A system of simultaneous equations

I'm currently stuck solving this set of equations. $$x(x+y+z)=4-yz$$ $$y(x+y+z)=9-zx$$ $$z(x+y+z)=25-xy$$

Here's what I've done so far:

By subtracting the second equation from the first, I got

$$(x-y)(x+y+z)=(4-yz)-(9-zx)=-5+z(x-y) \rightarrow (x-y)(x+y)=-5$$

Similarly, by subtracting the third equation from the second, I got

$$(y-z)(x+y+z)=(9-zx)-(25-xy)=-16+x(y-z) \rightarrow (y-z)(y+z)=-16$$

I've also tried adding the three equations together, and got

$$(x+y+z)^2=38-xy-yz-zx$$

Unfortunately, I don't think this equation helps.

WolframAlpha tells me that the answer is

$$(x,y,z)=(\mp\frac{89}{60},\pm\frac{161}{60},\pm\frac{289}{60})$$

but I'm more interested in how the answer is found. Thanks in advance!

Upon further inspection of the output that WolframAlpha gave me, I think I figured out how to solve this system.

Rearranging the first equation, we get

$$x(x+y+z)=4-yz \rightarrow x^2+xy+xz+yz=4 \rightarrow (x+y)(z+x)=4$$

Similarly, we get

$$(y+z)(x+y)=9$$

and

$$(z+x)(y+z)=25$$

From there, we get $$(x+y)(y+z)(z+x)=\pm30$$ and then from there get

$$x+y=\pm\frac{6}{5},y+z=\pm\frac{15}{2},z+x=\pm\frac{10}{3}$$

$$(x,y,z)=(\mp\frac{89}{60},\pm\frac{161}{60},\pm\frac{289}{60})$$
• upon reflection, the direction I proved for dimension $3$ will work for all larger dimension as well. It is misleading for dimension $2,$ though, a binary quadratic form factors if and only if the determinant of its Hessian matrix is minus a square; that is, the traditional "discriminant" as in the Quadratic Formula is a square. May 24 '15 at 20:06