A system of equations

Question

Given three equations $x^2+y^2+xy=a$, $y^2+z^2+yz=b$ and $x^2+z^2+xz=c$, how can I solve for $x,y$ and $z$ in terms of $a,b$ and $c$?

Answer 1

If we multiply second equation by $-1$ and add to first we have:

$x^2+y^2+xy-y^2-z^2-yz=(x+y+z)(x-z)=a-b$

Next do the same with other equation:

$(x+y+z)(y-z)=a-c$

$(x+y+z)(y-x)=b-c$

Now if $a \neq c$ we can divide first by second and get:

$\frac{x-z}{y-z}=\frac{a-b}{a-c}$, so:

$(x-z)(a-c)=(a-b)(y-z)$. It's linear.

Next do the same with the other equations and we get a system of three linear equation which is easy to solve. There are also cases $a=c$, $a=b$ or $b=c$, but it's easier than in general case.