How do you prove a statement in the form of "for all integers $x$, there is some integer $y$, such that $3 \mid x + y$". I believe the statement is true. I know you start by assuming $x$ is an integer, and you pick a $y$. Let's say $y = 3$. And next you need to prove that $3 \mid x + y$ in order to prove the statement is true.
I think the next step is to say $3k = x + y$ where $k$ is an integer. I'm just not sure where to go from here. My initial thought was that $3k = x + y$ where $k$ is an integer proves that $3$ divides $x + y$, but then what is the point of picking a value for $y$? And if I'm wrong here, how can I prove that $3 \mid x + y$ based on my assumption that $x$ is an integer and $y = 3$?

Let $y$ be dependent on $x$. $y$ can't be independent of $x$, this can be seen by trying a few values of $x$.
Given $x\in \mathbb{Z}$, let $y=3-x \in \mathbb{Z}$, then we have $x+y=3$. This would answer your question.
However, I would encourage you to practice more:

*

*Try to think of a different choice of $y$ as an exercise.

*Also try to think given an $x$, find all the possible choice of $y$.

