Solve the following system of equations - (3)Solve the following system of equationsSolve the system of equations…!System of symmetric equationsSolve the system of equations with one symmetrical equationHow can I solve this hard system of equations?How to solve system of equations involving square rootsHow to solve this kind of system of equations using software?Solve the system of nonlinear equationsSolve the following system of equations (1)Solve the following system of equations - (2).
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Solve the following system of equations - (3)
Solve the following system of equationsSolve the system of equations…!System of symmetric equationsSolve the system of equations with one symmetrical equationHow can I solve this hard system of equations?How to solve system of equations involving square rootsHow to solve this kind of system of equations using software?Solve the system of nonlinear equationsSolve the following system of equations (1)Solve the following system of equations - (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
asked 3 hours ago
Lê Thành ĐạtLê Thành Đạt
20210
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.
$endgroup$
|
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
asked 3 hours ago
Lê Thành ĐạtLê Thành Đạt
20210
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
3 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
3 hours ago
$begingroup$
But you forgot this to say.
$endgroup$
– Dr. Sonnhard Graubner
3 hours ago
$begingroup$
@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
3 hours ago
$begingroup$
Well.... uh.... (I'm sorry.)
$endgroup$
– Lê Thành Đạt
3 hours ago
|
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
asked 3 hours ago
Lê Thành ĐạtLê Thành Đạt
20210
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:
$$large
left{
beginalign*
3x^2 + xy - 4x + 2y - 2 = 0\
x(x + 1) + y(y + 1) = 4
endalign*
right.
$$
I tried writing the first equation as $(x - 2)(3x + y - 10) = -18$, but it didn't help.
systems-of-equations
systems-of-equations
asked 3 hours ago
Lê Thành ĐạtLê Thành Đạt
20210
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.
asked 3 hours ago
Lê Thành ĐạtLê Thành Đạt
20210
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.
edited 3 hours ago
Lê Thành Đạt
asked 3 hours ago
Lê Thành ĐạtLê Thành Đạt
20210
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.
asked 3 hours ago
Lê Thành ĐạtLê Thành Đạt
20210
asked 3 hours ago
Lê Thành ĐạtLê Thành Đạt
20210
20210
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.
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
3 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
3 hours ago
$begingroup$
But you forgot this to say.
$endgroup$
– Dr. Sonnhard Graubner
3 hours ago
$begingroup$
@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
3 hours ago
$begingroup$
Well.... uh.... (I'm sorry.)
$endgroup$
– Lê Thành Đạt
3 hours ago
|
show 4 more comments
$begingroup$
Only for $$yne 2$$
$endgroup$
– Dr. Sonnhard Graubner
3 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
3 hours ago
$begingroup$
But you forgot this to say.
$endgroup$
– Dr. Sonnhard Graubner
3 hours ago
$begingroup$
@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
3 hours ago
$begingroup$
Well.... uh.... (I'm sorry.)
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
Only for $$yne 2$$
$endgroup$
– Dr. Sonnhard Graubner
3 hours ago
$begingroup$
Only for $$yne 2$$
$endgroup$
– Dr. Sonnhard Graubner
3 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
3 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
3 hours ago
$begingroup$
But you forgot this to say.
$endgroup$
– Dr. Sonnhard Graubner
3 hours ago
$begingroup$
But you forgot this to say.
$endgroup$
– Dr. Sonnhard Graubner
3 hours ago
$begingroup$
@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
3 hours ago
$begingroup$
@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
3 hours ago
$begingroup$
Well.... uh.... (I'm sorry.)
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
Well.... uh.... (I'm sorry.)
$endgroup$
– Lê Thành Đạt
3 hours ago
|
show 4 more comments
4 Answers
4
active
oldest
votes
$begingroup$
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
$$(5x+4)(x-1)^3=0$$
answered 3 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
$begingroup$
Please wait. I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
add a comment |
$begingroup$
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
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$
answered 3 hours ago
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
$begingroup$
I apologize. I typed the question wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
@LêThànhĐạt And what is the correct question?
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
I just fixed it. Thanks for asking.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
@LêThànhĐạt I fixed my answer, too.
