SOLUTION: A bakery delivery truck leaves the bakery at 5:00 AM each morning on its 105-mile route. One day the driver gets a late start and does not leave the bakery until 5:30 AM To finish

Question 856956: A bakery delivery truck leaves the bakery at 5:00 AM each morning on its 105-mile route. One day the driver gets a late start and does not leave the bakery until 5:30 AM To finish her route on time the driver drives 5 miles per hour faster than usual. At what speed does she usually drive?
Answer by josgarithmetic(39630) (Show Source): You can put this solution on YOUR website!
r = normal speed for a trip
t = AMOUNT of time in hours, NOT a point on a time line

___________________speed__________time__________distance
NORMALLY___________r______________t_____________105
START LATE________r+5____________t-1/5__________105

When you understand the sense of the data table, you can continue to solve.

___________________speed__________time__________distance
NORMALLY___________r______________t_____________105=rt
START LATE________r+5____________t-1/5__________105=(r+5)(t-1/2)

The rest of the solution process of solving those two equations for r and t is only slightly tricky. The "late" equation will have a rt term which can be substituted using the "normally" equation. Another substitution for one of the variables may also be needed. You examine the data table and solve the system.

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