Solve |3x - 6| > 3
x > 3 and x <1
x > 3 or x > 9
x > 3 and x > 9
x > 3 or x < 1
1 < x < 3
2007-10-25 10:29:07 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
According to the definition of the absolute value:
If 3x - 6 >= 0 (thus x >= 2)
|3x - 6| = 3x - 6
For x>= 2:
|3x - 6| > 3
3x - 6 > 3 // + 6
3x > 9 // divide by 3
x > 3
If x<2 (which means 3x - 6 < 0)
-(3x - 6) > 3
-3x + 6 > 3 // - 6
-3x > - 3 // divide by -3
x < 1
Combine with the case that x>=2, and you'll get:
x > 3 or x < 1
2007-10-25 10:37:36 · answer #1 · answered by Amit Y 5 · 0⤊ 0⤋
-3 > 3x -6 >3
3 > 3x > 9
1 > x > 3
2007-10-25 10:33:22 · answer #2 · answered by roguetrader12002 4 · 0⤊ 0⤋
x > 3 or x < 1
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Ideas: Solve 3x-6 > 3 => x > 3, or -(3x-6) > 3 => x < 1
2007-10-25 10:38:29 · answer #3 · answered by sahsjing 7 · 0⤊ 0⤋
3x-6 >3
3x > -3
x> -1
-3x+6 < 3
-3x < -3
x>1
2007-10-25 10:37:36 · answer #4 · answered by Darkskinnyboy 6 · 0⤊ 0⤋