Solve 1+1/x>0

$$1 + \frac { 1 } { x } > 0$$

Solve for x

$$x\in \left(-\infty,-1\right)\cup \left(0,\infty\right)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

$$\frac{x}{x}+\frac{1}{x}>0$$

Since \frac{x}{x} and \frac{1}{x} have the same denominator, add them by adding their numerators.

$$\frac{x+1}{x}>0$$

For the quotient to be positive, x+1 and x have to be both negative or both positive. Consider the case when x+1 and x are both negative.

$$x+1<0 x<0$$

The solution satisfying both inequalities is x<-1.

$$x<-1$$

Consider the case when x+1 and x are both positive.

$$x>0 x+1>0$$

The solution satisfying both inequalities is x>0.

$$x>0$$

The final solution is the union of the obtained solutions.

$$x<-1\text{; }x>0$$