1. **State the problem:** Solve the system of simultaneous equations: $$x^2 = 4 - y$$ $$x = y + 2$$ 2. **Use substitution:** From the second equation, express $y$ in terms of $x$: $$y = x - 2$$ 3. **Substitute $y$ into the first equation:** $$x^2 = 4 - (x - 2)$$ Simplify the right side: $$x^2 = 4 - x + 2$$ $$x^2 = 6 - x$$ 4. **Rearrange to form a quadratic equation:** $$x^2 + x - 6 = 0$$ 5. **Factor the quadratic:** $$(x + 3)(x - 2) = 0$$ 6. **Solve for $x$:** $$x = -3 \quad \text{or} \quad x = 2$$ 7. **Find corresponding $y$ values using $y = x - 2$:** - For $x = -3$: $$y = -3 - 2 = -5$$ - For $x = 2$: $$y = 2 - 2 = 0$$ 8. **Final solutions:** $$(x, y) = (-3, -5) \quad \text{and} \quad (2, 0)$$