= Solution
Put $M=[E_m]$, $r=V_R[R]$ and $b=V_B[B]$. Saturated <CheR> supplies a constant <receptor methylation> flux, while <CheB> removes it with <Michaelis-Menten kinetics>:
$$
\dot M=r-\frac{bM}{K_B+M}.
$$
At a finite <steady state>, $r(K_B+M_0)=bM_0$, giving
$$
\boxed{M_0=\frac{K_BV_R[R]}{V_B[B]-V_R[R]}=\frac{K_Br}{b-r}.}
$$
It requires $0<r<b$ and $M_0$ below the total available <receptor> pool. The <steady state> is stable because $d\dot M/dM=-bK_B/(K_B+M)^2<0$. If $r\geq b$, the ideal equation has no finite positive <steady state>; actual methylatable sites would become exhausted.
Attractant binding reduces <receptor> signalling, reducing <CheA>-dependent <phosphorylation> of <CheY> and initially reducing tumbling frequency, so smooth runs become longer. A spatially uniform stimulus provides no preferred direction. <Perfect adaptation> means that after this transient the tumbling frequency returns exactly to its pre-step value despite continued attractant presence.
The equation above alone does not specify attractant occupancy or its influence on activity. If binding simply changes activity from $M$ to $fM$ with $0<f<1$, without changing either modifying flux, $M$ remains $M_0$ and the output remains $fM_0$: there is no adaptation. Thus the usual fall followed by recovery requires the additional feedback mechanism developed in <2/c> or <2/d>, rather than following from the stated <Michaelis-Menten kinetics> alone.
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