# Solve |x-1|+|x-2|+|x-3|>6

148 Points
14 years ago

Dear chilukuri

|x-1|+|x-2|+|x-3|>6

case 1        x>=3

|x-1|+|x-2|+|x-3|>6

or (x-1)+(x-2)+(x-3)>6

x>=4

case 2         2=<x <3

|x-1|+|x-2|+|x-3|>6

or

(x-1)+(x-2)+(-x+3)>6

x>=6

so it is not a solution

case 3   1=<x<2

|x-1|+|x-2|+|x-3|>6

or

(x-1)+(-x+2)+(-x+3)>6

x<=-2

so it is not a solution

case 4  x<1

|x-1|+|x-2|+|x-3|>6

or

(-x+1)+(-x+2)+(-x+3)>6

x<=0

so final solution

-∞<x≤0   and   4≤x<

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Akshay Mr
13 Points
5 years ago
The detailed answer of this question please give me I can't understand the way you Explain the answer

Rishi Sharma
3 years ago
Dear Student,

To get rid of the absolute value functions, we need to take 4 cases:
Case 1: x≥3or [3,∞)Here all the bars will open with a positive sign.
x−1+x−2+x−3≥6
3x≥12
x≥4
Case 2: 2≤x<3
or [2,3)
Here, only |x−3|
will open with a negative sign.
x−1+x−2−x+3≥6
x≥6
But this lies out of [2,3)
So no solution
Case 3: 1≤x<2 or [1,2)
Here only |x−1|
will open with a positive sign.
x−1−x+2−x+3≥6
−x≥2
x≤−2
Case 4: x<1(−∞,1)
Here all bars will open with a negative sign.
−x+1−x+2−x+3≥6
x≤0
Eliminating the redundant solutions, we get the final answer as:
x∈(−∞,0]∪[4,∞)
or x∈R−(0,4)

Thanks and Regards
Rishi Sharma
3 years ago
Dear Student,

To get rid of the absolute value functions, we need to take 4 cases:
Case 1: x≥3or [3,∞)Here all the bars will open with a positive sign.
x−1+x−2+x−3≥6
3x≥12
x≥4
Case 2: 2≤x<3
or [2,3)
Here, only |x−3|
will open with a negative sign.
x−1+x−2−x+3≥6
x≥6
But this lies out of [2,3)
So no solution
Case 3: 1≤x<2 or [1,2)
Here only |x−1|
will open with a positive sign.
x−1−x+2−x+3≥6
−x≥2
x≤−2
Case 4: x<1(−∞,1)
Here all bars will open with a negative sign.
−x+1−x+2−x+3≥6
x≤0
Eliminating the redundant solutions, we get the final answer as:
x∈(−∞,0]∪[4,∞)
or x∈R−(0,4)

Thanks and Regards