Simultaneous Equations Question

Marko1011

Hi,

I've been trying to solve these simultaneous equations, but I just can't. Any help would be greatly appreciated.

x + y + z = 5

1/x + 1/y + 1/z = 1/5

xy + xz + yz = -9

Yakuza

Multiply both sides of the second equation by xyz, you'll get
xyz/x + xyz/y + xyz/z = xyz/5
Simplify:
yz + xz + xy = xyz/5

From the 3rd equation, we now have -9 = xyz/5 or xyz = -45; that might be easier to work with than the sum of the reciprocals.

Marko1011

But that just brings me around in circles, does it not?

bnt

Well, you have now xyz = -45 (as Yakuza worked out), and you also have x + y + z = 5 (the first equation). Just looking at those, without any working, I can see three integers that make both equations true*. That doesn't tell me which is x, y, or z, but looking at the equations, it doesn't look like that matters anyway! It's kind-of a trick question.

*

-3, 3, 5

Marko1011

Well yes, I suppose you could solve it by inspection like that, but that method wouldn't work for more complicated equations. Is there a way to solve it algebraically?

bnt

The form of those equations is what sent me looking for non-analytical solutions. They can't be expressed in the standard ways that you'd use for more analytical methods e.g. inverting a matrix.

Also, the left sides are "symmetrical" (I don't know the correct word, if there is one), in that x, y, and z are totally interchangeable. You can see this in the results i.e. you can swap them around and the equations still work. This could send you on a wild goose chase if you tried to do it only with algebra.

In short, I think you're expected to solve this at least partly by inspection. If there's a fully algebraic method, I'd like to see it.

Marko1011

What about creating a cubic equation and letting the roots equal x, y and z?

ray giraffe

Marko1011 wrote: »

What about creating a cubic equation and letting the roots equal x, y and z?

Yes, that works!