Particular method to prove that if $\gcd(a,b,c)=1$ then there exists $n\in \mathbb{Z}$ such that :

Question

$\gcd(a,b+nc)=1$.

I tried to solve this by the analysis-synthesis method :

$\underline{Analysis}$ : Suppose that we have the existence of $n$ such that $\gcd(a,b+nc)=1$. So by Bachet-Bézout theorem we have the existence of $k,l\in \mathbb{Z}$ such that : $ak+(b+nc)l=1$.

Moreover we have by Bachet-Bézout generalized the existence of $u,v,w\in \mathbb{Z}$ such that : $au+bv+cw=1$.

By multiplying we obtain : $a^2ku+abkv+acwk+ablu+b^2lv+bclw+n(clau+clbv+c^2lw)=1$.

Now I want to obtain a special form for $n$ but I don't know what to develop.

$\underline{Synthesis}$ : By the expression of $n$ we found in the analysis part, we just have to prove that $n\in \mathbb{Z}$.

I know that this question is solved on this website but I ask for a particular method.

Thanks in advance !