Solve x^2+x-1/1-x>-x

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

Solve for x

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

Solution Steps

Add x to both sides.

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

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

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

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

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

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

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

Combine like terms in x^{2}+x-1+x-x^{2}.

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

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

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

This is false for any x.

$$x\in \emptyset$$

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

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

The solution satisfying both inequalities is x\in \left(\frac{1}{2},1\right).

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

The final solution is the union of the obtained solutions.

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