x <1/3
1-3x + 3x+1= 2
0= 2
no sol
when x< -1/3
1-3x - (3x+1) = 2
x = -1/3
so x = 1/3 or -1/3

(3x-1)+(3x+1)=2
3x-1+3x+1=2
6x=2
x=2/6
x=1/3