# Simultaneous Equations

1. Oct 1, 2012

### FeDeX_LaTeX

The problem:

"$$xy + yz + zx = 12$$
$$xyz = 2 + x + y + z$$

Find a solution of the above simultaneous equations, in which all of x, y and z are positive, and prove that it is the only such solution.

Show that a solution exists in which x, y and z are real and distinct."

I haven't really made much progress on this problem. I divided the first equation by the second and got;

$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{12}{2 + x + y + z}$$

or

$$2 + x + y + z = \frac{12}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$$

if we do this:

$$\frac{2 + x + y + z}{4} = \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$$

I notice that the LHS is the arithmetic mean of {2, x, y, z} and the RHS is the harmonic mean of {x, y, z}. They are equal iff x = y = z = 2, so one solution is x = 2, y = 2, z = 2. BUT it appears that they want x, y and z to be distinct.

Can anyone help me here?

Thanks.

EDIT:
Wait... on second thought, what I wrote is wrong. The two sets {2,x,y,z} and {x,y,z} aren't identical. Looks like I am back to square one.

Last edited: Oct 1, 2012
2. Oct 1, 2012

### ramsey2879

Since the solution is unique, you can try to make the problem simple by adding a relationship between x,y and z and see if the problem can still be solved. perhaps you may try x,y,z form an arithmetic sequence, etc. What makes you think z,y,z must be different? Looks like a valid solution, i.e. the one in which x thru z are each positive. In the second part, a negative number could be included.

Last edited: Oct 1, 2012
3. Oct 1, 2012

### pwsnafu

There are two different problems here, and you are conflating them

You want the unique positive solution. It is obvious that x=y=z=2 is the solution because you can substitute the values in the original. What you need to show is that unique implies symmetric.

The only positive solution is symmetric, so if you want distinct, at least one must be negative.