Edited: '0' is a real number as well

(1) x @ y = 1/x + 1/y
y @ x = 1/y + 1/x
Addition will give you the same value, except for x or y =0, when the equation becomes indeterminable. Sufficient

(2) x @ y = x - y
y @ x = y - x

x-y = y-x, only when x=y.

A IMO

the click here is @, it must be @ in all cases

st(1) x@y=(x+y)/xy, we quickly note this statement cannot be solved if x and/or y=0 and 0 is real value Sufficient to answer No
st(2) x@y=x-y, here we have no undefined solution areas so we can conclude Sufficient to answer Yes

d

gmatblood wrote:Is x @ y = y @ x, for all real values of x and y?

Stmt 1: x @ y = 1/x + 1/y
Stmt 2: x @ y = x - y

gmatblood wrote:Is x @ y = y @ x, for all real values of x and y?

Stmt 1: x @ y = 1/x + 1/y
Stmt 2: x @ y = x - y

IMO D

Whts OA?

gmatblood wrote:Is x @ y = y @ x, for all real values of x and y?

Stmt 1: x @ y = 1/x + 1/y
Stmt 2: x @ y = x - y

stmt 1: @ can be any mathematical expression here.

let's take few numbers here.
X=1 and Y=1

so, we have 1@1 = 2


from stmt 2: if I take same values then

1@1 = 0

--

If I combine the 2 statements then -- then I get

1@1 =2 and 1@1 = 0

two contradicting statements?

or am I missing something?

what's the source of this question?

Looks A, since first one will hold for all real values, whereas second one will fail at x=y=0.

What is the OA?

Thanks.