System of five equations in five variables?

Question

If someone could give a little help on this problem I'd really appreciate it. I've been trying to solve this for 2 hours.
yz=n(2y+2z)+4µ
xz=n(2x+2z)+4µ
xy+n(2x+2y)+4µ
2xz+2xy+2yz=1500
4x+4y+4z=200

Answer 1

Let's assume the first plus sign in the third equation is a typo, and should be an equal sign instead.

Then if you add the first three equations together and use the info in the other two, you get a simple relationship between n and mu.

Next, let's experiment. Set z = 0 and see what happens. Whoops. We get a contradiction. Ditto if we set x=y=z.

Well, let's set x = -y. Then z = 50. 200x - 2x^2 = 1500. The quadratic formula will lead you to solutions to that. Ditto if you set x = -z or y = -z.

I'm also tempted to see what we can learn about (x+y)(x+z)(y+z) and/or (x+y+1)(x+z+1)(y+z+1). Or about the intersections of the solution set with spheres of the form x^2+y^2+z^2=r^2.

Basically, I favor experimenting until you suddenly see what was supposed to be a blindingly obvious shortcut.