1. **State the problem:** Solve the system of equations using substitution: $$-x + y = 4$$ $$x^{2} + y = 3$$ 2. **Isolate one variable:** From the first equation, solve for $y$: $$-x + y = 4 \implies y = x + 4$$ 3. **Substitute into the second equation:** Replace $y$ in the second equation with $x + 4$: $$x^{2} + (x + 4) = 3$$ 4. **Simplify and solve for $x$:** $$x^{2} + x + 4 = 3$$ $$x^{2} + x + \cancel{4} = \cancel{3}$$ $$x^{2} + x + 1 = 0$$ 5. **Use the quadratic formula:** For $ax^{2} + bx + c = 0$, $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$. Here, $a=1$, $b=1$, $c=1$. $$x = \frac{-1 \pm \sqrt{1^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2}$$ 6. **Interpret the result:** Since the discriminant is negative ($-3$), there are no real solutions for $x$. 7. **Final answer:** The system has no real solutions; the solutions are complex: $$x = \frac{-1 \pm i\sqrt{3}}{2}$$ and corresponding $y$ values: $$y = x + 4 = \frac{-1 \pm i\sqrt{3}}{2} + 4 = \frac{7 \pm i\sqrt{3}}{2}$$