Solve 0leq2+x/x+1

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

Solve for x

$$x\in (-\infty,-2]\cup (-1,\infty)$$

Solution Steps

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$\frac{2+x}{x+1}\geq 0$$

For the quotient to be ≥0, x+2 and x+1 have to be both ≤0 or both ≥0, and x+1 cannot be zero. Consider the case when x+2\leq 0 and x+1 is negative.

$$x+2\leq 0 x+1<0$$

The solution satisfying both inequalities is x\leq -2.

$$x\leq -2$$

Consider the case when x+2\geq 0 and x+1 is positive.

$$x+2\geq 0 x+1>0$$

The solution satisfying both inequalities is x>-1.

$$x>-1$$

The final solution is the union of the obtained solutions.

$$x\leq -2\text{; }x>-1$$