R11. Rational Expressions: Distance
Warm-up. Clear the fractions of
a
b
= dc .
Solution. The LCD is bd. Multiply each term by the LCD.
c
a
=
a b d c bd
=
bd
b
d
da = cb
ad = bc Write the letters alphabetically.
This technique is often referred to as cross multiplying and results from the
techniques studied in this chapter. In other words, multiply the denominator
of one fraction by the numerator of the other fraction and set that equal to the
product of the other numerator and denominator.
We are more familiar with these problems than we might realize. For example,
if one drove at 70 miles per hour for 3 hours, most of us are comfortable that 210
miles were traveled. This resulted in taking the speed, or rate, and multiplying
it by the time.
Distance = 70mph × 3hours
One way to understand the formula is to remove the numbers from the problem
and examine the units. Units multiply the same as numbers. Thus,
Distance =
miles
miles
= miles
× hours = hours
×
hour
hour
This makes sense. If the units were otherwise, the problem would be incorrect.
The general formula is
d=r·t
where d is distance, r is the rate of travel, and t is time spent traveling.
Example. A car travels 180 miles in the same time a truck travels 120 miles.
If the car’s speed is 20 mph faster than the truck, find the speed of each vehicle.
This problem may seem confusing at first. Make a table to keep track of the
information and insert pieces as discovered.
d
Truck
Car
=
r
t
120 =
180 =
As information is uncovered, place the numbers in the table where they belong.
The problem indicates that both vehicles traveled for the same amount of time,
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but how much time is unknown. Let the amount of time traveled be t. This
gives the table,
d = r · t
Car
Truck
· t
· t
180 =
120 =
The last piece of information is the rate of the vehicles. It is described as the
car travels 20 mph faster than the truck. If the truck is traveling at a rate of 30
mph, then the car is traveling (30+20)mph= 50mph. If the truck is traveling 55
mph, then the car is traveling (55 + 20)mph= 75mph.1 . If the truck is traveling
r mph, then the car is traveling (r + 20)mph. Let r be the rate of the truck and
r + 20 be the rate of the car. The table then is completed as follows.
d
Car
Truck
r
· t
(r + 20)
r
· t
· t
=
180 =
120 =
Recall solving a simple problem like 6 = 2x. It is solved by dividing both sides
of the equation by 2. With this in mind, solve the equations for the Car and
Truck from the above table for t.
180 = (r + 20)t
180
Time for car:
=t
(r + 20)
120 = rt
120
=t
Time for truck:
r
Both equations equal t. Set each equation equal to the other and solve for r.
120
180
=
(r + 20)
r
180r = 120(r + 20) Cross multiply.
180r = 120r + 2400
60r = 2400
r = 40
The truck travels 40 mph and the car travels (40 + 20)mph= 60mph.
Example. A boat can travel 9 miles upstream in the same amount of time it
takes to travel 11 miles downstream. If the current of the river is 3 mph, find
the speed of the boat in still water.
1 Probably
not this fast, since this would be speeding.
2
This problem also utilizes the formula d = rt. Before writing the table, notice
that this problem also gives equal times of travel. Let the time traveled by the
boat be represented by t.
The current of the river affects the speed of the boat. When traveling upstream,
the boat is slowed by the current. When traveling downstream, the boat gets
and additional 3 mph of speed it otherwise would not have had. Let r be the
speed of the boat in still water. Then the rate of the boat going upstream is
r −3 and the rate of the boat going downstream is r +3. This gives the following
table.
d =
r
t
Upstream
Downstream
9 = (r − 3) t
11 = (r + 3) t
Solve each equation for t
9 = (r − 3)t
9
Time upstream:
=t
(r − 3)
11 = (r + 3)t
11
Time downstream:
=t
(r + 3)
Set the equations equal to each other and solve for r.
9
11
=
(r − 3)
(r + 3)
9(r + 3) = 11(r − 3) Cross multiply.
9r + 27 = 11r − 33 Distribute.
−2r + 27 = −33
−2r = −60
r = 30
The speed of the boat in still water is 30 mph.
0.1
Practice Problems
1. A boat can travel 17 miles downstream in the same time it takes it to
travel 13 miles to go upstream in the same river. If the boat’s speed is 45
mph in still water, what is the speed of the current?
2. The river’s current is 3 mph. A barge can go 6 miles upstream in the same
amount of time it takes it to go 10 miles downstream. Find the speed of
the barge in still water.
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3. A fisherman travels 15 miles downstream in the same amount of time it
takes him to go 10 miles upstream. If the current is 6 mph, what is the
speed of the boat in still water?
4. Sandy rows 12 miles downstream in the same amount of time it takes her
to row 5 miles upstream. If the speed of the current is 3 mph, what is the
speed of the boat in still water? (Round your answer to one decimal.)
5. A swimmer swims 8 miles upstream and 8 miles back in 3 hours. If his
speed in still water is 6 mph, find the speed of the current.
6. A yacht travels 75 miles upstream and 75 miles back in 4 hours. If the
speed of the current is 10 mph, find the speed of the yacht in still water.
7. A person drives 54 miles into town during the rush hour and then goes back
via the same route when the traffic is lighter, traveling 15 mph faster. If
he spent a total of 3 hours traveling, what was his speed in each direction?
8. A bicyclist training for competition rides 21 miles out and then comes back
through the same route but riding 20 mph slower. He rode his bicycle for
a total of 2 hours. What was his speed in each direction?
0.2
Solutions
1. 6 mph
4. 7.3 mph
7. 45 mph
2. 12 mph
5. 2 mph
8. 15 mph
3. 30 mph
6. 40 mph
12. mph
5. Obviously the time for the swimmer to swim upstream is more than the time
needed to swim downstream. Let t1 be the time to swim upstream and t2 be
the time to swim downstream. What is know about these times is the total trip
took 3 hours. This means that
t1 + t2 = 3.
As in earlier examples build a table based on the d = rt. Let x be the speed of
the current. This gives the following table.
d =
Upstream
Downstream
r
t
= (6 − x) t1
= (6 + x) t2
8
8
Solve each equation for t.
4
8 = (6 − x)t1
8
(
6−
x)t1
=
6−x
6 −
x
8
= t1
6−x
And for the second equation,
8 = (6 + x)t2
(
6+
x)t2
8
=
6+x
6+
x
8
= t2
6+x
Substitute t1 and t1 into the first equation and solve. Notice that the lowest
common denominator is (6 − x)(6 + x). Use this to clear fractions. This is
illustrated below.
t1 + t2 = 3
8
8
+
=3
6−x 6+x
8
8
+
· (6 − x)(6 + x) = 3 · (6 − x)(6 + x)
6−x 6+x
+ x) · 8 + 8 · (6 − x)(6
= 3 · (6 − x)(6 + x)
(6
−
x)(6
x)
+
6−
x 6+
x
8(6 + x) + 8(6 − x) = 3(6 − x)(6 + x)
(6 − x)(6 + x) ·
48 + 8x + 48 − 8x = 108 − 3x2
96 = 108 − 3x2
3x2 − 12 = 0
x2 − 12 = 0
(x + 2)(x − 2) = 0
x = −2, 2
Since speed is a positive number, the answer is that the speed of the current is
2 mph.
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