Solving system of equations

[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

[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.

 

[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

[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.

 

[ZaidAlyafey]

Spoiler

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

 

[anemone]

Hello.

Spoiler

[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]

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

[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.

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

 

[kaliprasad]

Hello.

Spoiler

[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.

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]

Spoiler

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

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.

 

[ZaidAlyafey]

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

Spoiler

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

is a solution. At first glance I thought of using the interesting formula

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

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.

 

[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.

The time difference is

My last post: 5:45 AM

Regards.