Let $f=f(x,y),g=g(x,y) \in \mathbb{C}[x,y]$, each of degree $\geq 1$, and $f,g$ are algebraically independent over $\mathbb{C}$ (= their Jacobian $\in \mathbb{C}[x,y]-\{0\}$).
In this question I have asked for a sufficient (and necessary) condition for $k(f,g)=k(x,y)$.
How to apply Jeremy Blanc's answer in the following two specific cases: $f=ax+by, g=cx+dy$, $ad-bc \neq 0$; $f=ux, g=vy+H(x)$, $uv \neq 0$, $H(x) \in \mathbb{C}[x]$.
Another answer to the above question can be found in the comments to this question; it would be nice if someone could help me in understanding and proving the claim in the comments that says:
" $s_1,s_2$ form birational generators if and only if for general $a,b \in \mathbb{C}$, $s_1−a,s_2−b$ have exactly one common zero". (Theorem 2.1 may be relevant. Perhaps the answer to this question can be generalized to the two variable case).
Thank you very much!