I have the following system of nonlinear equations:
$$\begin{align} \Delta t \omega x^{(\omega-2)/2} y^{\phi/2} &= r_L \tag{1}\label{eq1} \\ \Delta t \phi x^{\omega/2} y^{(\phi-2)/2} &= r_D \tag{2}\label{eq2} \end{align} $$
Where $\Delta t >0$ is a constant commom to both equations. I want to solve with respect to $x$ and $y$ without specifying any parameter ($\omega,\phi, r_L, r_D$). First I thought directly on the substitution method but it seemed not so easy.
So it cames on mind to divide $(1)$ by $(2)$ and isolate $y$ as a function of $x$ which leaves me to the following pair of equations
$$ \begin{align} y = \frac{\phi r_L}{\omega r_D} x \tag{3}\label{eq3} \\ \Delta t \phi x^{\omega/2} y^{(\phi-2)/2} = r_D \tag{4}\label{eq4} \end{align} $$
Then the next step was inserting $(3)$ in $(4)$, which gave me
$$ x = \left(\frac{\omega}{r_L}\right)^{\frac{2-\phi}{\omega+\phi-2}} \left(\frac{r_D}{\phi}\right)^\frac{\phi}{\omega+\phi-2} \frac{1}{(\Delta t)^{\frac{2}{\omega+\phi-2}}} $$
Finally, I isolated $x$ in $(3)$ and substituted in $(4)$ which gave me
$$ y = \left(\frac{r_D}{\phi}\right)^{\frac{2-\omega}{\omega+\phi-2}} \left(\frac{r_L}{\omega}\right)^\frac{\omega}{\omega+\phi-2} \frac{1}{(\Delta t)^{\frac{2}{\omega+\phi-2}}} $$
As important as the result: is my reasoning right? And is there any software that would give me the solutions? Thank you!
Start simplifying the equations
$$\Delta t \omega x^{(\omega-2)/2} y^{\phi/2} = r_L\quad \implies \quad x^{(\omega-2)/2} y^{\phi/2}=\frac {r_L}{\Delta t \omega }=a$$ $$\Delta t \phi x^{\omega/2} y^{(\phi-2)/2} = r_D \quad \implies \quad x^{\omega/2} y^{(\phi-2)/2}=\frac{r_D }{ \Delta t \phi}=b$$
Now, taking logarithms, let $$X=\log(x) \qquad Y=\log(y)\qquad A=\log(a) \qquad B=\log(b)$$ which gives $$\frac{\omega-2}2 X+\frac \phi 2 Y=A$$ $$\frac{\omega}2 X+\frac {\phi-2} 2 Y=B$$ which is more than simple.
When done, go back to the original variables.