Weird equation

$\frac{x + y + z}{x(y + z)}$ = 4 .............1

$\frac{x + y + z}{y(x + z)}$ = $\frac{9}{2}$ ...................2

$\frac{x + y + z}{z( x + y )}$ = $\frac{36}{5}$ .................3

so, 4xy + 4xz = $\frac{9}{2}$ (xy + yz)

z = $\frac{xy}{8x - 9y}$

solving 1 and 3 we get

4xz = y(5x - 9z)

using the value of z we get y = $\frac{2}{3}$ x

again if we put the value of z in equation 3 we get

40x^2 - 45y^2 = 36xy( x + y )

as y = $\frac{2}{3}$ x

we get x = $\frac{1}{2}$

so y = 2/3

an z = 1/6

so, 1/x + 1/y + 1/z = 2 + 3 + 6

Similar to Omar's, but simplier.

We will use the substitution $\frac{1}{x} = a, \frac{1}{y } = b, \frac{ 1}{z} = c$ . (where $abc \neq 0$ .)
Notice that $\frac{1}{y+z} = \frac{ \frac{1}{yz} } { \frac{1}{z} + \frac{1}{y} } = \frac{ bc } { b+c }$ .
Thus, the first equation is $a + \frac{ bc} { b+ c } = 4$ , or that $ab + bc + ca = 4 ( b + c )$ .
Similarly, the second equation is $ab + bc + ca = \frac{9}{2} ( c + a)$ ,
and the third equation is $ab + bc + ca = \frac{ 36}{5} ( a+b)$ .

Solving $4 (b+c) = \frac{ 9}{2} ( c + a ) = \frac{ 36}{5} ( a + b )$ , we obtain that $3a = 2b = c$ .
Substituting this into $ab + bc + ca = 4 (b+c)$ , we get that $a = 0 , 2$ . Since $a \neq 0$ , thus $a = 2$ . Then, $b = 3, c = 6$ .

Hence $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = a + b + c = 2 + 3 + 6 = 11$ .