1. **State the problem:** Solve the simultaneous equations for real numbers $x, y$: $$x^2 + 2xy - 8 = 0$$ $$\frac{2}{x} - \frac{2y}{x} = 1, \quad x \neq 0$$ 2. **Rewrite the second equation:** Multiply both sides by $x$ to clear the denominator: $$2 - 2y = x$$ 3. **Isolate $y$:** $$2 - x = 2y \implies y = \frac{2 - x}{2}$$ 4. **Substitute $y$ into the first equation:** $$x^2 + 2x \left( \frac{2 - x}{2} \right) - 8 = 0$$ 5. **Simplify:** $$x^2 + x(2 - x) - 8 = 0$$ $$x^2 + 2x - x^2 - 8 = 0$$ $$2x - 8 = 0$$ 6. **Solve for $x$:** $$2x = 8$$ $$x = 4$$ 7. **Find $y$ using $x=4$:** $$y = \frac{2 - 4}{2} = \frac{-2}{2} = -1$$ **Final solution:** $$\boxed{(x, y) = (4, -1)}$$