= Solution
Use $s=s_0u$, $c=e_0v$ and slow time $\boxed{\tau=k_1e_0t}$. Dividing the equations by their respective <concentration> and <time scales> gives
$$
\boxed{\epsilon=\frac{e_0}{s_0},\quad
K=\frac{k_{-1}+k_2}{k_1s_0},\quad\lambda=\frac{k_2}{k_1s_0},\quad
\eta=\frac{k_0}{k_1e_0}.}
$$
Thus $u'=\eta u-u+uv+(K-\lambda)v$ and $\epsilon v'=u-uv-Kv$. Notice that $K>\lambda>0$ and the physical <enzyme> fraction satisfies $0\leq v\leq1$. The initial values are $u(0)=1$, $v(0)=0$ for this particular normalization; the threshold analysis compares different initial <substrate> abundances using a fixed reference scale.
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