I'm not quite sure I understand my book's reasoning for the answer

Question

I have the following homework problem:

Does there exist a graph, $G$, with 28 edges and 12 vertices, each of degree 3 or 4?

First, my solution.

$$ \sum deg(v_i) = 2 \cdot |E| \\ |E| = 28 \Rightarrow 2 \cdot |E| = 2 \cdot 28 = 56 \\ \forall v_i \in V, deg(v_i) = 4 \\ \text{thus} \sum deg(v_i) = 4 \cdot 12 = 48 \neq 56 \\ \therefore \text{no such graph exists} \square $$

Now, the book's:
$$ \text{Suppose G has } a\, \text{vertices of degree 3 and } b\, \text{vertices of degree 4.} \\ \text{Then, } \sum deg\,v_i = 3a + 4b = 56 = 2\cdot |E| \, \text{and}\, a+b = 12. \\ \text{These equations imply}\, a < 0, \text{which cannot be. No such G exists.} $$

I'm not sure I see why the equations imply that $a < 0$. Is it essentially what I have or have I missed the mark?

Thanks for the help,
Andy

Answer 1

For the book solution, you have two equations: $3a+4b = 56$ and $a+b =12$, subtracting 4 multiples of the second from the first yields $-a = 8$, so $a<0$, which cannot be.

Your proof cannot assume all degrees are 4, some could be 3, so correct it to an inequality: $$\sum \deg v_i \leq 4 \cdot 12 = 48 < 56$$ which yields the desired contradiction.

Answer 2

$$3a+4b=56\\ a+b=12\implies b=12-a\\ \implies 3a+48-4a=56\\ \implies -a=8\\ \implies a=-8<0$$