Fdtxjr

"listening to me about as much as you're listening to this pole here"

Doomsday-clock for my fantasy planet

How would photo IDs work for shapeshifters?

Is Social Media Science Fiction?

A poker game description that does not feel gimmicky

Can I legally use front facing blue light in the UK?

What happens when a metallic dragon and a chromatic dragon mate?

Lied on resume at previous job

What do you call something that goes against the spirit of the law, but is legal when interpreting the law to the letter?

If a centaur druid Wild Shapes into a Giant Elk, do their Charge features stack?

Symmetry in quantum mechanics

When blogging recipes, how can I support both readers who want the narrative/journey and ones who want the printer-friendly recipe?

Are white and non-white police officers equally likely to kill black suspects?

How to manage monthly salary

Unbreakable Formation vs. Cry of the Carnarium

Is domain driven design an anti-SQL pattern?

Is this food a bread or a loaf?

What is the command to reset a PC without deleting any files

What to wear for invited talk in Canada

What is the meaning of "of trouble" in the following sentence?

Re-submission of rejected manuscript without informing co-authors

aging parents with no investments

Filling an area between two curves

How can I add custom success page

Solve the following system of equations - (3)

Solve the following system of equationsSolve the system of equations…!System of symmetric equationsHow can I solve this hard system of equations?Solving equations system: $xy+yz=a^2,xz+xy=b^2,yz+zx=c^2$System of two symmetric equationsSolve the system of nonlinear equationsSolve the following system of equations (1)Solve the following system of equations - (2).Solving linear system of equations by block

2

$begingroup$

Solve the following system of equations:
$$large
left{
beginalign*
3x^2 + xy - 4x + 2y - 2 = 0\
x(x + 1) + y(y + 1) = 4
endalign*
right.
$$

I tried writing the first equation as $(x - 2)(3x + y - 10) = -18$, but it didn't help.

systems-of-equations symmetric-polynomials

share|cite|improve this question

edited Apr 3 at 4:51

Harry Peter

5,51911439

asked Mar 22 at 16:14

Lê Thành ĐạtLê Thành Đạt

47313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Only for $$yne 2$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 22 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner For $xneq 2$, yes. Then we can say if $x=2$, $y$ must equal $14/6$ and this does not satisfy the second equation.
  $endgroup$
  – Infiaria
  Mar 22 at 16:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But you forgot this to say.
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 22 at 16:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner Good thing 4 other people already have answers. (Since OP updated the question, now $y=frac2+4x-3x^22+x$. For $xneq -2$, of course.)
  $endgroup$
  – Infiaria
  Mar 22 at 16:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well.... uh.... (I'm sorry.)
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:38
  

  
  
  
  

 | 
show 4 more comments

2

$begingroup$

Solve the following system of equations:
$$large
left{
beginalign*
3x^2 + xy - 4x + 2y - 2 = 0\
x(x + 1) + y(y + 1) = 4
endalign*
right.
$$

I tried writing the first equation as $(x - 2)(3x + y - 10) = -18$, but it didn't help.

systems-of-equations symmetric-polynomials

share|cite|improve this question

edited Apr 3 at 4:51

Harry Peter

5,51911439

asked Mar 22 at 16:14

Lê Thành ĐạtLê Thành Đạt

47313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Only for $$yne 2$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 22 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner For $xneq 2$, yes. Then we can say if $x=2$, $y$ must equal $14/6$ and this does not satisfy the second equation.
  $endgroup$
  – Infiaria
  Mar 22 at 16:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But you forgot this to say.
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 22 at 16:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner Good thing 4 other people already have answers. (Since OP updated the question, now $y=frac2+4x-3x^22+x$. For $xneq -2$, of course.)
  $endgroup$
  – Infiaria
  Mar 22 at 16:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well.... uh.... (I'm sorry.)
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:38
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Solve the following system of equations:
$$large
left{
beginalign*
3x^2 + xy - 4x + 2y - 2 = 0\
x(x + 1) + y(y + 1) = 4
endalign*
right.
$$

