= Solution
The conserved total is $e_0=[E](0)$ under the chosen initial conditions. Substitute $E=e_0-X-Y-E_i$ into the equations from the previous part. This closes the subsystem for $S,X,Y,E_i$; product can be recovered afterward by integrating $\dot P=k_3Y$.
Let $k_\Sigma=k_{-1}+k_2$ and $K_m=k_\Sigma/k_1$. The required <nondimensionalization> is completed by
$$
\boxed{\sigma=\frac{s_0}{K_m},\quad\rho=\frac{k_{-1}}{k_2},\quad
\varepsilon=\frac{e_0}{s_0+K_m},\quad
\psi=\frac{k_3+k_4}{k_\Sigma},\quad\phi=\frac{k_4}{k_\Sigma},\quad
T=\varepsilon k_\Sigma t.}
$$
Indeed $X=e_0\sigma x/(1+\sigma)$ and the free-enzyme fraction is $1-\sigma x/(1+\sigma)-y-e_i$. Substitution and division by the indicated time scale give
$$
\begin{aligned}
s_T&=-s[(1+\sigma)-\sigma x-(1+\sigma)y-(1+\sigma)e_i]+\frac\rho{1+\rho}x,\\
\varepsilon x_T&=s[(1+\sigma)-\sigma x-(1+\sigma)y-(1+\sigma)e_i]-x,\\
\varepsilon y_T&=\frac{\sigma}{(1+\sigma)(1+\rho)}x-\psi y,\\
\varepsilon(e_i)_T&=\phi y.
\end{aligned}
$$
For example $k_2/k_\Sigma=1/(1+\rho)$ explains the coefficient in the third equation. The small multiplier on the intermediate-species derivatives identifies their potentially fast adjustment relative to substrate depletion.
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