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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 14 Jun 2017, 05:59
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If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis


SOLUTION (OE) IS HERE.

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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 23 Jun 2017, 12:33
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OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.


(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.


Answer: E.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 14 Jun 2017, 06:49
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If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis
Show more


So the question basically asks if Y=0

(x^2 - y^2)^(1/2) = x + y

Squaring both sides

(x+y)(x-y) = (x+y)^2
x-y = x+y
y=0

(1) xy is not a square of an integer
Since \(-x \leq y \leq x\) and xy is not a square means x is not equal to y
But x+y > 0

(\(\sqrt{x^2 - y^2}\) is +ve) => x^2>y^2

If y = 0 then xy=0 which is a square of an integer hence y is not zero
and x cannot be 0 because y will be negative then as x>y and then x+y becomes negative but we know that x+y>0

Hence Sufficient

(2) Point (x,y) is not on x axis shows that y can be anything but 0. Sufficient

Hence both can solve
Answer D

Please correct me if i am wrong
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 14 Jun 2017, 11:16
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Bunuel
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis
Show more

As per question stem -
\(\sqrt{x^2 - y^2} = x + y\). squaring it we get

\(x^2 - y^2= x^2 + y^2 + 2xy\) solving this we get
\(y^2+xy=0\) OR y(y+x) = 0
So the question asks us IS y=0 or y=-x

Statement 1: xy is not a square, implies that "y" is not equal to 0 but y can be equal to -x. For e.g x=-2, y=2, then xy = -4, which is not a perfect square as square cannot be negative, or any other value e.g x=2 y=3. Hence insufficient

Statement 2: this implies that x & y are not 0. but x and y can be any number, so y may or may not be equal to -x e.g (2,3) or (2,-2). Hence insufficient

Combining 1 & 2: we know that y is not equal to 0 but "y" and "x" can be any other number. Hence insufficient.

Option E
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 14 Jun 2017, 11:52
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There are flaws in both solutions above, even in the one with correct answer.

P.S. Above question and similar question HERE are brand new questions by me and they seem to be quite tricky.
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If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis

As per question stem -
\(\sqrt{x^2 - y^2} = x + y\). squaring it we get

\(x^2 - y^2= x^2 + y^2 + 2xy\) solving this we get
\(y^2+xy=0\) OR y(y+x) = 0
So the question asks us IS y=0 or y=-x

Statement 1: xy is not a square, implies that "y" is not equal to 0 but y can be equal to -x. For e.g x=-2, y=2, then xy = -4, which is not a perfect square as square cannot be negative, or any other value e.g x=2 y=3. Hence insufficient

Statement 2: this implies that x & y are not 0. but x and y can be any number, so y may or may not be equal to -x e.g (2,3) or (2,-2). Hence insufficient

Combining 1 & 2: we know that y is not equal to 0 but "y" and "x" can be any other number. Hence insufficient.

Option E
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It is mentioned in the stem that -x<y<x. So y cannot be equal to -x.

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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 15 Jun 2017, 07:51
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Bunuel
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis

As per question stem -
\(\sqrt{x^2 - y^2} = x + y\). squaring it we get

\(x^2 - y^2= x^2 + y^2 + 2xy\) solving this we get
\(y^2+xy=0\) OR y(y+x) = 0
So the question asks us IS y=0 or y=-x

Statement 1: xy is not a square, implies that "y" is not equal to 0 but y can be equal to -x. For e.g x=-2, y=2, then xy = -4, which is not a perfect square as square cannot be negative, or any other value e.g x=2 y=3. Hence insufficient

Statement 2: this implies that x & y are not 0. but x and y can be any number, so y may or may not be equal to -x e.g (2,3) or (2,-2). Hence insufficient

Combining 1 & 2: we know that y is not equal to 0 but "y" and "x" can be any other number. Hence insufficient.

Option E
It is mentioned in the stem that -x<y<x. So y cannot be equal to -x.

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app
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It's less than or equal sign there: \(-x \leq y \leq x\), so y can equal to -x, as well as x. In fact -x=y=x=0 is also possible.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 15 Jun 2017, 18:53
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There are flaws in both solutions above, even in the one with correct answer.

P.S. Above question and similar question HERE are brand new questions by me and they seem to be quite tricky.
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Hi Bunuel,

Those questions are awesome. You tagged both question as GMAT Club tests. Have you add them in the tests? i.e should I face them if I take mock exam in GMAT Club tests?
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GMAT Focus 1: 735 Q90 V89 DI81
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 15 Jun 2017, 20:54
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Bunuel
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis
Show more

\(-x \leq y \leq x\) means x is positive or 0..
\(\sqrt{x^2 - y^2} = x + y\) ..
Square both sides..
\(x^2-y^2=(x+y)^2=x^2+y^2+2xy.........2y^2+2xy=0......2y(y+x) =0\)
So we have to find whether y=0 or x=-y...

I tells us that x is not equal to y..
But can be 0 or x=-y
Insuff
II tells us y is not equal to 0..
Insuff..

Combined
X can be = -y or not
Insufficient

E
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 15 Jun 2017, 22:09
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Bunuel
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis
Show more


\(\sqrt{x^2 - y^2} = x + y\) - > this result will come about when \(x^2 - y^2\) is either 0 or 1.
Let's jump on to the statement analysis.
1) xy is not a square of integer.

