Solving an Inequality Involving a Fraction 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?

First note that when you multiply by a negative number the inequality changes in sign. In general, if $\dfrac{a}{b} > 0$, we have $a>0, b>0$ or $a<0, b<0$. Hence, we get that
$$(x-1) > 0, \,\,\,\,\, (x+2) > 0 \text{ i.e. } x>1$$
or
$$(x-1) < 0, \,\,\,\,\, (x+2) < 0 \text{ i.e. }x < -2$$
If $\dfrac{x-1}{x+2} = 0$, then $x=1$. Hence, the solution is $$x \in (-\infty,-2) \cup [1,\infty)$$
