Two buses leave a station at the same time and travel in opposite directions. one bus travels 12 mi/hr faster than the other. if the two buses are 750 miles apart after 5 hours, what is the rate of each bus?

speed of the first bus is 81 miles per hour.

speed of the second bus is 69 miles per hour.

Step-by-step explanation:

Let x = the speed of the first bus

Let y = speed of the second bus

One bus travels 12 miles/hour faster than the other. Assuming that the first bus travels at a faster rate, then

x = y + 12

Recall that distance = speed × time.

The distance travelled by the first bus in 5 hours would be x × 5 = 5x

The distance travelled by the second bus in 5 hours would be y × 5 = 5y

the two buses are 750 miles apart after 5 hours. This means that total distance travelled by first bus and second bus in 5 hours is equal to 750 miles. Therefore,

5x + 5y = 750 - - - - - - - -- - 1

Substituting x = y + 12 into equation 1, it becomes

5(y+12) + 5y = 750

5y + 60 + 5y = 750

5y + 5y = 750 - 60

10y = 690

y = 690/10 = 69 miles per hour

x = y + 12 = 69 + 12

x = 81 miles per hour

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363