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Hi johncondon101, thanks for the question. The choice is kind of strategic. Either way will lead to the answer, but the algebra just looks a little tidier one way vs. the other, and the choice is more a matter of intuition. In problem #5 it made sense to have $f_2$ in the numerator since it was the unknown, and we ended up multiplying by the denominator ($f_1$ in that case) to isolate $f_2$ on one side of the equation. In this problem it's a bit less clear which choice to make, but since $m$ is the unknown and it will appear in both the numerator and denominator (eventually, after a couple of steps, which is why, again, setting things up in this case is more a matter of intuition), it's better to have it by itself in the denominator instead of adding to a term since we'll be multiplying by the denominator, and it's easier to multiply by a single factor.