1. Given that $x$ and $y$ are nonzero real numbers such that $x+\frac{1}{y}=10$ and $y+\frac{1}{x}=\frac{5}{12}$, find all possible values of $x$.
Answer: 4,6 OR 6,4 Let $z=\frac{1}{y}$. Then $x+z=10$ and $\frac{1}{x}+\frac{1}{z}=\frac{5}{12}$. Since $\frac{1}{x}+\frac{1}{z}=\frac{x+z}{x z}=\frac{10}{x z}$, we have $x z=24$. Thus, $x(10-x)=24$, so $x^{2}-10 x+24=(x-6)(x-4)=0$, whence $x=6$ or $x=4$.

Alternate solution: Clearing denominators gives $x y+1=10 y$ and $y x+1=\frac{5}{12} x$, so $x=24 y$. Thus we want to find all real (nonzero) $x$ such that $\frac{x^{2}}{24}+1=\frac{5}{12} x$ (and for any such $x, y=x / 24$ will satisfy the original system of equations). This factors as $(x-4)(x-6)=0$, so precisely $x=4,6$ work.