Solve x+3/x+1geq0

$$\frac { x + 3 } { x + 1 } \geq 0$$

Solve for x

$$x\in (-\infty,-3]\cup (-1,\infty)$$

Solution Steps

For the quotient to be ≥0, x+3 and x+1 have to be both ≤0 or both ≥0, and x+1 cannot be zero. Consider the case when x+3\leq 0 and x+1 is negative.

$$x+3\leq 0 x+1<0$$

The solution satisfying both inequalities is x\leq -3.

$$x\leq -3$$

Consider the case when x+3\geq 0 and x+1 is positive.

$$x+3\geq 0 x+1>0$$

The solution satisfying both inequalities is x>-1.

$$x>-1$$

The final solution is the union of the obtained solutions.

$$x\leq -3\text{; }x>-1$$