True or false: $\forall x \in \Bbb R,\exists y\in \Bbb R,y+x=x+y$ How would one tell if this is true or false.
1: $\forall x \in \Bbb R,\exists y\in \Bbb R,y+x=x+y$
2: $\exists x \in \Bbb R,\forall y \in \Bbb R,y+x=x+y$

For the first I think it would be true because if $x=8$ then
$y+8=8+y$ 
$y=2$
So to justify it would I say true because of commutative property.
 A: Both statements are true, regardless of our choices for $x, y$. $$y + x = x+y$$ is true for any arbitrary pair of real numbers $(x, y)$, and as you note, this is due to the commutativity of addition over the real numbers.
Note: In your first case, if $x = 8$, the statement holds regardless of the value of $y\in \mathbb R$, and not just when $y = 2$. 
A: Since $\mathbb{R}$ is a field, every element has an additive inverse which commutes with it. Hence there does exist $y\in\mathbb{R}$ such that $y+x=x+y$ for all $x\in\mathbb{R}$, namely $y=-x$.
Similarly for the second statement, there must exist an additive identity, namely $x=0$ such that $y+x=x+y$ for all $y\in\mathbb{R}$.
A: Both statements are true, but your example of $2+8 = 8+2$ doesn't prove either, it just shows that the statement
$\exists x \in \mathbb{R}, \,\exists y \in \mathbb{R},\,x+y = y+x$
is true; that is, that there exists some $x$ and $y$ that commute with one another.  (We say that $x$ and $y$ commute if $x+y = y+x$.) On the other hand the original statements


*

*$\forall x \in \mathbb{R}, \,\exists y \in \mathbb{R},\,x+y = y+x$

*$\exists x \in \mathbb{R}, \,\forall y \in \mathbb{R},\,x+y = y+x$


say respectively that


*

*For every $x$, there is a $y$ that commutes with it, and

*There is an $x$ such that every $y$ commutes with it.


Unfortunately this is not a very good example to clarify the meaning of $\forall$ versus $\exists$, because in $\mathbb{R}$ we have that every $x$ commutes with every $y$, so that all of the statements are trivially true.
