Solve 1-x/x<0

$$\frac{ 1-x }{ x } < 0$$

Solve for x

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

Solution Steps

For the quotient to be negative, 1-x and x have to be of the opposite signs. Consider the case when 1-x is positive and x is negative.

$$1-x>0 x<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when x is positive and 1-x is negative.

$$x>0 1-x<0$$

The solution satisfying both inequalities is x>1.

$$x>1$$

The final solution is the union of the obtained solutions.

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