# Linear Inequations-18

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**Solve |x - 1|/(x + 2) < 1**

**Solution:** We have,

|x - 1|/(x + 2) < 1

=>|x - 1|/(x + 2) – 1 < 0

=> [|x - 1| - (x + 2)]/(x + 2) < 0

Now, the following cases arise.

Case I

When x – 1 ≥ 0 i.e. x ≥ 1

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

Therefore, [|x - 1| - (x + 2)]/(x + 2) < 0

=>[(x - 1) – (x + 2)]/(x + 2) < 0

=> -3/(x + 2) < 0

=> x + 2 > 0 [since, a/b < 0 and a < 0 => b > 0]

=> x > -2

But, x ≥ 1. Therefore, x ≥ 1.

Thus, in this case the solution set of the given inequation is [1, ∞) ------------- (i)

Case II

When x – 1 < 0 i.e. x < 1

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

Therefore, [|x - 1| - (x + 2)]/(x + 2) < 0

[-(x - 1) – (x + 2)]/(x + 2) < 0

=> -(2x + 1)/(x + 2) < 0

=> (2x + 1)/(x + 2) > 0

=> x ∈ (-∞, -2) U (-1/2, ∞)

But, x < 1. Therefore, x ∈ (-∞, -2) U (-1/2, 1)

Thus, in this case the solution set of the given inequation is (-∞, -2) U (-1/2, 1)----(ii)

Combining (i) and (ii), we obtain that the solution set of the given inequation is (-∞, -2) U (-1/2, ∞)