System $\,x+y+z=1\,$ and $\,\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$

Question

I'm trying to solve the following system of equations:

I. $\,x+y+z=1$

II. $\,\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$

And an elegant solution just eludes me. It should be a rather easy problem, but I'm having slight problems solving it. I tried just brute force substituting it, but didn't seem to get anywhere with that approach either...

Answer 1

As pointed out in the comments, this is an indeterminate system of equations. However we can still solve for a general form of the solutions. Notice that $\dfrac1x+ \dfrac1y+ \dfrac1z=\dfrac{xy+yz+xz}{xyz}$, by Vieta's theorem for cubic equations we may construct an equation in t which has solutions $t= x,y,$ or $z$.

$t^3-t^2-a^2t+a^2=0\implies (t^2-a^2)(t-1)=0$ , where a is an arbitrary constant.

So $t=a, -a$ (these two solution exhibits the non-negativity restriction on $a^2$ if you're solving in $\Bbb R$), or $t=1$.

Set $x=a,y=-a, z=1$, and you have a set of solutions.

Edit: Made the substitution $A=-a^2$ as suggested by Yves in the comments.

Answer 2

$$\begin{cases}y+z=1-x,\\\dfrac{y+z}{yz}=1-\dfrac1x\end{cases}$$

implies

$$yz=-x.$$

Then

$$y^2+yz=(1-x)y$$

and

$$y^2-(1-x)y-x=0.$$

Solve for $y$ as a function of $x$ and you are done.

Answer 3

We have $$xy+xz+yz=xyz,$$ which gives $$(xy+xz+yz)(x+y+z)=xyz$$ or $$(x+y)(x+z)(y+z)=0.$$ Let $x+y=0$.

Thus, $z=1$ and we see that it gives solutions as $(t,-t,1)$, where $t\neq0$.

The cases $x+z=0$ and $y+z=0$ are similar.