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
It is not clear how the equation of degree 4 in y was reached...
$endgroup$
– NoChance
2 hours ago
|
show 2 more comments
$begingroup$
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.
answered 3 hours ago
Shamim AkhtarShamim Akhtar
707
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
3 hours ago
$begingroup$
You apologised in different sentences for all answers 😂😂
$endgroup$
– Shamim Akhtar
3 hours ago
1
$begingroup$
Yup. That makes me sound like an actual human.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
I am not sure you can get a solution this way.
$endgroup$
– NoChance
2 hours ago
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 $y = 1$.
answered 3 hours ago
Robert IsraelRobert Israel
328k23216469
$endgroup$
$begingroup$
Could you wait for me a little bit? I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$(5x+4)(x-1)^3=0$$
answered 3 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
$begingroup$
Please wait. I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
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
$$(5x+4)(x-1)^3=0$$
answered 3 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
$begingroup$
Please wait. I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
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
$$(5x+4)(x-1)^3=0$$
answered 3 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
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
$$(5x+4)(x-1)^3=0$$
answered 3 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
edited 3 hours ago
answered 3 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
answered 3 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
answered 3 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
78k42866
$begingroup$
Please wait. I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
add a comment |
$begingroup$
Please wait. I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
Please wait. I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
Please wait. I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
add a comment |
$begingroup$
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
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$
answered 3 hours ago
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
$begingroup$
I apologize. I typed the question wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
@LêThànhĐạt And what is the correct question?
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
I just fixed it. Thanks for asking.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
@LêThànhĐạt I fixed my answer, too.
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
It is not clear how the equation of degree 4 in y was reached...
$endgroup$
– NoChance
2 hours ago
|
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
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$
answered 3 hours ago
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
$begingroup$
I apologize. I typed the question wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
@LêThànhĐạt And what is the correct question?
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
I just fixed it. Thanks for asking.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
@LêThànhĐạt I fixed my answer, too.
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
It is not clear how the equation of degree 4 in y was reached...
$endgroup$
– NoChance
2 hours ago
|
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
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$
answered 3 hours ago
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
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
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$
answered 3 hours ago
Dietrich BurdeDietrich Burde
81.2k648106
edited 1 hour ago
answered 3 hours ago
Dietrich BurdeDietrich Burde
81.2k648106
answered 3 hours ago
Dietrich BurdeDietrich Burde
81.2k648106
answered 3 hours ago
Dietrich BurdeDietrich Burde
81.2k648106
81.2k648106
$begingroup$
I apologize. I typed the question wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
@LêThànhĐạt And what is the correct question?
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
I just fixed it. Thanks for asking.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
@LêThànhĐạt I fixed my answer, too.
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
It is not clear how the equation of degree 4 in y was reached...
$endgroup$
– NoChance
2 hours ago
|
show 2 more comments
$begingroup$
I apologize. I typed the question wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
@LêThànhĐạt And what is the correct question?
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
I just fixed it. Thanks for asking.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
@LêThànhĐạt I fixed my answer, too.
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
It is not clear how the equation of degree 4 in y was reached...
$endgroup$
– NoChance
2 hours ago
$begingroup$
I apologize. I typed the question wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
I apologize. I typed the question wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
@LêThànhĐạt And what is the correct question?
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
@LêThànhĐạt And what is the correct question?
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
I just fixed it. Thanks for asking.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
I just fixed it. Thanks for asking.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
@LêThànhĐạt I fixed my answer, too.
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
@LêThànhĐạt I fixed my answer, too.
$endgroup$
– Dietrich Burde
3 hours ago
$begingroup$
It is not clear how the equation of degree 4 in y was reached...
$endgroup$
– NoChance
2 hours ago
$begingroup$
It is not clear how the equation of degree 4 in y was reached...
$endgroup$
– NoChance
2 hours ago
|
show 2 more comments
$begingroup$
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.
answered 3 hours ago
Shamim AkhtarShamim Akhtar
707
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
3 hours ago
$begingroup$
You apologised in different sentences for all answers 😂😂
$endgroup$
– Shamim Akhtar
3 hours ago
1
$begingroup$
Yup. That makes me sound like an actual human.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
I am not sure you can get a solution this way.
