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0

$begingroup$

I am trying to solve under determined simultaneous non - linear equations, where the variables are multiplied, but the power of the variables is always 1, is there a formal way doing it?

For eg: x + y + z = 6, xyz = 6

I was previously using reduced row Echelon form, but now that is less useful, is there a counterpart of Echelon matrix form for the above kind of equations?

systems-of-equations

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13.1k41960

asked yesterday

VARUN.N RAOVARUN.N RAO

33

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RodrigodeAzevedo No powers greater than 1 are used, the variables are multiplied that's all
  $endgroup$
  – VARUN.N RAO
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then you're in the realm of algebraic geometry. Solution sets will be algebraic curves, surfaces, etc. You can parameterize them.
  $endgroup$
  – Rodrigo de Azevedo
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RodrigodeAzevedo is there a name for solving them formally? some method name or something on which I can read on
  $endgroup$
  – VARUN.N RAO
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm basically trying code a program, where if you give the equation it generates the family of possible answers, so I need something like a ready made formula
  $endgroup$
  – VARUN.N RAO
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't want to discourage you, but this is a really ambitious task in the general setting.
  $endgroup$
  – Yves Daoust
  yesterday
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

I am trying to solve under determined simultaneous non - linear equations, where the variables are multiplied, but the power of the variables is always 1, is there a formal way doing it?

For eg: x + y + z = 6, xyz = 6

I was previously using reduced row Echelon form, but now that is less useful, is there a counterpart of Echelon matrix form for the above kind of equations?

systems-of-equations

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13.1k41960

asked yesterday

VARUN.N RAOVARUN.N RAO

33

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RodrigodeAzevedo No powers greater than 1 are used, the variables are multiplied that's all
  $endgroup$
  – VARUN.N RAO
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then you're in the realm of algebraic geometry. Solution sets will be algebraic curves, surfaces, etc. You can parameterize them.
  $endgroup$
  – Rodrigo de Azevedo
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RodrigodeAzevedo is there a name for solving them formally? some method name or something on which I can read on
  $endgroup$
  – VARUN.N RAO
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm basically trying code a program, where if you give the equation it generates the family of possible answers, so I need something like a ready made formula
  $endgroup$
  – VARUN.N RAO
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't want to discourage you, but this is a really ambitious task in the general setting.
  $endgroup$
  – Yves Daoust
  yesterday
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I am trying to solve under determined simultaneous non - linear equations, where the variables are multiplied, but the power of the variables is always 1, is there a formal way doing it?

For eg: x + y + z = 6, xyz = 6

I was previously using reduced row Echelon form, but now that is less useful, is there a counterpart of Echelon matrix form for the above kind of equations?

systems-of-equations

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13.1k41960

asked yesterday

VARUN.N RAOVARUN.N RAO

33

$endgroup$

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13.1k41960

asked yesterday

VARUN.N RAOVARUN.N RAO

33

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13.1k41960

asked yesterday

VARUN.N RAOVARUN.N RAO

33

edited yesterday

Rodrigo de Azevedo

13.1k41960

edited yesterday

Rodrigo de Azevedo

13.1k41960

asked yesterday

VARUN.N RAOVARUN.N RAO

33

asked yesterday

VARUN.N RAOVARUN.N RAO

33

2 Answers
2

active

oldest

votes

0

$begingroup$

When you have too many unknowns, an option is to consider some of them as free parameters and solve for the remaining ones. In the given example, solve

$$begin{cases}x+y+z=6,\xyz=6end{cases}$$ where $z$ is assumed to be known.

We can eliminate $y$ by multiplying the first equation by $xz$,

$$begin{cases}x^2z+xyz+xz^2=6xz,\xyz=6end{cases}$$

which leaves us

$$x^2z+x(z^2-6z)+6=0,$$ quadratic in $x$.

This simple trick turns an indeterminate system into a determinate one, but the real problem of solving the determinate one remains.

share|cite|improve this answer

answered yesterday

Yves DaoustYves Daoust

130k676227

$endgroup$

add a comment | 

0

$begingroup$

Hint: With $$z=frac{6}{xy}$$ we get
$$x^2y+xy^2-6xy+6=0$$ or
$$y^2+y(6+x)+frac{6}{x}=0$$ so we get
$$y_{1,2}=-frac{6+x}{2}pmsqrt{left(frac{6+x}{2}right)^2-frac{6}{x}}$$
The solution is a curve.
And a Groebner bases is given by
$$6-6yz+y^2z+yz^2,x+y+z-6$$

share|cite|improve this answer

edited yesterday

answered yesterday

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.2k42866

$endgroup$

add a comment | 

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0

$begingroup$

When you have too many unknowns, an option is to consider some of them as free parameters and solve for the remaining ones. In the given example, solve

$$begin{cases}x+y+z=6,\xyz=6end{cases}$$ where $z$ is assumed to be known.

