Please help how to work the inequation (x-1)/(x-5)<0 I am studying for college finals and I cannot solve the inequation (x-1)/(x-5)<0 using the method the professor taught us, which I have to use in the exam.
This is another inequation using said method:

I noticed that the inequation I posted has a different symbol than the pic, but the first one is the only example of this type the professor gave us with the less than symbol. 
This is how I tried to solve it:

I appreciate any help you can give. Thanks in advance. 

Remember if $\frac ab < 0$ then $a$ and $b$ are "different signs".
So if $\frac {x-1}{x-5} < 0$ then either
1)  $x - 1 > 0$ and $x -5 < 0$
OR
2) $x-1 < 0$ and $x-5 > 0$.
In case 1) we have $x - 1> 0$ so $x > 1$ and $x-5 < 0$ so $x < 5$.  So $x$ is between $1$ and $5$ or $1 < x < 5$
In case 2) we have $x -1 < 0$ and $x< 1$ and $x-5> 0$ so $x > 5$. So $x$ is both less than $1$ and greater than $5$. That's impossible.
So Case 1: is the true case and $1 < x < 5$.
.....
Alternatively.  $1 < 5$ always and so $-5 < -1$ always, and $x-5 <x - 1$ always no matter what $x$ is.
So when we know that $x-1$ and $x-5$ are "different signs" we know that $x-5$ must be the negative one (because it is the smaller one) and $x -1$ must be the positive one because it is bigger.
So $x -5 < 0 < x-1$
So $x - 5 < 0$ and $x < 5$.  And $0 < x-1$ so $1 < x$.  So $1 < x < 5$.
