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

Question

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

Badiuddin askIITians.ismu Expert · Answer

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= 6 or (x-1)+(x-2)+(-x+3) > 6 x>=6 so it is not a solution case 3 1= 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 -∞

Akshay Mr · Answer

The detailed answer of this question please give me I can't understand the way you Explain the answer

Rishi Sharma · Answer

Dear Student,
Please find below the solution to your problem.

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 · Answer

Dear Student,
Please find below the solution to your problem.

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

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

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

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