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

Hi

In other words "Is \(x^2 > y^2\) ?

(1) \(x^2-y^2>(x - y)\)

\(x(x - 1) > y(y - 1)\)

We can have \(x>y>0\) -----> \(3>2>0\)

\(3*2>2*1\) --------> \(6^2 > 2^2\) Yes.

Or \(0<x<y\) ---------> \(0<\frac{1}{3}<\frac{1}{2}\)

\(\frac{1}{3}*(-\frac{2}{3}) "?" \frac{1}{2} * (-\frac{1}{2})\) -------> \(-\frac{2}{9} > -\frac{1}{4}\) ----------> \((\frac{1}{3})^2 < (\frac{1}{2})^2\) No. Insufficient.

(2) \(x^2-y^2<(x + y)\)

\(x(x - 1) < y(y + 1)\)

\(x=2, y=4\) ----> \(2*1 < 4*5\) -----> \(2^2 < 4^2\) No.

\(x=\frac{1}{2}, y=\frac{1}{3}\) ---> \(\frac{1}{2}*(-\frac{1}{2}) "?" \frac{1}{3}*\frac{4}{3}\) -----> \(-\frac{1}{4} < \frac{4}{9}\) -------> \((\frac{1}{2})^2 > (\frac{1}{4})^2\) Yes. Insufficient.

(1)&(2) \(x^2 - y^2 > x - y\) & \(x^2 - y^2 < x + y\)

or \(y^2 - x^2 < y - x\) & \(x^2 - y^2 < x + y\)

Adding we'll get: 0 < 2y ----> y>0. But we already have different answers when y is positive (x can be >0, <0, integer or a fraction). Insufficient.

Answer E.

What's the source of the question?

Here's how I solved this - please critique!

Is \(x^{2}\)-\(y^{2}\)>0?

Rewriting the stem: Is (x-y)(x+y)>0?

Statement 1) \(x^{2}\)-\(y^{2}\)>(x-y)

Rewriting: (x-y)(x+y)>(x-y)

If (x-y)>0
Dividing both sides by (x-y) we are left with (x+y)>1.

If (x-y)<0
Dividing both sides by (x-y) we are left with (x+y)<1.

Because we don't know if (x-y) is positive, negative or zero, Statement 1 is insufficient.


Statement 2) \(x^{2}\)-\(y^{2}\)<(x+y)

Rewriting: (x-y)(x+y)<(x+y)

If (x+y)>0
Dividing both sides by (x+y) we are left with (x-y)<1.

If (x+y)<0
Dividing both sides by (x+y) we are left with (x-y)>1.

Because we don't know if (x+y) is positive, negative or zero, Statement 2 is insufficient.

Statement 1 & 2) We are left with a bunch of possibilities and no way to come to 1 final answer; as such, "C" is insufficient.

E is the answer.

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.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________