# System of 3 equations

I am doing thermal calculation in electronics and when trying to device a general formula for equivalent system resistance to air flow of a part of real system, I ended with this system of three equations ($$x,y,z$$ are unknown; $$a, b, c$$ are positive parameters): $$\frac{1}{\sqrt x} + \frac{1}{\sqrt{y+z}} = a\\ \frac{1}{\sqrt y} + \frac{1}{\sqrt{x+z}} = b\\ \frac{1}{\sqrt z} + \frac{1}{\sqrt{x+y}} = c$$ There is no problem in solving $$x, y, z$$ in Excel solver, Matlab, or other. The problem is with analytical solution that can be useful in solving more complex systems elegantly. I was trying to do some substitutions, but with no success at all. So, question arises: is it possible at all to solve above system of equations for $$x, y, z$$ analytically?

• Mathematica is using more than a minute now for solving it and it still has found nothing, so I guess it has no analytical solution. Dec 18, 2013 at 22:29
• Could someone post the numeric solution? Perhaps the inverse symbolic calculator could help... Dec 18, 2013 at 22:53
• @anorton: His $a,b,c$ are free parameters. As such, his unknowns $x,y,z$ will vary. Feb 14, 2015 at 11:59

This is an old question, but for the benefit of those who'll come across this and left hanging.

Case 1: If $b=c$, then there is an analytic solution which involves an octic that factors into quartics over $\sqrt{2}$.

Let $x,y,z = p^2,\;q^2,\;r^2$ and, to avoid the octic, let $a=\sqrt{2}\,a$. We get,

\begin{aligned} \frac{1}{p}+\frac{1}{\sqrt{q^2+r^2}}\;&= \sqrt{2}\,a\\ \frac{1}{q}+\frac{1}{\sqrt{p^2+r^2}}\;&= b\\ \frac{1}{r}+\frac{1}{\sqrt{p^2+q^2}}\;&= b \end{aligned}

and the solution,

$$p =\frac{(r+2ar^2)\sqrt{2}}{4a^2r^2-1},\quad q = r$$

and $r$ is an appropriate root(s) of the quartic,

$$2 - 6 b r + b (8 a + 3 b) r^2 - 4 a b (2 a + b) r^3 + 4 a^2 b^2 r^4=0$$

Case 2: For general $a,b,c$, one has to solve a $32$-deg eqn. (I don't know if over an appropriate radical extension it will factor into quartics as well.)

This can be converted into a system of polynomial equations.

Let $u^2 = x$, $v^2 = y$, $w^2 = z$. Then set $s^2 = x+z$, $t^2 = x+y$, $r^2 = y+z$. The equations are now

$\frac{1}{u} + \frac{1}{r} = a$, $\frac{1}{v} + \frac{1}{s} = b$, $\frac{1}{w} + \frac{1}{t} = c$. Clear denominators and we have six polynomials in the six vars $r$, $s$, $t$, $u$, $v$, $w$, and three parameters $a$, $b$, $c$.

This system can be "solved" with the Dixon resultant in 0.9 secs. The resultant for u has degree 32, 1105 terms, and begins and ends

$16a^8b^16c^16u^32 - 64a^10b^14c^16u^32 + 96a^12b^12c^16u^32 … \\ - 65536ab^8u + 4096c^8 - 8192b^4c^4 + 4096b^8$

As another poster mentioned, if we assume $a = b$, $a = c$, or $b = c$, the polynomial factors a lot.

If you want $x$ instead of $u$, that will take one more easy resultant computation.