Problem - 3617
Let real numbers $x, y,$ and $z$ satisfy $$x+\frac{1}{y}=4\quad\text{,}\quad y+\frac{1}{z}=1\quad\text{,}\quad z +\frac{1}{x}=\frac{7}{3}$$ Find the value of $xyz$.
Let's construct an equation of $x$ by canceling $y$ and $z$.
$$4=x+\frac{1}{y}=x+\frac{1}{1-\frac{1}{z}}=x+\frac{1}{1-\frac{1}{\frac{7}{x}-\frac{1}{x}}}$$
This equation can be simplified to $$4x^2 - 12x+9=0\Leftrightarrow (2x-3)^2=0\implies x=\frac{3}{2}$$
It follows that $z=\frac{7}{x}-\frac{1}{x}=\frac{5}{3}$ and $y=1-\frac{1}{z}=\frac{2}{5}$.
$$\therefore\quad xyz=\frac{3}{2}\times\frac{2}{5}\times\frac{5}{x}=\boxed{1}$$