Mixing
How do I solve this algebra problem
Clash Royale CLAN TAG#URR8PPP
up vote
6
down vote
favorite
The question goes, solve in real number.
$x^5 - 5 x^3y - 5x^2 + 5xy^2 + 5y = 16 tag1$
$ y^5 + 5xy^3 + 5y^2 + 5x^2y + 5x = -57 tag2$
I tried simplifying the first equation to,
$$ x^5 + 5left[ left(xy+1 right) left( y - x^2 right) right] = 16 $$
and second equation to,
$$ y^5 + 5 left[ left(xy+1 right) left( y^2 + x right) right] = -57$$
I know not much efforts shown, but this is where I'm stuck. Any hints on where do I go from here?
Thanks
algebra-precalculus systems-of-equations nonlinear-system
edited Jul 28 at 18:45
Batominovski
23k22777
asked Jul 28 at 17:45
William
730214
 |Â
show 3 more comments
up vote
6
down vote
favorite
The question goes, solve in real number.
$x^5 - 5 x^3y - 5x^2 + 5xy^2 + 5y = 16 tag1$
$ y^5 + 5xy^3 + 5y^2 + 5x^2y + 5x = -57 tag2$
I tried simplifying the first equation to,
$$ x^5 + 5left[ left(xy+1 right) left( y - x^2 right) right] = 16 $$
and second equation to,
$$ y^5 + 5 left[ left(xy+1 right) left( y^2 + x right) right] = -57$$
I know not much efforts shown, but this is where I'm stuck. Any hints on where do I go from here?
Thanks
algebra-precalculus systems-of-equations nonlinear-system
edited Jul 28 at 18:45
Batominovski
23k22777
asked Jul 28 at 17:45
William
730214
2
Where is this problem from? Using a basic Bezout bound, there could be $25$ solutions.
– Michael Burr
Jul 28 at 17:53
@MichaelBurr I'm guessing this is contest/contest prep; there is likely no way to "solve for" the eventual solution, but uniqueness can be inferred from inequalities, and the numbers are pleasant. Anyway, in the third quadrant, the curves are apparently separated by the ray $y=-x,$ I imagine that observation can be firmed up. Then just the fourth quadrant remains
– Will Jagy
Jul 28 at 18:16
@i.m.soloveichik My man, what sorcery did you use?
– William
Jul 28 at 18:20
Oops I made an error--the answer given is incorrect and there may be more than one solution. I determined the real solutions to the resultant of the two polynomials.
– i. m. soloveichik
Jul 28 at 18:23
5
Apparently, $(x,y)=(1,-2)$ is the unique real solution.
– Batominovski
Jul 28 at 18:40
 |Â
show 3 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
The question goes, solve in real number.
$x^5 - 5 x^3y - 5x^2 + 5xy^2 + 5y = 16 tag1$
$ y^5 + 5xy^3 + 5y^2 + 5x^2y + 5x = -57 tag2$
I tried simplifying the first equation to,
$$ x^5 + 5left[ left(xy+1 right) left( y - x^2 right) right] = 16 $$
and second equation to,
$$ y^5 + 5 left[ left(xy+1 right) left( y^2 + x right) right] = -57$$
I know not much efforts shown, but this is where I'm stuck. Any hints on where do I go from here?
Thanks
algebra-precalculus systems-of-equations nonlinear-system
edited Jul 28 at 18:45
Batominovski
23k22777
asked Jul 28 at 17:45
William
730214
The question goes, solve in real number.
$x^5 - 5 x^3y - 5x^2 + 5xy^2 + 5y = 16 tag1$
$ y^5 + 5xy^3 + 5y^2 + 5x^2y + 5x = -57 tag2$
I tried simplifying the first equation to,
$$ x^5 + 5left[ left(xy+1 right) left( y - x^2 right) right] = 16 $$
and second equation to,
$$ y^5 + 5 left[ left(xy+1 right) left( y^2 + x right) right] = -57$$
I know not much efforts shown, but this is where I'm stuck. Any hints on where do I go from here?
Thanks
algebra-precalculus systems-of-equations nonlinear-system
edited Jul 28 at 18:45
Batominovski
23k22777
asked Jul 28 at 17:45
William
730214
edited Jul 28 at 18:45
Batominovski
23k22777
edited Jul 28 at 18:45
Batominovski
23k22777
edited Jul 28 at 18:45
Batominovski
23k22777
23k22777
asked Jul 28 at 17:45
William
730214
asked Jul 28 at 17:45
William
730214
asked Jul 28 at 17:45
William
730214
730214
2
Where is this problem from? Using a basic Bezout bound, there could be $25$ solutions.
