Is |x-y| > |x+y|?NandishSS wrote:Is |x−y|>|x+y| ?
(1) x^2−y^2=9
(2) x−y=2
OA: C
When each side of an inequality is an absolute value, we can square the inequality.
(x-y)² > (x+y)²
x² - 2xy + y² > x² + 2xy + y²
0 > 4xy
0 > xy.
The resulting inequality is true only if x and y have different signs.
Question stem, rephrased:
Do x and y have different signs?
Statement 1: x² - y² = 9
If x=5 and y=4, then x and y have the same sign.
If x=5 and y=-4, then x and y have different signs.
INSUFFICIENT.
Statement 2: x-y = 2
If x=3 and y=1, then x and y have the same sign.
If x=1 and y=-1, then x and y have different signs.
INSUFFICIENT.
Statements combined:
x² - y² = 9 can be rephrased as follows:
(x+y)(x-y) = 9.
Substituting x-y = 2 into the blue equation, we get:
(x+y)(2) = 9
x+y = 9/2.
Since we have two variables (x and y) and two distinct linear equations (x-y=2 and x+y=9/2), we can solve for the two variables, enabling us to determine whether x and y have different signs.
SUFFICIENT.
The correct answer is C.
