Bunuel wrote:
If xyz ≠ 0, is x(y + z) = 0?
(1) |y + z| = |y| + |z|
(2) |x + y| = |x| + |y|
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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Since xyz ≠ 0, we have x ≠ 0 and the question ask if y + z = 0 or not.
Condition 1)
Remind that |y+z| = |y|+|z| is equivalent to yz ≥ 0.
Since xyz ≠ 0, we have yz ≠ 0 and yz > 0.
Thus x(y+z) can not be zero since both y and z are positive or negative.
Since condition 1) yields a unique solution "No", it is sufficient.
Since 'no' is also a unique answer by CMT (Common Mistake Type) 1, both conditions are sufficient, when used together.
Condition 2)
If x = 1, y = 1, z = -1, then x(y+z) = 0 and the answer is "yes".
If x = 1, y = 1, z = 1, then x(y+z) = 2 one the answer is "no".
Since condition 2) does not yield a unique solution, it is not sufficient.
Therefore, A is the answer.