Answer :

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 >


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