Solve the system of equations in the set of real numbers:
$$\begin{cases}
\frac1x + \frac1{y+z} = \frac13 \\
\frac1y + \frac1{x+z} = \frac15 \\
\frac1z + \frac1{x+y} = \frac17
\end{cases}$$
I got:
$$\begin{cases}
3(x+y+z)=x(y+z) \\
5(x+y+z)=y(x+z) \\
7(x+y+z)=z(x+y)
\end{cases}$$
However, no matter how I continue from here, I always get $x=y=z=0$, which cannot be true; or I get a new system of equations, but still with 3 variables (which I cannot solve).
How can I solve this problem or how should I approach it?
Best Answer
Let $x+y+z=m$
Adding all the equations, we get,
$xy+yz+zx=\frac{3m+5m+7m}{2}=\frac{15m}{2}$
Subtracting each equation one by one from this, we get,
$xy=\frac{m}{2}$
$yz=\frac{9m}{2}$
$zx=\frac{5m}{2}$
Dividing by $ xyz$, we get, $$\frac{1}{x}:\frac{1}{y}:\frac{1}{z}=9:5:1$$ $$\Longrightarrow x:y:z=\frac{1}{9}:\frac{1}{5}:\frac{1}{1}$$ $$\Longrightarrow x:y:z=5:9:45$$ Now, let $x=5k, y=9k, z=45k$ and get the result.
Hope it is helpful