One example of elliptic fibration is obtained as follows:
Let $Z(F),Z(G)\subset\Bbb{P}^2$ be two non-singular cubics intersecting in distinct points $P_1,...,P_9$ and take the rational map \begin{align*} \varphi:\Bbb{P}^2&\to\Bbb{P}^1\\ P&\mapsto (F(P):G(P)) \end{align*}
If $p:X\to \Bbb{P}^2$ is the blow-up of $\Bbb{P}^2$ in $P_1,...,P_9$, then $\pi:=\varphi\circ p:X\to\Bbb{P}^1$ defines an elliptic fibration.
I'm trying to understand why almost all fibers of $\pi$ are elliptic curves.
Here's where I'm at: If $(a:b)\in\Bbb{P}^1$, we see that $\varphi^{-1}(a:b)=C\setminus\{P_1,...,P_9\}$ where $C:=Z(bF-aG)$.
I'm not sure how to prove this, but intuitively I'm convinced that $C$ is irreducible for almost all $(a:b)$.
Now if $\widetilde{C}\subset X$ the strict transform of $C$, we have $\pi^{-1}(a:b)=\widetilde{C}$, and we should be able to prove that $g(\widetilde{C})=1$. If $m_i$ is the multiplicity of $C$ in $P_i$, then $m_i\cdot m_F(P_i)\leq I(P_i, C\cap F)=I(P_i, F\cap G)=1$, so $m_i=1$ for all $i=1,...,9$. Therefore: $$\widetilde{C}^2=C^2-(m_1^2+...+m_9^2)=9-(1+...+1)=0$$
This seems relevant, but I don't know how to conclude that $g(\widetilde{C})=1$.