Solve each of the following in equations and represent the solution set on

Solution:

Given:

$|x+a|+|x|>3, x \in R$

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

$|x|=-x$ or $x$

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

Then,

$|x+a|+|x|>3 \rightarrow-(x+a)+(-x)>3$

$-x-a-x>3$

$-2 x-a>3$

Adding a on both the sides in above equation

$-2 x-a+a>3+a$

$-2 x>3+a$

Dividing both the sides by 2 in above equation

$\frac{-2 x}{2}>\frac{3+a}{2}$

$-x>\frac{3+a}{2}$

Multiplying both the sides by -1 in the above equation

$-x(-1)>\left(\frac{3+a}{2}\right)(-1)$

$x<-\left(\frac{3+a}{2}\right)$

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

Then,

$|x+a|+|x|>3 \rightarrow-(x+a)+x>3$

$-x-a+x>3$

$-a>3$

In this case no solution for x

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

Then,

$|x+a|+|x|>3 \rightarrow(x+a)+(-x)>3$

$x+a-x>3$

a > 3

In this case no solution for x.

Now when,

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

Then,

$|x+a|+|x|>3 \rightarrow(x+a)+(x)>3$

$x+a+x>3$

$2 x+a>3$

Subtracting a from both the sides in above equation

$2 x+a-a>3-a$

$2 x>3-a$

Dividing both the sides by 2 in above equation

$\frac{2 x}{2}>\frac{3-a}{2}$

$x>\frac{3-a}{2}$

Therefore,

$x<-\left(\frac{3+a}{2}\right)$ or $x>\left(\frac{3-a}{2}\right)$