Solve the following system of equations - (3)












2












$begingroup$



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




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






systems-of-equations







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Lê Thành Đạt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$












  • $begingroup$
    Only for $$yne 2$$
    $endgroup$
    – Dr. Sonnhard Graubner
    14 hours ago










  • $begingroup$
    @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
    14 hours ago










  • $begingroup$
    But you forgot this to say.
    $endgroup$
    – Dr. Sonnhard Graubner
    14 hours ago










  • $begingroup$
    @Dr.SonnhardGraubner Good thing 4 other people already have answers. (Since OP updated the question, now $y=frac{2+4x-3x^2}{2+x}$. For $xneq -2$, of course.)
    $endgroup$
    – Infiaria
    14 hours ago












  • $begingroup$
    Well.... uh.... (I'm sorry.)
    $endgroup$
    – Lê Thành Đạt
    14 hours ago
















2












$begingroup$



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




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






systems-of-equations







share|cite|improve this question









New contributor




Lê Thành Đạt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Only for $$yne 2$$
    $endgroup$
    – Dr. Sonnhard Graubner
    14 hours ago










  • $begingroup$
    @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
    14 hours ago










  • $begingroup$
    But you forgot this to say.
    $endgroup$
    – Dr. Sonnhard Graubner
    14 hours ago










  • $begingroup$
    @Dr.SonnhardGraubner Good thing 4 other people already have answers. (Since OP updated the question, now $y=frac{2+4x-3x^2}{2+x}$. For $xneq -2$, of course.)
    $endgroup$
    – Infiaria
    14 hours ago












  • $begingroup$
    Well.... uh.... (I'm sorry.)
    $endgroup$
    – Lê Thành Đạt
    14 hours ago














2












2








2





$begingroup$



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




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






systems-of-equations







share|cite|improve this question









New contributor




Lê Thành Đạt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





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




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






systems-of-equations




systems-of-equations






share|cite|improve this question









New contributor




Lê Thành Đạt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Lê Thành Đạt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 14 hours ago







Lê Thành Đạt













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asked 14 hours ago









Lê Thành ĐạtLê Thành Đạt

21310




21310




New contributor




Lê Thành Đạt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Lê Thành Đạt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Lê Thành Đạt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Only for $$yne 2$$
    $endgroup$
    – Dr. Sonnhard Graubner
    14 hours ago










  • $begingroup$
    @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
    14 hours ago










  • $begingroup$
    But you forgot this to say.
    $endgroup$
    – Dr. Sonnhard Graubner
    14 hours ago










  • $begingroup$
    @Dr.SonnhardGraubner Good thing 4 other people already have answers. (Since OP updated the question, now $y=frac{2+4x-3x^2}{2+x}$. For $xneq -2$, of course.)
    $endgroup$
    – Infiaria
    14 hours ago












  • $begingroup$
    Well.... uh.... (I'm sorry.)
    $endgroup$
    – Lê Thành Đạt
    14 hours ago


















  • $begingroup$
    Only for $$yne 2$$
    $endgroup$
    – Dr. Sonnhard Graubner
    14 hours ago










  • $begingroup$
    @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
    14 hours ago










  • $begingroup$
    But you forgot this to say.
    $endgroup$
    – Dr. Sonnhard Graubner
    14 hours ago










  • $begingroup$
    @Dr.SonnhardGraubner Good thing 4 other people already have answers. (Since OP updated the question, now $y=frac{2+4x-3x^2}{2+x}$. For $xneq -2$, of course.)
    $endgroup$
    – Infiaria
    14 hours ago












  • $begingroup$
    Well.... uh.... (I'm sorry.)
    $endgroup$
    – Lê Thành Đạt
    14 hours ago
















$begingroup$
Only for $$yne 2$$
$endgroup$
– Dr. Sonnhard Graubner
14 hours ago




$begingroup$
Only for $$yne 2$$
$endgroup$
– Dr. Sonnhard Graubner
14 hours ago












$begingroup$
@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
14 hours ago




$begingroup$
@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
14 hours ago












$begingroup$
But you forgot this to say.
$endgroup$
– Dr. Sonnhard Graubner
14 hours ago




