Probably the easiest way to solve these types of problems is to run scenarios.

Let x, which must be odd, = 1
Let y, which must be even, = 2

(A) 3(1)+2/1 = 3+1 = 4

So (A) could be even. That's your answer. No need to test further.

lheiannie07 wrote:If y is an even integer and x is an odd integer, which of the following expressions could be an even integer?

(A) 3x + y/2
(B) (x + y)/2
(C) x + y
(D) x/4 − y/2
(E) x^2 + y^2

Notice the problem is a "could be" problem, not a "must be" problem. Let's go through each answer choice, recalling that odd + odd = even, and odd + even = odd, and even + even = even.

A) 3x + y/2

3x is odd and y/2 could be odd (for example, if y = 6). Thus 3x + y/2 could be even.

The answer must be A. However, as an exercise, let' verify that none of the other four choices could be even. That is, they will never be even.

B) (x + y)/2

x + y is odd and thus (x + y)/2 is not even an integer (no pun intended).

C) x + y

x + y is odd.

D) x/4 - y/2

x/4 is not an integer and y/2 is an integer, but their difference will not be an integer.

E) x^2 + y^2

x^2 is odd and y^2 is even, but their sum will be odd. Recall here that odd x odd = odd and even x even = even.

Answer: A