Fractional Inequality $|\frac{2}{x-4}| > 1$

Question

I have tried like this $|\frac{2}{x-4}| > 1 \quad$ $\frac{2}{x-4} > 1$ Or. $\frac{2}{x-4} < -1$; $x-4 < 2$. Or. $4-x < 2$; $x>6$. Or $x>2$; But my text book says it is wrong and the correct answer is $(2,4) \cup (4,6)$

Answer 1

Hint: For $x\neq 4$ we get $$2>|x-4|$$ so we have to solve $2>x-4$ for $x>4$ or $2>-x+4$ for $x<4$. The solution is given by $$2<x<4$$ or $$4<x<6$$

Answer 2

We obtain $x\neq4$ and $$|x-4|<2$$ or $$-2<x-4<2$$ or $$2<x<6,$$ which with $x\neq4$ gives: $$(2,4)\cup(4,6).$$

Answer 3

Hint:

Take the reciprocals of both sides. The inequation becomes, taking into account its domain of validity, $$\biggl|\frac{x-4}2\biggr|<1\iff |x-4|<2\quad\textbf{ and }\quad x\ne 4$$ and interpret this inequality in terms of distance.

Answer 4

Since $|2/x-4|> 1$, so $|(x-4)/2| <1$ This gives $$-2< x-4 <2$$ and hence $2<x < 6$ But since $x\neq 4$ so $x \in (2,4) \cup (4,6)$

Answer 5

$x \not = 4;$

$\left | \dfrac{2}{4-x} \right |$ implies

1)$ \dfrac{2}{x-4} >1$, or

2) $\dfrac{2}{x-4}<-1$;

1) $x >4$ (why?) , and

$2 >x-4$, or $6 >x$,

Hence $x \in (4,6)$.

2) $x< 4$ (why?), and

$2 >(4-x)$, or $x>2$.

Hence $x \in (2,4)$.

Finally $x \in (2,4)\cup (4,6)$.