Solving the inequality $-1<\frac{2}{x}$ 
What is the range of $x$ for inequality $-1<\frac{2}{x}$ ?

My attempt is as follows:
$$-1<\frac{2}{x}$$
$$-1\cdot{x}<2$$
Multiplying by $-1$ on both sides, which changes sign, we get:
$$x>-2$$
So according to this resultant equation lets put some value of x into the first equation.
Let's take $x=-1$
$$-1<\frac{2}{-1}$$
$$-1<-2$$
I am not able to understand what is wrong here, why am I getting contradictory result when I put values of $x$ in the original question.
 A: For $x>0$ we get $$-x<2$$ so $$x>-2$$
For $$x<0$$ we get $$-x>2$$ so $$x<-2$$
A: The following step
$$-1<\frac{2}{x} \implies -1\cdot{x}<\frac{2}{x}\cdot{x}=2$$
is correct ony for $x>0$.
Otherwise we need to reverse the inequality sign.
A: $$-1<\dfrac2x\iff\dfrac{x+2}x>0$$
$\iff x(x+2)>0$
$\implies x>$max$(0,-2)$ 
Or $x<$min$(0,-2)$
A: We can multiply both sides by $x^2$ since $x^2 \geq 0$, so the sign won't change after multiplying both sides by $x^2$. 
$$\begin{aligned}
-1 &< \frac{2}{x}\\
-x^2 &< 2x\\
x^2 + 2x &> 0\\
x(x+2) &> 0\\
x > 0, &\quad x < -2
\end{aligned}$$
Therefore, the range of solutions is $x \in (-\infty, -2) \cup (0, \infty)$.
A: The simplest way: $$\frac{2}{x} >-1 \Rightarrow \frac{2}{x}+1>0 \Rightarrow \frac{x+2}{x} >0 \Rightarrow x(x+2)>0 \Rightarrow x<-2~or~x>0$$
A: When you go from 
$-1 < \frac{2}{x}$
to
$-x < 2$,
you're assuming that $x$ is positive since multiplication by a negative number flips the inequality sign. So when you have $x = -1$(or simply $x < 0$), you will have 
 $-x > 2$ instead of $-x < 2$.
A: THE GIVEN INEQUALITY HAS GOT RANGE AS X<-2 AND X>=0.
Reason for it is 
                          2/X >-1.
                         Implies two cases 1) if 2/X >0.
                                                               Implies X>=0
                                                           2)if -1<2/X<0.
                                                               Implies X<-2.
