If $a,b$ is an $R$-sequence, then $(ax-b)$ is prime

Question

If $R$ is an integral domain, $a, b\in R, a\neq 0$ and $\bar b$ is not a zero divisor in $R/(a)$. I'm trying to prove $(ax-b)\in R[x]$ is prime.

This question seems easy but I couldn't prove it, maybe there is some trick which I could solve this problem easily?

Thanks