Solve (3x-1)(x+1)geq0

$$( 3 x - 1 ) ( x + 1 ) \geq 0$$

Solve for x

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

Solution Steps

For the product to be ≥0, x+1 and 3x-1 have to be both ≤0 or both ≥0. Consider the case when x+1 and 3x-1 are both ≤0.

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

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

$$x\leq -1$$

Consider the case when x+1 and 3x-1 are both ≥0.

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

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.

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