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If x and y are integers such that x^2+2x+2y+4=2x^2+3x+y-2, we can dedu

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If x and y are integers such that x^2+2x+2y+4=2x^2+3x+y-2, we can dedu  [#permalink]

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New post 06 Feb 2016, 19:00
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A
B
C
D
E

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Question Stats:

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If x and y are integers such that x^2+2x+2y+4=2x^2+3x+y-2, we can deduce that y is

A. not an even
B. an even
C. not an odd
D. an odd
E. a prime


* A solution will be posted in two days.

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Re: If x and y are integers such that x^2+2x+2y+4=2x^2+3x+y-2, we can dedu  [#permalink]

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New post 07 Feb 2016, 00:45
x^2 + 2x + 2y + 4 = 2x^2 + 3x + y - 2
2(x + y + 2) = x^2 + 3x + y - 2
even = x^2 + 3x + y - 2
If x = even --> even + even + y - even = even --> y has to be even
If x = odd --> odd + odd + y - even = even --> y has to be even

Answer: B
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If x and y are integers such that x^2+2x+2y+4=2x^2+3x+y-2, we can dedu  [#permalink]

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New post 07 Feb 2016, 02:11
1. Given: \(x^2 + 2x + 2y + 4 = 2x^2 + 3x + y - 2\)
2. \(y + 4 = x^2 - x - 2\)
3. \(y = x^2 - x - 6\)
4. We know that even - even = even, and that odd - odd = even
5. We see \(x^2 - x\) in (3). This will always translate to one of the two statements: even - even or odd - odd. The result of either statement will be even.
6. We plug in (5) into our simplified formula: \(y = even - 6\). Since even - even = even, we know that y = even.

Therefore B, y must be even.
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Re: If x and y are integers such that x^2+2x+2y+4=2x^2+3x+y-2, we can dedu  [#permalink]

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New post 07 Feb 2016, 05:20
MathRevolution wrote:
If x and y are integers such that x^2+2x+2y+4=2x^2+3x+y-2, we can deduce that y is

A. not an even
B. an even
C. not an odd
D. an odd
E. a prime


* A solution will be posted in two days.

\(2x^2-x^2+3x-2x+y-2y-2-4=0\)
\(x^2+x-y-6=0\)

Let's assume x is even
Then even+even-y-even=even
even - y=even,
y is even

Let's assume x is odd
Then odd-odd-y-even=even
even-y-even=even
y is even

Y is even in any case
Answer: B
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Re: If x and y are integers such that x^2+2x+2y+4=2x^2+3x+y-2, we can dedu  [#permalink]

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New post 08 Feb 2016, 18:37
If x and y are integers such that x^2+2x+2y+4=2x^2+3x+y-2, we can deduce that y is

A. not an even
B. an even
C. not an odd
D. an odd
E. a prime


--> In x^2+2x+2y+4=2x^2+3x+y-2, y=x^2+x-6=x(x+1)-6 is derived. Since x(x+1) is multiplication of consecutive integers, it is always an even number. Then, y=even number-6=even number. Therefore, the answer is B.
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Re: If x and y are integers such that x^2+2x+2y+4=2x^2+3x+y-2, we can dedu  [#permalink]

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New post 24 Jun 2018, 21:52
MathRevolution wrote:
If x and y are integers such that x^2+2x+2y+4=2x^2+3x+y-2, we can deduce that y is

A. not an even
B. an even
C. not an odd
D. an odd
E. a prime


* A solution will be posted in two days.


Don't B and C mean the same thing. Given x and y are integers and if Y is not an odd integer then it must be an even integer! If an integer is not odd then it must be even , must it not ?

Unlike positive and negative where an integer need not be positive if it not negative ( e.g. zero ), even and odd have no intermediate category unless I am missing something .

according to wikipedia " An integer that is not an odd number is an even number."

https://simple.wikipedia.org/wiki/Odd_number.

If I am correct then the answer choices in this question are ambiguous . Kindly share your views too.
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Re: If x and y are integers such that x^2+2x+2y+4=2x^2+3x+y-2, we can dedu &nbs [#permalink] 24 Jun 2018, 21:52
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