Quantifier Order Matters? Does the quantifier order matter in this problem? I don't think it does.
(∃y in R)(∀x in R)(x+y=x) For some real number y and for all real numbers x, x plus y equals x. I assert this to be true. (I won't bother with the proof unless somebody really wants it)
I feel that, although the quantifiers are switched, the sentence is still true. Is this right?
(∀x in R)(∃y in R)(x+y=x) For all real numbers x there is a real number y such that x plus y = x
I also feel that these sentences are also all true, regardless of their order. Am I missing something here? 
(∃y in R)(∀x in R)(x+y=0) and (∀x in R)(∃y in R)(x+y=0). 
Plus these two sentences (∃y in R)(∀x in R)(xy=1) and (∀x in R)(∃y in R)(xy=1)

Just because these two sentences have the same truth-value (true for both) does not mean they are equivalent. Indeed, please do not think that just because the truth-value didn't change for this example, that they will never change in general, as in general, swapping quantifiers will change the meaning of the statement.
Indeed, your follow-up examples are a perfect case in point:
$\exists y \in R \: \forall x \in R \: x+y=0$ is false, (since there is no $y$ which has the property such that if you add it to any $x$, it will add to 0)
But
$\forall x \in R \: \exists y \in R \: x+y=0$ is true! (since for any $x$ we can pick $y =-x$)
