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For integers, x and y, x=y+2. From the table below, select the values

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goodyear2013
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For integers, x and y, x=y+2. From the table below, select the values [#permalink] New post   Updated on: 02 Oct 2023, 12:09
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For integers, \(x\) and \(y, x=y+2.\) From the table below, select the values that could correspond to \(x\) and \(x^2 − y^2.\)
\(x\) \(x^2\)- \( y^2\) Values
21
23
25
36
40
80
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\(x\): 21 \(x^2\)- \( y^2\): 80

Originally posted by goodyear2013 on 06 Jul 2014, 05:45.
Last edited by BottomJee on 02 Oct 2023, 12:09, edited 16 times in total.
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Re: For integers, x and y, x=y+2. From the table below, select the values [#permalink] New post   29 Oct 2019, 23:15
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Hi, is the question framed correctly?

When X=21, then Y=19 & (X^2-Y^2)= (441-361) = 80.

As per the given condition, with increasing X and Y, the value of (X^2-Y^2) will also increase. So there will be no value in the the table which will correspond to both X & (X^2-Y^2).
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Re: For integers, x and y, x=y+2. From the table below, select the values [#permalink] New post   08 Feb 2020, 23:57
rocky620 wrote:
Hi, is the question framed correctly?

When X=21, then Y=19 & (X^2-Y^2)= (441-361) = 80.

As per the given condition, with increasing X and Y, the value of (X^2-Y^2) will also increase. So there will be no value in the the table which will correspond to both X & (X^2-Y^2).


You're right. There's a problem with the values given or the equations given. I would change the values. Doesn't work
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Re: For integers, x and y, x=y+2. From the table below, select the values [#permalink] New post   25 Sep 2020, 02:53
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x²-y² = y²+2y+4-y²=2*(y+2)
thus x²-y² should be a multiple of 2=> 2*(y+2) = 36 or 40 or 80
when x²-y²= 2*(y+2)=36 => y=16 and x=18 reject this answer
when x²-y²= 2*(y+2)=40 => y=18 and x=20 reject this answer
when x²-y²=2*(y+2)=80 => y=38 and x=40 correct answer
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Re: For integers, x and y, x=y+2. From the table below, select the values [#permalink] New post   12 Oct 2020, 19:47
I got x=21,

If x=21, then y=19 and x^2 - y^2 = 80
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Re: For integers, x and y, x=y+2. From the table below, select the values [#permalink] New post   14 Mar 2024, 00:35
 
goodyear2013 wrote:
For integers, \(x\) and \(y, x=y+2.\) From the table below, select the values that could correspond to \(x\) and \(x^2 − y^2.\)


Given : x=y+2 ----> y = x-2
To find : x & (x^2 - y^2).

How can you solve it it in a faster way ?

Bring it in terms of X. Why ? Because , we have to find the value of x and we are given answer choices for value of x.
So,
(x^2 - y^2) = [x^2] - [(x-2)^2] = [x^2] - [x^2 + 4 - 4x] = x^2 - x^2 - 4 + 4x = 4x -4

Therefore, (x^2 - y^2) = 4x-4

Now as soon as you put , x=21 , 4x-4 = (21*4) - 4 = 80.

Hence , 21 and 80 are the answers respectively for x & (x^2 - y^2).

Hope it helps.
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Re: For integers, x and y, x=y+2. From the table below, select the values [#permalink]
14 Mar 2024, 00:35
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