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November 23, 2014

November 23, 2014

Posted by **dante** on Monday, September 12, 2011 at 7:40pm.

- math -
**MathMate**, Tuesday, September 13, 2011 at 8:31am2y+6,3y and y-4 are all integers and y-4 is the median of these intergers

=>

case 1:

2y+6≥y-4 AND 3y≤y-4

2y+6≥y-4 => y≥-10

AND

3y≤y-4 => y≤-2

=>

-10≤y≤-2

or case 2:

2y+6≤y-4 AND 3y≥y-4

solving similarly,

-2≤y≤-10

which is impossible.

Now you only have to figure out the invalid choice from your list.

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