Solving $\dfrac{x+2}{x}>0$ I want to find values of $x$ such that $\dfrac{x+2}{x}>0$ : $1+\dfrac{2}{x}=\dfrac{x+2}{x}>0 \implies \dfrac{2}{x}>-1 \implies \dfrac{1}{x}>\frac{-1}{2} \implies x<-2 $. But by intuition $x>0$ is also results in $\dfrac{x+2}{x}>0$. How possible that $x>0$ doesn't come from solving the original inequality? I mean, why $\dfrac{x+2}{x}>0$ results in $x<-2$ not the correct solution, i.e., $x<-2 \cup x>0$ ?
Thank you. 
 A: $$\frac{x+2}{x}>0\iff (x+2>0\wedge x>0)\lor(x+2<0\wedge x<0)\\\iff x\in(0,\infty)\cup(-\infty,-2)$$
A: There is a way to avoid dealing with a fraction. Note that $x \neq 0 \to x^2 > 0 \to \dfrac{x+2}{x} > 0 \iff \dfrac{x(x+2)}{x^2}> 0\iff x(x+2) > 0\iff x< -2 \cup x > 0$
A: Be careful when "inverting" fractions. You wrote 
\begin{equation}
\frac{1}{x}>-\frac{1}{2}\Rightarrow x<-2
\end{equation}
Here, you "inverted" the fraction and switched the inequality sign. You must remember that the inequality changes direction only if you multiply by a negative number; otherwise it stays the same. To multiply by $x$ (as you did) you must first decide if $x$ is positive or negative. So it should turn out this way:
\begin{equation}
\text{if $x>0$:}\qquad\frac{1}{x}>-\frac{1}{2}\Rightarrow 2>-x\Rightarrow -2<x
\end{equation}
which gives no further constraint, so $x>0$ is a solution, and
\begin{equation}
\text{if $x<0$:}\qquad\frac{1}{x}>-\frac{1}{2}\Rightarrow 2<-x\Rightarrow -2>x
\end{equation}
So the solution in this case is actually $x<-2$.
Combine the two solutions and you have the answer.
A: You are correct that your inequality is equal to $$\frac{1}{x}>-\frac12,$$
but that is not equal to $x<-2$. In fact, if $x$ is positive, then you multiply the inequality by $-2x$ on both sides and you get $-2 < x$, because if $a<b$ and $c$ is a negative number, then $ca>cb$ (the inequality turns around!). This means that if $x$ is positive, then the inequality is always true.
On the other hand, if $x$ is negative, then $-2x$ is a positive number, so miltiplying the inequality by $-2x$ yields $-2 > x$, meaning that $x\in(-2,0)$ is a valid solution.
A: You made a mistake when you reached $\frac{1}{x}>-\frac{1}{2}\Rightarrow x<-2$. You don't have the right to do it if you don't know if $x$ is positive or negative because you're inversing a term. Multipliying and inversing without mistakes needs for you to use positive numbers.
If $x>0$ then $\frac{1}{x}>0>-\frac{1}{2}$ so it's correct (your inequalities were equivalent except $x<-2$).
Now if $x<0$ then $-x>0$. $\frac{1}{x}>-\frac{1}{2}\Leftrightarrow 0<\frac{1}{-x}<\frac{1}{2}$ which is equivalent to $2<-x$ which is equivalent to $x<-2<0$.
Therefore the set of solution is $(0,\infty)\cup (-\infty, -2)$
A: There's an easier way: the signs of both dividend and divisor must be the same for the quotient to be positive (greater than 0). So,
$$\frac{x+2}{x}>0$$
implies:
Case 1: $$\begin{array}{rcl}x+2>0&\wedge& x>0\\x>-2&\wedge& x>0\\\Rightarrow&x>0&\end{array}$$
Case 2: $$\begin{array}{rcl}x+2<0&\wedge& x<0\\x<-2&\wedge& x<0\\\Rightarrow&x<-2&\end{array}$$
Both cases are independent, so the solution is: 
$$x<-2 \vee x>0 \Rightarrow x\in(-\infty,-2)\cup(0,\infty)$$
