The Time Difference Between Posts

[anemone]

Find all ordered triples $(x,\,y,\,z)$ that satisfy the following system of equations:

$xy+z=6$

$yz+x=6$

$zx+y=6$

 

[mente oscura]

Find all ordered triples $(x,\,y,\,z)$ that satisfy the following system of equations:

$xy+z=6$

$yz+x=6$

$zx+y=6$

Hello.

Spoiler

$$z=6-xy$$

$$6y-xy^2+x=6 \rightarrow{}6(y-1)=x(y^2-1)$$

$$6=x(y+1)$$(*)

$$6x-x^2y+y=6 \rightarrow{}6(x-1)=y(x^2-1)$$

$$6=y(x+1)$$(**)

For (*) and (**):

$$x(y+1)=y(x+1) \rightarrow{}x=y$$

Same:

$$x=6-yz$$

...

...

$$y=z$$

Conclusion:

$$x=y=z$$

Therefore:

$$xy+z=6 \rightarrow{} x^2+x-6=0$$

Resolving:

$$x=2 , \ and \ x=-3$$

Solution:

$$(2,2,2),(-3,-3,-3)$$

Regards.

 

[MarkFL]

There are also:

Spoiler

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

But have we found them all? ;)

 

[mente oscura]

There are also:

Spoiler

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

But have we found them all? ;)

What I can not deduct:

Spoiler

$$(1,1,5),(1,5,1),(5,1,1)$$

I I had lying down to sleep, and I've had to raise this issue again. now yes I, which is late.

Regards.

 

[alyafey22]

[sp]
Assume that $xyz=0$ then we get a contradiction. So it must be $xyz\neq 0$. Divide all equations by $xyz$.

$$\frac{1}{z}+\frac{1}{xy}=\frac{6}{xyz}
$$
$$\frac{1}{x}+\frac{1}{yz}=\frac{6}{xyz}
$$
$$\frac{1}{y}+\frac{1}{xz}=\frac{6}{xyz}
$$
so we get

$$\frac{1}{z}+\frac{1}{xy}=\frac{1}{x}+\frac{1}{yz}=\frac{1}{y}+\frac{1}{xz}$$

Consider

$$\frac{1}{z}+\frac{1}{xy}=\frac{1}{x}+\frac{1}{yz}$$

Then we have

$$\frac{1}{z}-\frac{1}{x}=\frac{1}{yz}-\frac{1}{xy}$$

$$\frac{1}{z}-\frac{1}{x}=\frac{1}{y}\left(\frac{1}{z}-\frac{1}{x} \right)$$

Case [1]

$$\frac{1}{x}-\frac{1}{z}=0$$

So $$x=y=z$$, Hence $$x^2+x=6$$ and we have the solutions $$(2,2,2),(-3,-3,-3)$$.

Case [2]

$$\frac{1}{x}-\frac{1}{z}\neq 0$$

Then $$y=1$$ so we have

$$\frac{1}{x}+\frac{1}{z}=1+\frac{1}{xz}$$

or $x(1-z)=1-z$ then it is immediate that either $$x=1$$ or $$z=1$$

By symmetry of solutions we have $$(1,1,5),(1,5,1),(5,1,1)$$

[/sp]

 

[anemone]

Hello.

Spoiler

$$z=6-xy$$

$$6y-xy^2+x=6 \rightarrow{}6(y-1)=x(y^2-1)$$

$$6=x(y+1)$$(*)

$$6x-x^2y+y=6 \rightarrow{}6(x-1)=y(x^2-1)$$

$$6=y(x+1)$$(**)

For (*) and (**):

$$x(y+1)=y(x+1) \rightarrow{}x=y$$

Same:

$$x=6-yz$$

...

...

$$y=z$$

Conclusion:

$$x=y=z$$

Therefore:

$$xy+z=6 \rightarrow{} x^2+x-6=0$$

Resolving:

$$x=2 , \ and \ x=-3$$

Solution:

$$(2,2,2),(-3,-3,-3)$$

There are also:

Spoiler

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

But have we found them all? ;)

Thanks MarkFL for pointing this out!

Regards.

What I can not deduct:

Spoiler

$$(1,1,5),(1,5,1),(5,1,1)$$

I I had lying down to sleep, and I've had to raise this issue again. now yes I, which is late.

Regards.

Thanks, mente for participating and yes, all those 5 solutions are the complete answer to this problem.

 

[kaliprasad]

Hello.

Spoiler

$$z=6-xy$$

$$6y-xy^2+x=6 \rightarrow{}6(y-1)=x(y^2-1)$$

$$6=x(y+1)$$(*)

$$6x-x^2y+y=6 \rightarrow{}6(x-1)=y(x^2-1)$$

$$6=y(x+1)$$(**)

For (*) and (**):

$$x(y+1)=y(x+1) \rightarrow{}x=y$$

Same:

$$x=6-yz$$

...

...

$$y=z$$

Conclusion:

$$x=y=z$$

Therefore:

$$xy+z=6 \rightarrow{} x^2+x-6=0$$

Resolving:

$$x=2 , \ and \ x=-3$$

Solution:

$$(2,2,2),(-3,-3,-3)$$

Regards.

Spoiler

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

 

[anemone]

[sp]
Assume that $xyz=0$ then we get a contradiction. So it must be $xyz\neq 0$. Divide all equations by $xyz$.

$$\frac{1}{z}+\frac{1}{xy}=\frac{6}{xyz}
$$
$$\frac{1}{x}+\frac{1}{yz}=\frac{6}{xyz}
$$
$$\frac{1}{y}+\frac{1}{xz}=\frac{6}{xyz}
$$
so we get

$$\frac{1}{z}+\frac{1}{xy}=\frac{1}{x}+\frac{1}{yz}=\frac{1}{y}+\frac{1}{xz}$$

Consider

$$\frac{1}{z}+\frac{1}{xy}=\frac{1}{x}+\frac{1}{yz}$$

Then we have

$$\frac{1}{z}-\frac{1}{x}=\frac{1}{yz}-\frac{1}{xy}$$

$$\frac{1}{z}-\frac{1}{x}=\frac{1}{y}\left(\frac{1}{z}-\frac{1}{x} \right)$$

Case [1]

$$\frac{1}{x}-\frac{1}{z}=0$$

So $$x=y=z$$, Hence $$x^2+x=6$$ and we have the solutions $$(2,2,2),(-3,-3,-3)$$.

Case [2]

$$\frac{1}{x}-\frac{1}{z}\neq 0$$

Then $$y=1$$ so we have

$$\frac{1}{x}+\frac{1}{z}=1+\frac{1}{xz}$$

or $x(1-z)=1-z$ then it is immediate that either $$x=1$$ or $$z=1$$

By symmetry of solutions we have $$(1,1,5),(1,5,1),(5,1,1)$$

[/sp]

Thanks to you too, Zaid for participating and showing us another way to solve the problem.

Spoiler

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

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. :)

 

[alyafey22]

I wouldn't have arrived to the solutions unless Mark pointed out the other possiblities. It wasn't quite easy to see that the permutations of [sp]$$(1,1,5)$$[/sp] is a solution. At first glance I thought of using the interesting formula

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

One interesting question is to look at all the solutions of

$$xy+z=a$$
$$xz+y=a$$
$$zy+x=a$$

where the domain is complex numbers.

 

[mente oscura]

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. :)

Hello. (Time)

The time difference is (Yawn)

My last post: 5:45 AM

Regards.