Algebraic approach"Is |x+y|>|x-y|?
a) |x|>|y|
b) |x-y|<|x|
Is |x+y| > |x-y|?
When there is absolute value notation on each side, we can square the inequality.
(x+y)² > (x-y)²
x² + 2xy + y² > x² - 2xy + y²
4xy > 0
xy > 0.
Question rephrased: Do x and y have the same sign?
Statement 1: |x| > |y|
Here, x and y could have the same sign or different signs.
INSUFFICIENT.
Statement 2: |x-y| < |x|
Squaring both sides, we get:
(x-y)² < x²
x² - 2xy + y² < x²
y² < 2xy
xy > y²/2.
Since the square of a value cannot be negative, y²/2 cannot be negative.
Thus, xy>0, implying that x and y have the same sign.
SUFFICIENT.
The correct answer is B.
Number line approach:
|x-y| = the distance between x and y.
|x+y| = the distance between x and -y.
|x| = the distance between x and 0.
Is |x-y| < |x+y|?
In words:
Is the distance between between x and y less than the distance between x and -y?
For x to be CLOSER TO Y than to -y, x and y must be TO THE SAME SIDE OF 0.
To illustrate:
x.....y..............0.............-y
.......y.....x.......0.............-y
As these examples show, when x and y are to the same side of 0, x is closer to y than to -y.
Question rephrased: Are x and y to the same side of 0?
Statement 1: |x| > |y|
x and y could be to the same of 0 or to opposite sides of 0.
INSUFFICIENT.
Statement 2: |x-y| < |x|
In words:
The distance between x and y is less than the distance between x and 0.
This will not be true if x and y are to opposite sides of 0.
To illustrate:
x.........0...........y
y.........0..........x
In each case here, the distance between x and y is GREATER than the distance between x and 0.
Thus, to satisfy statement 2, x and y must be TO THE SAME SIDE OF 0.
SUFFICIENT.
The correct answer is B.
