Hi there,

You're right to start off factoring from the first statement as you have: 2(x-y) = 1. From there we can divide both sides of the equation by 2 and get to x-y = 1/2, and then, adding y to both sides, we land at x = y + 1/2. All this tells us is that x is .5 greater than y, but that could be the case if they're both positive, or if they're both negative, or even if x is a just-above-zero positive and y is a just-below-zero negative. So Statement (1) alone is insufficient.

With statement (2), be careful! Remember that when you're dealing with an inequality, if you divide or multiply both sides by a negative number, you must flip the direction of the inequality sign. In this case, if we multiply through by y, we don't KNOW whether we're multiplying through by a positive or a negative number. So, we wind up with two possibilities: if y is positive, the statement becomes x>y, but if y is negative, the statement becomes precisely the opposite -- x<y. This alone isn't enough to tell us whether x and y are both positive -- though it IS enough to tell us that x and y have the same sign (if y is positive, x is "even more" positive; if y is negative, x is "even more" negative. Insufficient alone; now let's combine.

In combination, we know that x and y could both be positive, with x being 1/2 greater than y. Could it equally well be that x and y are both negative? Well, no, because if they're negative, x must be less than y, but we KNOW that x is .5 greater than y. So now the only possibility is indeed that they're both positive. So in combination, the statements are sufficient.

Statement 1) 2x-2y = 1, 2(x-y) = 1, x-y = 1/2, x = y + 1/2 or y = x - 1/2
Here's an alternative way to think about this statement if you have trouble determining where to go from here. Think about how the graph of this equation looks, y as a function of x. It's a line with a slope of 1 and a y-intercept of -1/2. Thus the plot will cross through quadrants I, III, and IV and you know x and y could both be positive, but also that y could be negative while x is positive or both could be negative. Insufficient.

Statement 2) As was pointed out, avoid multiplying or dividing by variables when the sign is unknown. In this case, x and y could both be positive or both be negative.

Whenever I see x/y > 0 or xy > 0, I think of this as "x and y are the same sign." I recommend memorizing this wording as it seems that this concept is heavily tested on the GMAT.

Statements 1 & 2) Combining and substituting does the trick here: with x = y + 1/2 we can substitute into the second inequality, (y + 1/2)/y > 1, (y + 1/2)/y - 1 > 0, (y + 1/2 - y)/y > 0, (1/2)/y > 0, y > 0. As I mentioned above, statement 2 says "x and y are the same sign," so x must be positive as well. The first statement also confirms this: if y is positive the graph will always be in the first quadrant and x will be positive also.