Faces within a planar embedding In a past assignment, I encountered the following problem:

Let $G$ be a $4$-regular connected graph with a planar embedding. Suppose that
every face of the embedding has degree in $\{ 3, 5 \}$.
Let $f_5$ be the number of faces of degree $5$. Determine the number of faces of degree $3$, call this
$f_3$, as a function of $f_5$. Prove your assertions.

My solution feels like I've missed out on some big point or there's some obvious mistake, but I can't work it out:

Let $n=f_3+f_5$ be the number of faces in $G$ and label its faces $g_1, ..., g_n$. Since $G$ is connected with a planar embedding, we know by the "faceshaking lemma" that $\sum^{n}_{i=1} \deg(g_i) = 2 |E(G)|$ and since $g_i \in \{ 3, 5 \} $, we have
$$
5f_5+3f_3 = 2|E(G)| \implies f_3 = \frac{2}{3}|E(G)|-\frac{5}{3}f_5.
$$

Thanks in advance for the help!
 A: The identity $$f_3 = \frac{2}{3}|E(G)|-\frac{5}{3}f_5$$ is true as far as it goes, but it is not a complete solution, because it does not enable us to determine $f_3$ knowing only $f_5$. We can go further, because this identity is not the only identity available to us here.
Let $f_3, f_5$ be the same variables as before, but let $n$ be the number of vertices instead (we'll need it) and let $m$ be the number of edges. Then we have the following equations:

*

*$n - m + (f_3 + f_5) = 2$: this is Euler's formula, true for any plane embedding of a connected graph.

*$5f_5 + 3f_3 = 2m$: this is the handshaking lemma applied to the dual of the plane embedding. (I am charmed by the term "faceshaking lemma".)

*$4n = 2m$: this is the ordinary handshaking lemma, since we know that the graph is $4$-regular.

We now have three linear equations in four unknowns; we cannot find their values, but we can solve for any three variables in terms of the fourth. To achieve our final goal in this question, we should solve for $n$, $m$, and $f_3$ in terms of $f_5$.
