Compund Inequality with Reciprocal $0<(2x-4)^{-1}<\frac{1}{2}$ I need help with this exercise: $0<(2x-4)^{-1}<\frac{1}{2}$.
$$0<(2x-4)^{-1}<\frac{1}{2}$$
$$0<\frac{1}{2x-4}<\frac{1}{2}$$
I separate this into two inequalities:

*

*$0<\frac{1}{2x-4}$


*$\frac{1}{2x-4}<\frac{1}{2}$
The first one I had no problem solving it.
$$0<\frac{1}{2x-4}$$
By the Reciprocal Property of inequalities
$$0<2x-4$$
$$4<2x$$
$$\frac{4}{2}<\frac{2x}{2}$$
$$2<x$$
$$x>2$$
Is the second that I have a problem with. Is there a way to solve it without using a sign chart/table? I do know how to do it with a sign chart/table. I just want to know if there is another way to do it that is not the sign table.
 A: As you have correctly concluded, one must have that $x > 2$.
Under such condition, $2x - 4 > 0$.
Hence we can multiply both sides by it so that we obtain an equivalent inequation:
\begin{align*}
\frac{1}{2x - 4} < \frac{1}{2} & \Longleftrightarrow 2 < 2x - 4\\\\
& \Longleftrightarrow 2x > 6\\\\
& \Longleftrightarrow x > 3
\end{align*}
So the solution set to such problem is given by $S = (3,+\infty)\cap(2,+\infty) = (3,+\infty)$.
Hopefully this helps!
A: $$0<(2x-4)^{-1}<\frac{1}{2}$$
$$2<(2x-4)<\infty$$
$$6<2x<\infty$$
$$3<x<\infty$$
A: If you had $\frac 1{2x-4} < \frac 12$ you'd have a difficulty because you don't know if $\frac 1{2x-4}$ is positive or negative.
!BUT!  you are GIVEN that $0 < \frac 1{2x-4}$.
There's no reason to separate it into two inequalities... just take it as a single inequality.
$0 < \frac 1{2x-4} < \frac 12 \iff \frac 1{2x-4}$ is positive and $\frac 1{2x-4} < \frac 12$.
So because $\frac 1{2x-4}$ is given to be positive we have reciprocal law $2x-4 > 2$.
That's all we need.
..... alternatively we should be comfortable with $0 < a < b\implies 0 < \frac 1b < \frac 1a$ is a common rule that out to be second nature.
A: \begin{align*}
0<(2x-4)^{-1}<\frac{1}{2}\\ \\
\implies 
0<\frac{1}{2x-4}<\frac{1}{2}\\ \\
\implies  0< 2< 2x-4\\ \\
\implies  4< (2)+4< (2x-4) + 4 \\ \\
\implies  4< 6< 2x \\
\implies 2< 3< x\\
\implies 3< x\\
\end{align*}