– Michael Burr
Jul 28 at 17:53
@MichaelBurr I'm guessing this is contest/contest prep; there is likely no way to "solve for" the eventual solution, but uniqueness can be inferred from inequalities, and the numbers are pleasant. Anyway, in the third quadrant, the curves are apparently separated by the ray $y=-x,$ I imagine that observation can be firmed up. Then just the fourth quadrant remains
– Will Jagy
Jul 28 at 18:16
@i.m.soloveichik My man, what sorcery did you use?
– William
Jul 28 at 18:20
Oops I made an error--the answer given is incorrect and there may be more than one solution. I determined the real solutions to the resultant of the two polynomials.
– i. m. soloveichik
Jul 28 at 18:23
5
Apparently, $(x,y)=(1,-2)$ is the unique real solution.
– Batominovski
Jul 28 at 18:40
 |Â
show 3 more comments
2
Where is this problem from? Using a basic Bezout bound, there could be $25$ solutions.
– Michael Burr
Jul 28 at 17:53
@MichaelBurr I'm guessing this is contest/contest prep; there is likely no way to "solve for" the eventual solution, but uniqueness can be inferred from inequalities, and the numbers are pleasant. Anyway, in the third quadrant, the curves are apparently separated by the ray $y=-x,$ I imagine that observation can be firmed up. Then just the fourth quadrant remains
– Will Jagy
Jul 28 at 18:16
@i.m.soloveichik My man, what sorcery did you use?
– William
Jul 28 at 18:20
Oops I made an error--the answer given is incorrect and there may be more than one solution. I determined the real solutions to the resultant of the two polynomials.
– i. m. soloveichik
Jul 28 at 18:23
5
Apparently, $(x,y)=(1,-2)$ is the unique real solution.
– Batominovski
Jul 28 at 18:40
2
2
Where is this problem from? Using a basic Bezout bound, there could be $25$ solutions.
– Michael Burr
Jul 28 at 17:53
Where is this problem from? Using a basic Bezout bound, there could be $25$ solutions.
– Michael Burr
Jul 28 at 17:53
@MichaelBurr I'm guessing this is contest/contest prep; there is likely no way to "solve for" the eventual solution, but uniqueness can be inferred from inequalities, and the numbers are pleasant. Anyway, in the third quadrant, the curves are apparently separated by the ray $y=-x,$ I imagine that observation can be firmed up. Then just the fourth quadrant remains
– Will Jagy
Jul 28 at 18:16
@MichaelBurr I'm guessing this is contest/contest prep; there is likely no way to "solve for" the eventual solution, but uniqueness can be inferred from inequalities, and the numbers are pleasant. Anyway, in the third quadrant, the curves are apparently separated by the ray $y=-x,$ I imagine that observation can be firmed up. Then just the fourth quadrant remains
– Will Jagy
Jul 28 at 18:16
@i.m.soloveichik My man, what sorcery did you use?
– William
Jul 28 at 18:20
@i.m.soloveichik My man, what sorcery did you use?
– William
Jul 28 at 18:20
Oops I made an error--the answer given is incorrect and there may be more than one solution. I determined the real solutions to the resultant of the two polynomials.
– i. m. soloveichik
Jul 28 at 18:23
Oops I made an error--the answer given is incorrect and there may be more than one solution. I determined the real solutions to the resultant of the two polynomials.
– i. m. soloveichik
Jul 28 at 18:23
5
5
Apparently, $(x,y)=(1,-2)$ is the unique real solution.
– Batominovski
Jul 28 at 18:40
Apparently, $(x,y)=(1,-2)$ is the unique real solution.
– Batominovski
Jul 28 at 18:40
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
0
down vote
By adding these two equations I ended up in a quite symmetric term
$$x^5+y^5+5[(1+xy)(y+x)(y-x+1)]~=~-41$$
I would also recommend to use the fact that you can write any $(x+y)^n$ only in terms of $xy$ and $(x+y)$ and the two $x^n,y^n$ like
$$(x+y)^3~=~x^3+y^3+3xy(x+y)$$
Maybe someone can proceed form there.