I tried writing the first equation as $(x - 2)(3x + y - 10) = -18$, but it didn't help.

systems-of-equations symmetric-polynomials

share|cite|improve this question

edited Apr 3 at 4:51

Harry Peter

5,51911439

asked Mar 22 at 16:14

Lê Thành ĐạtLê Thành Đạt

47313

$endgroup$

share|cite|improve this question

edited Apr 3 at 4:51

Harry Peter

5,51911439

asked Mar 22 at 16:14

Lê Thành ĐạtLê Thành Đạt

47313

share|cite|improve this question

edited Apr 3 at 4:51

Harry Peter

5,51911439

asked Mar 22 at 16:14

Lê Thành ĐạtLê Thành Đạt

47313

edited Apr 3 at 4:51

Harry Peter

5,51911439

edited Apr 3 at 4:51

Harry Peter

5,51911439

asked Mar 22 at 16:14

Lê Thành ĐạtLê Thành Đạt

47313

asked Mar 22 at 16:14

Lê Thành ĐạtLê Thành Đạt

47313

3 Answers
3

active

oldest

votes

3

$begingroup$

Solving the first equation for $y$ we get $$y=frac-3x^2+4x+22+x$$ for $$xneq -2$$
plugging this in the second equation we get after simplifications
$$(5x+4)(x-1)^3=0$$

share|cite|improve this answer

edited Mar 22 at 16:44

answered Mar 22 at 16:23

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please wait. I typed the problem wrong.
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:33
  

  
  
  
  

add a comment | 

2

$begingroup$

Substituting for the updated equation yields
$$
x=-frac45, ; y=-frac135
$$
or $(x,y)=(1,1)$. This is a very pleasant result, compared with the old one (with $-xy$ in the first equation instead of $xy$).

$$
x=frac5y^3 - 26y^2 - 24y + 9165,
$$
with
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$

share|cite|improve this answer

edited Mar 22 at 17:54

answered Mar 22 at 16:26

Dietrich BurdeDietrich Burde

81.9k649107

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I apologize. I typed the question wrong.
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LêThànhĐạt And what is the correct question?
  $endgroup$
  – Dietrich Burde
  Mar 22 at 16:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just fixed it. Thanks for asking.
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LêThànhĐạt I fixed my answer, too.
  $endgroup$
  – Dietrich Burde
  Mar 22 at 16:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not clear how the equation of degree 4 in y was reached...
  $endgroup$
  – NoChance
  Mar 22 at 16:57
  

  
  
  
  

 | 
show 2 more comments

0

$begingroup$

The resultant of $3,x^2-xy-4,x+2,y-2$ and $
x left( x+1 right) +y left( y+1 right) -4$ with respect to $y$ is
$$ 10,x^4-24,x^3-10,x^2+42,x-8$$
which is irreducible over the rationals. Its roots can be written in terms of
radicals, but they are far from pleasant. There are two real roots,
$x approx -1.287147510$ and $x approx 0.2049816008$, which correspond to
$y approx -2.469872787$ and $y approx 1.500750095$ respectively.

EDIT: For the corrected question, the resultant of $3,x^2+xy-4,x+2,y-2$ and $x left( x+1 right) +y left( y+1 right) -4$ with respect to $y$ is
$$ 10 x^4 - 22 x^3 + 6 x^2 + 14 x - 8 = 2 (5 x + 4) (x-1)^3$$
Thus we want $x = -4/5$, corresponding to $y = -13/5$, or $x = 1$, corresponding to $y = 1$.

share|cite|improve this answer

edited Mar 22 at 18:16

answered Mar 22 at 16:25

Robert IsraelRobert Israel

331k23220475

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you wait for me a little bit? I typed the problem wrong.
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:34
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158340%2fsolve-the-following-system-of-equations-3%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

