Given:

, x ϵ R.

Intervals of |x|:

|x| = -x, x<0

|x| = x, x ≥ 0

Domain of

|x| + 3 = 0

X = -3 or x = 3

Therefore,

-3 < x < 3

Now, combining intervals with domain:

x < -3, -3<x<0, 0 ≤ x <3, x >3

For x < -3

→

Now, subtracting from both the sides

Signs of x + 5:

x + 5 = 0 → x = -5 (Subtracting 5 from both the sides)

x + 5 > 0 → x > -5 (Subtracting 5 from both the sides)

x + 5 < 0 → x < -5 (Subtracting 5 from both the sides)

Signs of -2x - 6:

-2x - 6 = 0 → x = -3

(Adding 6 on both the sides, then multiplying both the sides by -1 and then dividing both the sides by 2)

-2x - 6 > 0 → x < -3

(Adding 6 on both the sides, then multiplying both the sides by -1 and then dividing both the sides by 2)

-2x - 6 < 0 → x > -3

(Adding 6 on both the sides, then multiplying both the sides by -1 and then dividing both the sides by 2)

Intervals satisfying the required condition: ≤ 0

x < -5, x= -5, x >-3

or

x ≤ -5 or x >-3

Similarly, for -3 < x < 0:

x ≤ -5 or x >-3

Merging overlapping intervals:

-3 < x < 0

For, 0 ≤ x < 3:

→

Subtracting from both the sides

Multiplying both the sides by 2

Signs of 5 – x:

5 – x = 0 → x = 5

(Subtracting 5 from both the sides and then dividing both sides

by -1)

5 – x > 0 → x < 5

(Subtracting 5 from both the sides and then multiplying both sides by -1)

5 – x < 0 → x > 5

(Subtracting 5 from both the sides and then multiplying both sides by -1)

Signs of x – 3:

x – 3 = 0 → x = 3 (Adding 3 to both the sides)

5 – x > 0 → x > 3 (Adding 3 to both the sides)

5 – x < 0 → x < 3 (Adding 3 to both the sides)

Intervals satisfying the condition: x ≤ 0

x < 3 or x = 5 or x > 5

or

x <3 and x ≥ 5

Similarly, for 0 ≤ x < 3:

x <3 and x ≥ 5

Merging overlapping intervals:

0 ≤ x < 3

Now, combining all the intervals satisfying condition: ≤ 0

x ≤ -5 or -3 < x < 0 or 0 ≤ x < 3 or x ≥ 5

Therefore

x є (-∞, -5] υ (-3, 3) υ [5, ∞)