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**emakarov** $\displaystyle \frac{x+1}{x}>0$ iff either x + 1 > 0 and x > 0, or x + 1 < 0 and x < 0. In the first case, x + 1 > 0 gives x > -1. However, we need *both* conditions to be true: x + 1 > 0 and x > 0. The second one is more restrictive (it implies the first one). Therefore, x + 1 > 0 and x > 0 is equivalent to x > 0. Now consider the second case similarly.

Edit: Dividing both sides of an inequality by x has to be done carefully. If you divide both sides of an equation, you just need to require that x ≠ 0. If you divide both sides of an inequality, you need to consider the cases x > 0 and x < 0 since dividing by a negative number flips the inequality. Thus, one inequality is split into two systems of two inequalities. In this particular problem, it is easier to use the fact that a / b > 0 iff a and b are either both positive or both negative.