Solving the first equation for $y$ we get $$y=frac-3x^2+4x+22+x$$ for $$xneq -2$$
plugging this in the second equation we get after simplifications
$$(5x+4)(x-1)^3=0$$

share|cite|improve this answer

edited Mar 22 at 16:44

answered Mar 22 at 16:23

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please wait. I typed the problem wrong.
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:33
  

  
  
  
  

add a comment | 

3

$begingroup$

Solving the first equation for $y$ we get $$y=frac-3x^2+4x+22+x$$ for $$xneq -2$$
plugging this in the second equation we get after simplifications
$$(5x+4)(x-1)^3=0$$

share|cite|improve this answer

edited Mar 22 at 16:44

answered Mar 22 at 16:23

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please wait. I typed the problem wrong.
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:33
  

  
  
  
  

add a comment | 

$begingroup$

Solving the first equation for $y$ we get $$y=frac-3x^2+4x+22+x$$ for $$xneq -2$$
plugging this in the second equation we get after simplifications
$$(5x+4)(x-1)^3=0$$

share|cite|improve this answer

edited Mar 22 at 16:44

answered Mar 22 at 16:23

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

$endgroup$

share|cite|improve this answer

edited Mar 22 at 16:44

answered Mar 22 at 16:23

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

answered Mar 22 at 16:23

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

answered Mar 22 at 16:23

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

2

$begingroup$

Substituting for the updated equation yields
$$
x=-frac45, ; y=-frac135
$$
or $(x,y)=(1,1)$. This is a very pleasant result, compared with the old one (with $-xy$ in the first equation instead of $xy$).

$$
x=frac5y^3 - 26y^2 - 24y + 9165,
$$
with
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$

share|cite|improve this answer

edited Mar 22 at 17:54

answered Mar 22 at 16:26

Dietrich BurdeDietrich Burde

81.9k649107

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I apologize. I typed the question wrong.
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LêThànhĐạt And what is the correct question?
  $endgroup$
  – Dietrich Burde
  Mar 22 at 16:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just fixed it. Thanks for asking.
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LêThànhĐạt I fixed my answer, too.
  $endgroup$
  – Dietrich Burde
  Mar 22 at 16:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not clear how the equation of degree 4 in y was reached...
  $endgroup$
  – NoChance
  Mar 22 at 16:57
  

  
  
  
  

 | 
show 2 more comments

2

$begingroup$

Substituting for the updated equation yields
$$
x=-frac45, ; y=-frac135
$$
or $(x,y)=(1,1)$. This is a very pleasant result, compared with the old one (with $-xy$ in the first equation instead of $xy$).

$$
x=frac5y^3 - 26y^2 - 24y + 9165,
$$
with
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$

share|cite|improve this answer

edited Mar 22 at 17:54

answered Mar 22 at 16:26

Dietrich BurdeDietrich Burde

81.9k649107

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I apologize. I typed the question wrong.
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LêThànhĐạt And what is the correct question?
  $endgroup$
  – Dietrich Burde
  Mar 22 at 16:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just fixed it. Thanks for asking.
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LêThànhĐạt I fixed my answer, too.
  $endgroup$
  – Dietrich Burde
  Mar 22 at 16:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not clear how the equation of degree 4 in y was reached...
  $endgroup$
  – NoChance
  Mar 22 at 16:57
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Substituting for the updated equation yields
$$
x=-frac45, ; y=-frac135
$$
or $(x,y)=(1,1)$. This is a very pleasant result, compared with the old one (with $-xy$ in the first equation instead of $xy$).

$$
x=frac5y^3 - 26y^2 - 24y + 9165,
$$
with
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$

share|cite|improve this answer

edited Mar 22 at 17:54

answered Mar 22 at 16:26

Dietrich BurdeDietrich Burde

81.9k649107

$endgroup$

share|cite|improve this answer

edited Mar 22 at 17:54

answered Mar 22 at 16:26

Dietrich BurdeDietrich Burde

81.9k649107

answered Mar 22 at 16:26

Dietrich BurdeDietrich Burde

81.9k649107

answered Mar 22 at 16:26

Dietrich BurdeDietrich Burde

81.9k649107

0

$begingroup$

The resultant of $3,x^2-xy-4,x+2,y-2$ and $
x left( x+1 right) +y left( y+1 right) -4$ with respect to $y$ is
$$ 10,x^4-24,x^3-10,x^2+42,x-8$$
which is irreducible over the rationals. Its roots can be written in terms of
radicals, but they are far from pleasant. There are two real roots,
$x approx -1.287147510$ and $x approx 0.2049816008$, which correspond to
$y approx -2.469872787$ and $y approx 1.500750095$ respectively.

