Can someone help with how to solev this? i cannot get the correct answer?

Question

Mark drove 150 mi to visit Sandra. Returning by a shorten route, he found that the trip was only 130 mi, but traffic slowed his speed by 6 miles. If the two trips took exactly the same time, what was his rate on the return trip?
Avail answers are: 40,39,36,25 Thanks!!

Thank you to those of u who truly answered! I appreciate your help

Answer 1

Distance = Rate * Time

D = RT

Solving for T:
T = D/R

Let R be the rate on the return trip:
Let R + 6 be the faster rate on the first leg.

Trip time to see Sandra:
t = 150 / (R + 6)

Return trip:
t = 130 / R

Because these are equivalent you can equate them:
150 / (R + 6) = 130 / R

Cross multiply:
150R = 130(R + 6)
150R = 130R + 780
20R = 780
R = 780 / 20
R = 39

On the trip out he traveled 45 mph, on the return 39 mph.
(He took 3 hours 20 minutes each way).

The answer is 39 mph on the return trip.

Answer 2

t = time, 150/t = rate on going trip, 130/t - 6 = rate on return trip

rate on going trip:
150/r = 130/(r - 6)
15(r - 6) = 13r
15r - 90 = 13r
2r = 90
r = 45

travel time (each going and return) in hours:
= 150/ 45
= 10/3 or 3 1/3

rate on return trip:
= 45 - 6
= 39

Answer: rate on return trip is 39 miles per hour:

Proof (equal time going and return):
150 mi / 45 mph = 130 mi / 39 mph
10/3 hours = 10/3 hours
3 1/3 hours = 3 1/3 hours

Answer 3

130/x = 150/(x + 6)
=> 13(x + 6) = 15x
=> 2x = 78
=> x = 39
39 miles an hour.

Answer 4

39

150x=130x+6
20x=6
x=6/20
x=3/10

130*3/10=


39.0