$begingroup$
But you forgot this to say.
$endgroup$
– Dr. Sonnhard Graubner
14 hours ago












$begingroup$
@Dr.SonnhardGraubner Good thing 4 other people already have answers. (Since OP updated the question, now $y=frac{2+4x-3x^2}{2+x}$. For $xneq -2$, of course.)
$endgroup$
– Infiaria
14 hours ago






$begingroup$
@Dr.SonnhardGraubner Good thing 4 other people already have answers. (Since OP updated the question, now $y=frac{2+4x-3x^2}{2+x}$. For $xneq -2$, of course.)
$endgroup$
– Infiaria
14 hours ago














$begingroup$
Well.... uh.... (I'm sorry.)
$endgroup$
– Lê Thành Đạt
14 hours ago




$begingroup$
Well.... uh.... (I'm sorry.)
$endgroup$
– Lê Thành Đạt
14 hours ago










4 Answers
4






active

oldest

votes


















3












$begingroup$

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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Please wait. I typed the problem wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago



















2












$begingroup$

Substituting for the updated equation yields
$$
x=-frac{4}{5}, ; y=-frac{13}{5}
$$
or $(x,y)=(1,1)$. This is a very pleasant result, compared with the old one (with $-xy$ in the first equation instead of $xy$).



$$
x=frac{5y^3 - 26y^2 - 24y + 91}{65},
$$
with
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I apologize. I typed the question wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    @LêThànhĐạt And what is the correct question?
    $endgroup$
    – Dietrich Burde
    14 hours ago










  • $begingroup$
    I just fixed it. Thanks for asking.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    @LêThànhĐạt I fixed my answer, too.
    $endgroup$
    – Dietrich Burde
    14 hours ago










  • $begingroup$
    It is not clear how the equation of degree 4 in y was reached...
    $endgroup$
    – NoChance
    14 hours ago



















0












$begingroup$

Multiply the first equation with $x$ , the right side is unaffected, multiply the second equation with $y$, and write in matrix form and use row reduction echleon form.






share|cite|improve this answer








New contributor




Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$









  • 1




    $begingroup$
    I typed the problem wrong. Sorry for the inconvenience.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    You apologised in different sentences for all answers 😂😂
    $endgroup$
    – Shamim Akhtar
    14 hours ago






  • 1




    $begingroup$
    Yup. That makes me sound like an actual human.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    I am not sure you can get a solution this way.
    $endgroup$
    – NoChance
    13 hours ago



















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











$endgroup$













  • $begingroup$
    Could you wait for me a little bit? I typed the problem wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago











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4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Please wait. I typed the problem wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago
















3












$begingroup$

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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Please wait. I typed the problem wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago














3












3








3





$begingroup$

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






share|cite|improve this answer











$endgroup$



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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 14 hours ago

























answered 14 hours ago









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78k42866




78k42866












  • $begingroup$
    Please wait. I typed the problem wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago


















  • $begingroup$
    Please wait. I typed the problem wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago
















$begingroup$
Please wait. I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
14 hours ago




$begingroup$
Please wait. I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
14 hours ago











2












$begingroup$

Substituting for the updated equation yields
$$
x=-frac{4}{5}, ; y=-frac{13}{5}
$$
or $(x,y)=(1,1)$. This is a very pleasant result, compared with the old one (with $-xy$ in the first equation instead of $xy$).



$$
x=frac{5y^3 - 26y^2 - 24y + 91}{65},
$$
with
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I apologize. I typed the question wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    @LêThànhĐạt And what is the correct question?
    $endgroup$
    – Dietrich Burde
    14 hours ago










  • $begingroup$
    I just fixed it. Thanks for asking.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    @LêThànhĐạt I fixed my answer, too.
    $endgroup$
    – Dietrich Burde
    14 hours ago










  • $begingroup$
    It is not clear how the equation of degree 4 in y was reached...
    $endgroup$
    – NoChance
    14 hours ago
















2












$begingroup$

Substituting for the updated equation yields
$$
x=-frac{4}{5}, ; y=-frac{13}{5}
$$
or $(x,y)=(1,1)$. This is a very pleasant result, compared with the old one (with $-xy$ in the first equation instead of $xy$).



