Is this true? An element which is not contained in a maximal ideal.

Question

$a \notin M$ : maximal ideal of a ring $R$ $\Rightarrow$ $M+Ra=R$

I tried some my attempts but nothing was useful, can anybody help me?

add: My attempt was, for any $r \in R$, to construct an element $m$ of $M$ such that for some $b \in R, m+ba=r$ but I failed. I forgot what a maximal ideal was.

Answer 1

Assuming that $R$ is commutative, then $M+Ra$ is an ideal. Besides, it contains $M$. But $M+Ra$ cannot be a proper ideal, since $M$ is a maximal ideal. Therefore, $M+Ra=R$.

Answer 2

If $M$ is a maximal ideal of $R$, with $a\not \in M$, then $M+Ra=\langle M\cup \{a\}\rangle$. That is, it is the ideal generated by $M$ and $\{a\}$. $M$ is maximal, so that since $M+Ra\supsetneq M$, we have that $M+Ra=R$.