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Bambi2021
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Is x^2 - y^2 > 0 ? 18 May 2021, 15:56
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

45% (medium)

Question Stats:

83% (02:06) correct 17% (01:32) wrong based on 6 sessions

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Found this on Veritas Prep.

Is \(x^2–y^2>0\) ?

1. \(x^2–y^2>(x–y)\)

2. \(x^2–y^2<(x+y)\)




I have a question here:
Show Spoiler

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.

OE:
Show Spoiler
Solution: E

Let’s start by rewriting the equation on the left of both inequalities. x2–y2
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; INSUFFICIENT. Statement (2) says that (x+y)(x-y) < (x+y), so we have the same issues; INSUFFICIENT. Together we can express the inequality as (x+y) > (x2–y2) > (x-y), but the statements remain INSUFFICIENT.

For example, if (x+y) = 5 and (x-y) = ½ then the inequality is true and the answer to the original question is “yes.” If (x+y) = ½ and (x-y) = -½, the inequality is true but the answer to the original question is “no.” (E).

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Is x^2 - y^2 > 0 ? 18 May 2021, 15:59
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Bambi2021
Found this on Veritas Prep.

Is \(x^2–y^2>0\) ?

1. \(x^2–y^2>(x–y)\)

2. \(x^2–y^2<(x+y)\)




I have a question here:
Show Spoiler

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.

OE:
Show Spoiler
Solution: E

Let’s start by rewriting the equation on the left of both inequalities. x2–y2
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; INSUFFICIENT. Statement (2) says that (x+y)(x-y) < (x+y), so we have the same issues; INSUFFICIENT. Together we can express the inequality as (x+y) > (x2–y2) > (x-y), but the statements remain INSUFFICIENT.

For example, if (x+y) = 5 and (x-y) = ½ then the inequality is true and the answer to the original question is “yes.” If (x+y) = ½ and (x-y) = -½, the inequality is true but the answer to the original question is “no.” (E).
Show more

Discussed here: is-x-2-y-236550.html?fl=similar

Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Bunuel
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113146 posts