$endgroup$
– NoChance
2 hours ago
add a comment |
$begingroup$
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.
answered 3 hours ago
Shamim AkhtarShamim Akhtar
707
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
3 hours ago
$begingroup$
You apologised in different sentences for all answers 😂😂
$endgroup$
– Shamim Akhtar
3 hours ago
1
$begingroup$
Yup. That makes me sound like an actual human.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
I am not sure you can get a solution this way.
$endgroup$
– NoChance
2 hours ago
add a comment |
$begingroup$
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.
answered 3 hours ago
Shamim AkhtarShamim Akhtar
707
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.
$endgroup$
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.
answered 3 hours ago
Shamim AkhtarShamim Akhtar
707
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.
answered 3 hours ago
Shamim AkhtarShamim Akhtar
707
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.
answered 3 hours ago
Shamim AkhtarShamim Akhtar
707
answered 3 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.
Check out our Code of Conduct.
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
3 hours ago
$begingroup$
You apologised in different sentences for all answers 😂😂
$endgroup$
– Shamim Akhtar
3 hours ago
1
$begingroup$
Yup. That makes me sound like an actual human.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
I am not sure you can get a solution this way.
$endgroup$
– NoChance
2 hours ago
add a comment |
1
$begingroup$
I typed the problem wrong. Sorry for the inconvenience.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
You apologised in different sentences for all answers 😂😂
$endgroup$
– Shamim Akhtar
3 hours ago
1
$begingroup$
Yup. That makes me sound like an actual human.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
I am not sure you can get a solution this way.
$endgroup$
– NoChance
2 hours ago
1
1
$begingroup$
I typed the problem wrong. Sorry for the inconvenience.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
I typed the problem wrong. Sorry for the inconvenience.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
You apologised in different sentences for all answers 😂😂
$endgroup$
– Shamim Akhtar
3 hours ago
$begingroup$
You apologised in different sentences for all answers 😂😂
$endgroup$
– Shamim Akhtar
3 hours ago
1
1
$begingroup$
Yup. That makes me sound like an actual human.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
Yup. That makes me sound like an actual human.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
I am not sure you can get a solution this way.
$endgroup$
– NoChance
2 hours ago
$begingroup$
I am not sure you can get a solution this way.
$endgroup$
– NoChance
2 hours ago
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 $y = 1$.
answered 3 hours ago
Robert IsraelRobert Israel
328k23216469
$endgroup$
$begingroup$
Could you wait for me a little bit? I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
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 $y = 1$.
answered 3 hours ago
Robert IsraelRobert Israel
328k23216469
$endgroup$
$begingroup$
Could you wait for me a little bit? I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
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 $y = 1$.
answered 3 hours ago
Robert IsraelRobert Israel
328k23216469
$endgroup$
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 $y = 1$.
answered 3 hours ago
Robert IsraelRobert Israel
328k23216469
edited 1 hour ago
answered 3 hours ago
Robert IsraelRobert Israel
328k23216469
answered 3 hours ago
Robert IsraelRobert Israel
328k23216469
answered 3 hours ago
Robert IsraelRobert Israel
328k23216469
328k23216469
$begingroup$
Could you wait for me a little bit? I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
add a comment |
$begingroup$
Could you wait for me a little bit? I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
Could you wait for me a little bit? I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
$begingroup$
Could you wait for me a little bit? I typed the problem wrong.
$endgroup$
– Lê Thành Đạt
3 hours ago
add a comment |
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.
Lê Thành Đạt is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Only for $$yne 2$$
$endgroup$
– Dr. Sonnhard Graubner
3 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
3 hours ago
$begingroup$
But you forgot this to say.
$endgroup$
– Dr. Sonnhard Graubner
3 hours ago
$begingroup$
@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
3 hours ago
$begingroup$
Well.... uh.... (I'm sorry.)
$endgroup$
– Lê Thành Đạt
3 hours ago