Solutions for inequality $\frac{1}{x} + \frac{1}{1-x} \gt 0$ How to find all real solutions for $$\frac{1}{x} + \frac{1}{1-x} \gt 0$$
I came up with $\frac{1}{x(1-x)} \gt 0$ implies $x(1-x)>0$ and finally ended with $0<x<1$ but the answer provided was $0<x<1$ or $x>1$. I tried sample values for $x>1$ but they don't satisfy the inequality.

It should be $\frac{1}{x(1-x)} \gt 0$ thus, $x(1-x)>0$ and so, $0 < x < 1$ is the only solution.
