<s>[INST] Steps for solving this inequality: $ x \geq \frac{6}{x - 1} $ Solve:
$$
x \geq \frac{6}{x - 1}
$$
I figure that I can't really multiply by both sides by $(x - 1)$ since I'm not sure if it will be positive or negative. So I multiplied by $(x - 1)^2$ to get:
$$
x^3 - 2x^2 -5x \geq -6
$$
but wasn't sure how to proceed to an answer.
 [/INST] $$
x \ge \frac{6}{x-1}
$$
Move everything to one side:
$$
x-\frac{6}{x-1} \ge 0
$$
The common denominator is $x-1$, so we have:
$$
\frac{x(x-1)}{x-1} - \frac{6}{x-1} \ge 0
$$
Simplify:
$$
\frac{x^2-x-6}{x-1}\ge 0
$$
Factor:
$$
\frac{(x-3)(x+2)}{x-1} \ge 0
$$
This changes signs at $-2$, $1$, and $3$.  It is positive if $x>3$, negative if $x$ is between $3$ and $1$, positive if $x$ is between $1$ and $-2$, and negative if $x<-2$.  It is zero when the numerator is $0$, i.e. when $x=-2$ or $x=3$.
So the solution is
$$
x\ge3\text{ or } -2\le x<1.
$$
</s>