Elementary Algebra Inequality question $$
\frac{1}{x-1} < -\frac{1}{x+2}
$$
(see this page in wolframalpha)
Ok, so I think the main problem is that I don't really know how to do these questions. What I tried to do was move $-1/(x + 2)$ to the LHS and then tried to get a common denominator. I ended up with 
$$
\frac{(x + 2) + (x -1)}{(x-1)(x+2)} < 0.
$$
So then I went 
$$
\frac{2x + 1}{(x-1)(x+2)}<0
$$
and got $x < -1/2$  and $x \ne 1$ and $x\ne -2$. Therefore, the answer should be $x$ is $(-\infty, -2)$ or $(-2, -1/2)$. But it's not.
 A: $\frac{1}{x-1}<-\frac{1}{x+2} \rightarrow \frac{1}{x-1}+\frac{1}{x+2}<0 \rightarrow$
$\frac{(x+2)+(x-1)}{(x+2)(x-1)} <0 \rightarrow \frac{(x+2)+(x-1)}{(x+2)(x-1)}((x+2)(x-1)^2<0  \rightarrow (x+2+x-1)(x+2)(x+1)<0 $
for $x$ different to $-2,1$. (-2 and 1 arent soultions cause they are undefined.)
using the fact that rational functions are continuous wherever they are defined and the intermediate value theorem:
$(x+2+x-1)(x+2)(x-1)=2(x+\frac{1}{2})(x+2)(x+1)<0$
Since the polynomial 2(x+\frac{3}{2})(x+2)(x+1) has three roots:(each with multiplicity 1)
$-\frac{1}{2},-1,-2$ and -1000 is a solution to the inequality we can know that the solution set is $(-\infty,-2)\cup (-1,-\frac{1}{2})$
A: When you have the form $\frac{a}{b} < 0$ you know there are only two options:
$a < 0$, $b>0$ and $a>0$, $b<0$, because $a$ and $b$ need to differ in sign for the fraction to be negative.
First case:
$$
2x+1 < 0 \implies x < -\frac12
\\
(x-1)(x+2) > 0 \implies x < ...
$$
and then repeat for the other case
A: Define $f(x)=\frac{1}{x-1}+\frac{1}{x+2}$.  As $f$ isn't defined at $x=1$ and $x=-2$ and $f(x)=0\iff x=1/2$ the sign of $f$ on the intervals $(-\infty,-2)$, $(-2,1/2)$, $(1/2,1)$ and $(1,\infty)$ can't change, because $f$ is continuous.
