# Elementary Algebra Inequality question

$$\frac{1}{x-1} < -\frac{1}{x+2}$$

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.

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A product of three numbers is negative when all three of them are negative, or only one is negative. – egreg Oct 17 '13 at 23:49
huh? What are you referring to? – Kat Oct 17 '13 at 23:53
You have the product of $2x+1$ with $(x-1)^{-1}$ and $(x+2)^{-1}$, don't you? – egreg Oct 17 '13 at 23:54
yeah. so then 2x + 1 becomes x < -1/2 – Kat Oct 17 '13 at 23:57
We will want to get rid of the denominator, by multiplying by $(x+2)9x-1)$. If $x\lt -2$, we will be multiplying by a positive, also when $x\gt 1$. When $-2\lt x\lt 1$ we will be multiplying by a negative, and the inequality is reversed. – André Nicolas Oct 18 '13 at 0:25

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

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Thanks for keeping it simple. – Kat Oct 18 '13 at 21:37

$\frac{1}{x-1}<-\frac{1}{x+2} \rightarrow \frac{1}{x-1}+\frac{1}{x+2}<0 \rightarrow$

$\frac{(x+2)+(x-1)}{(x+2)(x-1)} <0 \rightarrow \frac{(x+2)+(x-1)}{(x+2)(x-1)}((x+2)(x-1)^2<0 \rightarrow (x+2+x-1)(x+2)(x+1)<0$

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 $(-\infty,-2)\cup (-1,-\frac{1}{2})$

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how did you get $((x+2)(x+1))^2$? – Kat Oct 18 '13 at 0:48
i multiply both sides by that (its standard in this type of problems) its usefull because since squares of real numbers are non negative im preserving the sign, and therefore its also gonna be positive unless its 0. But it can only be 0 if x is a root of the polynomial. – Carry on Smiling Oct 18 '13 at 1:23
Unfortunately, there is an error on the first line, it should be $1/(x-1)$ instead of $1/(x+1)$. – Mikael Öhman Oct 18 '13 at 1:39
oops. fixing fail sorry guys – Carry on Smiling Oct 18 '13 at 12:03

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.

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