answered Jul 28 at 18:20
mrtaurho
703118
add a comment |Â
up vote
0
down vote
COMMENT.- It seems that Batominovski's comment about $(1,-2)$ is true. One has $$x^5+5(xy+1)(y-x^2)=16\y^5+5(xy+1)(y^2+x)=-57$$ From which
$$fracy^2+xy-x^2=fracy^5+57x^5-16qquad(1)$$
A necessary condition for solutions $(x,y)$ is equality $(1)$. However it is not suffisant because if not it would be an infinity of solutions. Making separately
$$fracy^2+xy-x^2=aqquad(2)$$ $$fracy^5+57x^5-16=aqquad(3)$$ we have in $(2)$ a conic, hyperbola, ellipse, circle ($a=1$) and two lines ($a=-1$) while in $(3)$ there is a quintic.The corresponding graphics are suggestive in order to search solutions. In any case the graphics of the given equations (in the attached figure green and black respectively) also give $(1,-2)$ as probably only real solution.
answered Jul 29 at 13:56
Piquito
17.3k31234
It's exactly the WA's solution! I have a solution of this problem, but it's a very ugly solution. Mysterious problem!
– Michael Rozenberg
Jul 29 at 14:11
There should be no doubt about the fact that the non-real solutions are quite "ugly". However, I think, it would be very good to expose them explicitly.
– Piquito
Jul 29 at 14:38
@Batominosvki: your solution is certainly formed by the roots of $$(apmsqrta^2-2ax^2-x+57)^5-2ax^5+32a=0$$ for a certain value of the parameter 2a (instead of a), is not it?
– Piquito
Jul 29 at 15:24
What software did you use to plot the curves?
– saulspatz
Jul 29 at 22:08
Just write the equations in desmos.com/calculator
– Piquito
Jul 30 at 1:24
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
By adding these two equations I ended up in a quite symmetric term
$$x^5+y^5+5[(1+xy)(y+x)(y-x+1)]~=~-41$$
I would also recommend to use the fact that you can write any $(x+y)^n$ only in terms of $xy$ and $(x+y)$ and the two $x^n,y^n$ like
$$(x+y)^3~=~x^3+y^3+3xy(x+y)$$
Maybe someone can proceed form there.
answered Jul 28 at 18:20
mrtaurho
703118
add a comment |Â
up vote
0
down vote
By adding these two equations I ended up in a quite symmetric term
$$x^5+y^5+5[(1+xy)(y+x)(y-x+1)]~=~-41$$
I would also recommend to use the fact that you can write any $(x+y)^n$ only in terms of $xy$ and $(x+y)$ and the two $x^n,y^n$ like
$$(x+y)^3~=~x^3+y^3+3xy(x+y)$$
Maybe someone can proceed form there.
answered Jul 28 at 18:20
mrtaurho
703118
add a comment |Â
up vote
0
down vote
up vote
0
down vote
By adding these two equations I ended up in a quite symmetric term
$$x^5+y^5+5[(1+xy)(y+x)(y-x+1)]~=~-41$$
I would also recommend to use the fact that you can write any $(x+y)^n$ only in terms of $xy$ and $(x+y)$ and the two $x^n,y^n$ like
$$(x+y)^3~=~x^3+y^3+3xy(x+y)$$
Maybe someone can proceed form there.
answered Jul 28 at 18:20
mrtaurho
703118
By adding these two equations I ended up in a quite symmetric term
$$x^5+y^5+5[(1+xy)(y+x)(y-x+1)]~=~-41$$
I would also recommend to use the fact that you can write any $(x+y)^n$ only in terms of $xy$ and $(x+y)$ and the two $x^n,y^n$ like
$$(x+y)^3~=~x^3+y^3+3xy(x+y)$$
Maybe someone can proceed form there.
answered Jul 28 at 18:20
mrtaurho
703118
edited Jul 28 at 19:21
answered Jul 28 at 18:20
mrtaurho
703118
answered Jul 28 at 18:20
mrtaurho
703118
answered Jul 28 at 18:20
mrtaurho
703118
703118
add a comment |Â
add a comment |Â
up vote
0
down vote
COMMENT.- It seems that Batominovski's comment about $(1,-2)$ is true. One has $$x^5+5(xy+1)(y-x^2)=16\y^5+5(xy+1)(y^2+x)=-57$$ From which
$$fracy^2+xy-x^2=fracy^5+57x^5-16qquad(1)$$
A necessary condition for solutions $(x,y)$ is equality $(1)$. However it is not suffisant because if not it would be an infinity of solutions. Making separately
$$fracy^2+xy-x^2=aqquad(2)$$ $$fracy^5+57x^5-16=aqquad(3)$$ we have in $(2)$ a conic, hyperbola, ellipse, circle ($a=1$) and two lines ($a=-1$) while in $(3)$ there is a quintic.The corresponding graphics are suggestive in order to search solutions. In any case the graphics of the given equations (in the attached figure green and black respectively) also give $(1,-2)$ as probably only real solution.
answered Jul 29 at 13:56
Piquito
17.3k31234
It's exactly the WA's solution! I have a solution of this problem, but it's a very ugly solution. Mysterious problem!
