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Absolute value question

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by cramya » Fri Mar 13, 2009 6:24 am
2x+y=12

x=12-y/2


y has to be even satisfying |y|<=12

12,10,8,6,4,2,0,-2,-4,-6,-8,-10,-12

13 ordered pairs
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by jfranco23 » Fri Mar 13, 2009 7:38 am
how did you get that conclusion?
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by DanaJ » Fri Mar 13, 2009 7:59 am
jfranco23: notice that you have the first condition:
2x + y = 12
Now, 2x is obviously even, and so is 12. This is why you have that:
(even number) + y = (even number), making y even as well.
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by sureshbala » Fri Mar 13, 2009 7:59 am
Given 2x+y = 12
and |y|<=12.
Hence -12<=y<=12.
So y can take 25 values from -12 to 12.

We have 2x= 12-y
i.e x = (12-y)/2= 6 - y/2
Hence x will be integer if y is divisible by 2.
From -12 to 12, there are 13 values which are divisible by 2.

Hence the answer is 13
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