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

Beginner2022-06-27Added 23 answers

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,

$m+n+1=1+4+1=6$

$\frac{x(29x-1)}{x(5-x)}=1$

$29x-1=5-x$

$x=\frac{1}{5}$

$y=\frac{29/5-1}{1/5}=24$

$z=\frac{1}{5-1/5}=\frac{5}{24}$

$z+\frac{1}{y}=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}$

Thus,

$m+n+1=1+4+1=6$

Garrett Black

Beginner2022-06-28Added 5 answers

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 $145Q=xyz(x+1)(y+1)(z+1)$ and hence $145Q=(x+1)(y+1)(z+1)$

Expanding, we have

$\begin{array}{rl}(x+1)(y+1)(z+1)& =xyz+xy+yz+xz+x+y+z+1\\ & =2+xy+yz+xz+x+y+z\\ & =2+5+29+Q\\ & =36+Q\end{array}$

and so $145Q=36+Q$, from which it follows that $Q=\frac{1}{4}$

$\begin{array}{rl}x+xy& =5\\ y+yz& =29\\ z+xz& =Q\end{array}$

where Q is the quantity we're trying to determine.

Multiplying these three equations together gives $145Q=xyz(x+1)(y+1)(z+1)$ and hence $145Q=(x+1)(y+1)(z+1)$

Expanding, we have

$\begin{array}{rl}(x+1)(y+1)(z+1)& =xyz+xy+yz+xz+x+y+z+1\\ & =2+xy+yz+xz+x+y+z\\ & =2+5+29+Q\\ & =36+Q\end{array}$

and so $145Q=36+Q$, from which it follows that $Q=\frac{1}{4}$

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