Here's the question you clicked on:
melbel
(x-4)/(2x+3) <=3
\[\frac{ x-4 }{ 2x+3 }\le3\]
My prof said that I can't multiply the 3 by the stuff in the denominator because the denominator contains an x. Since you don't know if x is going to be pos or negative, you can't determine whether or not you'll need to switch the sign. I have the answer, I just don't know how to solve it: \[(-\infty, -\frac{ 13 }{ 5 }]\cup(-\frac{ 3 }{ 2 },\infty)\]