1. **State the problem:** We have the system of equations: $$2x + 3y = -8$$ $$3y^2 - 8y = 2x + 10$$ We want to find the product of $x_1$ and $x_2$, where $(x_1,y_1)$ and $(x_2,y_2)$ are distinct solutions to this system. 2. **Express $x$ from the first equation:** From $$2x + 3y = -8$$, we isolate $x$: $$2x = -8 - 3y$$ $$x = \frac{-8 - 3y}{2}$$ 3. **Substitute $x$ into the second equation:** The second equation is: $$3y^2 - 8y = 2x + 10$$ Substitute $x$: $$3y^2 - 8y = 2 \times \frac{-8 - 3y}{2} + 10$$ Simplify the right side: $$3y^2 - 8y = -8 - 3y + 10$$ $$3y^2 - 8y = 2 - 3y$$ 4. **Bring all terms to one side:** $$3y^2 - 8y - 2 + 3y = 0$$ $$3y^2 - 5y - 2 = 0$$ 5. **Solve the quadratic for $y$:** Using the quadratic formula: $$y = \frac{5 \pm \sqrt{(-5)^2 - 4 \times 3 \times (-2)}}{2 \times 3}$$ $$y = \frac{5 \pm \sqrt{25 + 24}}{6}$$ $$y = \frac{5 \pm \sqrt{49}}{6}$$ $$y = \frac{5 \pm 7}{6}$$ So the two values of $y$ are: $$y_1 = \frac{5 + 7}{6} = 2$$ $$y_2 = \frac{5 - 7}{6} = -\frac{1}{3}$$ 6. **Find corresponding $x$ values:** Using $x = \frac{-8 - 3y}{2}$: For $y_1 = 2$: $$x_1 = \frac{-8 - 3 \times 2}{2} = \frac{-8 - 6}{2} = \frac{-14}{2} = -7$$ For $y_2 = -\frac{1}{3}$: $$x_2 = \frac{-8 - 3 \times (-\frac{1}{3})}{2} = \frac{-8 + 1}{2} = \frac{-7}{2} = -\frac{7}{2}$$ 7. **Calculate the product $x_1 x_2$:** $$x_1 x_2 = (-7) \times \left(-\frac{7}{2}\right) = \frac{49}{2}$$ **Final answer:** $$\boxed{\frac{49}{2}}$$