Math Help!?

Question

I don't want the answer I just need like an example that's similar to this problem.
So Deric and Andrew go on a bike ride. Deric goes 8mph and gets an one hour head start. Andrew follows behind at 10mph.How many hours will it be before Andrew catches up? So now I have;
8(x)
10(x-1) = 10x-10
What are the steps to finding the answer?

hey! I do pay attention, it just didn't make sense the first time genius! *sarcasm* you don't talk crap to people if they can't get it the first time steven b

thank-you everone who helped! you are all saviors!

Answer 1

First start by calling Deric's position x, Andrew's postion is y
and t is the time elapsed since Andrew started.
Since we want to find the point where Andrew and Deric are at the same point, we need to find out when x=y. That is, for what t is x=y.
So let's figure out an equation which describes the two riders' positions.
Deric's position----x=8mi +(8mi/hr)t (where t is in hours and the 8 miles added represents the head start).
Andrew's position--- y=(10mi/hr)t
Set these equations equal, so

8mi+(8mi/hr)t=(10mi/hr)t
8mi=(10mi/hr-8mi/hr)t
Now solve for t to find out WHEN they passed.
8mi=(2mi/hr)t
(8mi)/(2mi/hr)=t
t=4 hrs
Notice that( 2mi/hr)(4hrs)=8 miles
So how far down the road did they pass?
Well in 4 hrs, Andrew went (10mi/hr)(4hrs)=40 miles
In 4 hrs Deric went (8mi/hr)(4hrs)=32 miles, but he had an 8mi
head start, so after 4 hours, Deric was at the 40 mile mark as well.
So they passed 4 hours after Andrew started and had both traveled 40 miles.
The steps to take:
Write equations which decribe the positions of each rider in terms of the speed they are traveling and the time they are traveling. Remember (speed)(time)=distance traveled.
Solve for t, the time traveled. In these types of problems I like to start my t at the point when both parties are traveling simultaneously.
Put the t you calculate back into the (speed)(time) equation to find the distance traveled for each party.
Good luck.

Answer 2

Then you set them equal to each other
8x = 10x -10
10 = 2x
x = 5

Answer 3

in one hour Dreic would have gone 8 miles......so try setting it up like this

10x=8x+8 8 for the 8 mile head start


also to check you answer think about it logically....andrew is going 2MPH faster had has to make up 8 miles......he gains 2 miles per hour