Say we are dealing with a commutative ring that contains 1, and we'll call it $R$. Suppose we want to show that $I \subseteq R$ is an ideal $\iff$ $\forall x,y \in I$ and $r \in R$, that $x+ry \in I$.
Now, for showing ideals I know I must show that $ar$ and $ra$ are in $I$ for all $r \in R$. Would I have to show that each individual portion of $x+ry$ could be an ideal and that sum of ideals is also an ideal?
$(\impliedby)$: Take $r=1$ and you find that $I$ is closed under addition. If you take $x = 0$, you find that $ry \in I$ for any $r \in R$ and $y \in I$. It's also clear that $0 \in I$. This is enough to show that $I$ is an ideal.
$(\implies):$ If $I \subseteq R$ is an ideal, then by definition $ry \in I$ for any $r \in R$ and $y \in I$. Furthermore, $I$ is closed under addition. This implies that $x + ry \in I$ for any $x,y \in I$ and $r \in R$.