If x and y are nonzero integers, is (xˉ¹ + yˉ¹)ˉ¹ > [(xˉ¹)(yˉ¹)]ˉ¹ ?
(1) x = 2y
(2) x + y > 0
Work from the INSIDE OUT:
xˉ¹ + yˉ¹ = 1/x + 1/y = (x+y)/xy.
xˉ¹yˉ¹ = 1/(xy).
Substituting the resulting expressions back into the question stem, we get:
[ (x+y)/xy ]ˉ¹ > [ (1/xy) ]ˉ¹
xy/(x+y) > xy.
Question stem, rephrased:
Is xy/(x+y) > xy?
Statement 1: x=2y
Substituting x=2y into xy/(x+y) > xy, we get:
(2y*y)/(2y+y) > (2y*y)?
2y²/3y > 2y² ?
Since y is nonzero, 2y²>0.
Thus, we can safely divide each side by 2y².
The question stem then becomes:
1/(3y) > 1?
Since y is a nonzero integer, 1/(3y) is equal to a positive or negative fraction between -1 and 1.
Thus, the answer to the question stem is NO:
It is not possible that 1/(3y) > 1.
SUFFICIENT.
Statement 2: x+y > 0
Case 1: x=y=1
Substituting x=y=1 into xy/(x+y) > xy, we get:
(1*1)/(1+1) > 1*1
1/2 > 1.
NO.
Case 2: x=-1 and y=3
Substituting x=-1 and y=3 into xy/(x+y) > xy, we get:
(-1*3)/(-1+3) > -1*3
-3/2 > -3.
YES.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.
The correct answer is
A.
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