Is |x - y| > |x + y|?

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Is |x - y| > |x + y|?

by Vincen » Tue Nov 28, 2017 8:18 am
Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2

The OA is C.

I don't know how to use both statements together. Experts, may you show me how to do it? Thanks.

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by ErikaPrepScholar » Wed Nov 29, 2017 10:50 am
We know from the question stem that we want to find out information about x - y and x + y. We can express Statement 1 as:

(x - y) (x + y) = 9

Using Statement 2, we can substitute 2 in for x - y:

2 (x + y) = 9
x + y = 9/2

So |x - y| = |2| = 2, and |x + y| = |9/2| = 9/2

So |x - y| is NOT greater than |x + y|. Both statements together are sufficient.
Last edited by ErikaPrepScholar on Thu Nov 30, 2017 7:12 am, edited 1 time in total.
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by Jay@ManhattanReview » Wed Nov 29, 2017 8:43 pm
Vincen wrote:Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2

The OA is C.

I don't know how to use both statements together. Experts, may you show me how to do it? Thanks.
(1) x^2 - y^2 = 9

Case 1: Say x = 3 and y = 0, then 3^2 - 0^2 = 9. We see that |x - y| > |x + y| => |3 - 0| > |3 + 0| => 3 = 3. The answer is No.
Case 2: Say x = -4 and y = 1, then (-4)^2 - 1^2 = 9. We see that |x - y| > |x + y| => |-4 - 1| > |-4 + 1| => 5 > 3. The answer is Yes.
No unique answer. Insufficient.

(2) x - y = 2

Case 1: Say x = 2 and y = 0, then 2 - 0 = 2. We see that |x - y| > |x + y| => |2 - 0| > |2 + 0| => 2 = 2. The answer is No.
Case 2: Say x = 1 and y = -1, then 1 + 1 = 2. We see that |x - y| > |x + y| => |1 + 1| > |1 - 1| => 2 > 0. The answer is Yes.
No unique answer. Insufficient.

(1) and (2) combined together

From (2), we get that x = y + 2. Plugging in the value of x in (1), we get x^2 - y^2 = 9 => (y + 2)^2 - y^2 = 9 => (y+2+y).(y+2-y) = 9 => (2y+2).2 = 9 => y = 5/4. Thus, x = 13/4.

At x = 13/4 and y = 5/4, we see that we see that |x - y| > |x + y| => |13/4 - 5/4| > |13/4 + 5/4| => 2 < 9/2 . The answer is No. Sufficient,

The correct answer: C

Hope this helps!

-Jay
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absolute value

by GMATGuruNY » Thu Nov 30, 2017 7:37 am
Vincen wrote:Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2
Is |x-y| > |x+y|?
When each side of an inequality is enclosed in absolute value symbols, we can square the inequality:
(x-y)² > (x+y)²
x² - 2xy + y² > x² + 2xy + y²
0 > 4xy
xy < 0.
Question stem, rephrased:
Do x and y have DIFFERENT SIGNS?

Statement 1: x² - y² = 9
Perfect squares: 1, 4, 9, 16, 25...
The difference between the two perfect squares in blue is 9, implying that one solution for Statement 1 is as follows:
x² = 25, with the result that x = ±5.
y² = 16, with the result that y = ±4.
Since x and y could have the same sign or 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:
Statement 1 implies the following:
(x+y)(x-y) = 9.
Substituting x-y=2 into (x+y)(x-y) = 9, 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 both variables, allowing us to determine whether x and y have the same sign or different signs.
SUFFICIENT.

The correct answer is C.
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