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How do I solve this type of inequalities analytically? I know the answer is $ x<-1 $ and $ x> 0 $ but: $$\frac{1}{x} > -1 $$ $$1>-x $$ $$ x>-1 $$ Wat I'm doing wrong?

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    $\begingroup$ To the answerers: someone please address the '$x<-1$ and $x>0$' $\endgroup$ – Git Gud Jul 13 '13 at 9:52
  • $\begingroup$ @Git Gud: Done - I suppose this is worth pointing out, though perhaps it would be better be suited as a comment anyhow? $\endgroup$ – Sharkos Jul 13 '13 at 11:18
  • $\begingroup$ @Sharkos I think it's better in an answer. +1 $\endgroup$ – Git Gud Jul 13 '13 at 11:41
  • $\begingroup$ $$\begin{align} \dfrac{1}{x}>-1 &\iff -\dfrac1x<1 \\ \end{align}$$ Then make two separate cases, one where $x$ is positive and the other where $x$ is negative. $\endgroup$ – Hakim Mar 8 '14 at 17:01
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You can't multiply both sides of an inequality by an unknown (which is your first step). This is because when you multiply by a negative, you have to flip the inequality, but when you multiply by a positive, you don't. If you multiply by an unknown you don't know which to do, because you don't know its sign.

What you can do is separate the case into "Assuming $x > 0$..." and "Assuming $x < 0$...".

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  • $\begingroup$ This type of case analysis can get pretty cumbersome when the denominator contains more terms. $\endgroup$ – augurar Jul 15 '13 at 1:08
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When you multiply an inequality by a negative number the direction changes - for example $1<2$ but $-1>-2$. Since you don't know whether $x$ is positive or negative, you can't multiply the inequality by it.

You could, however, instead choose to look at the cases where $x$ is positive and where $x$ is negative separately. The argument you gave shows that it works for positive $x$ which are bigger than $-1$, which is just all positive $x$. This gives one of the two bits you wanted. Can you find the other?

Note: As Git Gud says in the comment, note that we would usually express the answer as $x<-1$ or $x>0$ because there is no real number $x$ satisfying the first and the second inequality - instead we are expressing two alternative solutions. Rigorously, $$1/x>-1 \text{ (and $x\neq 0$)}\iff (x<-1 \text{ or } x>0)$$

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Subtract $-1$ from both sides of the equation to get: $$ \frac{1+x}{x} > 0 $$

This inequality will be satisifed when $1+x$ and $\frac{1}{x}$ have the same sign, i.e. when $x>0$ or $x<-1$.

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As others have pointed out, you can't just multiply by $x$ since you don't know whether it's positive or negative. One way to get around this is to multiply by something which is definitely nonnegative. Which functions of $x$ have this property? Well, the simplest examples are squares. In this case, we can multiply by $x^2$ to get $$x>-x^2 \\x^2+x>0 \\ x(x+1) >0$$ and hopefully you can solve it from here.

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