(x-2)/(x+3) less than (x+1)/(x)
The Attempt at a Solution
I broke it up into cases.
When x+3 less than 0 and x less than 0
and then when both are positive, when one is positive and the other is negative and then the other way around.
I'm not sure if that's the right way though. I was thinking that maybe you can say:
(x+3)(x) less than 0
and a second case
(x+3)(x) greater than 0