1. **State the problem:** Solve the equation $$-(3x + 2) \cdot \left(x - \frac{2}{3}\right) = 0$$ for $x$. 2. **Recall the zero product property:** If a product of two factors equals zero, then at least one of the factors must be zero. That is, if $$A \cdot B = 0,$$ then either $$A = 0$$ or $$B = 0$$. 3. **Apply the zero product property:** Here, the two factors are $$-(3x + 2)$$ and $$\left(x - \frac{2}{3}\right)$$. Set each factor equal to zero: $$-(3x + 2) = 0$$ and $$x - \frac{2}{3} = 0$$ 4. **Solve the first equation:** $$-(3x + 2) = 0 \implies 3x + 2 = 0$$ Subtract 2 from both sides: $$3x + 2 - 2 = 0 - 2 \implies 3x = -2$$ Divide both sides by 3: $$\frac{\cancel{3}x}{\cancel{3}} = \frac{-2}{3} \implies x = -\frac{2}{3}$$ 5. **Solve the second equation:** $$x - \frac{2}{3} = 0$$ Add $$\frac{2}{3}$$ to both sides: $$x = \frac{2}{3}$$ 6. **Final solution:** The solutions to the equation are $$x = -\frac{2}{3} \quad \text{or} \quad x = \frac{2}{3}$$ These are the values of $x$ that make the original equation true.