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

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Manager
Joined: 23 Mar 2008
Posts: 214
2x+y=12 IyI<=12 For how many ordered pairs (x,y) that are  [#permalink]

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11 May 2008, 06:14
1
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
Joined: 10 Sep 2007
Posts: 892

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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.

Current Student
Joined: 28 Dec 2004
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Location: New York City
Schools: Wharton'11 HBS'12

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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
Joined: 27 Jun 2007
Posts: 187

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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)
Intern
Joined: 29 Apr 2008
Posts: 20

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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
Manager
Joined: 12 Feb 2008
Posts: 174

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13 May 2008, 16:28
13 here as well
oi oi
Non-Human User
Joined: 09 Sep 2013
Posts: 8509
Re: 2x+y=12 IyI<=12 For how many ordered pairs (x,y) that are  [#permalink]

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

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