If |x|×y+ 9 > 0, and x and y are integers, is x < 6?
(1) y is negative
(2) |y| < 1
I'm having trouble with the solution to part 1 of this problem. It reads:
(1) INSUFFICIENT: We know that |x|×y > -9 and that y is a negative integer. Suppose y = -1. Then |x|×(-1) > -9, which means |x| < 9 (since dividing by a negative number reverses the direction of the inequality). Thus x could be less than 6 (for example, x could equal 2), but does not have to be less than 6 (for example, x could equal 7).
I understand that you can simplify to |x|×(-1) > -9, but when you divide out the -1, I'm left with |x|x<9, not |x| < 9. What happens to the extra X?
Thank you