Solve |x+1|+|x|>3

We have, $|x+1|+|x|>3$

As, $|x+1|=\left\{\begin{array}{l}(x+1), x \geq-1 \\ -(x+1), x<-1\end{array}\right.$

and $|x|=\left\{\begin{array}{l}x, x \geq 0 \\ -x, x<0\end{array}\right.$

Case I: When $x<-1$,

$|x+1|+|x|>3$

$\Rightarrow-(x+1)-x>3$

$\Rightarrow-2 x-1>3$

$\Rightarrow-2 x>4$

$\Rightarrow x<\frac{4}{-2}$

$\Rightarrow x<-2$

So, $x \in(-\infty,-2)$

Case II : When $-1 \leq x<0$,

$|x+1|+|x|>3$

$\Rightarrow(x+1)-x>3$

$\Rightarrow 1>3$, which is not possible

So, $x \in \phi$

Case III : When $x \geq 0$,

$|x+1|+|x|>3$

$\Rightarrow(x+1)+x>3$

$\Rightarrow 2 x+1>3$

$\Rightarrow 2 x>2$

$\Rightarrow x>\frac{2}{2}$

$\Rightarrow x>1$

So, $x \in(1, \infty)$

From all the cases, we get

$x \in(-\infty,-2) \cup(1, \infty)$