1. **State the problem:** Solve the simultaneous equations: $$2x^2 + 3xy + 4y^2 = 6$$ $$3x + 4y = -2$$ 2. **Use substitution or elimination:** From the linear equation, express $x$ in terms of $y$: $$3x + 4y = -2 \implies 3x = -2 - 4y \implies x = \frac{-2 - 4y}{3}$$ 3. **Substitute $x$ into the quadratic equation:** $$2\left(\frac{-2 - 4y}{3}\right)^2 + 3\left(\frac{-2 - 4y}{3}\right)y + 4y^2 = 6$$ 4. **Simplify each term:** $$2 \cdot \frac{(-2 - 4y)^2}{9} + 3 \cdot \frac{-2 - 4y}{3} y + 4y^2 = 6$$ 5. **Multiply through by 9 to clear denominators:** $$9 \times \left(2 \cdot \frac{(-2 - 4y)^2}{9} + 3 \cdot \frac{-2 - 4y}{3} y + 4y^2\right) = 9 \times 6$$ $$2(-2 - 4y)^2 + 9(-2 - 4y) y + 36 y^2 = 54$$ 6. **Expand $(-2 - 4y)^2$:** $$(-2 - 4y)^2 = (-2)^2 + 2 \times (-2) \times (-4y) + (-4y)^2 = 4 + 16y + 16y^2$$ 7. **Substitute back:** $$2(4 + 16y + 16y^2) + 9(-2 - 4y) y + 36 y^2 = 54$$ 8. **Expand and simplify:** $$8 + 32 y + 32 y^2 - 18 y - 36 y^2 + 36 y^2 = 54$$ Combine like terms: $$8 + (32 y - 18 y) + (32 y^2 - 36 y^2 + 36 y^2) = 54$$ $$8 + 14 y + 32 y^2 = 54$$ 9. **Bring all terms to one side:** $$32 y^2 + 14 y + 8 - 54 = 0 \implies 32 y^2 + 14 y - 46 = 0$$ 10. **Simplify by dividing by 2:** $$\cancel{2} \times 16 y^2 + \cancel{2} \times 7 y - \cancel{2} \times 23 = 0 \implies 16 y^2 + 7 y - 23 = 0$$ 11. **Use quadratic formula:** $$y = \frac{-7 \pm \sqrt{7^2 - 4 \times 16 \times (-23)}}{2 \times 16} = \frac{-7 \pm \sqrt{49 + 1472}}{32} = \frac{-7 \pm \sqrt{1521}}{32}$$ $$\sqrt{1521} = 39$$ So, $$y = \frac{-7 \pm 39}{32}$$ 12. **Calculate the two values of $y$:** $$y_1 = \frac{-7 + 39}{32} = \frac{32}{32} = 1$$ $$y_2 = \frac{-7 - 39}{32} = \frac{-46}{32} = -\frac{23}{16}$$ 13. **Find corresponding $x$ values:** For $y = 1$: $$x = \frac{-2 - 4(1)}{3} = \frac{-2 - 4}{3} = \frac{-6}{3} = -2$$ For $y = -\frac{23}{16}$: $$x = \frac{-2 - 4 \times \left(-\frac{23}{16}\right)}{3} = \frac{-2 + \frac{92}{16}}{3} = \frac{-2 + 5.75}{3} = \frac{3.75}{3} = 1.25 = \frac{5}{4}$$ 14. **Final solutions:** $$\boxed{(x, y) = (-2, 1) \text{ or } \left(\frac{5}{4}, -\frac{23}{16}\right)}$$