EDIT: For the corrected question, the resultant of $3,x^2+xy-4,x+2,y-2$ and $x left( x+1 right) +y left( y+1 right) -4$ with respect to $y$ is
$$ 10 x^4 - 22 x^3 + 6 x^2 + 14 x - 8 = 2 (5 x + 4) (x-1)^3$$
Thus we want $x = -4/5$, corresponding to $y = -13/5$, or $x = 1$, corresponding to $y = 1$.

share|cite|improve this answer

edited Mar 22 at 18:16

answered Mar 22 at 16:25

Robert IsraelRobert Israel

331k23220475

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you wait for me a little bit? I typed the problem wrong.
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:34
  

  
  
  
  

add a comment | 

0

$begingroup$

The resultant of $3,x^2-xy-4,x+2,y-2$ and $
x left( x+1 right) +y left( y+1 right) -4$ with respect to $y$ is
$$ 10,x^4-24,x^3-10,x^2+42,x-8$$
which is irreducible over the rationals. Its roots can be written in terms of
radicals, but they are far from pleasant. There are two real roots,
$x approx -1.287147510$ and $x approx 0.2049816008$, which correspond to
$y approx -2.469872787$ and $y approx 1.500750095$ respectively.

EDIT: For the corrected question, the resultant of $3,x^2+xy-4,x+2,y-2$ and $x left( x+1 right) +y left( y+1 right) -4$ with respect to $y$ is
$$ 10 x^4 - 22 x^3 + 6 x^2 + 14 x - 8 = 2 (5 x + 4) (x-1)^3$$
Thus we want $x = -4/5$, corresponding to $y = -13/5$, or $x = 1$, corresponding to $y = 1$.

share|cite|improve this answer

edited Mar 22 at 18:16

answered Mar 22 at 16:25

Robert IsraelRobert Israel

331k23220475

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you wait for me a little bit? I typed the problem wrong.
  $endgroup$
  – Lê Thành Đạt
  Mar 22 at 16:34
  

  
  
  
  

add a comment | 

$begingroup$

The resultant of $3,x^2-xy-4,x+2,y-2$ and $
x left( x+1 right) +y left( y+1 right) -4$ with respect to $y$ is
$$ 10,x^4-24,x^3-10,x^2+42,x-8$$
which is irreducible over the rationals. Its roots can be written in terms of
radicals, but they are far from pleasant. There are two real roots,
$x approx -1.287147510$ and $x approx 0.2049816008$, which correspond to
$y approx -2.469872787$ and $y approx 1.500750095$ respectively.

EDIT: For the corrected question, the resultant of $3,x^2+xy-4,x+2,y-2$ and $x left( x+1 right) +y left( y+1 right) -4$ with respect to $y$ is
$$ 10 x^4 - 22 x^3 + 6 x^2 + 14 x - 8 = 2 (5 x + 4) (x-1)^3$$
Thus we want $x = -4/5$, corresponding to $y = -13/5$, or $x = 1$, corresponding to $y = 1$.

share|cite|improve this answer

edited Mar 22 at 18:16

answered Mar 22 at 16:25

Robert IsraelRobert Israel

331k23220475

$endgroup$

share|cite|improve this answer

edited Mar 22 at 18:16

answered Mar 22 at 16:25

Robert IsraelRobert Israel

331k23220475

answered Mar 22 at 16:25

Robert IsraelRobert Israel

331k23220475

answered Mar 22 at 16:25

Robert IsraelRobert Israel

331k23220475