3-regular connected planar graph

Let $G$ be a 3-regular connected planar graph with a planar embedding where each face has degree either 4 or 6 and each vertex is incident with exactly one face of degree 4. Determine the number of vertices, edges and faces of degree 4 and 6.

Using handshake lemmas and Euler Formula, I've come up with the following for $E$ edges and $n$ vertices:

$2E=3n$

$2E=4F_4+6F_6$

$n-E+F_4+F_6=2$

I'm missing an equation because I'm not sure how to use the restriction where each vertex is incident with one face of degree 4. Any help?

Solution 1:

Consider the edges adjacent to any vertex $v\in V(G)$. We know the vertex is incident to only one face of degree 4, thus 2 of the three edges adjacent to it form part of the length of a face of degree 4. Thus, two thirds of all edges lie on the border of a face of length 4, and we have:

$f_4=\frac{1}{4}*\frac{2}{3} *e=\frac{1}{6}e$

Using this fact and the rest of your equations, you should be able to get all the values you need.