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2x+y=12 IyI<=12 For how many ordered pairs (x,y) that are

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Manager
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2x+y=12 IyI<=12 For how many ordered pairs (x,y) that are [#permalink]

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New post 11 May 2008, 06:14
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2x+y=12
IyI<=12

For how many ordered pairs (x,y) that are solutions of the system above are x and y both integers?

a) 7
b)10
c)12
d)13
e) 14

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Director
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Re: solutions [#permalink]

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New post 11 May 2008, 07:15
|y|<=12 implies -12<=y<=12
Since we are interested only in integer values so y can only be -12, -11, -10, -9....,9, 10, 11, 12

For 2x+y=12 and x also to be integer, the old value of y cannot hold true. As that will produce x with fractional values.

So the number of possible values for even Y is the answer, which is 13.

Answer D.
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Re: solutions [#permalink]

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New post 12 May 2008, 09:05
i too get D..

2x+y=12

|y|<=12..

we are looking at integers ONLY..

range of y -12<=y<=12..25 possible values..

however 2X is always even..so Even+Even=Even.. we cannot have any Y which is ODD..

therefore the possible values of y are -12, -10, -8, -6, -4, -2, 0 2, 4, 6, 8, 10, 12..13 in all..
Manager
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Re: solutions [#permalink]

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New post 12 May 2008, 10:16
puma wrote:
2x+y=12
IyI<=12

For how many ordered pairs (x,y) that are solutions of the system above are x and y both integers?

a) 7
b)10
c)12
d)13
e) 14


D.

(0,12)
(1,10)
(2,8)
(3,6)
(4,4)
(5,2)
(6,0)
(7,-2)
(8,-4)
(9,-6)
(10,-8)
(11,-10)
(12,-12)
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Re: solutions [#permalink]

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New post 12 May 2008, 10:45
13
x = (12-y)/2 for x to be integer y has to be even
|y| <=12 means -12<=y<=12 ..total 13 even values
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Re: solutions [#permalink]

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New post 13 May 2008, 16:28
13 here as well
oi oi
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Re: 2x+y=12 IyI<=12 For how many ordered pairs (x,y) that are [#permalink]

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New post 02 Dec 2017, 02:47
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Re: 2x+y=12 IyI<=12 For how many ordered pairs (x,y) that are   [#permalink] 02 Dec 2017, 02:47
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