Solve a nonlinear system of equations in 3 variables

Question

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.

Answer 1

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

$$z=-2.$$

Answer 2

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 $x=yz-y-z$.

Answer 3

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 $$x=5,y=-1,z=-2$$