1.

Solve the system of equations in R.(|x + 2| – x)/x < 2

Answer»

Given as

(|x + 2| – x)/x < 2

Now, let us rewrite the equation as

|x + 2|/x – x/x < 2

|x + 2|/x – 1 < 2

On adding 1 on both sides, we get

|x + 2|/x – 1 + 1 < 2 + 1

|x + 2|/x < 3

On subtracting 3 on both sides, we get

|x + 2|/x – 3 < 3 – 3

|x + 2|/x – 3 < 0

So, clearly it states, x ≠ 2 therefore two case arise:

Case 1: x + 2 > 0

x > –2

Now, in this case |x + 2| = x + 2

x + 2/x – 3 < 0

(x + 2 – 3x)/x < 0

– (2x – 2)/x < 0

(2x – 2)/x < 0

Now, let us consider only the numerators, we get

2x – 2 > 0

x > 1

x ϵ (1, ∞) ….(1)

Case 2: x + 2 < 0

x < –2

In this case, |x + 2| = – (x + 2)

-(x + 2)/x – 3 < 0

(-x – 2 – 3x)/x < 0

– (4x + 2)/x < 0

(4x + 2)/x < 0

Then, let us consider only the numerators, we get

4x + 2 > 0

x > – 1/2

But x < -2

From the denominator we have,

x ∈ (–∞ , 0) …(2)

From (1) and (2)

∴ x ∈ (–∞ , 0) ⋃ (1, ∞)


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