Solve (3x+2)(x-1)leq0

$$( 3 x + 2 ) ( x - 1 ) \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-\frac{2}{3},1\end{bmatrix}$$

Solution Steps

For the product to be ≤0, one of the values 3x+2 and x-1 has to be ≥0 and the other has to be ≤0. Consider the case when 3x+2\geq 0 and x-1\leq 0.

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

The solution satisfying both inequalities is x\in \left[-\frac{2}{3},1\right].

$$x\in \begin{bmatrix}-\frac{2}{3},1\end{bmatrix}$$

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

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

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{2}{3},1\end{bmatrix}$$