Is x > 0 ?
(1) xy + y = y
(2) xy + x = x
Answer was given as E, but why couldn't A 1 be sufficient?
Is my algebra incorrect if I factor out Y?
xy + y = y
y(x+1)=y
(x+1)=1
x=0
sufficient
this was a kaplan question categorized as difficult
Given explanation:
The question stem does not give any information about x, so let's look at the statements.
Statement (1): insufficient. Let's subtract y from both sides of the equation xy + y = y. Then xy = 0. When the product of a group of numbers is 0, at least one of the numbers must be 0. So either x = 0 or y = 0. If y = 0 and x is any positive number, (for example, x = 7), then xy = 0, and x > 0; so statement (1) is true, and the answer to the question is "yes." If x = 0, it does not matter what y equals because xy = 0, but x is not greater than 0; so statement (1) is true, but the answer to the question is "no." Since there can be more than one answer to the question, Statement (1) is insufficient. We can eliminate choices (A) and (D).
Note that when dealing with the equation xy = 0, you cannot simply divide both sides by y to find that x = 0, because it might have been the case that y = 0, and you cannot divide by 0.