SOLUTION: AN ACCOUNTANT RIDES A BUS PART OF THE WAY TO WORK EVERY DAY AND WALK THE REST OF THE WAY. THE BUS AVERAGES 40MPH AND THE ACCOUNTANT WALKS AT A SPEED OF 3MPH.THE DISTANT FROM HOME

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Question 117631: AN ACCOUNTANT RIDES A BUS PART OF THE WAY TO WORK EVERY DAY AND WALK THE REST OF THE WAY. THE BUS AVERAGES 40MPH AND THE ACCOUNTANT WALKS AT A SPEED OF 3MPH.THE DISTANT FROM HOME TO WORK IS 22MI AND THE TOTAL TIME FOR THE TRIP IS
3 HOURS. FIND HOW FAR THE ACCOUNTANT WALKS.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Rate x Time = Distance
Distance = Rate/Time
Let's call Bus Time and Bus Distance, T1 and D1.
Let's call walk time and walk distance, T2 and D2.
1.D1%2BD2=22 Total distance is 22 miles.
2.T1%2BT2=3 Total time is 3 hours.
3.D1%2FT1=40 Rate of the bus is 40 mph.
4.D2%2FT2=3 Rate of the accountant walking is 3 mph.
From 3,
3.D1%2FT1=40
D1=40T1
and from 4,
4.D2%2FT2=3
D2=3T2
You can substitute these values into 1.
1.D1%2BD2=22
40T1%2B3T2=22
3T2=22-40T1
T2=%2822-40T1%29%2F3
You can substitute this value for T2 into equation 2 and solve for T1.
2.T1%2BT2=3
T1%2B%2822-40T1%29%2F3=3
3T1%2B22-40T1=9
22-37T1=9
-37T1=-13
T1=13%2F37
From 2,
T1%2BT2=3
13%2F37%2BT2=3
T2=3-14%2F37
T2=111%2F37-13%2F37
T2=98%2F37
From 3,
D1=40T1
D1=40%2813%2F37%29
D1=520%2F37
From 4,
D2=3T2
D2=3%2898%2F37%29
D2=294%2F37
So in approximate terms, the accountant rides the bus 14.1 miles for 0.35 hrs(21 minutes).
Then walks 2.65 hrs (2 hrs, 39 minutes) for a distance of 7.9 miles.
Total distance is 14.1+7.9 or 22 miles.
Total time is 0.35+2.65 or 3 hours.