Solve x/1-x-1<0

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

Solve for x

$$x\in \left(-\infty,\frac{1}{2}\right)\cup \left(1,\infty\right)$$

Solution Steps

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

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

Since \frac{x}{1-x} and \frac{1-x}{1-x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x-\left(1-x\right)}{1-x}<0$$

Do the multiplications in x-\left(1-x\right).

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

Combine like terms in x-1+x.

$$\frac{2x-1}{1-x}<0$$

For the quotient to be negative, 2x-1 and 1-x have to be of the opposite signs. Consider the case when 2x-1 is positive and 1-x is negative.

$$2x-1>0 1-x<0$$

The solution satisfying both inequalities is x>1.

$$x>1$$

Consider the case when 1-x is positive and 2x-1 is negative.

$$1-x>0 2x-1<0$$

The solution satisfying both inequalities is x<\frac{1}{2}.

$$x<\frac{1}{2}$$

The final solution is the union of the obtained solutions.

$$x>1\text{; }x<\frac{1}{2}$$