1. **State the problem:** Solve the system of equations for $x$ and $y$: $$\begin{cases} x^2 + xy - y^2 + 3x = -4 \\ 2x^2 + 2xy - 2y^2 + 5x - y = -6 \end{cases}$$ 2. **Rewrite the system:** $$\begin{cases} x^2 + xy - y^2 + 3x + 4 = 0 \\ 2x^2 + 2xy - 2y^2 + 5x - y + 6 = 0 \end{cases}$$ 3. **Divide the second equation by 2** to simplify coefficients: $$x^2 + xy - y^2 + \frac{5}{2}x - \frac{y}{2} + 3 = 0$$ 4. **Subtract the first equation from this simplified second equation**: $$\left(x^2 + xy - y^2 + \frac{5}{2}x - \frac{y}{2} + 3\right) - \left(x^2 + xy - y^2 + 3x + 4\right) = 0$$ Simplify: $$\frac{5}{2}x - \frac{y}{2} + 3 - 3x - 4 = 0$$ $$\left(\frac{5}{2}x - 3x\right) - \frac{y}{2} + (3 - 4) = 0$$ $$-\frac{1}{2}x - \frac{y}{2} - 1 = 0$$ Multiply both sides by $-2$: $$x + y + 2 = 0$$ 5. **Express $y$ in terms of $x$:** $$y = -x - 2$$ 6. **Substitute $y$ into the first original equation:** $$x^2 + x(-x - 2) - (-x - 2)^2 + 3x = -4$$ Simplify step-by-step: - $x^2 + x(-x - 2) = x^2 - x^2 - 2x = -2x$ - $(-x - 2)^2 = (x + 2)^2 = x^2 + 4x + 4$ So the equation becomes: $$-2x - (x^2 + 4x + 4) + 3x = -4$$ Simplify: $$-2x - x^2 - 4x - 4 + 3x = -4$$ $$-x^2 - 3x - 4 = -4$$ 7. **Add 4 to both sides:** $$-x^2 - 3x = 0$$ 8. **Multiply both sides by -1:** $$x^2 + 3x = 0$$ 9. **Factorize:** $$x(x + 3) = 0$$ 10. **Solve for $x$:** $$x = 0 \quad \text{or} \quad x = -3$$ 11. **Find corresponding $y$ values using $y = -x - 2$:** - If $x=0$, then $y = -0 - 2 = -2$ - If $x=-3$, then $y = -(-3) - 2 = 3 - 2 = 1$ 12. **Final solutions:** $$\boxed{(0, -2) \quad \text{and} \quad (-3, 1)}$$