sorry, just started to see graphs, but how can you proof that,
If G = (V,A) a connected graph, plannar, with p verices and q edges.
Proof that...
$q\leq 3p -6$
Any help or orientation would be deeply appreciated. Thank you!
(I made a drawing where p = 3, but I doubt this is the correct answer...):
Hint: Assume that $G$ is a maximal planar graph. By maximality of $G$, we can not add more edges without destroying planarity. Observe that each face in this maximal planar graph is a triangle. Otherwise, we could always add diagonal edges (convince yourself that this is true; draw a few planar graphs). Now, denote $p$ as the number of vertices, $q$ as the number of edges and $f$ as the number of faces in $G$. And every edge bounds two faces, hence $2q=3f$. Use Euler's formula $p-q+f=2$, and the other observations and you quickly get the upper bound.