Solve a nonlinear system of equations in 3 variables I need to solve this system of equations
$$\frac 1x+\frac{1}{y+z}=-\frac2{15}$$
$$\frac 1y+\frac{1}{x+z}=-\frac2{3}$$
$$\frac 1z+\frac{1}{x+y}=-\frac1{4}$$
I've tried to express $x$ in terms of $y$, then $y$ in terms of $z$. But this leads to nothing good. I think I should use some matrix method, but I'm not sure what exactly I have to do. Need help here.
 A: It is a linear algebra problem ! So multiply the first by $x(y+z)$ to get 
$$x+y+z=-\frac{2}{15}x(x+z)$$ similar for all the rest. Let $a=x+y+z$.
Then you have the linear system
$$\begin{pmatrix}\frac{2}{15}&\frac{2}{15}&0\\
\frac{2}{3}&0&\frac{2}{3}\\
0&\frac{1}{4}&\frac{1}{4}
\end{pmatrix}\begin{pmatrix}xy\\xz\\yz\end{pmatrix}=\begin{pmatrix}-1\\-1\\-1\end{pmatrix}a$$
Solving this by row reduction we get 
$$2xy=-5a$$
$$xz=-5a$$
$$yz=a$$
Subsituting and canceling we have 
$$x=-5y$$
$$z=2y$$ 
from which we see by adding those equations that $$a:=x+y+z=-2y$$
And then we substitute again to have 
$$yz=-2y$$ and thus 
$$z=-2.$$
A: The solution is given by
$$
(x,y,z)=(5,-1,-2)
$$
This follows by multiplying with the common denominator, which gives three polynomial equations in $x,y,z$, which can be easily solved using resultants. The first polynomial equation, for example, is $x(2y + 2z + 15) + 15(y + z)=0$. One of the resultant equations is, for example, $yz - x - y - z=0$, so that we can substitute $x=yz-y-z$.
A: setting $$x+y=a,x+z=b,y+z=c$$ then we have $$\frac{a+b+c}{2}-c=\frac{a+b-c}{2}=x$$ etc. Can you finish?
the solution is given by $$x=5,y=-1,z=-2$$
