1. **State the problem:** Solve the simultaneous equations: $$a^2 - ab + b^2 = 3 \quad (D)$$ $$a + 2b + 1 = 0 \quad (E)$$ 2. **Express one variable in terms of the other:** From equation (E), solve for $a$: $$a + 2b + 1 = 0 \implies a = -2b - 1$$ 3. **Substitute into equation (D):** Replace $a$ in equation (D) with $-2b - 1$: $$(-2b - 1)^2 - (-2b - 1) b + b^2 = 3$$ 4. **Expand and simplify:** $$(-2b - 1)^2 = 4b^2 + 4b + 1$$ $$- (-2b - 1) b = 2b^2 + b$$ So equation (D) becomes: $$4b^2 + 4b + 1 + 2b^2 + b + b^2 = 3$$ Combine like terms: $$4b^2 + 2b^2 + b^2 + 4b + b + 1 = 3$$ $$7b^2 + 5b + 1 = 3$$ 5. **Bring all terms to one side:** $$7b^2 + 5b + 1 - 3 = 0$$ $$7b^2 + 5b - 2 = 0$$ 6. **Solve quadratic equation:** Use the quadratic formula: $$b = \frac{-5 \pm \sqrt{5^2 - 4 \times 7 \times (-2)}}{2 \times 7} = \frac{-5 \pm \sqrt{25 + 56}}{14} = \frac{-5 \pm \sqrt{81}}{14}$$ $$b = \frac{-5 \pm 9}{14}$$ Two solutions for $b$: $$b_1 = \frac{-5 + 9}{14} = \frac{4}{14} = \frac{2}{7}$$ $$b_2 = \frac{-5 - 9}{14} = \frac{-14}{14} = -1$$ 7. **Find corresponding $a$ values:** Using $a = -2b - 1$: For $b = \frac{2}{7}$: $$a = -2 \times \frac{2}{7} - 1 = -\frac{4}{7} - 1 = -\frac{4}{7} - \frac{7}{7} = -\frac{11}{7}$$ For $b = -1$: $$a = -2 \times (-1) - 1 = 2 - 1 = 1$$ 8. **Final solutions:** $$\boxed{\left(a, b\right) = \left(-\frac{11}{7}, \frac{2}{7}\right) \text{ or } (1, -1)}$$