1. **State the problem:** We are given the system of equations: $$-2 - x = \frac{y^2 - 4y + 10}{11}$$ $$-11x + 3y = 62$$ We need to find the product of the $x$-values of the two distinct solutions $(x_1, y_1)$ and $(x_2, y_2)$, i.e., find $x_1 \cdot x_2$. 2. **Rewrite the first equation:** Multiply both sides by 11 to clear the denominator: $$11(-2 - x) = y^2 - 4y + 10$$ Simplify the left side: $$-22 - 11x = y^2 - 4y + 10$$ Bring all terms to one side: $$y^2 - 4y + 10 + 11x + 22 = 0$$ Simplify constants: $$y^2 - 4y + 11x + 32 = 0$$ 3. **Express $x$ from the second equation:** $$-11x + 3y = 62 \implies -11x = 62 - 3y \implies x = \frac{3y - 62}{11}$$ 4. **Substitute $x$ into the first equation:** $$y^2 - 4y + 11 \left( \frac{3y - 62}{11} \right) + 32 = 0$$ Simplify: $$y^2 - 4y + 3y - 62 + 32 = 0$$ Combine like terms: $$y^2 - y - 30 = 0$$ 5. **Solve the quadratic equation for $y$:** Use the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=-1$, $c=-30$: $$y = \frac{1 \pm \sqrt{1 + 120}}{2} = \frac{1 \pm \sqrt{121}}{2} = \frac{1 \pm 11}{2}$$ So, $$y_1 = \frac{1 + 11}{2} = 6, \quad y_2 = \frac{1 - 11}{2} = -5$$ 6. **Find corresponding $x$ values:** For $y_1 = 6$: $$x_1 = \frac{3(6) - 62}{11} = \frac{18 - 62}{11} = \frac{-44}{11} = -4$$ For $y_2 = -5$: $$x_2 = \frac{3(-5) - 62}{11} = \frac{-15 - 62}{11} = \frac{-77}{11} = -7$$ 7. **Calculate the product $x_1 \cdot x_2$:** $$x_1 \cdot x_2 = (-4) \times (-7) = 28$$ **Final answer:** $$\boxed{28}$$