lets evaluate the second statement: |y|<=12 ---> -12<=y<=12, thus, y can have integer values within this range. Now, we need to think about x.

Lets consider the first statement 2x+y=12 ----> 2x=(12-y), now for x to be an integer, (12-y) should be even (so that it gets divided by 2) ---> that means y should be even integer. There are 13 even values for y in the range -12<=y<=12, thus, the answer is D

thank you so much. don't we have check each even values of x and y in this range to see if their sum is 12

pankajks2010 wrote:lets evaluate the second statement: |y|<=12 ---> -12<=y<=12, thus, y can have integer values within this range. Now, we need to think about x.

Lets consider the first statement 2x+y=12 ----> 2x=(12-y), now for x to be an integer, (12-y) should be even (so that it gets divided by 2) ---> that means y should be even integer. There are 13 even values for y in the range -12<=y<=12, thus, the answer is D

one need not check individually, because for every value of y that you would put in the equation, you would get a corresponding value of x which would satisfy the equation. So, its just about finding the values of y which fit the bill.

For example: lets take y=-12, if we substitute this in the equation: 2x+y=12; 2x-12=12; 2x=24;x=12
similarly, for every possible value of y, we would get a corresponding x, which would satisfy the equation. Hope this helps!!