I forgot how to do inequalities with a rational function. The question I have is:

((3x-2)/(x-1))<1

Here is how I would do this in a formal way:

Would like to have a zero on the right side, so
((3x-2)/(x-1))<1
((3x-2)/(x-1))-1 < 0
(3x-2 - (x-1))/(x-1) < 0
(2x - 1)/(x-1) < 0

"critical values" are x = 1/2 and x = 1

This splits a number line into 3 parts,
1. less than 1/2
2. between 1/2 and 1
3. greater than 1

pick an arbitrary value of x in each part.

case1: x < 1/2
how about x = 0, so (0-1)/(0-1) < 0
1 < 0 , which is false , so all values of x < 1/2 don't work

case2:
x between 1/2 and 1, say x = .75
then (1.5 - 1)/(.75-1) < 0
.5/-.25 < 0 , any negative < 0, so YES

case 3: x > 1, let's pick x = 5
(10-1)(5-1) < 0 , NO

So all values between 1/2 and 1 will work,
BUT if x = 1, we would be dividing by zero,

final answer:
1/2 ≤ x < 1