Chapter 8 linear appliCaTionS 297

50 45 45

4

45 45

5 45

4

1

5

45

4

9

4

tt

tt

t

t

t

=+

−−

=

=⋅

=

We now know the time, but the problem asked for the distance. The dis-
tance from City A to City B is given by d = 50t, so d=^50 ()^94 ==^225211221. The
cities are^11221 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 – ^1560 would
represent the number of hours the semi spent traveling from City A to City
B, and the equation to solve is^50 ()tt  – ^1560 = ^45.

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 –  605 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 dt  =  (^12) ()–  605. As these distances are equal, we have that
10 td =  =  (^12) ()t –  605.
10 12 5
60
10 12 1
12
10 12 1
12
tt
tt
tt
t
=−



=−




=−
− −−
−=−
=−
−

12
21
1
2
1
2
t
t
t
t
The distance from home to the library is dt^ =  =  ()^12 = 10^5 miles.^10