1. We are given the system of equations: $$x + 2y = 6$$ $$y = -x^2 - x + 6$$ We need to find the solution set $(x_1, y_1), (x_2, y_2)$ where $y_1 < y_2$, and then calculate $4(y_1 - x_1)$. 2. First, express $y$ from the linear equation: $$x + 2y = 6 \implies 2y = 6 - x \implies y = \frac{6 - x}{2}$$ 3. Substitute this expression for $y$ into the quadratic equation: $$\frac{6 - x}{2} = -x^2 - x + 6$$ 4. Multiply both sides by 2 to clear the denominator: $$6 - x = 2(-x^2 - x + 6) = -2x^2 - 2x + 12$$ 5. Bring all terms to one side: $$6 - x + 2x^2 + 2x - 12 = 0$$ Simplify: $$2x^2 + ( - x + 2x ) + (6 - 12) = 0$$ $$2x^2 + x - 6 = 0$$ 6. Divide the entire equation by $\cancel{2}$ to simplify: $$\cancel{2}x^2 + \frac{1}{\cancel{2}}x - \frac{6}{\cancel{2}} = 0 \implies x^2 + \frac{1}{2}x - 3 = 0$$ 7. Solve the quadratic equation $x^2 + \frac{1}{2}x - 3 = 0$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=\frac{1}{2}$, and $c=-3$. Calculate the discriminant: $$\Delta = \left(\frac{1}{2}\right)^2 - 4(1)(-3) = \frac{1}{4} + 12 = \frac{49}{4}$$ 8. Find the roots: $$x = \frac{-\frac{1}{2} \pm \sqrt{\frac{49}{4}}}{2} = \frac{-\frac{1}{2} \pm \frac{7}{2}}{2}$$ Calculate each root: - For $+$: $$x_2 = \frac{-\frac{1}{2} + \frac{7}{2}}{2} = \frac{3}{2}$$ - For $-$: $$x_1 = \frac{-\frac{1}{2} - \frac{7}{2}}{2} = \frac{-4}{2} = -2$$ 9. Find corresponding $y$ values using $y = \frac{6 - x}{2}$: - For $x_1 = -2$: $$y_1 = \frac{6 - (-2)}{2} = \frac{8}{2} = 4$$ - For $x_2 = \frac{3}{2}$: $$y_2 = \frac{6 - \frac{3}{2}}{2} = \frac{\frac{12}{2} - \frac{3}{2}}{2} = \frac{\frac{9}{2}}{2} = \frac{9}{4} = 2.25$$ 10. Check the condition $y_1 < y_2$: We have $y_1 = 4$ and $y_2 = 2.25$, but $4 > 2.25$, so swap the points: $(x_1, y_1) = \left(\frac{3}{2}, \frac{9}{4}\right)$ and $(x_2, y_2) = (-2, 4)$. 11. Calculate $4(y_1 - x_1)$: $$4\left(\frac{9}{4} - \frac{3}{2}\right) = 4\left(\frac{9}{4} - \frac{6}{4}\right) = 4 \times \frac{3}{4} = 3$$ **Final answer:** 3