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- #1

- Feb 14, 2012

- 3,888

$xy+z=6$

$yz+x=6$

$zx+y=6$

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,888

$xy+z=6$

$yz+x=6$

$zx+y=6$

- Nov 29, 2013

- 172

Hello.

$xy+z=6$

$yz+x=6$

$zx+y=6$

[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.

- Admin
- #3

- Nov 29, 2013

- 172

What I can not deduct:There are also:

(1,1,5) and (5,1,1)

But have we found them all?

[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.

- Jan 17, 2013

- 1,667

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 \(\displaystyle (1,1,5),(1,5,1),(5,1,1)\)

- Thread starter
- Admin
- #6

- Feb 14, 2012

- 3,888

Hello.

[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]

ThanksThere are also:

(1,1,5) and (5,1,1)

But have we found them all?MarkFLfor pointing this out!

Regards.

Thanks,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.

- Mar 31, 2013

- 1,346

Hello.

[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.

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

- Thread starter
- Admin
- #8

- Feb 14, 2012

- 3,888

Thanks to you too,

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,

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

- Jan 17, 2013

- 1,667

\(\displaystyle (1,1,5)\)

\(\displaystyle x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)+3xyz\).

One interesting question is to look at all the solutions of

\(\displaystyle xy+z=a\)

\(\displaystyle xz+y=a\)

\(\displaystyle zy+x=a\)

where the domain is complex numbers.

- Nov 29, 2013

- 172

Hello.Thanks to you too,Zaidfor 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.

The time difference is

My last post: 5:45 AM

Regards.