Solve 1/x+1-1>0

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

Solve for x

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

Solution Steps

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

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

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

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

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

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

Combine like terms in 1-x-1.

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

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

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

This is false for any x.

$$x\in \emptyset$$

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

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

The solution satisfying both inequalities is x\in \left(-1,0\right).

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

The final solution is the union of the obtained solutions.

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