Find the set of the real numbers $x$ satisfying the inequalities $|x+4|<|2x-1|$ and $|x|+|x+1|<3$ 
Find the set of the real numbers $x$ satisfying the given inequality and sketch them on the real line:
1) $|x+4|<|2x-1|$



*

*If $x<-4$, we have $x+4<0$ and $2x-1<0$, Then $|x+4|=-x-4$ and $|2x-1|=1-2x$
hence, in this case:
$|x+4|<|2x-1| \iff -x-4<1-2x \iff x<5$.
and we have that, $(-\infty,4)\cap (-\infty,5)=(-\infty,4)$.
So, for any $x \in (-\infty,-4)$, $|x+4|<|2x-1|$.


*

*If $-4\leq x <\frac{1}{2}$, we have that $x+4\geq 0$, and $2x-1<0$, so $|x+4+x+4$ and $|2x-1|=1-2x$
hence, in this case:
$|x+4|<|2x-1| \iff x+4<1-2x \iff 3x<-3 \iff x<-1$.
Thus,  for any $x \in [-4,\frac{1}{2}) \cap (-\infty,-1)=[-4,-1)$, $|x+4|<|2x-1|$.


*

*If $x\geq \frac{1}{2}$, we have $x+4>0$, and $2x-1\geq 0$, then $|x+4|=x+4$ and $|2x-1|=2x-1$
Then, in this case
$|x+4|<|2x-1| \iff x+4<2x-1 \iff x>5$.
So, for any $x \in [\frac{1}{2},\infty) \cap (5,\infty)=(5,\infty)$, $|x+4|<|2x-1|$.
Therefore, The solution set of the inequality is $(-\infty,-4) \cup [-4,-1) \cup (5,\infty)=(-\infty,-1) \cup (5,\infty)$.

2) $|x|+|x+1|<3$



*

*If $x<-1$, we have that $x<0$ and $x+1<0$, Then from the Absolute value definition $|x|=-x$ and $|x+1|=-x-1$.


Thus, $|x|+|x+1|=-x-x-1=-2x-1$.
Hence in this case, $|x|+|x+1|<3 \iff -2x-1 < 3 \iff 2x>-4 \iff x>-2$
So, for any $x \in (-2,-1)$, $|x|+|x+1|<3$.


*

*If $-1 \leq x <0$, we have that $|x|=-x$ and $|x+1|=x+1$
we have, $|x|+|x+1|=-x+x+1=1<3$.
Hence, for any $x \in [-1,0)$, $|x|+|x+1|<3$.


*

*If $x \geq 0$, we have that $|x|=x$ and $|x+1|=x+1$
Then in this case, $|x|+|x+1|=x+x+1=2x+1<3 \iff 2x<2 \iff x<1$.
Thus, for any $x \in [0,1)$, $|x|+|x+1|<3$.
Therefore, The solution set of the inequality is $(-2,-1) \cup [-1,0) \cup
[0,1)=(-2,1)$
is it true, please?

Problem 1. 
If you square it you get $$x^2+8x+16<4x^2-4x+1$$
so $$ 3x^2-12x-15>0\implies 3(x-5)(x+1)>0$$
so $x\in(-\infty,-1)\cup (5,\infty)$

And a solution to the second problem you solved is correct.
