1. **State the problem:** Find the values of $x$, $y$, and $z$ such that the vector $(x y, x + y, z - 1)$ equals $(4, 2, 3)$. 2. **Set up equations:** Since the vectors are equal, their corresponding components must be equal: $$x y = 4$$ $$x + y = 2$$ $$z - 1 = 3$$ 3. **Solve for $z$:** From the third equation, $$z - 1 = 3 \implies z = 4$$ 4. **Solve for $x$ and $y$:** From the second equation, $$x + y = 2$$ From the first equation, $$x y = 4$$ 5. **Use substitution or quadratic approach:** Let $y = 2 - x$ from the second equation. Substitute into the first: $$x(2 - x) = 4$$ $$2x - x^2 = 4$$ $$-x^2 + 2x - 4 = 0$$ Multiply both sides by $-1$: $$x^2 - 2x + 4 = 0$$ 6. **Solve quadratic equation:** The quadratic is $$x^2 - 2x + 4 = 0$$ Calculate the discriminant: $$\Delta = (-2)^2 - 4 \times 1 \times 4 = 4 - 16 = -12$$ Since $\Delta < 0$, there are no real solutions for $x$ and $y$. 7. **Interpretation:** The system has no real solutions for $x$ and $y$, but complex solutions exist. 8. **Find complex solutions:** Using quadratic formula, $$x = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3}$$ Then, $$y = 2 - x = 1 \mp i\sqrt{3}$$ **Final answer:** $$x = 1 \pm i\sqrt{3}, \quad y = 1 \mp i\sqrt{3}, \quad z = 4$$