This question was uploaded on 23/12/21 on social media accounts.
Solution:
By doing partial fraction on all three equations:
\[\frac{x+y}{xyz} = \frac1{yz}+\frac1{xz} = \frac{-1}4 \rightarrow(1)\]
\[\frac{y+z}{xyz} = \frac1{xz}+\frac1{xy} = \frac{-1}{24} \rightarrow(2)\]
\[\frac{z+x}{xyz} = \frac1{xy}+\frac1{yz} = \frac1{24} \rightarrow(3)\]
Now add all three equations:
\[2\left(\frac1{xy}+\frac1{yz}+\frac1{zx}\right) = \frac{-1}4\]
\[\Rightarrow \frac1{xy}+\frac1{yz}+\frac1{zx} = \frac{-1}8 \rightarrow(4)\]
(4) - (1):
\[\frac1{xy} = \frac18\]
\[\Rightarrow xy = 8 \rightarrow(5)\]
(4) - (2):
\[\frac1{yz} = \frac{-1}{12}\]
\[\Rightarrow yz = -12 \rightarrow(6)\]
(4) - (3):
\[\frac1{zx} = \frac{-1}6\]
\[\Rightarrow zx = -6 \rightarrow(7)\]
Substituting values of y, z from equations (5), (7) in equation (6):
We will get:
\[\frac8y\cdot\frac{-6}z = -12\]
\[\Rightarrow x^2 = 4\]
\[\Rightarrow x=2ㅤorㅤx=-2\]
Substituting this value in equations (5) and (7):
We will get:
\[y=4ㅤorㅤy=-4\]
\[z=-3ㅤorㅤz=3\]
We get:
\[(x,y,z) = (2,4,-3)\]
or
\[(x,y,z) = (-2,-4,3)\]
