How to get rid of the absolute value in $|x + 2| > -4$? How to get rid of the absolute value in $|x + 2| > -4$? I know that $|x|> a$ is equivalent to $x > a$ or $x < -a$, but I think that this definition is not applicable here? am I right ?
 A: Since the absolute value is always greater than or equal to zero, this statement is true for all values of $x$.
To further answer your question, we can write this as
$$x + 2 > -4 \mbox { or } x + 2 < 4$$
Again, since $x+2$ is either at least $-4$ or at most $4$, this is true for all $x$.
A: You can use the definition of the absolute value: 
$$|x| = \left\{ \begin{array}{cc} x & x\geq 0\\
-x & x < 0 
\end{array}\right..$$
So for your case you have: 
$$|x+2| = \left\{ \begin{array}{cc} x+2 & x\geq -2\\
-(x+2) & x < -2 
\end{array}\right..$$
A: The equivalence
$$
|x| > a
\iff x > a\ \text{or}\ x<-a
$$
holds for every $a\in\mathbb{R}$. It's usually only stated for $a\ge 0$ because we have $|x|\ge 0$ for every $x\in\mathbb{R}$.
To see why it remains true, suppose $a<0$. Then the right hand side of the equivalence above becomes
$$
x\in(a,\infty)\cup(-\infty,-a)=\mathbb{R},
$$
which is the same set of solutions for the left hand side.
A: Just draw a number line and think about $|x+2|>-4$ when can this happen? there can be two cases 1) when $x+2$ is positive then from number line and the inequality $x+2 $ is in the right side of $-4$ or $x+2 > -4$ or $x > -6$
2) when $x+2$ is negative then from number line nd the inequality this can happen when $x+2$ is in the left of $-4$ or $x+2 < -4$ or $x < -6$ . 
Concepts about absolute values/modulus simplifies when you consider number line.
So your initial thought was correct right!. 
