Example of reduction of number of parameters

Consider equation

$\dot r = - (rv)'$ ;   $r \dot v = -Ar' - B v ^ S$

Where $r=r(x,t)$   and   $v=v(x,t)$ ; $A$, $B$, $S$ are parameters (positive real numbers).
As usually, prime $'$ differentiates with respect to the first (primary) argument,
and dot $\dot{}$ does the same with respect to the last argument (as dot appears at the end of the sentence).
Having no smart idea, I define

$X=\alpha x$ ;   $T=\beta t$

Let

$R=R(X,T)=r(x,t)$ ;   $V=V(X,T)=\gamma v(x,t)$

Then

$R' = \alpha r'$ ;   $V'=\alpha\gamma v'$

$\dot R = \beta \dot r$ ;   $\dot V=\beta\gamma\dot v$

$\dot R = - \beta (R V/\gamma)'$ ;   $R \dot V \beta/\gamma = -(A/\alpha) R' - B (V/\gamma)^S$

$\dot R = - \frac{\beta}{\gamma} (R'V+ R V')$ ;   $R \dot V = - \frac{A \gamma}{\alpha\beta} R' - \frac{B}{\beta\gamma^{S-1}} V^{S}$

I set $\gamma=\beta$ ; then

$\dot R = - R'V- R V'$ ;   $R \dot V = - \frac{A}{\alpha} R' - \frac{B}{\beta^{S}} V^{S}$

I set $\alpha=A$   and   $\beta=B^{1/S}$ ;   then

$\dot R = - R'V- R V'$ ;   $R \dot V = - R' - V^{S}$

Compared to the initial equations, two parameters are excluded.
In this sense, the exact solution of the initial system of equations is acheived:

$r(x,t) = R(x/A,\ t/\beta)$ ;   $v(x,t)=B^{-1/S}\ V(x/A,\ t/\beta)$