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

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

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

• for more general: given a quadratic form in three variables with integer coefficients, if it does factor into the product of two linear forms as in your answer, the discriminant (determinant of the Hessian matrix of second partials) is definitely zero. I suspect the reverse is true (discriminant zero implies factors), maybe it is in one of my books. May 24 '15 at 19:56
• 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