Question 403223
three key relationships for any vehicle in this case are:

(a) distance  = (speed)(time)
(b) speed = distance /time
(c) time = distance /speed


for the car:
distance = d = 15 miles
speed = v 
time = t 


for the tractor:
distance = D = 10 miles
speed = v - 15
time = T 


tractor time = car time, so


T = t


and using key relationship (c), we have:


10/(v-15) = 15/v


we can keep the equivalence if we multiply each side by the product of the denominators


10v = 15(v-15)
that is the same as

10v = 15v - 225


solving for v:


-5v = -225

5v = 225

v = 225/5

v = 45 miles per hour
V = v - 15 = 30 miles per hour


the rate of the tractor is 30 miles per hour