I am trying to understand how the behavior between the absolute value(s) and the inequality sign differ between equations of the form:
\begin{equation} |x| < a \tag{1} \label{1} \end{equation}
for some $a \in \mathbb{R}$ and
\begin{equation} |f(x)| < |g(x)| \tag{2} \label{2} \end{equation}
for $f : D_1 \subseteq\mathbb{R} \to \mathbb{R}$ and $g : D_2 \subseteq\mathbb{R}\to\mathbb{R}$.
Specifically, I know that when one solves an equation like eq. $\ref{1}$, you do the following:
- Solve $x < a$.
- Solve $x > -a$.
- Solve eq. $\ref{1}$ by taking the intersection of the solution found in steps 1. and 2.
I think of step 2. as first writing $-x < a$ (to make the LHS positive for the case when $x < 0$), and then multiplying both sides by $-1$, thereby flipping the inequality sign to get $x > -a$.
To solve eq. $\ref{2}$, you take each case separately, as shown in, for example, this question. However, in each of these cases, as far as I know, you do not switch the inequality sign like in step 2. for solving eq. $\ref{1}$.
Therefore, my question is why, conceptually, is this done for solving eq. $\ref{1}$, but not done in each case (i.e. when the argument of one of the absolute values is negative) when solving eq. $\ref{2}$.
Thanks in advance. I have been avoiding absolute values + inequalities for most of my career in mathematics; it's time I face them.