Solve xleft(x-1right)/x+1>x

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

Solve for x

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

Solution Steps

Use the distributive property to multiply x by x-1.

$$\frac{x^{2}-x}{x+1}>x$$

Subtract x from both sides.

$$\frac{x^{2}-x}{x+1}-x>0$$

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

$$\frac{x^{2}-x}{x+1}-\frac{x\left(x+1\right)}{x+1}>0$$

Since \frac{x^{2}-x}{x+1} and \frac{x\left(x+1\right)}{x+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{2}-x-x\left(x+1\right)}{x+1}>0$$

Do the multiplications in x^{2}-x-x\left(x+1\right).

$$\frac{x^{2}-x-x^{2}-x}{x+1}>0$$

Combine like terms in x^{2}-x-x^{2}-x.

$$\frac{-2x}{x+1}>0$$

For the quotient to be positive, -2x and x+1 have to be both negative or both positive. Consider the case when -2x and x+1 are both negative.

$$-2x<0 x+1<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when -2x and x+1 are both positive.

$$x+1>0 -2x>0$$

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.

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