$G_1$ is graph on the set of vertices $\{1,2,3,4,5,6,7,8\}$, $G_1$ vertice chromatic number is 5.
$G_2$ is graph on the set of vertices $\{7,8,9,10,11,12,13,14,15,17,18,19,20\}$, $G_2$ vertice chromatic number is 7.
we know $G_1$ and $G_2$ have an edge between vertice 7 and vertice 8. (we already used 2 differnt colors on them both).
$G$ is graph on the set of vertices $\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}$, that it's set of edges is the union of the edges set of $G_1$ with the edges set of $G_2$.
the vertice chromatic number of $G$ is: 5 / 7 / 12 / we can't know without further information
prove the answer.
As I understand, no vertice of $G_1$ can be connected to a vertice of $G_2$, except vertices 7 and 8, if we use 7 different colors on $G_2$, with vertice 7 and vertice 8 counted as 2 of them, and we use 5 colors on $G_1$ (with vertice 7 and vertice 8, counted as the same colors we used to color $G_2$ ) $G$ must have a vertice chromatic number of 7.
too simple in my opinion, I'm already waiting for my mistake.
I think I got it, can someone approve?
after the union to $G$, we copy the colors of $G_2$ to 7,8,9,10,11,12,13,14,15,16,17,18,19,20, obviously, $G$ edge chromatic number is 7 and can't be less.
we go through the edges 1,2,3,4,5,6,7,8
we look for an edge that doesn't connect to a different edge. (there must be one, because if only one vertice was missing in $G_1$, our chromatic number of $G_1$ was 7, but it's 5).
once we find a missing vertice, we color both it's edges in the same color.
we go through edges 1-6, and color each edge that doesn't have a color, with the colors of $G_2$ that we didn't use in $G_1$ (there should be 4-5 of them).