Binomial Expansion plus Algebra!

\(\dbinom{9}{k}=4\dbinom{7}{k-1}\\ \text{There's probably some identity out there that would make this easy}\\ \text{but nothing really jumped out at me on the wiki page, brute force isn't too bad}\\\)

\(\dfrac{9!}{k!(9-k)!}=4\dfrac{7!}{(k-1)!(7-(k-1))!} = 4\dfrac{7!}{(k-1)!(8-k)!} \)

\(\dfrac{(k-1)!}{k!}\dfrac{(8-k)!}{(9-k)!} = \dfrac{4\cdot 7!}{9!}\)

\(\text{let's look at each of these terms}\\ \dfrac{(k-1)!}{k!} = \dfrac 1 k\\ \dfrac{(8-k)!}{(9-k)!} = \dfrac{1}{9-k}\\ 4\dfrac{7!}{9!} = \dfrac{4}{9\cdot 8} = \dfrac{1}{18}\\ \dfrac{1}{k}\dfrac{1}{9-k}=\dfrac{1}{18}\\ \)

\(\text{take the reciprocal of both sides and with a bit of algebra}\\ -k^2+9k = 18\\ k^2-9k+18 = 0\\ (k-6)(k-3)=0\\ k=3\text{ or }k=6 \)