Mixing
Solution to a two-equations system.
$begingroup$
Consider the following system of two equations:
begin{cases}
delta = phi x^{phi-1}y^{1-phi} \ tag{1}
z = (1-phi)x^{phi}y^{-phi}
end{cases}
With $phi$ $in$ $(0,1)$.
To find the values $(x,y)$ that solve the system, I solve for $y$ in the first equation and obtain:
begin{equation}
y = Big(frac{delta}{phi}Big)^{frac{1}{1-phi}}xtag{2}
end{equation}
I then plug it in the second equation and obtain:
begin{equation}
z = (1-phi) Big(frac{delta}{phi}Big)^{frac{phi}{phi-1}} tag{3}
end{equation}
Where the unknowns cancel out. I have two related questions:
a) When equation (3) is satisfied, any combination of $(x,y)$ is a solution to the system. Correct?
b) When equation (3) is not satisfied, a solution does not exist. Correct?
algebra-precalculus proof-verification systems-of-equations
asked Jan 28 at 13:32
TeconTecon
174
$endgroup$
add a comment |
$begingroup$
Consider the following system of two equations:
begin{cases}
delta = phi x^{phi-1}y^{1-phi} \ tag{1}
z = (1-phi)x^{phi}y^{-phi}
end{cases}
With $phi$ $in$ $(0,1)$.
To find the values $(x,y)$ that solve the system, I solve for $y$ in the first equation and obtain:
begin{equation}
y = Big(frac{delta}{phi}Big)^{frac{1}{1-phi}}xtag{2}
end{equation}
I then plug it in the second equation and obtain:
begin{equation}
z = (1-phi) Big(frac{delta}{phi}Big)^{frac{phi}{phi-1}} tag{3}
end{equation}
Where the unknowns cancel out. I have two related questions:
a) When equation (3) is satisfied, any combination of $(x,y)$ is a solution to the system. Correct?
b) When equation (3) is not satisfied, a solution does not exist. Correct?
algebra-precalculus proof-verification systems-of-equations
asked Jan 28 at 13:32
TeconTecon
174
$endgroup$
add a comment |
$begingroup$
Consider the following system of two equations:
begin{cases}
delta = phi x^{phi-1}y^{1-phi} \ tag{1}
z = (1-phi)x^{phi}y^{-phi}
end{cases}
With $phi$ $in$ $(0,1)$.
To find the values $(x,y)$ that solve the system, I solve for $y$ in the first equation and obtain:
begin{equation}
y = Big(frac{delta}{phi}Big)^{frac{1}{1-phi}}xtag{2}
end{equation}
I then plug it in the second equation and obtain:
begin{equation}
z = (1-phi) Big(frac{delta}{phi}Big)^{frac{phi}{phi-1}} tag{3}
end{equation}
Where the unknowns cancel out. I have two related questions:
a) When equation (3) is satisfied, any combination of $(x,y)$ is a solution to the system. Correct?
b) When equation (3) is not satisfied, a solution does not exist. Correct?
algebra-precalculus proof-verification systems-of-equations
asked Jan 28 at 13:32
TeconTecon
174
$endgroup$
Consider the following system of two equations:
begin{cases}
delta = phi x^{phi-1}y^{1-phi} \ tag{1}
z = (1-phi)x^{phi}y^{-phi}
end{cases}
With $phi$ $in$ $(0,1)$.
To find the values $(x,y)$ that solve the system, I solve for $y$ in the first equation and obtain:
begin{equation}
y = Big(frac{delta}{phi}Big)^{frac{1}{1-phi}}xtag{2}
end{equation}
I then plug it in the second equation and obtain:
begin{equation}
z = (1-phi) Big(frac{delta}{phi}Big)^{frac{phi}{phi-1}} tag{3}
end{equation}
Where the unknowns cancel out. I have two related questions:
a) When equation (3) is satisfied, any combination of $(x,y)$ is a solution to the system. Correct?
b) When equation (3) is not satisfied, a solution does not exist. Correct?
algebra-precalculus proof-verification systems-of-equations
algebra-precalculus proof-verification systems-of-equations
asked Jan 28 at 13:32
TeconTecon
174
asked Jan 28 at 13:32
TeconTecon
174
edited Jan 28 at 14:32
Tecon
asked Jan 28 at 13:32
TeconTecon
174
asked Jan 28 at 13:32
TeconTecon
174
asked Jan 28 at 13:32
TeconTecon
174
174
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
a) no.
b) yes.
