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?


2 Answers 2


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.]

  • $\begingroup$ @Tecon: what do you think ? $\endgroup$
    – user65203
    Jan 28, 2019 at 14:59
  • $\begingroup$ @Tecon $\phi$ in $(0,1)$ has nothing special. $\endgroup$
    – user65203
    Jan 28, 2019 at 18:40

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 $\phi=1$.

  • $\begingroup$ After your comment I edited the question. $\phi$ is actually bounded between 0 and 1. With the extremes excluded. $\endgroup$
    – Tecon
    Jan 28, 2019 at 13:56
  • $\begingroup$ I don't see why the solution needs that restriction on $\phi$. Can you, please, explain? $\endgroup$
    – Tecon
    Jan 28, 2019 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. $\endgroup$ Jan 28, 2019 at 14:49
  • $\begingroup$ Ah ok. My bad I misread your comment. Thanks. $\endgroup$
    – Tecon
    Jan 28, 2019 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.