Algebra Demystified 2nd Ed

434 algebra De mystif ieD

3. Let r represent the boat’s speed in still water. The average speed down-
   stream is r + 4 and the average speed upstream is r − 4. The boat was in
   the water a total of 2 hours. The distance traveled in each direction is
   15 miles. The time the boat traveled downstream is^15
   r +  4

hours, and it

traveled upstream^15

r –  4

hours. The time the boat traveled upstream

plus the time it traveled downstream equals 2 hours. The equation to

solve is^15

4

15

4

2

rr

+ 

+ 

– 

= . The LCD is (r + 4)(r − 4).

()() ()() ()()

(

rr

r

rr

r

rr

r

+−

+

++ −

−

4415 =+−

4

4415

4

244

15 −−+ += +−

−++= −

4154244

15 60 15 60 212 6

)()[()()]

(

rrr

rrr ))

() ()

30 232

02 30 32

1

2

0 1

2

23032

0

2

2

2

rr

rr

rr

r

=−

=−−

=−−

=^221516

0161

−−

=− +

r

()rr()

r − 16 = 0 r + 1 = 0 (This does not lead to a solution.)

r = 16

The boat’s average speed in still water is 16 mph.

4. Let r represent the plane’s average speed without the wind. The plane’s
   average speed from Denver to Indianapolis is r + 20, and the plane’s
   average speed from Indianapolis to Denver is r − 20. The total time in
   flight is 512 hours and the distance between Denver and Indianapolis is
   1000 miles. The time in the air from Denver to Indianapolis is^1000
   r +  20
   hours and the time in the air from Indianapolis to Denver is^1000
   r –  20

hours.

The time in the air from Denver to Indianapolis plus the time in the air

from Indianapolis to Denver is 5.5 hours. The equation to solve is

1000

20

1000

20

55

rr

 .

+ 

+ 

– 

= . The LCD is (r + 20)(r − 20).

()rr() ()() (

r

rr

r

+− r

+

++ −

−

20 201000 =+

20

20 201000

20

20 ))( )( .)

() ().[()(

r

rrrr

−

−+ += −

20 55

1000 20 1000 20 55 20 ++ 20 )]^