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Solving a linear system of reciprocals.

Trouble Solving a system of 3 equationsSolving a system of equations with variables in denominator.Solving system of three quadratic equationsSolving system of non-linear equations.Solving a combined system of linear and bilinear equationsHow to solve the system of non-linear equations?Solving a system of equations $(x+y-z=1), (x^2+y^2-z^2=5-2xy), (x^3+y^3-z^3=43-3xy)$Easier solution for system of equations.Solving in terms of z , three variable two equation systemSolve the following system of equations - (4).

3

$begingroup$

Solve for $begincasesfrac1x +frac1y+frac1z=0\frac4x +frac3y+frac2z=5\frac3x +frac2y+frac4z=-4endcases$

I turn the equations into $begincasesyz+xz+xy=0\4yz+3xz+2xy=5xyz\3yz+2xz+4xy=-4xyzendcases$

Not sure if I am doing fine

systems-of-equations cubic-equations rational-functions symmetric-polynomials

share|cite|improve this question

edited May 3 at 5:38

Harry Peter

5,55411439

asked Apr 28 at 15:43

DavidDavid

764

$endgroup$

add a comment | 

3

$begingroup$

Solve for $begincasesfrac1x +frac1y+frac1z=0\frac4x +frac3y+frac2z=5\frac3x +frac2y+frac4z=-4endcases$

I turn the equations into $begincasesyz+xz+xy=0\4yz+3xz+2xy=5xyz\3yz+2xz+4xy=-4xyzendcases$

Not sure if I am doing fine

systems-of-equations cubic-equations rational-functions symmetric-polynomials

share|cite|improve this question

edited May 3 at 5:38

Harry Peter

5,55411439

asked Apr 28 at 15:43

DavidDavid

764

$endgroup$

add a comment | 

$begingroup$

Solve for $begincasesfrac1x +frac1y+frac1z=0\frac4x +frac3y+frac2z=5\frac3x +frac2y+frac4z=-4endcases$

I turn the equations into $begincasesyz+xz+xy=0\4yz+3xz+2xy=5xyz\3yz+2xz+4xy=-4xyzendcases$

Not sure if I am doing fine

systems-of-equations cubic-equations rational-functions symmetric-polynomials

share|cite|improve this question

edited May 3 at 5:38

Harry Peter

5,55411439

asked Apr 28 at 15:43

DavidDavid

764

$endgroup$

share|cite|improve this question

edited May 3 at 5:38

Harry Peter

5,55411439

asked Apr 28 at 15:43

DavidDavid

764

share|cite|improve this question

edited May 3 at 5:38

Harry Peter

5,55411439

asked Apr 28 at 15:43

DavidDavid

764

edited May 3 at 5:38

Harry Peter

5,55411439

edited May 3 at 5:38

Harry Peter

5,55411439

asked Apr 28 at 15:43

DavidDavid

764

asked Apr 28 at 15:43

DavidDavid

764

4 Answers
4

active

oldest

votes

8

$begingroup$

If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begincasest +u+v=0\4t +3u+2v=5\3t +2u+4v=-4endcases$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.

share|cite|improve this answer

answered Apr 28 at 15:50

Cameron BuieCameron Buie

88k773162

$endgroup$

add a comment | 

8

$begingroup$

It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.

share|cite|improve this answer

answered Apr 28 at 15:45

Matt SamuelMatt Samuel

40k63870

$endgroup$

add a comment | 

2

$begingroup$

how about this, let:
$$X=frac1x,,Y=frac1y,,Z=frac1z$$
and it is much easier to solve the following:
$$beginpmatrix
1&1&1\
4&3&2\
3&2&4
endpmatrix
beginpmatrix
X\
Y\
Z
endpmatrix=
beginpmatrix
0\5\-4
endpmatrix
$$

