1

How does one go about solving the roots for the following equation

$$x+y+z=xyz$$

There simply to many variables. Anyone have an idea ?

Community
  • 1
jessica
  • 1,002

2 Answers2

2

There are infinitely many solutions. One way to show this is to solve for one of the variables. For example, $$ \begin{align*} & x+y+z=xyz \\ \Rightarrow& z-xyz=-x-y \\ \Rightarrow& z(1-xy)=-x-y \\ \Rightarrow&z=\frac{-x-y}{1-xy} \\ & \,\,\,\,=\frac{x+y}{xy-1}. \end{align*} $$

Choose any $x,y$ so that $xy-1 \neq 0$. For example, Let $x=2$ and $y=1$. Then $z=3$, and by inspection you will find that $x+y+z=xyz$ holds. You could write the solution up as the set of all ordered triples $$\left\{\left( x,y, \frac{x+y}{xy-1} \right)\bigg| \, x,y \in \mathbb{R} \wedge xy \neq 1\right\}.$$

Additional Comment:

We introduced a possibility for division by zero into the solution set when we divided by $1-xy$ to get the result $z$, and this forced us to eliminate any values $x,y$ so that $xy=1$. We could have solved for one of the other two variables and encountered a similar restriction. As it turns out, it is not possible for the product of any two of $x,y,z$ to equal $1$ as was dutifully demonstrated by Will Jagy in comments.

Without loss of generality, we may assume that $xy=1$. Then $$x+y+z=z \Rightarrow x+y=0 \Rightarrow x=-y.$$ This is a contradiction.

J. W. Perry
  • 5,447
  • 3
  • 22
  • 28
  • What makes you say that there aren't any solutions with $xy=1$? Probably there were, but you lost them when you divided by $xy-1$. – Trevor Wilson Nov 15 '13 at 06:17
  • 1
    Assume $xy=1,$ then $x+y+z=z,$ so $x+y=0,$ then $y=-x;$ we get a contradiction. – Will Jagy Nov 15 '13 at 06:38
  • 1
    @WillJagy That is an excellent additional statement. I am trying to think of how to write this more compactly at the moment, and this shows that the product of no two values can equal $1$. – J. W. Perry Nov 15 '13 at 06:42
  • 1
    @WillJagy Oh, I see! Serves me right for being lazy and saying "probably". – Trevor Wilson Nov 15 '13 at 06:56
  • @TrevorWilson Yeah, me too, and I am thinking of deleting those last two sets. I think the first covers the set of all solutions. Thanks though for making me think. I have never done a rollback, but it might be in order. – J. W. Perry Nov 15 '13 at 06:58
  • 1
    You could roll it back, but you might want to leave some sort of comment about why your solution set isn't asymmetric even though it appears to be at first glance (so no one else makes the same mistake I did.) – Trevor Wilson Nov 15 '13 at 07:00
  • @TrevorWilson I think I will carefully back it up by hand. Keep an eye out for the record. – J. W. Perry Nov 15 '13 at 07:02
1

If we fix one of the variables, we get a hyperbola in that plane. So, for example, fixing any $z = z_0,$ this is your relationship: $$ \left(x - \frac{1}{z_0} \right) \left(y - \frac{1}{z_0} \right) = \; 1 + \frac{1}{z_0^2} $$ Makes me think the surface could be connected.

Indeed, as $|z| \rightarrow \infty,$ the curve approaches the fixed curve $xy=1.$

Meanwhile, the surface is smooth. The gradient of the function $xyz-x-y-z$ is only the zero vector at $(1,1,1)$ and $(-1,-1,-1)$ but those are not part of the surface.

And, in the plane $x+y+z = 0,$ the surface contains the entirety of three lines that meet at $60^\circ$ angles, these being the intersections with the planes $x=0$ or $y=0$ or $z=0.$

EDIT: the surface actually has three connected components, sometimes called sheets in the context of hyperboloids. One is in the positive octant $x,y,z > 0,$ with $xy>1, xz>1, yz > 1$ in addition. The nearest point to the origin is $(\sqrt 3,\sqrt 3,\sqrt 3 ),$ and the thing gets very close to the walls as $x+y+z$ increases.

The middle sheet goes through the origin and, near there, is a monkey saddle.

The third sheet is in the negative octant.

About the middle sheet: you may recall that the real part of $(x+yi)^3$ is $x^3 - 3 x y^2.$ The graph $z =x^3 - 3 x y^2 $ is, near the origin, a monkey saddle, room for two legs and a tail for a monkey sitting on a horse. If we rotate coordinates by $$ x = \frac{u}{\sqrt 3 } - \frac{v}{\sqrt 2 } - \frac{w}{\sqrt 3 }, $$ $$ y = \frac{u}{\sqrt 3 } + \frac{v}{\sqrt 2 } - \frac{w}{\sqrt 3 }, $$ $$ z = \frac{u}{\sqrt 3 } + \frac{2w}{\sqrt 3 }, $$

the surface becomes $$ \color{magenta}{ u \; \left( 1 + \frac{v^2}{6} + \frac{w^2}{6} - \frac{u^2}{9} \right) = \frac{1}{9 \sqrt 2} \; \left( w^3 - 3 w \, v^2 \right).} $$ So, say within the ball $u^2 + v^2 + w^2 < 1,$ the multiplier of $u$ on the left is quite close to $1,$ and we have a monkey saddle.

Took a bit of work to confirm: given constants $a+b+c = 0,$ and looking at $$ x=a+t, y = b+t, z = c+t, $$ there are exactly three distinct real values of $t$ that solve $xyz=x+y+z.$

Will Jagy
  • 139,541