Solve 2-x/x+1+2leq0

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

Solve for x

$$x\in [-4,-1)$$

Solution Steps

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

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

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.

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

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

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

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

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

For the quotient to be ≤0, one of the values x+4 and x+1 has to be ≥0, the other has to be ≤0, and x+1 cannot be zero. Consider the case when x+4\geq 0 and x+1 is negative.

$$x+4\geq 0 x+1<0$$

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

$$x\in [-4,-1)$$

Consider the case when x+4\leq 0 and x+1 is positive.

$$x+4\leq 0 x+1>0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in [-4,-1)$$