Solve the system of equations in the set of real numbers. 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?

We know that via your equations, $$\frac{15}2(x+y+z)=xy+yz+xz$$Hence, $$xy=\frac12(x+y+z)$$$$yz=\frac92(x+y+z)$$$$xz=\frac52(x+y+z)$$So, assuming $x+y+z\neq0$, $z=9x$, $z=5y$. Try using this to move forward!
