Solve the following system of equations - (2).How do we solve the system of equations?18-1-2013Solve the...
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Solve the following system of equations - (2).
How do we solve the system of equations?18-1-2013Solve the following system of equationsHow to solve this system of equations.Solve the non-linear system of equationsSolve the system of nonlinear equationsHow do you solve the following nonlinear equations?System of equations with special substitutionIdeas to solve the following system of equationsSolve the following system of equations (1)Solve the following system of equations - (3)
$begingroup$
(a) Solve the following system of equations:
$$begin{cases}
x(x + 1) + dfrac{1}{y}left(dfrac{1}{y} + 1right) = 4\
dfrac1{y^2}(x+1) + x^2left(dfrac{1}{y} + 1right) = 4
end{cases}$$
(b) Solve the following system of equations:
$$begin{cases}
x(x + 1) + dfrac{1}{y}left(dfrac{1}{y} + 1right) = 4\
dfrac{1}{y}(x^2 + 1) + xleft(dfrac{1}{y^2} + 1right) = 4
end{cases}
$$
This problem comes from a competition I participated this morning (19/3/2019). And the problems were hard. Out of 11 problems, I only did 7. That counts this one.
contest-math systems-of-equations rational-functions
edited 2 days ago
Harry Peter
5,50911439
asked Mar 19 at 4:48
Lê Thành ĐạtLê Thành Đạt
37013
$endgroup$
|
show 2 more comments
$begingroup$
(a) Solve the following system of equations:
$$begin{cases}
x(x + 1) + dfrac{1}{y}left(dfrac{1}{y} + 1right) = 4\
dfrac1{y^2}(x+1) + x^2left(dfrac{1}{y} + 1right) = 4
end{cases}$$
(b) Solve the following system of equations:
$$begin{cases}
x(x + 1) + dfrac{1}{y}left(dfrac{1}{y} + 1right) = 4\
dfrac{1}{y}(x^2 + 1) + xleft(dfrac{1}{y^2} + 1right) = 4
end{cases}
$$
This problem comes from a competition I participated this morning (19/3/2019). And the problems were hard. Out of 11 problems, I only did 7. That counts this one.
contest-math systems-of-equations rational-functions
edited 2 days ago
Harry Peter
5,50911439
asked Mar 19 at 4:48
Lê Thành ĐạtLê Thành Đạt
37013
$endgroup$
$begingroup$
So you managed to solve this problem (it seems like it, from your last sentence)?
$endgroup$
– Minus One-Twelfth
Mar 19 at 4:53
$begingroup$
I did manage to solve it. I just want to see another, shorter solution.
$endgroup$
– Lê Thành Đạt
Mar 19 at 4:53
$begingroup$
Ah I see. Congratulations!
$endgroup$
– Minus One-Twelfth
Mar 19 at 4:54
1
$begingroup$
Thanks! The next question will be the one that I couldn't do. It's geometry.
$endgroup$
– Lê Thành Đạt
Mar 19 at 4:56
2
$begingroup$
How can we know whether we have a shorter solution than yours, if we don't know yours?
$endgroup$
– Gerry Myerson
Mar 19 at 5:32
|
show 2 more comments
$begingroup$
(a) Solve the following system of equations:
$$begin{cases}
x(x + 1) + dfrac{1}{y}left(dfrac{1}{y} + 1right) = 4\
dfrac1{y^2}(x+1) + x^2left(dfrac{1}{y} + 1right) = 4
end{cases}$$
(b) Solve the following system of equations:
$$begin{cases}
x(x + 1) + dfrac{1}{y}left(dfrac{1}{y} + 1right) = 4\
dfrac{1}{y}(x^2 + 1) + xleft(dfrac{1}{y^2} + 1right) = 4
end{cases}
$$
This problem comes from a competition I participated this morning (19/3/2019). And the problems were hard. Out of 11 problems, I only did 7. That counts this one.
