# Solution to a two-equations system.

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:

$$$$y = \Big(\frac{\delta}{\phi}\Big)^{\frac{1}{1-\phi}}x\tag{2}$$$$

I then plug it in the second equation and obtain:

$$$$z = (1-\phi) \Big(\frac{\delta}{\phi}\Big)^{\frac{\phi}{\phi-1}} \tag{3}$$$$

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?

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.]

• @Tecon: what do you think ? – Yves Daoust Jan 28 '19 at 14:59
• @Tecon $\phi$ in $(0,1)$ has nothing special. – Yves Daoust Jan 28 '19 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 $$\phi=1$$.

• After your comment I edited the question. $\phi$ is actually bounded between 0 and 1. With the extremes excluded. – Tecon Jan 28 '19 at 13:56
• I don't see why the solution needs that restriction on $\phi$. Can you, please, explain? – Tecon Jan 28 '19 at 14:38
• Plug in $\phi=0$ and $\phi=1$ and you see what happens: The equations doesn't depend on $x$ and $y$ anymore. – Mundron Schmidt Jan 28 '19 at 14:49
• Ah ok. My bad I misread your comment. Thanks. – Tecon Jan 28 '19 at 14:52