Adding another reciprocal

I don't know is there an easier method to find $z + \frac{1}{y}$ , but this below is my method $\begin{aligned} \large x + \frac{1}{z} & \large = 5 \\ \large xz+ 1 & \large = 5z \\ \large 1 & \large = z{\left(5-x \right)} \\ \large xyz & \large = z{\left(5-x \right)} \\ \large \therefore y& \large = \frac{5}{x} - 1 \\ \end{aligned}$

Putting it into the third equation gives us $\begin{aligned} \large \frac{5}{x} - 1 + \frac{1}{x} = 29 \\ \large \frac{6}{x} = 30 \\ \large \therefore x = \frac{1}{5} \\ \end{aligned}$

Substitute $x$ into all equations gives us $y = 24, z=\frac{5}{24}$

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

So $m + n = 1 + 4 = 5$

By rearranging the equations we get:

$x+xy=x(y+1)=5$

$y+yz=y(z+1)=29$

$z+zx=z(x+1)=\frac{m}{n}$

Multiplying the equations:

$xyz(x+1)(y+1)(z+1)=(5)(29)(\frac{m}{n})$

Expanding:

$xyz(xyz+xy+xz+yz+x+y+z+1)=(5)(29)(\frac{m}{n})=145(\frac{m}{n})$

Substitute $xyz=1$ as well as $xy=\frac{1}{z}, xz=\frac{1}{y}, yz=\frac{1}{x}$ :

$1(1+(x+\frac { 1 }{ z } )+(y+\frac { 1 }{ x } )+(z+\frac { 1 }{ y } )+1)=145(\frac { m }{ n } )$

$1+5+29+\frac { m }{ n } +1=145(\frac { m }{ n } )$

Solving for $\frac{m}{n}$ yields $\frac{m}{n}=\frac{1}{4}$ . So $1+4=5$

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

$\begin{aligned} (1)\times yz: \quad xyz + \frac {yz}z & = 5yz \\ 1 + y & = 5yz & \small \color{#3D99F6} \text{Divide both sides by }y. \\ \frac 1y + 1 & = 5z \\ \frac 1y & = 5z - 1 & \small \color{#3D99F6} \text{Both sides add }z. \\ z + \frac 1y & = 6z - 1 & ...(3) \end{aligned}$

$\begin{aligned} (2)\times \frac 1{yz}: \quad \frac y{yz} + \frac 1{xyz} & = \frac {29}{yz} \\ \frac 1z + 1 & = \frac {29}{yz} \\ \frac {1+z}z & = \frac {29}{yz} \\ 1 + z & = \frac {29}y & \small \color{#3D99F6} \text{Both sides add }29z. \\ 1 + 30z & = 29 \left(z + \frac 1y\right) & ...(4) \end{aligned}$

We note that $29(3) = (4)$ :

$\begin{aligned} \implies 29(6z-1) & = 1 + 30z \\ 174z - 29 & = 1 + 30 z \\ 144 z & = 30 \\ \implies z & = \frac 5{24} \\ (3): \quad z + \frac 1y & = 6z - 1 \\ & = 6 \times \frac 5{24} - 1 \\ & = \frac 14 \end{aligned}$

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

First of all, in the first equation,

$\begin{aligned} & x + \dfrac 1z = 5 \\ \implies & \dfrac{zx+1}{z} = 5 \\ \implies & \dfrac{\frac 1y + 1}{\frac{1}{xy}} = 5 \\ \implies & (1+y)x = 5 \\ \implies & x = \dfrac{5}{1+y} \end{aligned}$

Substituting $x$ with this expression in the next equation, we obtain

$\begin{aligned} & y + \dfrac 1x = 29 \\ \implies & y + \dfrac{1+y}{5} = 29 \\ \implies & y = 24 \end{aligned}$

So,

$x = \dfrac{5}{1+y} = \dfrac{5}{25} = \dfrac 15$

And,

$\begin{aligned} & x + \dfrac 1z = 5 \\ \implies & \dfrac 15 + \dfrac 1z = 5 \\ \implies & \dfrac 1z = \dfrac{24}{5} \\ \implies & z = \dfrac{5}{24} \end{aligned}$

Finally,

$z + \dfrac 1x = \dfrac{5}{24} + \dfrac{1}{24} = \dfrac{6}{24} = \dfrac 14 = \dfrac mn$

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

Multiplying the equations, we get

$\left(x+\frac{1}{z}\right)\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)=$

$xyz+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xyz}=$

$1+x+y+z+\frac{xy+yz+xz}{xyz}+1=$

$x+y+z+xy+yz+xz+2=$

$25\cdot 5\cdot \frac{m}{n}$

Adding the equations, we get:

$\left(x+\frac{1}{z}\right)+\left(y+\frac{1}{x}\right)+\left(z+\frac{1}{y}\right)=$

$x+y+z+\frac{xy+yz+xz}{xyz}=$

$x+y+z+xy+yz+xz=$

$29+5+\frac{m}{n}$

Letting $x+y+z+xy+xz+yz=a$ , we get:

$a+2=145\frac{m}{n}$

$a=34+\frac{m}{n}$

Subtracting the equations,

$2=144\frac{m}{n}-34$

$144\frac{m}{n}=36$

$\frac{m}{n}=\frac{1}{4}$

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

Note the following identity.

$(x+\frac {1}{z})(y+\frac {1}{x})(z+\frac {1}{y})-(x+\frac {1}{z})-(y+\frac {1}{x})-(z+\frac {1}{y})=xyz+\frac {1}{xyz}$

Thus if $z+\frac {1}{y}=A$ then $(5)(29)A-5-29-A=2$ . Solving, $A=\frac {1}{4}$ and the answer is $1+4=5$ .

By multiplying the first two "reciprocal-summation" identities: $(x+\frac{1}{z}=5)\times (y+\frac{1}{x}=29)$

Gives $y(x+\frac{1}{z})+1+\frac{1}{xz}=145$

But, we know that $xyz=1$ , i.e., $\frac{1}{xz}=y$ , and $x+\frac{1}{z}=5$ .

Therefore, by substituting in the above expression, we conclude that $6y=144$ or $y=24$ and, by the previous identities: $x=\frac{1}{5}$ ; and $z=\frac{5}{24}$ .

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

So, $m+n=1+4=5$ .

First we need to substitute the value of x in terms of y and z and then substitute the value of x and the find the answer