Solve x+1/x+3geq0

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

Solve for x

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

Solution Steps

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

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

The solution satisfying both inequalities is x<-3.

$$x<-3$$

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

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

The solution satisfying both inequalities is x\geq -1.

$$x\geq -1$$

The final solution is the union of the obtained solutions.

$$x<-3\text{; }x\geq -1$$