There are some distance/rate problems for which there are three unknowns. You must reduce the number of unknowns to one. The clues on how to do so are given in the problem.
Examples
A semi-truck
traveled from City A to City B at 50 mph. On the return trip, it averaged only
45 mph and took 15 minutes longer. How far is it from City A to City B?
There are three unknowns-the distance
between City A and City B, the time spent traveling from City A to City B, and
the time spent traveling from City B to City A. We must eliminate two of these unknowns.
Let t represent the number of hours spent on the trip from City A to City B. We
know that it took 15 minutes longer traveling from City B to City A (the return
trip), so t + 15/60 represents the number of hours traveling
from City B to City A. We also know that the distance from City A to City B is
the same as from City B to City A. Let d represent the distance between the two
cities. We now have the following two equations.
From City A to City B: From City B to City A:
D = 50t d
= 45(t + 15/60)
But if the
distance between them is the same, then 50t = Distance from City A to City B is
equal to the distance from City B to City
. Therefore,
We now know
the time, but the problem asked for the distance. The distance from City A to
City B is given by d = 50t, so 
. The cities are 112 ½ miles
apart.
Another
approach to this problem would be to let t represent the number of hours the semi
spent traveling from City B to City A. Then t _ 15 60 would represent the
number of hours the semi spent traveling from City A to City B. The equation to
solve would be 50(t – 15/60) = 45t.
Kaye rode
her bike to the library. The return trip took 5 minutes less. If she rode to
the library at the rate of 10 mph and home from the library at the rate of 12
mph, how far is her house from the library?
Again there are three unknowns—the distance
between Kaye’s house and the library, the time spent riding to the library and
the time spent riding home. Let t represent the number of hours spent riding to
the library. She spent 5 minutes less riding home, so t _ 5
60
represents the number of hours spent riding home. Let d represent the distance between
Kaye’s house and the library.
The trip to the library is given by d = 10t,
and the trip home is given by d = 12(t – 5/60). As these
distances are equal, we have that 10t = d = 12(t – 5/60).
The distance
from home to the library is d = 10t = 10 (½) = 5 miles.
Practice
1. Terry,
a marathon runner, ran from her house to the high school then back. The return
trip took 5 minutes longer. If her speed was 10 mph to the high school and 9
mph to her house, how far is Terry’s house from the high school?
2. Because
of heavy morning traffic, Toni spent 18 minutes more driving to work than
driving home. If she averaged 30 mph on her drive to work and 45 mph on her
drive home, how far is her home from her work?
3. Leo
walked his grandson to school. If he averaged 3 mph on the way to school and 5
mph on his way home, and if it took 16 minutes longer to get to school, how far
is it between his home and his grandson’s school?
Solutions
1. Let
t represent the number of hours spent running from home to school. Then t + 5/60
represents the number of hours spent running from school to home. The distance
to school is given by d = 10t, and the distance home is given by d = 9(t + 5/60).
The distance between home and school is 
miles.
2. Let
t represent the number of hours Toni spent driving to work. Then t – 18/60
represents the number of hours driving home. The distance from home to
work is given by d = 30t, and the distance from work to home is given by d = 45(t
– 18/60).
The distance from Toni’s home and work is 
miles.
3. Let
t represent the number of hours Leo spent walking his grandson to school. Then
t – 16/60 represents the number of hours Leo spent
walking home. The distance from home to school is given by d = 3t, and the
distance from school to home is given by d = 5(t – 16/60).
The
distance from home to school is 
miles.
In the above examples and practice problems, the number for t was substituted in the first distance equation to get d. It does not matter which equation you used to find d, you should get the same value. If you do not, then you have made an error somewhere.
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