Suppose that $x, y, z$ are three positive numbers that satisfy the equation $xyz=1, x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$.
Matthew Harrington 
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?
$\endgroup$2 Answers
$\begingroup$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 $Q=\frac{1}{4}$.
$\endgroup$ 0 $\begingroup$Just substitute the values of $y$ and $z$ in $xyz=1$.$$\frac{x(29x-1)}{x(5-x)}=1$$$$29x-1=5-x$$$$x=\frac{1}{5}$$$$y=\frac{29/5-1}{1/5}=24$$$$z=\frac{1}{5-1/5}=\frac{5}{24}$$$$z+\frac{1}{y}=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}$$Thus, $$m+n+1=1+4+1=6$$
$\endgroup$ 6
