E + S → ES → E + P
E + I = EI

In the equal state,

[ET] = [E] + [ES] + [EI]
= [E] + [ES] +[E][I]/K_I from (2)
= [E] + [E][I]/K_i +[ES]
= [E](1 + I/K_i) + [ES]
= K_m[ES]/[S] (1 + I/K_i) + [ES] from (4)
E_T = [ES](K_m(1 + I/K_i)/[S] + 1)

For enzyme-catalysed reactions,

Divide (7) by (6).

Substituting (8) in (5), we get,

 

V_max/V = K_m(1 + I/K_i)/[S] + 1
V_max/V = K_m(1 + I/K_i) + [S]/[S]
V_max[S] = V(K_m(1 + I/K_i)+[S]

Where

 = K_m(1 + I/K_I)

This is the M–M equation for competitive inhibitor.

From the equation, it is clear that when an enzyme is inhibited competitively,

1. V_max does not change at all
2. K_m value increases as the concentration of I increases.