Posted by dante on Monday, September 12, 2011 at 7:40pm.
2y+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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