For a simple graph $G = (V,E)$ we know that there are $10$ edges, $|E|=10$. And we know that two vertices have a degree of $4$, and the rest have a degree of $3$. I know how to solve this, I'm just not sure how to do it formally; my current work is a mess, could somebody show how it would be done without all mess?
Question: Find |V|
My solution
Let $n = |V|,\quad |U|= |E|-1$
$a\cdot n = \sum_{v\in V}deg(v) = 2\cdot|U| = 18$
$a = 2 \implies 2n = 18 \iff n = 9$
$a=3\implies 3n=18\iff n= 6$
$a=4\not\mid18$
And after this, there wouldn't be any more solutions for $n$ unless loops and multiedges were allowed.
We want to solve for $n$, the number of vertices in $G$. Note that $2$ vertices have degree $4$, so $n - 2$ vertices have degree $3$. Thus, by the handshaking lemma, we have that: \begin{align*} \sum_{v \in V} \deg(v) &= 2|E| \\ 2(4) + (n - 2)(3) &= 2(10) \\ 8 + (3n - 6) &= 20 \\ 3n &= 18 \\ n &= 6 \end{align*}