If xyz = 1, x + \(\frac{1}{y}\) = 5, y + \(\frac{1}{x}\) = 29 find (z + \(\frac{1}{y}\))

\(\frac{1}{4}\)

Note the result:

if x + \(\frac{1}{y}\) = m

y + \(\frac{1}{z}\) = n

z + \(\frac{1}{x}\) = p

then (m×n×p) - (m+n+p) = xyz + \(\frac{1}{xyz}\)

So,

ATQ,

m = 5 and n = 29, xyz = 1, put these values in result:

(5 × 29 × p) - (5 + 29 + p) = xyz + \(\frac{1}{xyz}\)

145 p - (34 + p) = xyz + \(\frac{1}{xyz}\)

145 p - 34 - p = 1+1

145 p - p = 36

p (145 - 1) = 36

p = \(\frac{36}{144}\)

p = \(\frac{1}{4}\)