Suppose that $x, y, z$ are three positive numbers that satisfy the equation $xyz=1, x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$. 
Suppose that $x, y, z$ are three positive numbers that satisfy the equation $xyz=1, x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$. Then $z+\frac{1}{y}=\frac{m}{n}$, where $m$ and $n$ are coprime. Find $m+n+1$.

I tried this, $$y=29-\frac{1}{x}$$ $$y=\frac{29x-1}{x}$$
And, $$x+\frac{1}{z}=5$$ $$z=\frac{1}{5-x}$$
What can I do next?

Another solution: we have $\frac{1}{x}=yz$, $\frac{1}{y}=xz$, and $\frac{1}{z}=xy$. So
\begin{align*}
x+xy&=5\\
y+yz&=29\\
z+xz&=Q
\end{align*}
where Q is the quantity we're trying to determine.
Multiplying these three equations together gives $145Q=xyz(x+1)(y+1)(z+1)$ and hence $145Q=(x+1)(y+1)(z+1)$.
Expanding, we have
\begin{align*}
(x+1)(y+1)(z+1)&=xyz+xy+yz+xz+x+y+z+1\\
&=2+xy+yz+xz+x+y+z\\
&=2+5+29+Q\\
&=36+Q
\end{align*}
and so $145Q=36+Q$, from which it follows that $Q=\frac{1}{4}$.