$$
x=frac{5y^3 - 26y^2 - 24y + 91}{65},
$$
with
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I apologize. I typed the question wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    @LêThànhĐạt And what is the correct question?
    $endgroup$
    – Dietrich Burde
    14 hours ago










  • $begingroup$
    I just fixed it. Thanks for asking.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    @LêThànhĐạt I fixed my answer, too.
    $endgroup$
    – Dietrich Burde
    14 hours ago










  • $begingroup$
    It is not clear how the equation of degree 4 in y was reached...
    $endgroup$
    – NoChance
    14 hours ago














2












2








2





$begingroup$

Substituting for the updated equation yields
$$
x=-frac{4}{5}, ; y=-frac{13}{5}
$$
or $(x,y)=(1,1)$. This is a very pleasant result, compared with the old one (with $-xy$ in the first equation instead of $xy$).



$$
x=frac{5y^3 - 26y^2 - 24y + 91}{65},
$$
with
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$






share|cite|improve this answer











$endgroup$



Substituting for the updated equation yields
$$
x=-frac{4}{5}, ; y=-frac{13}{5}
$$
or $(x,y)=(1,1)$. This is a very pleasant result, compared with the old one (with $-xy$ in the first equation instead of $xy$).



$$
x=frac{5y^3 - 26y^2 - 24y + 91}{65},
$$
with
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 13 hours ago

























answered 14 hours ago









Dietrich BurdeDietrich Burde

81.2k648106




81.2k648106












  • $begingroup$
    I apologize. I typed the question wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    @LêThànhĐạt And what is the correct question?
    $endgroup$
    – Dietrich Burde
    14 hours ago










  • $begingroup$
    I just fixed it. Thanks for asking.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    @LêThànhĐạt I fixed my answer, too.
    $endgroup$
    – Dietrich Burde
    14 hours ago










  • $begingroup$
    It is not clear how the equation of degree 4 in y was reached...
    $endgroup$
    – NoChance
    14 hours ago


















  • $begingroup$
    I apologize. I typed the question wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    @LêThànhĐạt And what is the correct question?
    $endgroup$
    – Dietrich Burde
    14 hours ago










  • $begingroup$
    I just fixed it. Thanks for asking.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    @LêThànhĐạt I fixed my answer, too.
    $endgroup$
    – Dietrich Burde
    14 hours ago










  • $begingroup$
    It is not clear how the equation of degree 4 in y was reached...
    $endgroup$
    – NoChance
    14 hours ago
















$begingroup$
I apologize. I typed the question wrong.
$endgroup$
– Lê Thành Đạt
14 hours ago




$begingroup$
I apologize. I typed the question wrong.
$endgroup$
– Lê Thành Đạt
14 hours ago












$begingroup$
@LêThànhĐạt And what is the correct question?
$endgroup$
– Dietrich Burde
14 hours ago




$begingroup$
@LêThànhĐạt And what is the correct question?
$endgroup$
– Dietrich Burde
14 hours ago












$begingroup$
I just fixed it. Thanks for asking.
$endgroup$
– Lê Thành Đạt
14 hours ago




$begingroup$
I just fixed it. Thanks for asking.
$endgroup$
– Lê Thành Đạt
14 hours ago












$begingroup$
@LêThànhĐạt I fixed my answer, too.
$endgroup$
– Dietrich Burde
14 hours ago




$begingroup$
@LêThànhĐạt I fixed my answer, too.
$endgroup$
– Dietrich Burde
14 hours ago












$begingroup$
It is not clear how the equation of degree 4 in y was reached...
$endgroup$
– NoChance
14 hours ago




$begingroup$
It is not clear how the equation of degree 4 in y was reached...
$endgroup$
– NoChance
14 hours ago











0












$begingroup$

Multiply the first equation with $x$ , the right side is unaffected, multiply the second equation with $y$, and write in matrix form and use row reduction echleon form.






share|cite|improve this answer








New contributor




Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$









  • 1




    $begingroup$
    I typed the problem wrong. Sorry for the inconvenience.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    You apologised in different sentences for all answers 😂😂
    $endgroup$
    – Shamim Akhtar
    14 hours ago






