MHB What are the solutions for x, y, and z in this equation?

[Albert1]

$\dfrac {1}{x}+\dfrac{1}{y+z}=\dfrac {1}{3}$

$\dfrac {1}{y}+\dfrac{1}{z+x}=\dfrac {1}{4}$

$\dfrac {1}{z}+\dfrac{1}{x+y}=\dfrac {1}{5}$

please find $x,y,z$

 

[RLBrown]

by inspection

Spoiler

x=11/3, y=11/2 and z=11.

 

[Albert1]

by inspection

Spoiler

x=11/3, y=11/2 and z=11.

your answer is correct , how to inspect ?

 

[Albert1]

$\dfrac {1}{x}+\dfrac{1}{y+z}=\dfrac {1}{3}$

$\dfrac {1}{y}+\dfrac{1}{z+x}=\dfrac {1}{4}$

$\dfrac {1}{z}+\dfrac{1}{x+y}=\dfrac {1}{5}$

please find $x,y,z$

hint :find x:y:z

 

[Farmtalk]

Albert, I've been working on this problem on and off between work breaks for two days now. Could you post one more hint? I just have something not quite right but I'm not afraid to admit that! :p

 

[Albert1]

hint :find x:y:z

more hint:

Spoiler

$(xy+xz) : (yz+xy) : (xz+yz)=3:4:5$
$xy:xz:yz=?:?:?$

 

[Albert1]

more hint:

Spoiler

$(xy+xz) : (yz+xy) : (xz+yz)=3:4:5$
$xy:xz:yz=?:?:?$

solution of others

Spoiler

from $(xy+xz) : (yz+xy) : (xz+yz)=3:4:5$
we get $xy:xz:yz=1:2:3---(1)$
$(1)\times \dfrac {1}{xyz}$
we have $\dfrac {1}{x}:\dfrac {1}{y}:\dfrac {1}{z}=3:2:1$
$\therefore x:y:z=2:3:6---(2)$
let :$x=2k,y=3k,z=6k$
and from :$\dfrac {1}{x}+\dfrac {1}{y+z}=\dfrac {1}{3},\,\,\leftrightarrow k=\dfrac {11}{6}$
and $x=\dfrac {11}{3},y=\dfrac {11}{2},z=11$