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

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

Solve for x

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

Solution Steps

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

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

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

$$x\leq -\frac{1}{3}$$

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

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

The solution satisfying both inequalities is x\geq 1.

$$x\geq 1$$

The final solution is the union of the obtained solutions.

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