Another AIME problem

Using the given system we have that

$\left(x + \dfrac{1}{z}\right)\left(y + \dfrac{1}{x}\right)\left(z + \dfrac{1}{y}\right) = 5 \times 29 \times \left(z + \dfrac{1}{y}\right) = 145\left(z + \dfrac{1}{y}\right)$ .

Expanding and rearranging, we then find that

$xyz + x + y + z + \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} + \dfrac{1}{xyz} = 145\left(z + \dfrac{1}{y}\right) \Longrightarrow$

$xyz + \dfrac{1}{xyz} + \left(x + \dfrac{1}{z}\right) + \left(y + \dfrac{1}{x}\right) + \left(z + \dfrac{1}{y}\right) = 145\left(z + \dfrac{1}{y}\right) \Longrightarrow$

$1 + 1 + 5 + 29 = (145 - 1)\left(z + \dfrac{1}{y}\right) \Longrightarrow 36 = 144\left(z + \dfrac{1}{y}\right) \Longrightarrow$

$z + \dfrac{1}{y} = \dfrac{36}{144} = \dfrac{1}{4} \Longrightarrow a + b = 1 + 4 = \boxed{5}$ .

Comment: Not that it is required, but it can be determined that $(x,y,z) = \left(\dfrac{1}{5}, 24, \dfrac{5}{24}\right)$ .

$\begin{cases} xyz = 1 &...(1) \\ x + \dfrac 1x = 5 &...(2) \\ y + \dfrac 1x = 29 &...(3) \end{cases}$

$\begin{aligned} (2): \quad x + \frac 1z & = 5 & \small \color{#3D99F6} \text{Multiply both sides by }yz \\ {\color{#3D99F6}xyz} + \frac {yz}z & = 5yz & \small \color{#3D99F6} \text{Note that }(1): \ xyz = 1 \\ \implies {\color{#3D99F6}1} + y & = 5yz \ \ \ ...(4) \end{aligned}$

$\begin{aligned} (2): \quad y + \frac {\color{#3D99F6}1}x & = 29 \\ y + \frac {\color{#3D99F6}xyz}x & = 29 \\ y + yz & = 29 \ \ \ ...(5) & \small \color{#3D99F6} \text{Multiply both sides by }5 \\ 5y + {\color{#3D99F6}5yz} & = 145 & \small \color{#3D99F6} \text{Note that }(4): \ 5yz = 1+y \\ 5y + {\color{#3D99F6}1+y} & = 145 \\ \implies y & = 24 \ \ \ ...(6) \end{aligned}$

$\begin{aligned} (5): \quad y + yz & = 29 & \small \color{#3D99F6} \text{Divide both sides by }y \\ 1 + z & = \frac {29}y \\ \implies z + \frac 1y & = \frac {30}{\color{#3D99F6}y} - 1 & \small \color{#3D99F6} \text{Note that }(6): \ y = 24 \\ & = \frac {30}{\color{#3D99F6}24} - 1 \\ & = \frac 14 \end{aligned}$

$\implies m + n = 1 + 4 = \boxed{5}$

from the second eqs. z = 1/(5 – x),
from the third eqs. 1/y =x/(29x – 1), or y = (29x – 1)/x from the first eqs. xyz =(29x – 1)/(5 – x) = 1, x= 1/5, y=24,z=5/24 So z + 1/y = m/n = ¼ , m+n = 5