Existence of solution to linear congruence

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Given that $a$ and $b$ are relatively prime positive integers, and that $a$ is relatively prime to all the following primes: $p_1,p_2,...,p_n$ then the following congruence:$${a+bx}\equiv 1\pmod{ p_1\cdot p_2\cdot p_3 \cdot...\cdot p_n}$$ Has solution in x, why can we claim this?

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It is solvable $\iff \gcd(n,b)\mid 1\!-\!a,\,$ where $n = \prod p_i.\,$ Indeed

$\qquad\qquad \begin{align} \!\bmod n\!:\ \exists x\!:\ bx\equiv&\, 1\!-\!a\\[.3em] \iff \exists x,y\!:\ ny+bx =&\, 1\!-\!a\\[.3em] \iff \gcd(n,\,b)\,\ \mid\ &\ 1\!-\!a\ \ {\rm by\ Bezout}\end{align}$