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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}}x\tag{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?

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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,

$$t=\frac{z\phi}{\delta(1-\phi)}.$$ But for this solution to be correct, if needs to satisfy both equations, and a compatibility condition must be fulfilled. For instance, by eliminating $t$,

$$\left(\frac\delta\phi\right)^\phi=\left(\frac z{1-\phi}\right)^{\phi-1},$$ 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.]

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  • $\begingroup$ @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
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Be careful, because the solution needs the restriction $\phi\notin\{0,1\}$. But you are right, that or $\phi\notin\{0,1\}$ 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 $\phi\notin\{0,1\}$. But if $(3)$ leads to some contradition, then there are no solutions $(x,y)$ such that $(2)$ holds. Hence, there are no solutions if $\phi\notin\{0,1\}$.

Finally, you still have to check the cases $\phi=0$ and $\phi=1$.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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