kylexy wrote:
If |x|*y+ 9 > 0, and x and y are integers, is x < 6?
(1) \(y<0\).
(2) \(|y|\leq{1}\)
I can move around \(|x|\) without effecting the inequality since
absolute value will always return a positive number, correct? I re-wrote \(|x|*y+ 9>0\)
to \(y>\frac{-9}{|x|}\) before I went ahead. I think this makes statement 1 obsolete so I am starting to doubt my reasoning... I did get the right answer though... Help?
1) Doesn't provide us with any new information since
we already know \(y\) has to be a some negative number from the new equation. We know \(y\) is some integer.
Thus \(|x|\) has to be less than or equal to 9 and must be a factor of 9. Limits possible choices for \(x\) to {-9. -3, -1, 1, 3, 9}. Insufficient.
2) This equation can be understood as 'the distance between \(y\) and the origin is less than or equal to 1'. Thus possible value for \(y\) are {-1, 0, 1}.
\(y\) cannot equal 0 since the numerator is already defined as -9. 1 for \(y\) is also impossible since \(y\) has to be a negative number. So \(y\) is -1 and \(x\) can either be 9 or -9. Insufficient.
1+2) Pretty much the same reasoning as statement 2.
Am I over looking some fact? Even if this is correct I think I like Bunuel's solution better. Just plug handful of numbers to test yes/no possibilities. Far more quick and effective.
1. The absolute value is always more than or equal to zero, thus \(|x|\geq{0}\), not \(|x|>0\). Since |x| can be zero you cannot write \(y>\frac{-9}{|x|}\) from \(|x|*y+ 9 > 0\). We cannot divide by zero.
2. Even if we had \(y>\frac{-9}{|x|}\) it still does not mean that y must be a negative number: \(y>\frac{-9}{|x|}=negative\) --> y is greater than some negative number (-9/|x|), it can be negative as well as positive.
3. y being integer does not mean that -9/|x| must be an integer (|x| must not be a factor of 9). Consider x=5 and y=-1.
4. For (2) y can be -1, 0, or 1. We don't know for (2) that y must be negative.