Given: |x| + |y| = 32, x ≠0, y ≠0ruplun wrote:If x and y are non-zero integers and |x| + |y| = 32, what is xy?
(1) -4x - 12y = 0
(2) |x| - |y| = 16
Implies any one of the four,
(1) For x > 0, y > 0 => x + y = 32
(2) For x < 0, y > 0 => -x + y = 32
(3) For x > 0, y < 0 => x - y = 32
(4) For x < 0, y < 0 => -x - y = 32
Statement 1: -4x - 12y = 0
Implies 4x + 12y = 0. Now we have to solve this equation with any one of the four possible equation of the above. Remember each of the combination is a possible result unless contradicts the solution range. Let's solve the four equation pairs.
(1) For x > 0, y > 0 => 4x + 12y cannot be equal to zero as x and y are nonzero. No possible solution.
(2) For x < 0, y > 0 => -4x + 4y = 128 and 4x + 12y = 0 => x = -24, y = 8
(3) For x > 0, y < 0 => 4x - 4y = 128 and 4x + 12y = 0 => x = 24, y = -8
(4) For x < 0, y < 0 => 4x + 12y cannot be equal to zero as x and y are nonzero. No possible solution.
Only two possible solutions are: (x = 24, y = -8) and (x = -24, y = 8). In both cases xy = -(24*8)
Sufficient.
Statement 2: |x| - |y| = 16
Solving this equation with the given one, i.e. |x| + |y| = 32, we get x = ±24 and y = ±8. Thus xy can be (8*24) or -(8*24)
Not sufficient.
The correct answer A.