contest-math systems-of-equations rational-functions
edited 2 days ago
Harry Peter
5,50911439
asked Mar 19 at 4:48
Lê Thành ĐạtLê Thành Đạt
37013
$endgroup$
(a) Solve the following system of equations:
$$begin{cases}
x(x + 1) + dfrac{1}{y}left(dfrac{1}{y} + 1right) = 4\
dfrac1{y^2}(x+1) + x^2left(dfrac{1}{y} + 1right) = 4
end{cases}$$
(b) Solve the following system of equations:
$$begin{cases}
x(x + 1) + dfrac{1}{y}left(dfrac{1}{y} + 1right) = 4\
dfrac{1}{y}(x^2 + 1) + xleft(dfrac{1}{y^2} + 1right) = 4
end{cases}
$$
This problem comes from a competition I participated this morning (19/3/2019). And the problems were hard. Out of 11 problems, I only did 7. That counts this one.
contest-math systems-of-equations rational-functions
contest-math systems-of-equations rational-functions
edited 2 days ago
Harry Peter
5,50911439
asked Mar 19 at 4:48
Lê Thành ĐạtLê Thành Đạt
37013
edited 2 days ago
Harry Peter
5,50911439
asked Mar 19 at 4:48
Lê Thành ĐạtLê Thành Đạt
37013
edited 2 days ago
Harry Peter
5,50911439
edited 2 days ago
Harry Peter
5,50911439
edited 2 days ago
Harry Peter
5,50911439
5,50911439
asked Mar 19 at 4:48
Lê Thành ĐạtLê Thành Đạt
37013
asked Mar 19 at 4:48
Lê Thành ĐạtLê Thành Đạt
37013
asked Mar 19 at 4:48
Lê Thành ĐạtLê Thành Đạt
37013
37013
$begingroup$
So you managed to solve this problem (it seems like it, from your last sentence)?
$endgroup$
– Minus One-Twelfth
Mar 19 at 4:53
$begingroup$
I did manage to solve it. I just want to see another, shorter solution.
$endgroup$
– Lê Thành Đạt
Mar 19 at 4:53
$begingroup$
Ah I see. Congratulations!
$endgroup$
– Minus One-Twelfth
Mar 19 at 4:54
1
$begingroup$
Thanks! The next question will be the one that I couldn't do. It's geometry.
$endgroup$
– Lê Thành Đạt
Mar 19 at 4:56
2
$begingroup$
How can we know whether we have a shorter solution than yours, if we don't know yours?
$endgroup$
– Gerry Myerson
Mar 19 at 5:32
|
show 2 more comments
$begingroup$
So you managed to solve this problem (it seems like it, from your last sentence)?
$endgroup$
– Minus One-Twelfth
Mar 19 at 4:53
$begingroup$
I did manage to solve it. I just want to see another, shorter solution.
$endgroup$
– Lê Thành Đạt
Mar 19 at 4:53
$begingroup$
Ah I see. Congratulations!
$endgroup$
– Minus One-Twelfth
Mar 19 at 4:54
1
$begingroup$
Thanks! The next question will be the one that I couldn't do. It's geometry.
$endgroup$
– Lê Thành Đạt
Mar 19 at 4:56
2
$begingroup$
How can we know whether we have a shorter solution than yours, if we don't know yours?
$endgroup$
– Gerry Myerson
Mar 19 at 5:32
$begingroup$
So you managed to solve this problem (it seems like it, from your last sentence)?
$endgroup$
– Minus One-Twelfth
Mar 19 at 4:53
$begingroup$
So you managed to solve this problem (it seems like it, from your last sentence)?
$endgroup$
– Minus One-Twelfth
Mar 19 at 4:53
$begingroup$
I did manage to solve it. I just want to see another, shorter solution.
$endgroup$
– Lê Thành Đạt
Mar 19 at 4:53
$begingroup$
I did manage to solve it. I just want to see another, shorter solution.
$endgroup$
– Lê Thành Đạt
Mar 19 at 4:53
$begingroup$
Ah I see. Congratulations!
$endgroup$
– Minus One-Twelfth
Mar 19 at 4:54
$begingroup$
Ah I see. Congratulations!
