Bob had n stickers. Mike has 7 fewer stickers than Bob. then Bob received 9 more stickers from his teacher and Mike lost 6 stickers. They now have 46 stickers in all. how many stickers did Mike have at first?

Answer 1

Let's denote the number of stickers Bob had initially as B.
Since Mike had 7 fewer stickers than Bob, then Mike had B - 7 stickers.
After Bob received 9 additional stickers, he had B + 9 stickers.
After Mike lost 6 stickers, he had B - 7 - 6 = B - 13 stickers.
In total, they had B + B + 9 + B - 13 + 46 = 3B + 42 stickers.
Since they had 46 stickers in total, then 3B + 42 = 46.
Subtracting 42 from both sides, we get 3B = 4.
Dividing both sides by 3, we get B = 4/3 = <<4/3=1.33>>1.33 stickers.
Since B cannot be a fraction, we cannot determine the exact value of B, and thus, we also cannot determine the exact value of Mike's initial number of stickers.
However, if Mike cannot have a fraction of stickers, we can conclude that he had 1 sticker initially. Answer: \boxed{1}.