Mission2012 wrote:Is x2-y2>0?
1> x2-y2>(x-y)
2> x2-y2<(x+y)
An alternate approach is to plug in a value of y and solve for x.
Statement 1: x²-y² > x-y
If y=1, we get:
x² - 1¹ > x - 1
x² > x
Either x<0 or x>1.
If x=-1 and y=1, then x²-y² = 1-1 = 0.
In this case, x²-y²=0.
If x=2 and y=1, then x²-y² = 4-1 = 3.
In this case, x²-y²>0.
INSUFFICIENT.
Statement 2: x²-y² < x+y
If y=1, we get:
x² - 1¹ < x+1
x² - x - 2 < 0.
(x+1)(x-2) < 0.
Here, the CRITICAL POINTS are x=-1 and x=2.
These are the values where (x+1)(x-2) = 0.
To determine the range(s) where (x-2)(x+1) < 0, test ONE VALUE TO EACH SIDE of x=-1 and x=2.
If we plug x=-2, x=0, and x=3 into (x-2)(x+1) < 0, only x=0 works.
Since only the tested value BETWEEN -1 AND 2 works, (x-2)(x+1) < 0 only when -1<x<2.
To save time, consider values for x that also the constraint in statement 1 that x<0 or x>1.
If x=-1/2 and y=1, then x²-y² = 1/4 - 1 = -3/4.
In this case, x²-y²<0.
If x=3/2 and y=1, then x²-y² = 9/4 - 1 = 5/4.
In this case, x²-y²>0.
INSUFFICIENT.
Statements combined:
Both statements are satisfied by x=-1/2 and y=1.
In this case, x²-y²<0.
Both statements are satisfied by x=3/2 and y=1.
In this case, x²-y²>0.
INSUFFICIENT.
The correct answer is
E.
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