= Solution
<Law of mass action>[Mass action] gives
$$
\dot S=-k_1SE+k_{-1}C,\quad\dot E=-k_1SE+(k_{-1}+k_2)C,\quad
\dot C=k_1SE-(k_{-1}+k_2)C,\quad\dot P=k_2C.
$$
Thus $E+C=E_0$ and $S+C+P=S_0$ are conserved. Eliminating free <enzyme> yields the stated two-equation system.
With $s=S/S_0$, $c=C/E_0$ and $\tau=t/T$, substitution gives
$$
\boxed{\alpha=k_1E_0T,\quad\beta=\frac{k_{-1}E_0T}{S_0},\quad\gamma=k_1S_0T,\quad\delta=(k_{-1}+k_2)T.}
$$
The <dimensionless> initial data are $\boxed{s(0)=1,\quad c(0)=0}$. The parameter named $\delta$ here is an <enzyme> kinetic coefficient and is unrelated to the loss parameter used in the later blood-cell question.
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