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$x+\frac{1}{z}=4 \quad z+\frac{1}{y}=\frac{7}{3}$ $y+\frac{1}{x}=1 \quad xyz=?$

Let the given equations be:

1. $x+\frac{1}{z}=4$
2. $z+\frac{1}{y}=\frac{7}{3}$
3. $y+\frac{1}{x}=1$

We want to find the value of $xyz$.

There are two main approaches to solve this system.

Method 1: Substitution

1. From equation (1), express $x$ in terms of $z$: $x = 4 - \frac{1}{z} = \frac{4z-1}{z}$

2. From equation (3), express $y$ in terms of $x$: $y = 1 - \frac{1}{x} = \frac{x-1}{x}$

3. Substitute the expression for $x$ from step 1 into the expression for $y$ from step 2: $y = \frac{\frac{4z-1}{z} - 1}{\frac{4z-1}{z}} = \frac{\frac{4z-1-z}{z}}{\frac{4z-1}{z}} = \frac{3z-1}{4z-1}$

4. Now we have $y$ in terms of $z$. Substitute this into equation (2): $z + \frac{1}{y} = \frac{7}{3}$ $z + \frac{4z-1}{3z-1} = \frac{7}{3}$

5. Solve this equation for $z$. Multiply both sides by $3(3z-1)$ to eliminate denominators: $3z(3z-1) + 3(4z-1) = 7(3z-1)$ $9z^2 - 3z + 12z - 3 = 21z - 7$ $9z^2 + 9z - 3 = 21z - 7$ $9z^2 - 12z + 4 = 0$

6. This is a perfect square trinomial: $(3z-2)^2 = 0$. So, $3z-2 = 0 \Rightarrow z = \frac{2}{3}$.

7. Now substitute the value of $z$ back to find $x$ and $y$: $x = \frac{4(\frac{2}{3})-1}{\frac{2}{3}} = \frac{\frac{8}{3}-1}{\frac{2}{3}} = \frac{\frac{5}{3}}{\frac{2}{3}} = \frac{5}{2}$ $y = \frac{3(\frac{2}{3})-1}{4(\frac{2}{3})-1} = \frac{2-1}{\frac{8}{3}-1} = \frac{1}{\frac{5}{3}} = \frac{3}{5}$

8. Finally, calculate $xyz$: $xyz = \left(\frac{5}{2