I have the following inequality:
$ \frac{x-1}{x+2} \geq 0.$
I solved it pretty fast:
$\begin{align} \frac{x-1}{x+2} +1 & \geq 1\\\\ \left(\frac{x-1}{x+2} + 1\right)\cdot(x+2) & \geq 1 \cdot (x+2)\\\\ x-1 + 1\cdot(x+2) & \geq 1\cdot (x+2)\\\\ 2x + 1 & \geq x+2\\\\ x + 1 & \geq 2\\\\ x & \geq 1 \end{align}$
But that is not the only solution, the other solution is $x < -2$. How do I get to this solution?