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Basically, when the slow step involves an intermediate you use the fast step to solve for the concentration of that intermediate and use that substitution to write the rate law of the overall reaction. You can look at the example in the course reader for reference, but try looking at is as a substitution to make it easier.
You know you would have to use the pre-equilibrium approach when Step 1 is fast and Step 2 is slow. This is because the intermediate will be included in the rate law equation. That is not allowed so you would need to use substitution to replace the intermediate concentration.
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