An article on Khan-Academy, states the following:
The overall reaction rate is determined by the rates of the steps up to (and including) the rate-determining step
Here, it says that rate is determined by steps before and including that of RDS. But from what I know the rate is determined by one single elementary reaction, which is the slowest, that is, the RDS. So why does it include steps previous to RDS?
(I would appreciate an example of this)
The interpretation of the statement depends on exactly what is meant by "overall rate."
The rate-determining step corresponds to the largest free energy difference from a transition state to any preceding stable state (reactant or intermediate).
In the example above, the biggest forward barrier is from the reactant to the second transition state, so the second step is rate-determining. However, the barrier, as identified above, is comprised of both the first and second step, so they both figure into the overall rate of the reaction, as determined by the slow step.
EDIT:
Based on comments, I think a clarification is in order. Note that the above conclusion is not true in the general case. You could make the indicated claim for the depicted reaction specifically, but not, for example, in the case below:
The second step here is still rate-determining because the transition state is involved in the largest barrier. However, the biggest barrier is from the intermediate to the second transition state. In this example, I would not say that the overall rate depends on the first step.
It is perhaps worth noting that the 'rate determining step' is an approximation.
It is technically possible to model any reacting system without it, by using numerical methods.
It is even possible to model certain systems fully analytically.
Take for instance a sequence of 2 consecutive irreversible reactions:
$\ce{A ->[k_A] B ->[k_B] C}$
which is described by:
$\frac{dA}{dt}=-k_A A$
$\frac{dB}{dt}=+k_A A-k_B B$
$\frac{dC}{dt}=+k_B B$
This system of differential equations has an analytical solution. Assuming that only $\ce A$ is present at $t=0$, at a concentration $\ce{A(0)}$:
$A(t) = A(0) e^{-k_A t}$
$B(t) = \frac {k_A A(0)}{k_B-k_A} [e^{-k_A t}-e^{-k_B t}]$
$C(t) = \frac {A(0)}{k_B-k_A} [k_B (1-e^{-k_A t}) -k_A (1-e^{-k_B t})]$
Suppose that the first reaction is 100 times faster than the second ($k_A = 1, k_B = 0.01$).
If you plot the 3 concentrations for $t \in [0,100]$, you will see $A$ disappear quite fast and $B$ reach its maximum when $C$ is still quite low.
In other words, you can assume that the second step is the 'rate determining' one, i.e. assume that all $A$ almost immediately turns into $B$, so $B(0) \approx A(0)$, and model only the second and third reaction:
$B(t) \approx A(0) e^{-k_B t}$
$C(t) \approx A(0) [1-e^{-k_B t}]$
The same result can be reached directly from the exact analytical solution, by setting $k_A >> k_B$:
$B(t) \approx \frac {k_A A(0)}{-k_A} [-e^{-k_B t}] = A(0) e^{-k_B t}$
$C(t) = \frac {A(0)}{k_B-k_A} [k_B (1-e^{-k_A t}) -k_A (1-e^{-k_B t})] \approx \frac {A(0)}{-k_A} [-k_A (1-e^{-k_B t})] = A(0) [1-e^{-k_B t}]$
Suppose instead that the first reaction is 100 times slower than the second ($k_A = 0.01, k_B = 1$), and plot the 3 concentrations for $t \in [0,1000]$.
You will see $A$ disappear according to a rather typical first order decay, $C$ form equally typically, whereas $B$ remains barely visible at the bottom of the plot (in some contexts $B$ would be called a 'stationary intermediate').
You can assume that the first step is the 'rate determining' one, i.e. assume that $\frac {dB}{dt} \approx 0$, and consequently $\frac {dC}{dt} = k_B B \approx k_A A$, and model only the first and third reaction:
$A(t) = A(0) e^{-k_A t}$
$C(t) \approx A(0) [1-e^{-k_A t}]$
Here, too, the same result can be reached directly from the exact analytical solution, by setting this time $k_A << k_B$:
$A(t) = A(0) e^{-k_A t}$
$C(t) = \frac {A(0)}{k_B-k_A} [k_B (1-e^{-k_A t}) -k_A (1-e^{-k_B t})] \approx \frac {A(0)}{k_B} [k_B (1-e^{-k_A t})] = A(0) [1-e^{-k_A t}]$
In both these 'extreme' cases, the approximate curves described by the 'rate determining step' approximation fit quite closely the exact curves, but there is nothing better than the latter, for a more accurate account of the system.
Also considering that one does not know a priori if a given approximation holds true or not for a particular system.
It is sufficient to set $k_A = 1, k_B = 1.01$, i.e. have two consecutive reactions of comparable rate, and plot the exact concentrations for $t \in [0,10]$, to see that $C(t)$ is not really well described by either approximate rds solution.
Luckily these days there are plenty of excellent software solutions allowing to model systems of any complexity via very accurate numerical methods.
