Is x^2 - y^2 > 0? (1) x^2 - y^2 > x - y (2) x^2 - y^2 < x + y

Notice that x^2-y^2 factors to (x+y)(x-y) for this reason, and all of the confusion that can occur from writing so much, missing a sign etc, I like to define a =(x+y), b=(x-y)

then the question becomes ab>0 Yes/NO?

(1) ab>b if b>0, a>1. if b <0, a<1. Suppose b=1,a=2 then ab>0 Yes. Suppose b=-1/2 a=1/2, then ab<0 NO.

Since we get a Yes and a No. (1) is NS

(2) ab<a if a>0,b <1. if a<0, b>1. Suppose a=1, b=1/2, then ab>0 Yes. Suppose a=-1 b=2, then ab<0 No Since we get a Yes and a No NS

(1) and (2) if b>0,a>1, if b<0, a<1, if b>0, a>1, if b<0, a <1, if a>0, b<1, if a<0, b>1. Suppose a=1/2, b=-1 then ab<0 No. Suppose a=1/2, b=1/2, then ab>0 YES. Since we get a Yes and a NO (1) and (2) are NS E.

Thanks for the explanation
I dont think we can divide by x-y. since its no where mentioned that x is not equal to y
similarly for the next given statement x can be equal to -y.

Hi ocelot22

Nice solution. However, I would take your concept in simplicity by the following:

ab>b
ab<a
-----------Subtract the inequalities as different signs

ab -ab > b -a

0 > b -a

a > b....Form here it is easy to judge quickly

If a = 10 & b= 4...Then answer is Yes

If a = -10 & b = 4....Then answer No

Answer: E

hazelnut
Can you please post Veritas solution

Hi, I answered E using the following approach.
(1)
\(x^2-y^2>(x - y)\)
\((x + y)(x - y)>(x - y)\)
\((x + y)>1\)
Insufficient.

(2)
\(x^2-y^2<(x + y)\)
\((x + y)(x - y)<(x + y)\)
\((x - y)<1\)
Insufficient.

Together both statements are still insufficient as the possibility for (x - y) can range from positive fraction < 1 to unlimited negatives.
Let me know if that makes sense.

So Is (x+y)(x-y)>0 ? Yes/No
Let (x+y)= m , (x-y)= n
True ,when m and n have the same sign (+ve•+ve or -ve•-ve)
Question becomes Is (m•n) > 0 ?

(1)(x+y)(x-y)> (x-y)
m•n > n. Let m =2 ,n=3
2•3 > 3 , so m•n >0
But , Let m= 1/2 ,n = -2
1/2•-2 > -2 , but m•n <0
(Not Sufficient)

(2) (x+y)(x-y) < (x+y)
m•n < m. Let m =2 ,n=3
2•3 < 2 , but m•n >0
But, Let m=-1/2 ,n=2
-1/2•2 < -1/2 , so m•n <0
(not Sufficient)

(1+2) so in (1) we had m+ve,n+ve
.: (x+y)(x-y)>0 , Also m+ve ,n-ve
.: (x+y)(x-y) <0
Again, in (2) we had m+ve,n+ve
.: (x+y)(x-y)>0 ,Also m-ve,n+ve
.: (x+y)(x-y)<0
Clearly from the above, we have two overlap of answers:
x^2-y^2 >0 and x^2-y^2 <0

So E

Posted from my mobile device

\(x^2-y^2 > 0\)
\(x^2 > y^2\)
\(|x| > |y|\)

Question stem, rephrased:
Is |x| > |y|?

Let y=2.
Plugging y=2 into each statement, we get:
Statement 1: \(x^2-4 > x-2\) --> \(x^2-x>2\)
Statement 2: \(x^2-4 < x+2\) --> \(x^2-x < 6\)
Statements combined:
\(2 < x^2-x < 6\)

Since \(2^2-2 = 2\) and \(3^2-3=6\), both statements are satisfied if x is BETWEEN 2 AND 3:
\(x = 2.5\)
In this case, |x| > |y|, so the answer to the rephrased question stem is YES.
Since \((-2)^2-(-2)=6\) and \((-1)^2-(-1)=2\), both statements are satisfied if x is BETWEEN -2 AND -1:
\(x = -1.5\)
In this case, |x| < |y|, so the answer to the rephrased question stem is NO.
Since the answer is YES in the first case but NO in the second case, the two statements combined are INSUFFICIENT.

Sry Bunuel for starting a new thread with this question. It was hard to search for.


Here is my issue:

Statement I gives: (x+y)(x−y)>(x−y)

Now, why can't I just delete (x-y) from both sides? I mean divide (x-y) by (x-y). It doesnt matter if its positive or negative, RHS will still be 1. So I should get:

x+y>1

The same idea with statement 2 and combining the statements gives:

x+y>1>x−y

However, in the official explanation, they get: (x+y)>(x^2–y^2)>(x−y)

I still manage to get the correct answer. I just dont really get why my approach is wrong according to the official explanation from Veritas Prep:

"Let’s start by rewriting the equation on the left of both inequalities. x^2–y^2 is the difference of squares, so it can be expressed as (x+y)(x-y). Statement (1) thus says that (x+y)(x-y) > (x-y), so (x+y) > 1 if (x-y) is positive and (x+y) < 1 if (x-y) is negative."


As for the bolded part of the OE: Why does it matter whether (x-y) is positive or negative? I mean:

Let x+y = a, x-y = b

ab > b

If only b is negative, then I see why we have to flip the sign. But we dont know whether a is positive or negative, so we shouldnt be able to flip the sign just by assuming b is negative.

The inequality given is true and it only holds when a and b have the same sign. So either both are negative or both are positive and we don't have to flip any signs.

So - How can ab become a<1 if b is negative? I mean this doesnt hold anymore.

a = 2, b = -2 --> -4>2 - but this is not what we are told. b can only be negative if a is also negative. So ab>b can never become ab<1.

1)
Factored out to
(x+y)(x-y)>(x-y)
Divide both sides by (x-y), noticing (x-y) can't be 0, and we have
(x+y)>1 if x-y>0
(x+y)<1 if x-y<0
x^2-y^2 in the first case is always positive
x^2-y^2 in the second case is negative for 1>x+y>0 and positive for 0>x+y
Insufficient

2)
(x+y)(x-y)<(x+y)
Divide both sides by (x+y), noticing (x+y) can't be 0, and we have
(x-y)<1 if x+y>0
(x-y)>1 if x+y<0
x^2-y^2 in the first case is positive for 1>x-y>0 and negative for 0>x-y
x^2-y^2 in the second case is always negative
Insufficient

Combining the statements is tedious but straightforward.
Take from statement 1)
(x+y)>1 if x-y>0
Fold this into statement 2) (the other part of statement 2 doesn't matter since we have (x+y)>1 from statement 1)
(x-y)<1 if x+y>0

And we have for, given (x+y)>1 and x-y>0, 1>x-y>0, and x^2-y^2 is positive.

Take from statement 1)
(x+y)<1 if x-y<0
Combine with statement 2)
Notice both matter
(x-y)<1 if x+y>0
(x-y)>1 if x+y<0

For 0<(x+y)<1 and (x-y)<0, x^2-y^2<0. For 0<(x+y)<1 and 0<(x-y)<1, x^2-y^2>0.
No need to test other cases.

Insufficient.

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