Quintin Stafford

2022-06-26

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 $m+n+1$

grcalia1

Just substitute the values of y and z in $xyz=1$
$\frac{x\left(29x-1\right)}{x\left(5-x\right)}=1$
$29x-1=5-x$
$x=\frac{1}{5}$
$y=\frac{29/5-1}{1/5}=24$
$z=\frac{1}{5-1/5}=\frac{5}{24}$
$z+\frac{1}{y}=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}$
Thus,
$m+n+1=1+4+1=6$

Garrett Black

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

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