296 algebra De mystif ieD
(^2201) hours, or 2 hours 3 minutes 60 1
()⋅ 20 = ^3 , after the Hewitts leave
Dallas, the Smiths and Hewitts will pass each other. In other words, at
8:20, the Smiths and Hewitts will pass each other.
In the following problems, we have information on how long a trip and
return trip are and use this information to find an unknown distance.
EXAMPLE
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?
We have 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 num-
ber 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 + ^1560
represents the amount of time 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 dt=+ (^45) ()^1560
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 A = (^45) ()t + ^1560. Therefore,
50 45 15
60
50 45 1
4
tt
tt
=+
=+
EXAMPLE
A semi-truck traveled from City A to City B at 50 mph. On the return trip, it