share|cite|improve this answer

answered Apr 28 at 15:51

Henry LeeHenry Lee

2,249319

$endgroup$

add a comment | 

1

$begingroup$

Observe that the system's matrix of the reciprocal of the unknowns is:

$$A=beginpmatrix1&1&1&0\
4&3&2&5\
3&2&4&-4endpmatrixstackrelR_2-4R_1,,R_3-3R_1longrightarrowbeginpmatrix1&1&1&0\
0&-1&-2&5\
0&-1&1&-4endpmatrixstackrelR_3-R_2longrightarrow$$

$$beginpmatrix1&1&1&0\
0&-1&-2&5\
0&0&3&-9endpmatrix$$

Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$

share|cite|improve this answer

edited Apr 28 at 18:43

answered Apr 28 at 15:51

DonAntonioDonAntonio

180k1496234

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  There's an error at the 1st step: the third row should be $;0;-1; colorred+1;-4$.
  $endgroup$
  – Bernard
  Apr 28 at 15:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Bernard Good catch, edited. Thanks.
  $endgroup$
  – DonAntonio
  Apr 28 at 18:43
  

  
  
  
  

add a comment | 

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8

$begingroup$

If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begincasest +u+v=0\4t +3u+2v=5\3t +2u+4v=-4endcases$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.

share|cite|improve this answer

answered Apr 28 at 15:50

Cameron BuieCameron Buie

88k773162

$endgroup$

add a comment | 

8

$begingroup$

If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begincasest +u+v=0\4t +3u+2v=5\3t +2u+4v=-4endcases$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.

share|cite|improve this answer

answered Apr 28 at 15:50

Cameron BuieCameron Buie

88k773162

$endgroup$

add a comment | 

$begingroup$

If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begincasest +u+v=0\4t +3u+2v=5\3t +2u+4v=-4endcases$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.

share|cite|improve this answer

answered Apr 28 at 15:50

Cameron BuieCameron Buie

88k773162

$endgroup$

share|cite|improve this answer

answered Apr 28 at 15:50

Cameron BuieCameron Buie

88k773162

answered Apr 28 at 15:50

Cameron BuieCameron Buie

88k773162

answered Apr 28 at 15:50

Cameron BuieCameron Buie

88k773162

8

$begingroup$

It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.

share|cite|improve this answer

answered Apr 28 at 15:45

Matt SamuelMatt Samuel

40k63870

$endgroup$

add a comment | 

8

$begingroup$

It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.

share|cite|improve this answer

answered Apr 28 at 15:45

Matt SamuelMatt Samuel

40k63870

$endgroup$

add a comment | 

$begingroup$

It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.

share|cite|improve this answer

answered Apr 28 at 15:45

Matt SamuelMatt Samuel

40k63870

$endgroup$

share|cite|improve this answer

answered Apr 28 at 15:45

Matt SamuelMatt Samuel

40k63870

answered Apr 28 at 15:45

Matt SamuelMatt Samuel

40k63870

answered Apr 28 at 15:45

Matt SamuelMatt Samuel

40k63870

2

$begingroup$

how about this, let:
$$X=frac1x,,Y=frac1y,,Z=frac1z$$
and it is much easier to solve the following:
$$beginpmatrix
1&1&1\
4&3&2\
3&2&4
endpmatrix
beginpmatrix
X\
Y\
Z
endpmatrix=
beginpmatrix
0\5\-4
endpmatrix
$$

share|cite|improve this answer

answered Apr 28 at 15:51

Henry LeeHenry Lee

2,249319

$endgroup$

add a comment | 

2

$begingroup$

how about this, let:
$$X=frac1x,,Y=frac1y,,Z=frac1z$$
and it is much easier to solve the following:
$$beginpmatrix
1&1&1\
4&3&2\
3&2&4
endpmatrix
beginpmatrix
X\
Y\
Z
endpmatrix=
beginpmatrix
0\5\-4
endpmatrix
$$

