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?
 A: 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 $$x \in (-\infty,-2) \cup [1,\infty)$$
A: The $=0$ part is easy. So let's worry about the $\gt 0$ part.
Our expression can only change sign when the top changes sign, or when the bottom changes sign. Thus the only "sign change" candidates are $x=1$ and $x=-2$.
It follows that our function is uniform in sign in $(-\infty,-2)$, also in $(-2,1)$, also in $(1,\infty)$.
In each of these regions, sind a test point: any point will do.
For the region $(-\infty,-2)$, use say the test point $x=-12$. Our function is $\frac{-13}{-10}$ at this point, positive. So our function is positive in all of $(-\infty,-2)$.
For the interval $(-2,1)$, use the test point $x=0$. At $0$, our function is clearly negative, so it is negative in the whole interval. 
Finally, use a test point in $(1,\infty)$ to conclude our function is positive in that interval. 
A: A geometric idea applied to algebra: multiply the inequality by a positive quantity, so that you won't have problems with the inequality sign's direction:
$$\frac{x-1}{x+2}\geq 0\Longleftrightarrow \frac{x-1}{x+2}\cdot(x+2)^2\geq 0\cdot(x+2)^2\Longrightarrow (x-1)(x+2)\geq 0$$
The left side is a parabola that opens upwards and vanishes at $\,x=-2\,,\,1\,$ , and from the geometric picture we can see the parabola lies above the $\,x-$axis, which is what we want, precisely whenever $\,x<-2\,\,\;\vee\;\,\,x>1\,$.
Finally, we just add the point $\,x=1\,$ to the above as we had a weak inequality. 