The system is "degenerate" in the sense that the equations are both in terms of $dfrac xy$. If we set $t:=dfrac xy$,
$$begin{cases}delta=phi t^{phi-1},\z=(1-phi)t^phi.end{cases}$$
This is a system of two equations in a single unknown. An immediate solution is obtained by taking the ratio,
$$t=frac{zphi}{delta(1-phi)}.$$ But for this solution to be correct, if needs to satisfy both equations, and a compatibility condition must be fulfilled. For instance, by eliminating $t$,
$$left(fracdeltaphiright)^phi=left(frac z{1-phi}right)^{phi-1},$$ which is analogous to your equation 3).
But, when the system is compatible, the solutions in $x,y$ are $y=tx$, where $x$ is arbitrary.
[Not discussing the singular cases.]
answered Jan 28 at 13:47
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
@Tecon: what do you think ?
$endgroup$
– Yves Daoust
Jan 28 at 14:59
$begingroup$
@Tecon $phi$ in $(0,1)$ has nothing special.
$endgroup$
– Yves Daoust
Jan 28 at 18:40
add a comment |
$begingroup$
Be careful, because the solution needs the restriction $phinotin{0,1}$.
But you are right, that or $phinotin{0,1}$ the second equation doesn't depend on $x$ or $y$. But that doesn't mean that any $(x,y)$ solves the whole system. You just don't get further restrictions on the solution of $(x,y)$. So, all $(x,y)$ such that $(2)$ holds are your solution for the case $phinotin{0,1}$.
But if $(3)$ leads to some contradition, then there are no solutions $(x,y)$ such that $(2)$ holds. Hence, there are no solutions if $phinotin{0,1}$.
Finally, you still have to check the cases $phi=0$ and $phi=1$.
answered Jan 28 at 13:45
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
$begingroup$
After your comment I edited the question. $phi$ is actually bounded between 0 and 1. With the extremes excluded.
$endgroup$
– Tecon
Jan 28 at 13:56
$begingroup$
I don't see why the solution needs that restriction on $phi$. Can you, please, explain?
$endgroup$
– Tecon
Jan 28 at 14:38
1
$begingroup$
Plug in $phi=0$ and $phi=1$ and you see what happens: The equations doesn't depend on $x$ and $y$ anymore.
$endgroup$
– Mundron Schmidt
Jan 28 at 14:49
$begingroup$
Ah ok. My bad I misread your comment. Thanks.
$endgroup$
– Tecon
Jan 28 at 14:52
add a comment |
Your Answer
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
a) no.
b) yes.
The system is "degenerate" in the sense that the equations are both in terms of $dfrac xy$. If we set $t:=dfrac xy$,
$$begin{cases}delta=phi t^{phi-1},\z=(1-phi)t^phi.end{cases}$$
This is a system of two equations in a single unknown. An immediate solution is obtained by taking the ratio,
$$t=frac{zphi}{delta(1-phi)}.$$ But for this solution to be correct, if needs to satisfy both equations, and a compatibility condition must be fulfilled. For instance, by eliminating $t$,
$$left(fracdeltaphiright)^phi=left(frac z{1-phi}right)^{phi-1},$$ which is analogous to your equation 3).
But, when the system is compatible, the solutions in $x,y$ are $y=tx$, where $x$ is arbitrary.
[Not discussing the singular cases.]
answered Jan 28 at 13:47
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
@Tecon: what do you think ?
