Jon recently drove to visit his parents who live 270 270 miles away. On his way there his average speed was 24 24 miles per hour faster than on his way home (he ran into some bad weather). If Jon spent a total of 12 12 hours driving, find the two rates.

He drove there at 60 mph, and he drove back at 36 mph.

Step-by-step explanation:

one way distance = d = 270 miles

average speed on way back = s

average speed on the way there = s + 24

time driving there = t

time driving back = 12 - t

average speed = distance/time

distance = speed * time

going there:

270 = (s + 24)t

270 = st + 24t

going back

270 = s(12 - t)

270 = 12s - st

We have a system of equations:

270 = st + 24t

270 = 12s - st

Solve the first equation for t.

t(s + 24) = 270

t = 270/(s + 24)

Substitute in the second equation.

270 = 12s - s[270/(s + 24)]

270 = 12s - 270s/(s + 24)

Multiply both sides by s + 24.

270s + 6480 = 12s^2 + 288s - 270s

12s^2 - 252s - 6480 = 0

Divide both sides by 12.

s^2 - 21s - 540 = 0

(s - 36)(s + 15) = 0

s = 36 or s = -15

The average speed cannot be negative, so we discard the solution s = -15.

s = 36

s + 24 = 60

He drove there at 60 mph, and he drove back at 36 mph.

answer: a

step-by-step explanation: