Propostion 1.6 on Atiyah's commutative algebra text

Question

Let $A$ be a ring (commutative with 1) and $M$ a maximal ideal of $A$. Let $x \in A-M $. Since $M$ is maximal, the ideal generated by $x$ and $M$ is $(1)$, i.e. the entire ring.

I don't understand how that is true. If $x$ were a unit, certainly. However, $x$ could be contained in some other maximal ideal and is not guaranteed to be a unit. So how can we be sure that $x$ and $M$ can generate $A$?

Answer 1

The ideal generated by $x$ and $M$ strictly contains $M$ (as $x \notin M$ by assumption). Then, since $M$ is maximal, by definition, any ideal containing $M$ must be the whole ring; thus, $(x,M)$ is the entire ring (as dezdichado said in the comments).

Answer 2

$\mathscr M\subset A$ a maximal ideal implies that for any ideal $\mathscr J\supsetneq\mathscr M$, $\mathscr J=(1)$.

But if $x\in A\setminus \mathscr M$, then $(x,\mathscr M)\supsetneq \mathscr M$.