OG (10th ed) - DS - #233

[Saurabh Malpani]

Is (x-y)/((x+y) > 1

a) x>0
b) y<0

I am not sure how B is wrong.

My way to solve the problem:

(x-y)>(x+y)

x-y>x+y ---Cancelling X on both the sides.

-y>y
0>y

which is B) so it's B the answer???

[StaceyKoprince]

When you multiple or divide by a negative number in an inequality, the sign switches. In this case, when you multiply by (x+y) to move it to the right side of the equation, you don't know whether (x+y) represents a positive number or a negative number - so the problem splits into two parts:

x-y > x+y IF (x+y) is positive
x-y < x+y IF (x+y) is negative

Statement 1: x>0 this is not enough to tell me whether (x+y) is positive or negative, so both paths are still open. The two paths (the two inequalities above) directly contradict each other, so I can't answer the question.

Statement 2: y<0 this is also not enough to tell me whether (x+y) is positive or negative, so both paths are still open. The two paths (the two inequalities above) directly contradict each other, so I can't answer the question.

Statements 1+2: x>0 and y<0 this is still not enough to tell me whether (x+y) is positive or negative, so both paths are still open. The two paths (the two inequalities above) directly contradict each other, so I can't answer the question.

So the right answer is E.