Solve x-3/x+4geq0

$$\frac { x - 3 } { x + 4 } \geq 0$$

Solve for x

$$x\in (-\infty,-4)\cup [3,\infty)$$

Solution Steps

For the quotient to be ≥0, x-3 and x+4 have to be both ≤0 or both ≥0, and x+4 cannot be zero. Consider the case when x-3\leq 0 and x+4 is negative.

$$x-3\leq 0 x+4<0$$

The solution satisfying both inequalities is x<-4.

$$x<-4$$

Consider the case when x-3\geq 0 and x+4 is positive.

$$x-3\geq 0 x+4>0$$

The solution satisfying both inequalities is x\geq 3.

$$x\geq 3$$

The final solution is the union of the obtained solutions.

$$x<-4\text{; }x\geq 3$$