Solve 3x-1/x+2geq0

$$\frac { 3 x - 1 } { x + 2 } \geq 0$$

Solve for x

$$x\in (-\infty,-2)\cup [\frac{1}{3},\infty)$$

Solution Steps

For the quotient to be ≥0, 3x-1 and x+2 have to be both ≤0 or both ≥0, and x+2 cannot be zero. Consider the case when 3x-1\leq 0 and x+2 is negative.

$$3x-1\leq 0 x+2<0$$

The solution satisfying both inequalities is x<-2.

$$x<-2$$

Consider the case when 3x-1\geq 0 and x+2 is positive.

$$3x-1\geq 0 x+2>0$$

The solution satisfying both inequalities is x\geq \frac{1}{3}.

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

The final solution is the union of the obtained solutions.

$$x<-2\text{; }x\geq \frac{1}{3}$$