1. [3] Find all ordered pairs of real numbers $(x, y)$ such that $x^{2} y=3$ and $x+x y=4$.
Answer: $(1,3),\left(3, \frac{1}{3}\right) \quad$ Multiplying the second equation by $x$ gives
$$
x^{2}+x^{2} y=4 x,
$$
and substituting our known value of $x^{2} y$ gives the quadratic
$$
x^{2}-4 x+3=0,
$$
so $x=1$ or $x=3$. Hence, we obtain the solutions $(x, y)=(1,3),(3,1 / 3)$.