Solving $|a| < |b|$ I apologize if this question in general, but I've been having trouble finding solutions as Google discards absolute value signs and inequality symbols.
I am looking for a way to eliminate absolute value functions in $|a| < |b|$.
I can solve $|a| < b$ and $|a| > b$, but I am unsure what method / combination of methods to use to eliminate absolute value signs from both sides.
Thank you!
An example problem:
$$|x + 2| < |x - 4|$$
 A: There are different approaches; one is to look at the zeroes of the expressions inside the absolute values, and split up $\mathbb R$ into intervals accordingly. In your example $|x + 2| < |x - 4|$, the points of interest are at $x=-2$ and $x=4$. You can therefore consider three cases:
1. If $x \in (-\infty,-2)$, then $x+2 < 0$ and $x-4 < 0$, so the absolute values will reverse the signs of both. This gives:
$$\begin{align} -(x+2) &< -(x-4) \\
x+2 &> x-4 \\
2 &> -4
\end{align}$$
This is true for all $x$ in the interval.
2. If $x \in [-2,4)$, then $x+2 \geq 0$ so its sign is unaffected by the absolute value, but $x-4 <0$ so its sign will be reversed:
$$\begin{align} x+2 &< -(x-4) \\
2x+2 &< 4 \\
x &< 1
\end{align}$$
Combining this last inequality with the assumption that $x \in [-2,4)$, we see that any $x$ in $[-2,1)$ is valid.
3. Finally, if $x \in [4,\infty)$, neither expression's sign is reversed:
$$\begin{align} x+2 &< x-4 \\
2 &< -4
\end{align}$$
This is false for all $x$ in the interval.
Putting all the information together from the above three cases, we have $x \in (-\infty, 1)$.

Note: as some other contributors have mentioned, there are simpler ways to deal with your problem, such as viewing it geometrically. The method that I have shown above is more useful when the expressions are more complicated or when you have several absolute values; for example, an inequality like $3 |x^2-1|+|x-2|+|x^2-3x| > 5$.
A: We can see $|x-a|$ as a distance point $x$ from point $a$.
Now, the question with the above "definition" would be like:
For which $x$ values distance point $x$ from $-2$ be less than distance point $x$ from $4$?
Clearly, by drawing it maybe, you can observe that the answer is for all $x<1$.
A: We have $|a| \lt |b|\,$ if any of these is true:
(i) $\,a$ and $b$ are $\gt 0$ and $a \lt b$
(ii) $a\lt 0$ and $b\ge 0$ and $-a \lt b$
(iii) $b \lt 0$ and $a \gt 0$ and $a \lt -b\,$
(iv) $\,a\lt 0$ and $b\lt 0$ and $-a\lt -b$. We can rewrite this as $b \lt a$.
Four cases! Not surprising, since eliminating a single absolute value sign often involves breaking up the problem into $2$ cases.
Sometimes, one can exploit the simpler $|a| \lt| b|\,$ iff $\,a^2\lt b^2$. But squaring expressions generally makes them substantially messier. 
Added: With your new sample problem, squaring happens to work nicely. We have $|x+2| \lt |x-4|$ iff $(x+2)^2 \lt (x-4)^2$. Expand. We are looking at the inequality
$$x^2+4x+4 \lt x^2-8x+16.$$
The $x^2$ cancel, and after minor algebra we get the equivalent inequality $12x \lt 12$, or equivalently $x\lt 1$.  The squaring strategy works well for any inequality of the form $|ax+b| \lt |cx+d|$.  
But the best approach for this particular problem is geometric. Draw a number line, with $-2$ and $4$ on it. Our inequality says that we are closer to $-2$ than we are to $4$. The number $1$ is halfway between $-2$ and $4$, so we must be to the left of $1$. 
A: We can divide by $|b|$ to get $|a/b|<1$. Let $x=a/b$ then $|x|<1$ so $-1<x<1$. 
Now multiply through by $b$. 
If $b>0$  then $-b<a<b$. 
If $b<0$ then $-b>a>b$
A: $|a| < |b| \iff -|b| < a < |b| \iff 
\begin{cases}
   -b < a < b & \text{If $b > 0$} \\
   \text{No solution.} &\text{If $b=0$} \\
   b < a < -b & \text{If $b < 0$}
\end{cases}$
E. G. $|x + 2| < |x - 4|$
\begin{align}
   x > 4 &\implies-x+4 < x+2 < x-4 \\
         &\implies -2x+4 < 2 < -4 \\
         &\implies \text{No solution.}
\end{align}
\begin{align}
   x > 4 &\implies -x+4 < x+2 < x-4 \\
         &\implies  4 < 2 < 2x-4 \\
         &\implies x < 1
\end{align}
\begin{align}
   x < 4 &\implies x-4 < x+2 < -x+4 \\
         &\implies  -4 < 2 < -2x+4 \\
         &\implies x < 1
\end{align}
Hence $x \in (-\infty, 1)$