$endgroup$
– Minus One-Twelfth
Mar 19 at 4:54
1
1
$begingroup$
Thanks! The next question will be the one that I couldn't do. It's geometry.
$endgroup$
– Lê Thành Đạt
Mar 19 at 4:56
$begingroup$
Thanks! The next question will be the one that I couldn't do. It's geometry.
$endgroup$
– Lê Thành Đạt
Mar 19 at 4:56
2
2
$begingroup$
How can we know whether we have a shorter solution than yours, if we don't know yours?
$endgroup$
– Gerry Myerson
Mar 19 at 5:32
$begingroup$
How can we know whether we have a shorter solution than yours, if we don't know yours?
$endgroup$
– Gerry Myerson
Mar 19 at 5:32
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
We will substitute $$y=tx$$ in the second equation, then we get
$$frac{x}{t^2x^2}+frac{1}{t^2x^2}+frac{x^2}{tx}+x^2-4=0$$ and this is
$$(tx^2+2tx+x+1)(tx^2-2tx+1)=0$$ so we get
$$tx^2+2tx+x+1=0$$
or
$$tx^2-2tx+1=0$$
which can be solved for $t$.
answered Mar 20 at 16:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
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1 Answer
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$begingroup$
We will substitute $$y=tx$$ in the second equation, then we get
$$frac{x}{t^2x^2}+frac{1}{t^2x^2}+frac{x^2}{tx}+x^2-4=0$$ and this is
$$(tx^2+2tx+x+1)(tx^2-2tx+1)=0$$ so we get
$$tx^2+2tx+x+1=0$$
or
$$tx^2-2tx+1=0$$
which can be solved for $t$.
answered Mar 20 at 16:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
We will substitute $$y=tx$$ in the second equation, then we get
$$frac{x}{t^2x^2}+frac{1}{t^2x^2}+frac{x^2}{tx}+x^2-4=0$$ and this is
$$(tx^2+2tx+x+1)(tx^2-2tx+1)=0$$ so we get
$$tx^2+2tx+x+1=0$$
or
$$tx^2-2tx+1=0$$
which can be solved for $t$.
answered Mar 20 at 16:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
We will substitute $$y=tx$$ in the second equation, then we get
$$frac{x}{t^2x^2}+frac{1}{t^2x^2}+frac{x^2}{tx}+x^2-4=0$$ and this is
$$(tx^2+2tx+x+1)(tx^2-2tx+1)=0$$ so we get
$$tx^2+2tx+x+1=0$$
or
$$tx^2-2tx+1=0$$
which can be solved for $t$.
answered Mar 20 at 16:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
We will substitute $$y=tx$$ in the second equation, then we get
$$frac{x}{t^2x^2}+frac{1}{t^2x^2}+frac{x^2}{tx}+x^2-4=0$$ and this is
$$(tx^2+2tx+x+1)(tx^2-2tx+1)=0$$ so we get
$$tx^2+2tx+x+1=0$$
or
$$tx^2-2tx+1=0$$
which can be solved for $t$.
answered Mar 20 at 16:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Mar 20 at 16:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Mar 20 at 16:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Mar 20 at 16:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
78.4k42867
add a comment |
add a comment |
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$begingroup$
So you managed to solve this problem (it seems like it, from your last sentence)?
$endgroup$
– Minus One-Twelfth
Mar 19 at 4:53
$begingroup$
I did manage to solve it. I just want to see another, shorter solution.
$endgroup$
– Lê Thành Đạt
Mar 19 at 4:53
$begingroup$
Ah I see. Congratulations!
$endgroup$
– Minus One-Twelfth
Mar 19 at 4:54
1
$begingroup$
Thanks! The next question will be the one that I couldn't do. It's geometry.
$endgroup$
– Lê Thành Đạt
Mar 19 at 4:56
2
$begingroup$
How can we know whether we have a shorter solution than yours, if we don't know yours?
$endgroup$
– Gerry Myerson
Mar 19 at 5:32