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Solving system of equations

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  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684
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

Well-known member
Nov 29, 2013
172
Find all ordered triples $(x,\,y,\,z)$ that satisfy the following system of equations:

$xy+z=6$

$yz+x=6$

$zx+y=6$
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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
There are also:

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


But have we found them all? ;)
 

mente oscura

Well-known member
Nov 29, 2013
172
There are also:

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


But have we found them all? ;)
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.
 

ZaidAlyafey

Well-known member
MHB Math Helper
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 \(\displaystyle (1,1,5),(1,5,1),(5,1,1)\)

 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684
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]



There are also:

(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:

[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

Well-known member
Mar 31, 2013
1,309
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.
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
 
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  • #8

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684

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.

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

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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
\(\displaystyle (1,1,5)\)
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

Well-known member
Nov 29, 2013
172
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.:eek: (Time)

The time difference is (Yawn)

My last post: 5:45 AM

Regards.