- Thread starter
- Admin
- #1
- Feb 14, 2012
- 3,816
Find all ordered triples $(x,\,y,\,z)$ that satisfy the following system of equations:
$xy+z=6$
$yz+x=6$
$zx+y=6$
$xy+z=6$
$yz+x=6$
$zx+y=6$
Hello.Find all ordered triples $(x,\,y,\,z)$ that satisfy the following system of equations:
$xy+z=6$
$yz+x=6$
$zx+y=6$
What I can not deduct:There are also:
(1,1,5) and (5,1,1)
But have we found them all?![]()
Hello.
[tex]z=6-xy[/tex]
[tex]6y-xy^2+x=6 \rightarrow{}6(y-1)=x(y^2-1)[/tex]
[tex]6=x(y+1)[/tex](*)
[tex]6x-x^2y+y=6 \rightarrow{}6(x-1)=y(x^2-1)[/tex]
[tex]6=y(x+1)[/tex](**)
For (*) and (**):
[tex]x(y+1)=y(x+1) \rightarrow{}x=y[/tex]
Same:
[tex]x=6-yz[/tex]
...
...
[tex]y=z[/tex]
Conclusion:
[tex]x=y=z[/tex]
Therefore:
[tex]xy+z=6 \rightarrow{} x^2+x-6=0[/tex]
Resolving:
[tex]x=2 , \ and \ x=-3[/tex]
Solution:
[tex](2,2,2),(-3,-3,-3)[/tex]
Thanks MarkFL for pointing this out!There are also:
(1,1,5) and (5,1,1)
But have we found them all?![]()
Regards.
Thanks, mente for participating and yes, all those 5 solutions are the complete answer to this problem.What I can not deduct:
[tex](1,1,5),(1,5,1),(5,1,1)[/tex]
I I had lying down to sleep, and I've had to raise this issue again. now yes I, which is late.
Regards.
Hello.
[tex]z=6-xy[/tex]
[tex]6y-xy^2+x=6 \rightarrow{}6(y-1)=x(y^2-1)[/tex]
[tex]6=x(y+1)[/tex](*)
[tex]6x-x^2y+y=6 \rightarrow{}6(x-1)=y(x^2-1)[/tex]
[tex]6=y(x+1)[/tex](**)
For (*) and (**):
[tex]x(y+1)=y(x+1) \rightarrow{}x=y[/tex]
Same:
[tex]x=6-yz[/tex]
...
...
[tex]y=z[/tex]
Conclusion:
[tex]x=y=z[/tex]
Therefore:
[tex]xy+z=6 \rightarrow{} x^2+x-6=0[/tex]
Resolving:
[tex]x=2 , \ and \ x=-3[/tex]
Solution:
[tex](2,2,2),(-3,-3,-3)[/tex]
Regards.
Thanks to you too, Zaid for participating and showing us another way to solve the problem.
Assume that $xyz=0$ then we get a contradiction. So it must be $xyz\neq 0$. Divide all equations by $xyz$.
\(\displaystyle \frac{1}{z}+\frac{1}{xy}=\frac{6}{xyz}
\)
\(\displaystyle \frac{1}{x}+\frac{1}{yz}=\frac{6}{xyz}
\)
\(\displaystyle \frac{1}{y}+\frac{1}{xz}=\frac{6}{xyz}
\)
so we get
\(\displaystyle \frac{1}{z}+\frac{1}{xy}=\frac{1}{x}+\frac{1}{yz}=\frac{1}{y}+\frac{1}{xz}\)
Consider
\(\displaystyle \frac{1}{z}+\frac{1}{xy}=\frac{1}{x}+\frac{1}{yz}\)
Then we have
\(\displaystyle \frac{1}{z}-\frac{1}{x}=\frac{1}{yz}-\frac{1}{xy}\)
\(\displaystyle \frac{1}{z}-\frac{1}{x}=\frac{1}{y}\left(\frac{1}{z}-\frac{1}{x} \right)\)
Case [1]
\(\displaystyle \frac{1}{x}-\frac{1}{z}=0\)
So \(\displaystyle x=y=z\), Hence \(\displaystyle x^2+x=6\) and we have the solutions \(\displaystyle (2,2,2),(-3,-3,-3)\).
Case [2]
\(\displaystyle \frac{1}{x}-\frac{1}{z}\neq 0\)
Then \(\displaystyle y=1\) so we have
\(\displaystyle \frac{1}{x}+\frac{1}{z}=1+\frac{1}{xz}\)
or $x(1-z)=1-z$ then it is immediate that either \(\displaystyle x=1\) or \(\displaystyle z=1\)
By symmetry of solutions we have \(\displaystyle (1,1,5),(1,5,1),(5,1,1)\)
Yes, you're absolutely right, kaliprasad! I think mente oscura realized it, but it's just that he was too tired and yet wanted to solve some challenge problems on his way to go to bed.the case that was left out was y = 1
y =1 gives x + z = 6, xz = 5 solving it we get
y = 1 , x = 5, z = 1
y = 1 ,z = 5, x = 1
similarly we can find one more x = 5, y = 1, z = 1
actually 2 solutions for y = 1 , 2 for z = 1 and 2 for x = 1 but that is 3 solutions in duplicate
Hello.Thanks to you too, Zaid for participating and showing us another way to solve the problem.
Yes, you're absolutely right, kaliprasad! I think mente oscura realized it, but it's just that he was too tired and yet wanted to solve some challenge problems on his way to go to bed.![]()