Given:

|x + a| + |x| > 3, x ϵ R.

|x + a| = -(x + a) or (x + a)

|x| = -x or x

When |x + a| = -(x + a) and |x| = -x

Then,

|x + a| + |x| > 3 → -(x + a) + (-x) > 3

-x -a – x > 3

-2x – a > 3

Adding a on both the sides in above equation

-2x -a + a> 3 + a

-2x > 3 + a

Dividing both the sides by 2 in above equation

Multiplying both the sides by -1 in the above equation

x <

Now when, |x + a| = -(x + a) and |x| = x

Then,

|x + a| + |x| > 3 → -(x + a) + x > 3

-x -a + x > 3

– a > 3

In this case no solution for x.

Now when, |x + a| = (x + a) and |x| = -x

Then,

|x + a| + |x| > 3 → (x + a) + (-x) > 3

x + a - x > 3

a > 3

In this case no solution for x.

Now when,

|x + a| = (x + a) and |x| = x

Then,

|x + a| + |x| > 3 → (x + a) + (x) > 3

x + a + x > 3

2x + a > 3

Subtracting a from both the sides in above equation

2x + a – a > 3 – a

2x > 3 – a

Dividing both the sides by 2 in above equation

x >

Therefore,

x < or x >