How do I solve a double absolute value inequality? How do I solve this?

|x-2| > |x-4|

Do I split the inequality up into two as so?

-(x-2) > x-4
x-2 > -(x-4)

 A: You are given:$$|x-2|\gt|x-4|$$Since both sides have an absolute value, we can just square both sides and remove the absolute value symbol (since the square of any number is always non-negative). So we get:$$(x-2)^2\gt(x-4)^2$$$$\therefore x^2-4x+4\gt x^2-8x+16$$$$\therefore -4x+4\gt-8x+16$$$$\therefore 4x\gt12$$$$\therefore x\gt3$$
A: I recommend Henning Makholm's idea of sketching the two functions.
But aside from that, it is also useful to build the intuition that in one dimension, $|x-a|$, for any $a$, is simply the (positive) distance that $x$ is from $a$ on the number line.  So, if $|x-2| > |x-4|$, that simply means that $x$ is further from $2$ than it is from $4$.  For what values of $x$ is that the case?
A: For the rigorous non-squaring approach, we have
$$\begin{align}x\lt 2&: -x+2\gt -x+4\\
2\le x\lt 4&: x-2\gt -x+4\\
4\le x&: x-2\gt x-4\end{align}$$
We immediately see that the first and third cases are resolved: there are no solutions for $x\lt 2$, and all $x\ge 4$ are solutions.  So the second case remains, and we get $2x\gt 6\to x\gt 3$.  Putting all of this together, we arrive at the simple solution of $x\gt 3$.
A: With an inequality where there are 2 absolute value equations on either side of the inequality i.e. $|ax+c| > |bx +d|$.
All you have to do is make both positive, and then a separate inequality where 1 absolute value is negative.
Both Positive:
$$|x-2| > |x-4|$$
$$x-2 > x-4$$
$$0>-2$$
The $x$'s cancel out, so this equation is no use to us.
One is negative:
$$|x-2| > |x-4|$$
$$x-2 > -(x-4)$$
$$x-2 > 4-x$$
$$2x > 6$$
$$\therefore x>3$$
So you do end up "splitting" it into 2 inequalities, but you only make 1 of them negative.