$endgroup$
– Yves Daoust
Jan 28 at 14:59
$begingroup$
@Tecon $phi$ in $(0,1)$ has nothing special.
$endgroup$
– Yves Daoust
Jan 28 at 18:40
add a comment |
$begingroup$
a) no.
b) yes.
The system is "degenerate" in the sense that the equations are both in terms of $dfrac xy$. If we set $t:=dfrac xy$,
$$begin{cases}delta=phi t^{phi-1},\z=(1-phi)t^phi.end{cases}$$
This is a system of two equations in a single unknown. An immediate solution is obtained by taking the ratio,
$$t=frac{zphi}{delta(1-phi)}.$$ But for this solution to be correct, if needs to satisfy both equations, and a compatibility condition must be fulfilled. For instance, by eliminating $t$,
$$left(fracdeltaphiright)^phi=left(frac z{1-phi}right)^{phi-1},$$ which is analogous to your equation 3).
But, when the system is compatible, the solutions in $x,y$ are $y=tx$, where $x$ is arbitrary.
[Not discussing the singular cases.]
answered Jan 28 at 13:47
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
@Tecon: what do you think ?
$endgroup$
– Yves Daoust
Jan 28 at 14:59
$begingroup$
@Tecon $phi$ in $(0,1)$ has nothing special.
$endgroup$
– Yves Daoust
Jan 28 at 18:40
add a comment |
$begingroup$
a) no.
b) yes.
The system is "degenerate" in the sense that the equations are both in terms of $dfrac xy$. If we set $t:=dfrac xy$,
$$begin{cases}delta=phi t^{phi-1},\z=(1-phi)t^phi.end{cases}$$
This is a system of two equations in a single unknown. An immediate solution is obtained by taking the ratio,
$$t=frac{zphi}{delta(1-phi)}.$$ But for this solution to be correct, if needs to satisfy both equations, and a compatibility condition must be fulfilled. For instance, by eliminating $t$,
$$left(fracdeltaphiright)^phi=left(frac z{1-phi}right)^{phi-1},$$ which is analogous to your equation 3).
But, when the system is compatible, the solutions in $x,y$ are $y=tx$, where $x$ is arbitrary.
[Not discussing the singular cases.]
answered Jan 28 at 13:47
Yves DaoustYves Daoust
131k676229
$endgroup$
a) no.
b) yes.
The system is "degenerate" in the sense that the equations are both in terms of $dfrac xy$. If we set $t:=dfrac xy$,
$$begin{cases}delta=phi t^{phi-1},\z=(1-phi)t^phi.end{cases}$$
This is a system of two equations in a single unknown. An immediate solution is obtained by taking the ratio,
$$t=frac{zphi}{delta(1-phi)}.$$ But for this solution to be correct, if needs to satisfy both equations, and a compatibility condition must be fulfilled. For instance, by eliminating $t$,
$$left(fracdeltaphiright)^phi=left(frac z{1-phi}right)^{phi-1},$$ which is analogous to your equation 3).
But, when the system is compatible, the solutions in $x,y$ are $y=tx$, where $x$ is arbitrary.
[Not discussing the singular cases.]
answered Jan 28 at 13:47
Yves DaoustYves Daoust
131k676229
answered Jan 28 at 13:47
Yves DaoustYves Daoust
131k676229
answered Jan 28 at 13:47
Yves DaoustYves Daoust
131k676229
answered Jan 28 at 13:47
Yves DaoustYves Daoust
131k676229
131k676229
$begingroup$
@Tecon: what do you think ?
$endgroup$
– Yves Daoust
Jan 28 at 14:59
$begingroup$
@Tecon $phi$ in $(0,1)$ has nothing special.
$endgroup$
– Yves Daoust
Jan 28 at 18:40
add a comment |
$begingroup$
@Tecon: what do you think ?
