= Solution
Write concentrations as $E,S,X,Y,E_i,P$ in this calculation. The <law of mass action> assigns the forward bimolecular rate $k_1ES$ and the first-order rates $k_{-1}X,k_2X,k_3Y,k_4Y$. Each species gains the rates of reactions producing it and loses those consuming it:
$$
\boxed{\begin{aligned}
\dot E&=-k_1ES+k_{-1}X+k_3Y,&\dot S&=-k_1ES+k_{-1}X,\\
\dot X&=k_1ES-(k_{-1}+k_2)X,&\dot Y&=k_2X-(k_3+k_4)Y,\\
\dot E_i&=k_4Y,&\dot P&=k_3Y.
\end{aligned}}
$$
Adding the equations for all enzyme-bearing species cancels every reaction rate, so $\boxed{E+X+Y+E_i=e_0}$. This is conservation of the total enzyme-derived material, including the inactivated form. It does not assert conservation of active enzyme: <mechanism-based inhibition> transfers some of that conserved material irreversibly into $E_i$.
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