Justin recently drove to visit his parents who live 240 miles away. on his way there his average speed was 11 miles per hour faster than on his way home (he ran into some bad weather). if justin spent a total of 8 hours driving, find the two rates.

Rate when he drove to visit his parents = 55 miles per hour

Rate when he was  coming home = 55 + 11 = 66 per hour

Step-by-step explanation:

The journey can be categorized into two phase:

when Justin was visiting his parents and when he was returning home.

let's compute the values of when he was returning home.

distance = 240 miles

let the rate = y

rate = distance/time

time = distance/rate

time = 240/y

let's compute the values of when he drove to visit his parents

distance = 240 miles

According to the question the rate was 11 miles per hour faster than on his way home. it can be represented as

rate = y + 11

rate = distance/time

time = distance/rate

time = 240/y + 11

The total time spent travelling is 8 hrs therefore,

240/y + 240/(y + 11) = 8

multiply through by y (y + 11)

240/y × y(y + 11)  + 240/(y + 11) × y(y + 11) = 8 × y × (y + 11)

240 (y + 11) + 240 y = 8y(y + 11)

240y + 2640 + 240y = 8y² + 88y

480y - 88y - 8y² + 2640 = 0

392y - 8y² + 2640 = 0

8y² - 392y - 2640 = 0

divide through by 8

y² - 49  - 330  = 0

(y - 55) (y + 6) = 0

Therefore y = 55 . Note y cannot be -6.

Rate when he drove to visit his parents = 55 miles per hour

Rate when he was  coming home = 55 + 11 = 66 per hour

answer: 1 2 3 6

step-by-step explanation:

1x6=6 2x3=6

55? i think maybe

step-by-step explanation: