I am stuck on problem involving absolute values and inequalities

Question

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.

Answer 1

$\dfrac{x-2}{x-4}-\dfrac{x+2}x=\dfrac8{(x-4)x}>0 \iff (x-4)x=(x-2)^2-4>0\iff|x-2|>2$

Answer 2

1. move all terms to the left. you will get

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

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

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

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

$x\lt 0 or x\gt 4\Leftrightarrow |x-2|>2(\therefore)$

Answer 3

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, $$ |x-2| > 2 \text{.} $$