To what extent can I square both sides of an absolute equation? I am working on some absolute equation problems like the following: 
$$\begin{align}
& {|x-4|} \lt 1 \\
& 1 \le |x| \le 4 \\
& |x+3| = |2x+1|
\end{align}$$
Now, for both of these equations, I simply squared both sides to get rid of the absolute and then continued solving from there. Now my question is: when can I not do this and what is the alternative if I can't?
Thanks a bunch!
 A: Order of a relation is preserved, when you apply a strictly monotonously increasing function (for $\leq, \geq$ you can drop the strictly).
$$f: x \mapsto x^2$$
is strictly monotonously increasing on $[0, \infty)$. so you can square whenever all expressions are guaranteed to be $\geq 0$.
A: The intuition here is to notice that the squaring function $f(x) = x^2$ is monotonically increasingly for non-negative real numbers, namely if $x<y$ and if both $x$ and $y$ are non-negative, then $x^2 < y^2$.  To understand what I mean, think of the graph of $f$ over the positive real axis.  It starts at $0$ and gets bigger and bigger.
On the other hand, suppose that $x<y$ but that $x$ is negative, then it is not necessarily true that $x^2 < y^2$.  Consider, for example, $x=-2$ and $y=1$.  Similarly, if both $x$ and $y$ are negative, then $x<y$ implies that $x^2>y^2$.  Think of $x=-2$ and $y=-1$ for example. These facts follow because the squaring function is a decreasing function for negative real numbers.
How does this affect inequalities with absolute values?  Well, consider, for example, $|x|<|y|$.  Since both $|x|$ and $|y|$ are non-negative, we can square both sides of this inequality to obtain $|x|^2<|y|^2$.
In all of your examples, all sides of all inequalities are non-negative, so you can square without reversing inequalities.  You would just have to be careful if some side of an inequality is negative.
A: The following facts hold because integers, rationals, algebraic numbers, and real numbers are all "totally ordered rings", which basically means that arithmetic plays well with order:
Fact 1: If $p\le q$ and $r\ge0$, then $pr\le qr$ and $rp\le rq$.
Fact 2: If $p<q$ and $r>0$, then $pr<qr$ and $rp<rq$.
(Note that these conclusions are really the same for numbers because of the commutative law, but there are situations where they are not).
Suppose $a\ge0$ and $b\ge0$.
If $a\le b$, then $a\cdot a\le a\cdot b$ and $a\cdot b\le b\cdot b$, so $a^2=a\cdot a\le b\cdot b$.
If $a<b$, then $b>0$, so $a^2=a\cdot a \le a \cdot b < b \cdot b=b^2$.
It turns out that we can reverse these implications using the fact that $a\le b \iff \lnot (b < a)$ and the other direction.
