1. **State the problem:** Find $x$, $y$, and $z$ such that the vector $(x y, x + y, z - 1) = (4, 2, 3)$. 2. **Use the definition of equality of vectors:** Corresponding components must be equal, so: $$x y = 4$$ $$x + y = 2$$ $$z - 1 = 3$$ 3. **Solve the system:** From the third equation, $$z - 1 = 3 \implies z = 4$$ 4. From the first two equations: $$x y = 4$$ $$x + y = 2$$ 5. Express $y$ from the second equation: $$y = 2 - x$$ 6. Substitute into the first equation: $$x (2 - x) = 4$$ $$2x - x^2 = 4$$ 7. Rearrange to standard quadratic form: $$-x^2 + 2x - 4 = 0$$ Multiply both sides by $-1$: $$x^2 - 2x + 4 = 0$$ 8. Calculate the discriminant: $$\Delta = (-2)^2 - 4 \times 1 \times 4 = 4 - 16 = -12$$ 9. Since the discriminant is negative, there are no real solutions for $x$ and $y$. **Conclusion:** The system has no real solutions for $x$ and $y$, but $z = 4$. **Note:** The original solution $x=3$, $y=1$, $z=4$ does not satisfy the first equation $x y = 4$ because $3 \times 1 = 3 \neq 4$. Hence, no real $(x,y,z)$ satisfy all three equations simultaneously.