Question 8070
You can use the distance formula: d = rt  (distance = rate X time)

For the outbound trip:
 d1 = (r1)(t1)
d1 = (r1)(5 hrs)

For the return trip:

d2 = (r2)(t2)
d2 = (r2)(4 hrs)

But, d1 = d2 (Same distance outbound as return)
And, r2 = r1+5 (Return speed was 5 mph more than the outbound speed)

So we can write: d1 = d2 or;

(r1)(5) = (r1+5)(4)  Simplify and solve for r1.
5(r1) = 4(r1) + 20
r1 = 20 mph  Outbound speed.
r2 = r1+5 mph = (20+5) mph = 25 mph Return speed.