# Linear Inequations-17

Solve |x - 1| + |x - 2| ≥ 4

Solution: On the LHS of the given inequation there are two terms both containing modulus. By equating the expressions within the modulus to zero, we get x = 1,2 as critical points. These points divide real line in three parts, namely (-∞ +1], [1, 2]and [2, ∞).

So, we consider the following three cases:

CASE I

When -∞ < x < 1.

In this case, we have |x - 1| = -(x - 1) and |x - 2| = -(x - 2)

Therefore, |x - 1| + |x - 2| ≥ 4

=>-(x - 1) – (x - 2) ≥ 4

=> -2x + 3 ≥ 4

=> -2x ≥ 1

=> x ≤ -1/2

But, -∞ < x < 1. Therefore, in this case the solution set of the given inequatin is (-∞, -1/2]-----(i)

Case II

When 1 ≤ x < 2

In this case, we have |x - 1| = (x - 1) and |x - 2| = -(x - 2)

Therefore, |x - 1| + |x - 2| ≥ 4

=>x – 1 – (x - 2) ≥ 4

=> 1 ≥ 4, which is an absurd result.

So, the given inequaiton has no solution for x ∈ [1,2).

Case III

When x ≥ 2.

In this case, we have |x - 1| = x – 1 and |x - 2| = x - 2

Therefore, |x - 1| + |x - 2| ≥ 4

=>x – 1 + x – 2 ≥ 4

=> 2x – 3 ≥ 4

=> 2x ≥ 7

x ≥ 7/2

But, x ≥ 2. Therefore, in this case the solution set of the given inequation is

[7/2, ∞]------------------------------------------------------------------------------------(ii)

Combining (i) abd (ii), we obtain that the solution set of the given inequation is

(-∞, -1/2] U [7/2, ∞)