  • 1




    $begingroup$
    Yup. That makes me sound like an actual human.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    I am not sure you can get a solution this way.
    $endgroup$
    – NoChance
    13 hours ago
















0












$begingroup$

Multiply the first equation with $x$ , the right side is unaffected, multiply the second equation with $y$, and write in matrix form and use row reduction echleon form.






share|cite|improve this answer








New contributor




Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$









  • 1




    $begingroup$
    I typed the problem wrong. Sorry for the inconvenience.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    You apologised in different sentences for all answers 😂😂
    $endgroup$
    – Shamim Akhtar
    14 hours ago






  • 1




    $begingroup$
    Yup. That makes me sound like an actual human.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    I am not sure you can get a solution this way.
    $endgroup$
    – NoChance
    13 hours ago














0












0








0





$begingroup$

Multiply the first equation with $x$ , the right side is unaffected, multiply the second equation with $y$, and write in matrix form and use row reduction echleon form.






share|cite|improve this answer








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Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$



Multiply the first equation with $x$ , the right side is unaffected, multiply the second equation with $y$, and write in matrix form and use row reduction echleon form.







share|cite|improve this answer








New contributor




Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this answer



share|cite|improve this answer






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answered 14 hours ago









Shamim AkhtarShamim Akhtar

707




707




New contributor




Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    I typed the problem wrong. Sorry for the inconvenience.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    You apologised in different sentences for all answers 😂😂
    $endgroup$
    – Shamim Akhtar
    14 hours ago






  • 1




    $begingroup$
    Yup. That makes me sound like an actual human.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    I am not sure you can get a solution this way.
    $endgroup$
    – NoChance
    13 hours ago














  • 1




    $begingroup$
    I typed the problem wrong. Sorry for the inconvenience.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    You apologised in different sentences for all answers 😂😂
    $endgroup$
    – Shamim Akhtar
    14 hours ago






  • 1




    $begingroup$
    Yup. That makes me sound like an actual human.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago










  • $begingroup$
    I am not sure you can get a solution this way.
    $endgroup$
    – NoChance
    13 hours ago








1




1




$begingroup$
I typed the problem wrong. Sorry for the inconvenience.
$endgroup$
– Lê Thành Đạt
14 hours ago




$begingroup$
I typed the problem wrong. Sorry for the inconvenience.
$endgroup$
– Lê Thành Đạt
14 hours ago












$begingroup$
You apologised in different sentences for all answers 😂😂
$endgroup$
– Shamim Akhtar
14 hours ago




$begingroup$
You apologised in different sentences for all answers 😂😂
$endgroup$
– Shamim Akhtar
14 hours ago




1




1




$begingroup$
Yup. That makes me sound like an actual human.
$endgroup$
– Lê Thành Đạt
14 hours ago




$begingroup$
Yup. That makes me sound like an actual human.
$endgroup$
– Lê Thành Đạt
14 hours ago












$begingroup$
I am not sure you can get a solution this way.
$endgroup$
– NoChance
13 hours ago




$begingroup$
I am not sure you can get a solution this way.
$endgroup$
– NoChance
13 hours ago











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











$endgroup$













  • $begingroup$
    Could you wait for me a little bit? I typed the problem wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago
















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











$endgroup$













  • $begingroup$
    Could you wait for me a little bit? I typed the problem wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago














0












0








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











$endgroup$



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














share|cite|improve this answer



share|cite|improve this answer








edited 12 hours ago

























answered 14 hours ago









Robert IsraelRobert Israel

328k23216470




328k23216470












  • $begingroup$
    Could you wait for me a little bit? I typed the problem wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago


















  • $begingroup$
    Could you wait for me a little bit? I typed the problem wrong.
    $endgroup$
    – Lê Thành Đạt
    14 hours ago
















$begingroup$
Could you wait for me a little bit? I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
14 hours ago




$begingroup$
Could you wait for me a little bit? I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
14 hours ago










Lê Thành Đạt is a new contributor. Be nice, and check out our Code of Conduct.










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Lê Thành Đạt is a new contributor. Be nice, and check out our Code of Conduct.













Lê Thành Đạt is a new contributor. Be nice, and check out our Code of Conduct.












Lê Thành Đạt is a new contributor. Be nice, and check out our Code of Conduct.
















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