# K’s Semi-solution to “Hard Problem”

September 19th, 2009 by k

I can’t think well after lunch, so I don’t have the solution. But I know you have to have the travelling times of each individual equal each other. And they have to share the bike. x=distance girl rides on bike, y=distance of boy on bike, z=distance of dog on bike.

$x+y+z = 38$ $16(38-x) + 6x = 15(38-y) + 5y = 10 (38-x) +4z$

I think I’m on the right track ,but if not, well, there’s a reason I’m not trying out for math team. Ok, people are coming over to work on the English project now.

Edit/Ben: You’re thinking on the right track with systems of linear equations, with three variables. However, I don’t see why you set $x+y+z = 38$, because $x+y+z$ is the sum of the time the girl, boy, and dog ride the bike, and that shouldn’t have anything to do with the distance they travel. As for the second equation, why are the three equal?

Continue >.< or leave it to Hanchan when you guys finish with your projects.

Edit/2: On second thought I shouldn’t really barge into your post, I should write a comment ^_^