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 as ordered pairs:** $$\left(\frac{-1 + i\sqrt{3}}{2}, \frac{7 + i\sqrt{3}}{2}\right) \quad \text{and} \quad \left(\frac{-1 - i\sqrt{3}}{2}, \frac{7 - i\sqrt{3}}{2}\right)$$