Solving system of equations (3 unknowns, 3 equations) So, I've been trying to solve this question but to no avail. The system of equations are as follows:
1) $x+ \frac{1}{y}=4$
2) $y+ \frac{1}{z}=1$
3) $z + \frac{1}{x}=\frac{7}{3}$
Attempt:
Using equation 1, we can rewrite it as $\frac{1}{y}=4-x \equiv y=\frac{1}{4-x}$
. Using $y=\frac{1}{4-x}$, we can substitute into equation 2 to get $\frac{1}{4-x}+\frac{1}{z}=1$ (eq. 2')
Together with equation 3, we can eliminate z by inversing equation 3 to be:
$x + \frac{1}{z}=\frac{3}{7}$
Now, we subtract both (eq. 2') and the inversed equation to get 
$\frac{1}{4-x} - x = \frac{4}{7}$
However, this is going no where. The final answer should be x=$\frac{3}{2}$, y = $\frac{2}{5}$ and z = $\frac{5}{3}$.
Any tips or help would be greatly appreciated!

\begin{eqnarray*}
x+ \frac{1}{1- \cfrac{1}{z}} =4
\end{eqnarray*}
\begin{eqnarray*}
x+ \frac{1}{1- \cfrac{1}{\frac{7}3-\frac{1}{x}}} =4
\end{eqnarray*}
\begin{eqnarray*}
x+ \frac{1}{1- \cfrac{3x}{7x-3}} =4
\end{eqnarray*}
\begin{eqnarray*}
x+ \frac{7x-3}{4x-3} =4
\end{eqnarray*}
& after a little algebra $(2x-3)^2=0$.
