<s>[INST] How to do $\lvert x - 1 \rvert < \lvert x-3 \rvert$ 
Solve for $x$ such that $$\lvert x - 1 \rvert < \lvert x-3 \rvert$$

I understand that the question is essentially saying what are the possible values for $x$, centered at $1$ with a distance of $x-3$ on either side of $x$. But how does that work? 
 [/INST] Several ways.
There are $4$ possibilities $x -1$ can be $<$ or $\ge 0$ and $x-3$ can be $<$ or $\ge 0$.
Or in other words $x$ can be $<$ or $\ge 1$ or $x$ can be $< $ or $\ge 3$. These are four possibilities but some may be contradictory (We can't have have $x < 1$ and $x \ge 3$ both) or redundant (if $x\ge 3$ then $x > 1$ is obviously also true.
So this breaks down into three cases.  $x < 1; 1 \le x < 3; x\ge 3$.
Case 1:  $x < 1$.  Then $|x-1| = 1-x$ and $|x-3| = 3-x$.
So $1 -x < 3-x\implies 1 < 3$ ... which is always true.  This means if $x< 1$ then $x$ an be any value less than $1$.
Case 2:  $1 \le x < 3$ so $|x-1| = x-1$ and $|x-3| = 3-x$.
So $x-1 < 3-x \implies 2x < 4\implies x < 2$.  This means if $1 \le x < 3$ then we know further than $x < 2$ and that $2\le x < 3$ is impossible.
Case 3:  $x \ge 3$ then $|x-1| = x-1$ and $|x-3| = x-3$ and we can $x-1 < x-3\implies -1 < -3$.  That is simply impossible.  So we know that $x \ge 3$ is not possible.
So we know 1) $x$ could be any value $< 1$.  2) $x$ could be any value so that $\le x < 2$ but $x$ can not be $2 \le x < 3$.  And 3) $x \ge 3$ is impossible.
Combining those three results we get .... $x < 2$.  And any value$x < 2$ will work.
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Just for kicks and giggles:
$|m| < k$ means $-k < m < k$.
So we have $-|x-3| < x-1 < |x-3|$.  
If $x \ge 3$ then that is $-x + 3 < x - 1 < x-3$ so $4 < 2x < 2x - 2$  and $-2x + 4 < 0 < -2$ which is impossible.
So $x < 3$ and $x -3 < x - 1 < -x + 3$ so $2x -3 < 2x - 1 <  3$ and so $2x < 4$ and $x < 2$.
So $x < 2$.
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And cute....
$\sqrt {k^2} = |k|$ and if $0 \le a$ and $0 \le b$ then $a^2 < b^2 \iff a < b$.
So $\sqrt {(x-1)^2} = |x-1| < |x - 3| = \sqrt{(x-2)^2} \iff $
$(x-1)^2 < (x-3)^2 \iff$
$x^2 -2x + 1 < x^2 - 6x + 9 \iff$
$4x < 8 \iff$
$x < 2$.
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