I am stuck on problem involving absolute values and inequalities The problem states to solve for $x$, and then to write the answer with absolute value notation.
The problem is: $(x-2)/(x-4) > (x+2)/(x)$
The correct answer is   abs$(x-2) > 2$
(abs stands for absolute value).
Can someone explain to me how to arrive at this answer?  Thank you.
 A: *

*move all terms to the left. you will get


$\frac{x-2}{x-4}-\frac{x+2}{x}\gt 0$


*common denominator; turn the left side into a rational expression. you will get


$\frac{8}{x\cdot (x-4)}\gt 0$


*then study the quadratic at the denominator. you will get the final answer


$x\lt 0 or x\gt 4\Leftrightarrow |x-2|>2(\therefore)$
A: $\dfrac{x-2}{x-4}-\dfrac{x+2}x=\dfrac8{(x-4)x}>0 \iff (x-4)x=(x-2)^2-4>0\iff|x-2|>2$
A: First, the given inequality is invalid when $x = 0$ or $x = 4$, since it contains undefined expressions in either of these cases.
Suppose $x \neq 0$ and $x \neq 4$.  Then we may multiply both sides of the inequality by $x(x-4)$ to clear denominators.  We should be careful: for some values of $x$, in particular $0 < x < 4$, we are multiplying both sides of the inequality by a negative number, so the sense of the inequality reverses; for other values of $x$, we are multiplying by a positive number, so the sense does not reverse.  In particular, we obtain \begin{align*}
(x(x-2) > (x+2)(x-4) &\text{ and } (x < 0 \text{ or } 4 < x))  \text{ or }\\
(x(x-2) < (x+2)(x-4) &\text{ and } 0<x<4)  \text{.}
\end{align*}
This is written in the form "(case 1) or (case 2)".  In the first case, we expand and then subtract terms to get
$$  x^2 - 2x > x^2 - 2 x - 8  \text{ , so}$$
$$ 8 > 0 \text{,} $$
which is always true.  Doing the same to the second case, we obtain
$$ 8 < 0 \text{,} $$
which is never true.  So we have simplified to 
\begin{align*}
(\text{True} &\text{ and } (x < 0 \text{ or } 4 < x))  \text{ or }\\
(\text{False} &\text{ and } 0<x<4)  \text{.}
\end{align*}
"True or (anything)" is always true and "False and (anything)" is always false, so actually we have 
$$ x < 0 \text{ or } 4 < x \text{.}  $$
(The next steps are very clear if you plot the current solution set on a number line.)  The midpoint of $0$ and $4$ is $2$, so we subtract $2$ from both equations to reveal
$$ x-2 < -2 \text{ or } 2 < x-2 \text{,}  $$
or what is the same thing,
$$  |x-2| > 2  \text{.}  $$ 