$endgroup$
– Yves Daoust
Jan 28 at 14:59
$begingroup$
@Tecon $phi$ in $(0,1)$ has nothing special.
$endgroup$
– Yves Daoust
Jan 28 at 18:40
$begingroup$
@Tecon: what do you think ?
$endgroup$
– Yves Daoust
Jan 28 at 14:59
$begingroup$
@Tecon: what do you think ?
$endgroup$
– Yves Daoust
Jan 28 at 14:59
$begingroup$
@Tecon $phi$ in $(0,1)$ has nothing special.
$endgroup$
– Yves Daoust
Jan 28 at 18:40
$begingroup$
@Tecon $phi$ in $(0,1)$ has nothing special.
$endgroup$
– Yves Daoust
Jan 28 at 18:40
add a comment |
$begingroup$
Be careful, because the solution needs the restriction $phinotin{0,1}$.
But you are right, that or $phinotin{0,1}$ the second equation doesn't depend on $x$ or $y$. But that doesn't mean that any $(x,y)$ solves the whole system. You just don't get further restrictions on the solution of $(x,y)$. So, all $(x,y)$ such that $(2)$ holds are your solution for the case $phinotin{0,1}$.
But if $(3)$ leads to some contradition, then there are no solutions $(x,y)$ such that $(2)$ holds. Hence, there are no solutions if $phinotin{0,1}$.
Finally, you still have to check the cases $phi=0$ and $phi=1$.
answered Jan 28 at 13:45
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
$begingroup$
After your comment I edited the question. $phi$ is actually bounded between 0 and 1. With the extremes excluded.
$endgroup$
– Tecon
Jan 28 at 13:56
$begingroup$
I don't see why the solution needs that restriction on $phi$. Can you, please, explain?
$endgroup$
– Tecon
Jan 28 at 14:38
1
$begingroup$
Plug in $phi=0$ and $phi=1$ and you see what happens: The equations doesn't depend on $x$ and $y$ anymore.
$endgroup$
– Mundron Schmidt
Jan 28 at 14:49
$begingroup$
Ah ok. My bad I misread your comment. Thanks.
$endgroup$
– Tecon
Jan 28 at 14:52
add a comment |
$begingroup$
Be careful, because the solution needs the restriction $phinotin{0,1}$.
But you are right, that or $phinotin{0,1}$ the second equation doesn't depend on $x$ or $y$. But that doesn't mean that any $(x,y)$ solves the whole system. You just don't get further restrictions on the solution of $(x,y)$. So, all $(x,y)$ such that $(2)$ holds are your solution for the case $phinotin{0,1}$.
But if $(3)$ leads to some contradition, then there are no solutions $(x,y)$ such that $(2)$ holds. Hence, there are no solutions if $phinotin{0,1}$.
Finally, you still have to check the cases $phi=0$ and $phi=1$.
answered Jan 28 at 13:45
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
$begingroup$
After your comment I edited the question. $phi$ is actually bounded between 0 and 1. With the extremes excluded.
$endgroup$
– Tecon
Jan 28 at 13:56
$begingroup$
I don't see why the solution needs that restriction on $phi$. Can you, please, explain?
$endgroup$
– Tecon
Jan 28 at 14:38
1
$begingroup$
Plug in $phi=0$ and $phi=1$ and you see what happens: The equations doesn't depend on $x$ and $y$ anymore.
$endgroup$
– Mundron Schmidt
Jan 28 at 14:49
$begingroup$
Ah ok. My bad I misread your comment. Thanks.
$endgroup$
– Tecon
Jan 28 at 14:52
add a comment |
$begingroup$
Be careful, because the solution needs the restriction $phinotin{0,1}$.
But you are right, that or $phinotin{0,1}$ the second equation doesn't depend on $x$ or $y$. But that doesn't mean that any $(x,y)$ solves the whole system. You just don't get further restrictions on the solution of $(x,y)$. So, all $(x,y)$ such that $(2)$ holds are your solution for the case $phinotin{0,1}$.