– Michael Rozenberg
Jul 29 at 14:11
There should be no doubt about the fact that the non-real solutions are quite "ugly". However, I think, it would be very good to expose them explicitly.
– Piquito
Jul 29 at 14:38
@Batominosvki: your solution is certainly formed by the roots of $$(apmsqrta^2-2ax^2-x+57)^5-2ax^5+32a=0$$ for a certain value of the parameter 2a (instead of a), is not it?
– Piquito
Jul 29 at 15:24
What software did you use to plot the curves?
– saulspatz
Jul 29 at 22:08
Just write the equations in desmos.com/calculator
– Piquito
Jul 30 at 1:24
add a comment |Â
up vote
0
down vote
COMMENT.- It seems that Batominovski's comment about $(1,-2)$ is true. One has $$x^5+5(xy+1)(y-x^2)=16\y^5+5(xy+1)(y^2+x)=-57$$ From which
$$fracy^2+xy-x^2=fracy^5+57x^5-16qquad(1)$$
A necessary condition for solutions $(x,y)$ is equality $(1)$. However it is not suffisant because if not it would be an infinity of solutions. Making separately
$$fracy^2+xy-x^2=aqquad(2)$$ $$fracy^5+57x^5-16=aqquad(3)$$ we have in $(2)$ a conic, hyperbola, ellipse, circle ($a=1$) and two lines ($a=-1$) while in $(3)$ there is a quintic.The corresponding graphics are suggestive in order to search solutions. In any case the graphics of the given equations (in the attached figure green and black respectively) also give $(1,-2)$ as probably only real solution.
answered Jul 29 at 13:56
Piquito
17.3k31234
It's exactly the WA's solution! I have a solution of this problem, but it's a very ugly solution. Mysterious problem!
– Michael Rozenberg
Jul 29 at 14:11
There should be no doubt about the fact that the non-real solutions are quite "ugly". However, I think, it would be very good to expose them explicitly.
– Piquito
Jul 29 at 14:38
@Batominosvki: your solution is certainly formed by the roots of $$(apmsqrta^2-2ax^2-x+57)^5-2ax^5+32a=0$$ for a certain value of the parameter 2a (instead of a), is not it?
– Piquito
Jul 29 at 15:24
What software did you use to plot the curves?
– saulspatz
Jul 29 at 22:08
Just write the equations in desmos.com/calculator
– Piquito
Jul 30 at 1:24
add a comment |Â
up vote
0
down vote
up vote
0
down vote
COMMENT.- It seems that Batominovski's comment about $(1,-2)$ is true. One has $$x^5+5(xy+1)(y-x^2)=16\y^5+5(xy+1)(y^2+x)=-57$$ From which
$$fracy^2+xy-x^2=fracy^5+57x^5-16qquad(1)$$
A necessary condition for solutions $(x,y)$ is equality $(1)$. However it is not suffisant because if not it would be an infinity of solutions. Making separately
$$fracy^2+xy-x^2=aqquad(2)$$ $$fracy^5+57x^5-16=aqquad(3)$$ we have in $(2)$ a conic, hyperbola, ellipse, circle ($a=1$) and two lines ($a=-1$) while in $(3)$ there is a quintic.The corresponding graphics are suggestive in order to search solutions. In any case the graphics of the given equations (in the attached figure green and black respectively) also give $(1,-2)$ as probably only real solution.
answered Jul 29 at 13:56
Piquito
17.3k31234
COMMENT.- It seems that Batominovski's comment about $(1,-2)$ is true. One has $$x^5+5(xy+1)(y-x^2)=16\y^5+5(xy+1)(y^2+x)=-57$$ From which
$$fracy^2+xy-x^2=fracy^5+57x^5-16qquad(1)$$
A necessary condition for solutions $(x,y)$ is equality $(1)$. However it is not suffisant because if not it would be an infinity of solutions. Making separately
$$fracy^2+xy-x^2=aqquad(2)$$ $$fracy^5+57x^5-16=aqquad(3)$$ we have in $(2)$ a conic, hyperbola, ellipse, circle ($a=1$) and two lines ($a=-1$) while in $(3)$ there is a quintic.The corresponding graphics are suggestive in order to search solutions. In any case the graphics of the given equations (in the attached figure green and black respectively) also give $(1,-2)$ as probably only real solution.
answered Jul 29 at 13:56
Piquito
17.3k31234
answered Jul 29 at 13:56
Piquito
17.3k31234
answered Jul 29 at 13:56
Piquito
17.3k31234
answered Jul 29 at 13:56
Piquito
17.3k31234
17.3k31234
It's exactly the WA's solution! I have a solution of this problem, but it's a very ugly solution. Mysterious problem!
