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