Question

Please Please help me with this  problem

Answer 1

The work become much simpler if you do it in a table. Hope this helps!!

Answer 2

Assuming that both cyclists are traveling AWAY from each other with the same starting point, consider the following:

Let us denote the speed of the first bicyclist by $s$ (measured in miles per hour). Then, the speed of the second bicyclist must be $s-5$ (miles per hour).

To verify this, assign an arbitrary number, say 17 mph, to the first bicyclist. Since the first bicyclist is always 5 mph faster than the second, we must have 17 - 5 = 12 mph as the second speed.

At the 0th hour,
- first bicyclist has traveled 0 miles
- second bicyclist has traveled 0 miles
- distance between both = 0 miles

At the 1st hour,
- first bicyclist has traveled $1\times s=s$ miles
- second bicyclist has traveled $-1\times(s-5)=-s+5$ miles (note the negative sign here. Since they are traveling in opposite directions, they must have opposite signs of each other in distances traveled)
- distance between both = $s-(-s+5)$ = $s+s-5$ = $2s-5$ miles

At the second hour,
- first bicyclist has traveled $2s$ miles
- second bicyclist has traveled $-2(s-5)=-2s+10$ miles
- distance between both = $2s-(-2s+10)=2s+2s-10=4s-10$ miles

According to the question statement, both of them are 70 miles apart at the second hour, i.e.
                                                $4s-10=70$
such that solving this gives:
     $4s=70+10$
     $4s=80$
     $s= \frac{80}{4}$
     $s=20$ mph

Therefore, their rates are as follows:
- first bicyclist = 20 mph
- second bicyclist = 20 - 5 = 15 mph