We can eliminate $y$ by multiplying the first equation by $xz$,

$$begin{cases}x^2z+xyz+xz^2=6xz,\xyz=6end{cases}$$

which leaves us

$$x^2z+x(z^2-6z)+6=0,$$ quadratic in $x$.

This simple trick turns an indeterminate system into a determinate one, but the real problem of solving the determinate one remains.

share|cite|improve this answer

answered yesterday

Yves DaoustYves Daoust

130k676227

$endgroup$

add a comment | 

0

$begingroup$

When you have too many unknowns, an option is to consider some of them as free parameters and solve for the remaining ones. In the given example, solve

$$begin{cases}x+y+z=6,\xyz=6end{cases}$$ where $z$ is assumed to be known.

We can eliminate $y$ by multiplying the first equation by $xz$,

$$begin{cases}x^2z+xyz+xz^2=6xz,\xyz=6end{cases}$$

which leaves us

$$x^2z+x(z^2-6z)+6=0,$$ quadratic in $x$.

This simple trick turns an indeterminate system into a determinate one, but the real problem of solving the determinate one remains.

share|cite|improve this answer

answered yesterday

Yves DaoustYves Daoust

130k676227

$endgroup$

add a comment | 

$begingroup$

When you have too many unknowns, an option is to consider some of them as free parameters and solve for the remaining ones. In the given example, solve

$$begin{cases}x+y+z=6,\xyz=6end{cases}$$ where $z$ is assumed to be known.

We can eliminate $y$ by multiplying the first equation by $xz$,

$$begin{cases}x^2z+xyz+xz^2=6xz,\xyz=6end{cases}$$

which leaves us

$$x^2z+x(z^2-6z)+6=0,$$ quadratic in $x$.

This simple trick turns an indeterminate system into a determinate one, but the real problem of solving the determinate one remains.

share|cite|improve this answer

answered yesterday

Yves DaoustYves Daoust

130k676227

$endgroup$

share|cite|improve this answer

answered yesterday

Yves DaoustYves Daoust

130k676227

answered yesterday

Yves DaoustYves Daoust

130k676227

answered yesterday

Yves DaoustYves Daoust

130k676227

0

$begingroup$

Hint: With $$z=frac{6}{xy}$$ we get
$$x^2y+xy^2-6xy+6=0$$ or
$$y^2+y(6+x)+frac{6}{x}=0$$ so we get
$$y_{1,2}=-frac{6+x}{2}pmsqrt{left(frac{6+x}{2}right)^2-frac{6}{x}}$$
The solution is a curve.
And a Groebner bases is given by
$$6-6yz+y^2z+yz^2,x+y+z-6$$

share|cite|improve this answer

edited yesterday

answered yesterday

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.2k42866

$endgroup$

add a comment | 

0

$begingroup$

Hint: With $$z=frac{6}{xy}$$ we get
$$x^2y+xy^2-6xy+6=0$$ or
$$y^2+y(6+x)+frac{6}{x}=0$$ so we get
$$y_{1,2}=-frac{6+x}{2}pmsqrt{left(frac{6+x}{2}right)^2-frac{6}{x}}$$
The solution is a curve.
And a Groebner bases is given by
$$6-6yz+y^2z+yz^2,x+y+z-6$$

share|cite|improve this answer

edited yesterday

answered yesterday

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.2k42866

$endgroup$

add a comment | 

$begingroup$

Hint: With $$z=frac{6}{xy}$$ we get
$$x^2y+xy^2-6xy+6=0$$ or
$$y^2+y(6+x)+frac{6}{x}=0$$ so we get
$$y_{1,2}=-frac{6+x}{2}pmsqrt{left(frac{6+x}{2}right)^2-frac{6}{x}}$$
The solution is a curve.
And a Groebner bases is given by
$$6-6yz+y^2z+yz^2,x+y+z-6$$

share|cite|improve this answer

edited yesterday

answered yesterday

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.2k42866

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.2k42866

answered yesterday

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.2k42866

answered yesterday

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.2k42866