Solve 2-x/x+1+2<0

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

Solve for x

$$x\in \left(-4,-1\right)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{x+1}{x+1}.

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

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

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

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

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

Combine like terms in 2-x+2x+2.

$$\frac{4+x}{x+1}<0$$

For the quotient to be negative, x+4 and x+1 have to be of the opposite signs. Consider the case when x+4 is positive and x+1 is negative.

$$x+4>0 x+1<0$$

The solution satisfying both inequalities is x\in \left(-4,-1\right).

$$x\in \left(-4,-1\right)$$

Consider the case when x+1 is positive and x+4 is negative.

$$x+1>0 x+4<0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(-4,-1\right)$$