Simultaneous Equations

  • #1
FeDeX_LaTeX
Gold Member
437
13

Main Question or Discussion Point

The problem:

"[tex]xy + yz + zx = 12[/tex]
[tex]xyz = 2 + x + y + z[/tex]

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;

[tex]\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{12}{2 + x + y + z}[/tex]

or

[tex]2 + x + y + z = \frac{12}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}[/tex]

if we do this:

[tex]\frac{2 + x + y + z}{4} = \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}[/tex]

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:

Answers and Replies

  • #2
841
0
The problem:

"[tex]xy + yz + zx = 12[/tex]
[tex]xyz = 2 + x + y + z[/tex]

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;
[tex]\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{12}{2 + x + y + z}[/tex]


[tex]2 + x + y + z = \frac{12}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}[/tex]

if we do this:

[tex]\frac{2 + x + y + z}{4} = \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}[/tex]

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.
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:
  • #3
pwsnafu
Science Advisor
1,080
85
There are two different problems here, and you are conflating them

The problem:

"[tex]xy + yz + zx = 12[/tex]
[tex]xyz = 2 + x + y + z[/tex]

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.
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.

Show that a solution exists in which x, y and z are real and distinct."
The only positive solution is symmetric, so if you want distinct, at least one must be negative.
 

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