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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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 $y=frac{2+4x-3x^2}{2+x}$. 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
















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











$endgroup$












  • $begingroup$
    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 $y=frac{2+4x-3x^2}{2+x}$. 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














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











$endgroup$





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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 22 at 16:34







Lê Thành Đạt

















asked Mar 22 at 16:14









Lê Thành ĐạtLê Thành Đạt

35813




35813












  • $begingroup$
    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 $y=frac{2+4x-3x^2}{2+x}$. 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


















  • $begingroup$
    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 $y=frac{2+4x-3x^2}{2+x}$. 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
















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




$begingroup$
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




$begingroup$
@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




$begingroup$
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 $y=frac{2+4x-3x^2}{2+x}$. For $xneq -2$, of course.)
$endgroup$
– Infiaria
Mar 22 at 16:38






$begingroup$
@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
Mar 22 at 16:38














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




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










3 Answers
3






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
    Mar 22 at 16:33



















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
    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



















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
    Mar 22 at 16:34












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3 Answers
3






active

oldest

votes








3 Answers
3






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
    Mar 22 at 16:33
















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
    Mar 22 at 16:33














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 Mar 22 at 16:44

























answered Mar 22 at 16:23









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867




78.4k42867












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


















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
















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




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











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
    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
















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
    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














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 Mar 22 at 17:54

























answered Mar 22 at 16:26









Dietrich BurdeDietrich Burde

81.5k648106




81.5k648106












  • $begingroup$
    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


















  • $begingroup$
    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
















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




$begingroup$
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




$begingroup$
@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




$begingroup$
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




$begingroup$
@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




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











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
    Mar 22 at 16:34
















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
    Mar 22 at 16:34














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 Mar 22 at 18:16

























answered Mar 22 at 16:25









Robert IsraelRobert Israel

330k23219473




330k23219473












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


















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
















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




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


















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