在google上找到了很好的解答：

There are 4 cases that will satisfy xy + z = odd
(i) x = even, y = even, z = odd
(ii) x = even, y = odd, z = odd
(iii) x = odd, y = even, z = odd
(iv) x = odd, y = odd, z = even

is x even?

Statement 1:

xy + xz = even
x(y + z) = even

case (a): x = odd, (y + z) = even. this is not possible since this will not satisfy any of the cases(i) - (iv).

case (b): x = even, (y + z) = odd/even. this satisfies cases (i) and (ii).

so, since only case b is valid, x must be even. sufficient.

Statement 2:

y + xz = odd

case (c): y = odd, xz = even. this satisfies cases (ii) and (iv) where x could either odd or even. insufficient.
case (d): y = even, xz = odd. this satisfies case (iii), where x is odd.

so, since this statement doesn't provide any restriction on x, it is not sufficient.

Choose A.