1. If $a$ and $b$ satisfy the equations $a+\frac{1}{b}=4$ and $\frac{1}{a}+b=\frac{16}{15}$, determine the product of all possible values of $a b$.
Answer: $\square$
We multiply $a+\frac{1}{b}=4$ and $\frac{1}{a}+b=\frac{16}{15}$ to get $2+a b+\frac{1}{a b}=\frac{64}{15} \Longleftrightarrow(a b)^{2}-\frac{34}{15}(a b)+1=0$. Since $\left(\frac{34}{14}\right)^{2}-4 \cdot 1 \cdot 1>0$, there are two roots $a b$, so the product of both possible values of $a b$ is 1 by Vieta's.