But if $(3)$ leads to some contradition, then there are no solutions $(x,y)$ such that $(2)$ holds. Hence, there are no solutions if $phinotin{0,1}$.
Finally, you still have to check the cases $phi=0$ and $phi=1$.
answered Jan 28 at 13:45
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
Be careful, because the solution needs the restriction $phinotin{0,1}$.
But you are right, that or $phinotin{0,1}$ the second equation doesn't depend on $x$ or $y$. But that doesn't mean that any $(x,y)$ solves the whole system. You just don't get further restrictions on the solution of $(x,y)$. So, all $(x,y)$ such that $(2)$ holds are your solution for the case $phinotin{0,1}$.
But if $(3)$ leads to some contradition, then there are no solutions $(x,y)$ such that $(2)$ holds. Hence, there are no solutions if $phinotin{0,1}$.
Finally, you still have to check the cases $phi=0$ and $phi=1$.
answered Jan 28 at 13:45
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 28 at 13:45
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 28 at 13:45
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 28 at 13:45
Mundron SchmidtMundron Schmidt
7,5042729
7,5042729
$begingroup$
After your comment I edited the question. $phi$ is actually bounded between 0 and 1. With the extremes excluded.
$endgroup$
– Tecon
Jan 28 at 13:56
$begingroup$
I don't see why the solution needs that restriction on $phi$. Can you, please, explain?
$endgroup$
– Tecon
Jan 28 at 14:38
1
$begingroup$
Plug in $phi=0$ and $phi=1$ and you see what happens: The equations doesn't depend on $x$ and $y$ anymore.
$endgroup$
– Mundron Schmidt
Jan 28 at 14:49
$begingroup$
Ah ok. My bad I misread your comment. Thanks.
$endgroup$
– Tecon
Jan 28 at 14:52
add a comment |
$begingroup$
After your comment I edited the question. $phi$ is actually bounded between 0 and 1. With the extremes excluded.
$endgroup$
– Tecon
Jan 28 at 13:56
$begingroup$
I don't see why the solution needs that restriction on $phi$. Can you, please, explain?
$endgroup$
– Tecon
Jan 28 at 14:38
1
$begingroup$
Plug in $phi=0$ and $phi=1$ and you see what happens: The equations doesn't depend on $x$ and $y$ anymore.
$endgroup$
– Mundron Schmidt
Jan 28 at 14:49
$begingroup$
Ah ok. My bad I misread your comment. Thanks.
$endgroup$
– Tecon
Jan 28 at 14:52
$begingroup$
After your comment I edited the question. $phi$ is actually bounded between 0 and 1. With the extremes excluded.
$endgroup$
– Tecon
Jan 28 at 13:56
$begingroup$
After your comment I edited the question. $phi$ is actually bounded between 0 and 1. With the extremes excluded.
$endgroup$
– Tecon
Jan 28 at 13:56
$begingroup$
I don't see why the solution needs that restriction on $phi$. Can you, please, explain?
$endgroup$
– Tecon
Jan 28 at 14:38
$begingroup$
I don't see why the solution needs that restriction on $phi$. Can you, please, explain?
$endgroup$
– Tecon
Jan 28 at 14:38
1
1
$begingroup$
Plug in $phi=0$ and $phi=1$ and you see what happens: The equations doesn't depend on $x$ and $y$ anymore.
$endgroup$
– Mundron Schmidt
Jan 28 at 14:49
$begingroup$
Plug in $phi=0$ and $phi=1$ and you see what happens: The equations doesn't depend on $x$ and $y$ anymore.
$endgroup$
– Mundron Schmidt
Jan 28 at 14:49
$begingroup$
Ah ok. My bad I misread your comment. Thanks.
$endgroup$
– Tecon
Jan 28 at 14:52
$begingroup$
Ah ok. My bad I misread your comment. Thanks.
$endgroup$
– Tecon
Jan 28 at 14:52
add a comment |
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