In Smith - Introduction to Algebraic Geometry page 334 it is claimed that definition of a maximal ideal $M$ is equivalent to saying that for any $a \in R \setminus M$, $M + (a) = R$. How this equivalence is true? :
For any $M$ let alone $M$ being maximal, $M + (a) \subset R$ is true, obviously. For the converse if $a \in R \setminus M$ then $(a,M)$, the ideal generated by $a$ and $M$, equals $R$. But how to prove that for any $s \in R \setminus M$, $s = c_1a + \sum_{k=2}^n c_k x_k$ for $x_k \in M$, esp when $c_1$ may not be a unit?
How $M + (a) = R$ implies $M$ being a maximal?
You're on the right track! Here are some hints:
For $1$, recall that $M + (a) \supsetneq M$ when $a \not \in M$. Since $M$ is maximal amongst proper ideals, and $M + (a)$ is an ideal, do you see why $M + (a) = R$?
For $2$, I'd suggest contradiction. Say $M$ is not maximal. Then $M \subsetneq N$ for some proper ideal $N$. Can you see how to find an element $a$ so that $M + (a) \neq R$?
I hope this helps ^_^