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?