I'm trying to solve the following inequality: $\tfrac{3x}{x^2+2}≥\tfrac{1}{x-1}$.
When I solve to get a zero on one side I do the following: $\tfrac{3x}{x^2+2}≥\tfrac{1}{x-1}\Leftrightarrow \tfrac{3x(x-1)}{x^2+2}-1≥0$, but when I check online it says I have to do $\tfrac{3x}{x^2+2}≥\tfrac{1}{x-1}\Leftrightarrow \tfrac{3x}{x^2+2}-\tfrac{1}{x-1}≥0$.
Could someone explain why I can't do the first method?
You cannot multiply by $x-1$ and preserve the inequality as this quantity could be negative (in particular when $x<1$), which would flip the inequality. If you really wanted to do this, you'd have to treat the cases $x<1$ and $x\geq 1$ separately, and adjust the inequality accordingly. Because of this the second method is preferable.