This means x and y are either different integers or same integers with different signs or they both are 0. Taking all the 3 scenarios,
When x=2 and y=-2, \(\sqrt{x^2 - y^2} = x + y\) is true.
When x=3 and y=2, \(\sqrt{x^2 - y^2} = x + y\) is not true.
When x=0 and y=0, \(\sqrt{x^2 - y^2} = x + y\) is true.

Hence insufficient.


2) Point (x, y) is NOT on x-axis

This implies y is not 0, but does not tell anything about x. Hence this statement in itself is insufficient.


Combining Stmt 1 and Stmt 2: Below two scenarios still hold :

When x=2 and y=-2, \(\sqrt{x^2 - y^2} = x + y\) is true.
When x=3 and y=2, \(\sqrt{x^2 - y^2} = x + y\) is not true.

Hence E.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 16 Jun 2017, 02:46
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Bunuel
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis

\(-x \leq y \leq x\) means x is positive or 0..
\(\sqrt{x^2 - y^2} = x + y\) ..
Square both sides..
\(x^2-y^2=(x+y)^2=x^2+y^2+2xy.........2y^2+2xy=0......2y(y+x) =0\)
So we have to find whether x=0 or x=-y...

I tells us that x is not equal to y..
But can be 0 or x=-y
Insuff
II tells us y is not equal to 0..
Insuff..

Combined
X can be = -y or not
Insufficient

E
Show more

I think you mean either y=o or x =-y

Also I have a question if y = 0, is not xy = 0 which is square of an integer zero??
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 21 Jun 2017, 01:35
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Statement 1:
if y = 0, answer is yes
if y is not 0, answer is no

Statement 1 only tells us that x is not equal to y

Insuff

Statement 2:
tells us that y is not equal to 0
but ignores the possibility of x = y or x = -y

Insuff.

Combining - we know that neither x = y and nor y is 0
however x = -y is still not known

hence E
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 21 Jun 2017, 01:50
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Bunuel - please provide detailed solution for the answer when OA is revealed,
This question is an absolute nightmare
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 23 Jun 2017, 12:34
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If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis


SOLUTION (OE) IS HERE.
Show more

Check fresh tricky GMAT CLUB question.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 23 Jun 2017, 13:34
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Bunuel

OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.


(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.


Answer: E.
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Kudos to you Bunuel

it is really spot on. It is high quality question and answer. But do you plan to add in the GMATCLub tests to experience it in mock test?

I hope so

Thanks
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 23 Jun 2017, 13:38
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Bunuel

OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.


(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.


Answer: E.

Kudos to you Bunuel

it is really spot on. It is high quality question and answer. But do you plan to add in the GMATCLub tests to experience it in mock test?

I hope so

Thanks
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Yes, all new questions will be added to the GMAT Club Tests.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 21 Sep 2017, 01:24
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Bunuel

OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.


(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.


Answer: E.
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I get the answer as B. Can you tell me where I might be making a mistake here:
(b) Point (x, y) is NOT on x-axis
\(\sqrt{x^2 - y^2} = x + y\)
=> \(\sqrt{(x-y)(x+y)} = (x+y)\)
Dividing both sides by \(\sqrt{(x+y)}\), we get
\(\sqrt{(x-y)} = \sqrt{(x+y)}\)

If y = 0 , then x-y = x+y, otherwise they are not equal.
Since y != 0(since point (x,y) is not on x axis), x-y not equal to to x+y, hence \(\sqrt{(x-y)} != \sqrt{(x+y)}\) . So, went with B
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 21 Sep 2017, 01:29
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Bunuel

OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.


(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.


Answer: E.

I get the answer as B. Can you tell me where I might be making a mistake here:
(b) Point (x, y) is NOT on x-axis
\(\sqrt{x^2 - y^2} = x + y\)
=> \(\sqrt{(x-y)(x+y)} = (x+y)\)
Dividing both sides by \(\sqrt{(x+y)}\), we get
\(\sqrt{(x-y)} = \sqrt{(x+y)}\)

If y = 0 , then x-y = x+y, otherwise they are not equal.
Since y != 0(since point (x,y) is not on x axis), x-y not equal to to x+y, hence \(\sqrt{(x-y)} != \sqrt{(x+y)}\) . So, went with B
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Please re-read the solution, paying attention to the highlighted part.

As for your solution, you cannot divide by \(\sqrt{(x+y)}\) because it can be 0 for x = -y and we cannot divide by 0.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 14 Apr 2021, 07:42
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E is the answer. Because multiple interpretations are possible.
Eg. in addition to Bunuel 's first answer, we also have 2y(x+y) = 0. From the question we know x+y can be taken as 2x since </=. then 4xy=0 => xy=0. then statements are not satisfied.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 20 Jul 2021, 15:38
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Bunuel

OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.


(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.


Answer: E.
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Hi Bunuel,

Is it as a rule I must never do the following?

Squaring both sides
(x+y)(x-y) = (x+y)^2
x-y = x+y
y=0

Should we always expand the equation and solve? I do not understand why we cannot cancel out (x+y) on both sides.. I understand the answer but the logic behind this one is not clear to me. Please help.
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