Suppose T is a tree with degree sequence 5, 5, 5, 4, 4, 3, 3 plus several 2’s and 1’s. Find the number of leaves of T.
I currently have made this much progress. Let $x$ be the number of degree $2$ vertices and $y$ be the number of degree $1$ vertices (leaves). Then $$|V| = 7 + x + y \implies |E| = 7 + x + y - 1 = 6 + x + y$$ and $$2|E| = 5(3)+4(2)+3(2)+2(x)+1(y) \implies |E| = 29/2 + x + y/2$$ so $6 + x + y = 29/2 + x + y/2$, eventually you get $y = 17$.
But from here, I don't know how to get $x$. Do I have the right approach or is there something else I'm missing?
Thanks in Advance!
You are not missing anything.
All you needed to do is solve for the number of leaves, which is $y$ in your notation. The number of degree $2$ vertices, which is $x$, cannot be determined from this information alone - but you also don't need to know it to solve for the number of leaves.
In fact, you can construct a tree with degree sequence $5,5,5,4,4,3,3$ followed by $y=17$ ones, in there are no vertices of degree $2$. From there, you can take any edge and subdivide it into a path of arbitrary length, which will give you a tree with degree sequence $5,5,5,4,4,3,3$, followed by any number of twos you like, followed by $17$ ones. So there exist trees in which $x$ is any nonnegative integer you want it to be.