share|cite|improve this answer

answered Apr 28 at 15:51

Henry LeeHenry Lee

2,249319

$endgroup$

add a comment | 

$begingroup$

how about this, let:
$$X=frac1x,,Y=frac1y,,Z=frac1z$$
and it is much easier to solve the following:
$$beginpmatrix
1&1&1\
4&3&2\
3&2&4
endpmatrix
beginpmatrix
X\
Y\
Z
endpmatrix=
beginpmatrix
0\5\-4
endpmatrix
$$

share|cite|improve this answer

answered Apr 28 at 15:51

Henry LeeHenry Lee

2,249319

$endgroup$

share|cite|improve this answer

answered Apr 28 at 15:51

Henry LeeHenry Lee

2,249319

answered Apr 28 at 15:51

Henry LeeHenry Lee

2,249319

answered Apr 28 at 15:51

Henry LeeHenry Lee

2,249319

1

$begingroup$

Observe that the system's matrix of the reciprocal of the unknowns is:

$$A=beginpmatrix1&1&1&0\
4&3&2&5\
3&2&4&-4endpmatrixstackrelR_2-4R_1,,R_3-3R_1longrightarrowbeginpmatrix1&1&1&0\
0&-1&-2&5\
0&-1&1&-4endpmatrixstackrelR_3-R_2longrightarrow$$

$$beginpmatrix1&1&1&0\
0&-1&-2&5\
0&0&3&-9endpmatrix$$

Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$

share|cite|improve this answer

edited Apr 28 at 18:43

answered Apr 28 at 15:51

DonAntonioDonAntonio

180k1496234

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  There's an error at the 1st step: the third row should be $;0;-1; colorred+1;-4$.
  $endgroup$
  – Bernard
  Apr 28 at 15:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Bernard Good catch, edited. Thanks.
  $endgroup$
  – DonAntonio
  Apr 28 at 18:43
  

  
  
  
  

add a comment | 

1

$begingroup$

Observe that the system's matrix of the reciprocal of the unknowns is:

$$A=beginpmatrix1&1&1&0\
4&3&2&5\
3&2&4&-4endpmatrixstackrelR_2-4R_1,,R_3-3R_1longrightarrowbeginpmatrix1&1&1&0\
0&-1&-2&5\
0&-1&1&-4endpmatrixstackrelR_3-R_2longrightarrow$$

$$beginpmatrix1&1&1&0\
0&-1&-2&5\
0&0&3&-9endpmatrix$$

Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$

share|cite|improve this answer

edited Apr 28 at 18:43

answered Apr 28 at 15:51

DonAntonioDonAntonio

180k1496234

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  There's an error at the 1st step: the third row should be $;0;-1; colorred+1;-4$.
  $endgroup$
  – Bernard
  Apr 28 at 15:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Bernard Good catch, edited. Thanks.
  $endgroup$
  – DonAntonio
  Apr 28 at 18:43
  

  
  
  
  

add a comment | 

$begingroup$

Observe that the system's matrix of the reciprocal of the unknowns is:

$$A=beginpmatrix1&1&1&0\
4&3&2&5\
3&2&4&-4endpmatrixstackrelR_2-4R_1,,R_3-3R_1longrightarrowbeginpmatrix1&1&1&0\
0&-1&-2&5\
0&-1&1&-4endpmatrixstackrelR_3-R_2longrightarrow$$

$$beginpmatrix1&1&1&0\
0&-1&-2&5\
0&0&3&-9endpmatrix$$

Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$

share|cite|improve this answer

edited Apr 28 at 18:43

answered Apr 28 at 15:51

DonAntonioDonAntonio

180k1496234

$endgroup$

share|cite|improve this answer

edited Apr 28 at 18:43

answered Apr 28 at 15:51

DonAntonioDonAntonio

180k1496234

answered Apr 28 at 15:51

DonAntonioDonAntonio

180k1496234

answered Apr 28 at 15:51

DonAntonioDonAntonio

180k1496234