– Michael Rozenberg
Jul 29 at 14:11
There should be no doubt about the fact that the non-real solutions are quite "ugly". However, I think, it would be very good to expose them explicitly.
– Piquito
Jul 29 at 14:38
@Batominosvki: your solution is certainly formed by the roots of $$(apmsqrta^2-2ax^2-x+57)^5-2ax^5+32a=0$$ for a certain value of the parameter 2a (instead of a), is not it?
– Piquito
Jul 29 at 15:24
What software did you use to plot the curves?
– saulspatz
Jul 29 at 22:08
Just write the equations in desmos.com/calculator
– Piquito
Jul 30 at 1:24
add a comment |Â
It's exactly the WA's solution! I have a solution of this problem, but it's a very ugly solution. Mysterious problem!
– Michael Rozenberg
Jul 29 at 14:11
There should be no doubt about the fact that the non-real solutions are quite "ugly". However, I think, it would be very good to expose them explicitly.
– Piquito
Jul 29 at 14:38
@Batominosvki: your solution is certainly formed by the roots of $$(apmsqrta^2-2ax^2-x+57)^5-2ax^5+32a=0$$ for a certain value of the parameter 2a (instead of a), is not it?
– Piquito
Jul 29 at 15:24
What software did you use to plot the curves?
– saulspatz
Jul 29 at 22:08
Just write the equations in desmos.com/calculator
– Piquito
Jul 30 at 1:24
It's exactly the WA's solution! I have a solution of this problem, but it's a very ugly solution. Mysterious problem!
– Michael Rozenberg
Jul 29 at 14:11
It's exactly the WA's solution! I have a solution of this problem, but it's a very ugly solution. Mysterious problem!
– Michael Rozenberg
Jul 29 at 14:11
There should be no doubt about the fact that the non-real solutions are quite "ugly". However, I think, it would be very good to expose them explicitly.
– Piquito
Jul 29 at 14:38
There should be no doubt about the fact that the non-real solutions are quite "ugly". However, I think, it would be very good to expose them explicitly.
– Piquito
Jul 29 at 14:38
@Batominosvki: your solution is certainly formed by the roots of $$(apmsqrta^2-2ax^2-x+57)^5-2ax^5+32a=0$$ for a certain value of the parameter 2a (instead of a), is not it?
– Piquito
Jul 29 at 15:24
@Batominosvki: your solution is certainly formed by the roots of $$(apmsqrta^2-2ax^2-x+57)^5-2ax^5+32a=0$$ for a certain value of the parameter 2a (instead of a), is not it?
– Piquito
Jul 29 at 15:24
What software did you use to plot the curves?
– saulspatz
Jul 29 at 22:08
What software did you use to plot the curves?
– saulspatz
Jul 29 at 22:08
Just write the equations in desmos.com/calculator
– Piquito
Jul 30 at 1:24
Just write the equations in desmos.com/calculator
– Piquito
Jul 30 at 1:24
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865436%2fhow-do-i-solve-this-algebra-problem%23new-answer', 'question_page');
);
Post as a guest
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
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
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
Where is this problem from? Using a basic Bezout bound, there could be $25$ solutions.
– Michael Burr
Jul 28 at 17:53
@MichaelBurr I'm guessing this is contest/contest prep; there is likely no way to "solve for" the eventual solution, but uniqueness can be inferred from inequalities, and the numbers are pleasant. Anyway, in the third quadrant, the curves are apparently separated by the ray $y=-x,$ I imagine that observation can be firmed up. Then just the fourth quadrant remains
– Will Jagy
Jul 28 at 18:16
@i.m.soloveichik My man, what sorcery did you use?
– William
Jul 28 at 18:20
Oops I made an error--the answer given is incorrect and there may be more than one solution. I determined the real solutions to the resultant of the two polynomials.
– i. m. soloveichik
Jul 28 at 18:23
5
Apparently, $(x,y)=(1,-2)$ is the unique real solution.
– Batominovski
Jul 28 at 18:40