Prepared by Doron Shahar
Word problems with Linear and Quadratic Equations
Prepared by Doron Shahar
Warm-up
Write each description as a mathematical statement.
 Page 9 #1, 3 more than twice a number
 Page 9 #2, The sum of a number and 16 is three times the
number.
 1.2.1 The sum of three consecutive integers is 78.
What is the smallest of the three integers?
 1.5.1 The sum of the square of a number and the square of 7
more than the number is 169. What is the number?
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Page 9 #1
Write “3 more than twice a number” as a mathematical statement.
Variable: Let x be the number.
Expression:
2x  3
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Page 9 #2
Write “the sum of a number and 16 is three times the number”
as a mathematical statement.
Variable: Let x be the number.
Equation:
x 16  3x
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1.2.1
The sum of three consecutive integers is 78.
What is the smallest of the three integers?
Want:
The smallest of the three integers.
Variable: Let x be the smallest of the three integers.
Examples of consecutive integers:
1, 2, 3
Equation:
17, 18 ,19
25, 26, 27
x, x+1, x+2
( x)  ( x  1)  ( x  2)  78
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1.5.1
The sum of the square of a number and the square of 7 more than
the number is 169. What is the number?
Want:
The number.
Variable: Let x be the number.
Equation: square of ( x )  square of
square of
(7 more than x)  169
( x)  square of ( x  7)  169
x  ( x  7)  169
2
2
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Approach to word problems
Solving word problems involves two key steps:
1. Constructing the equations you are to solve.
2. Solving the equations you found.
The word problems in sections 1.2 and 1.5 will lead to linear and
quadratic equations, which we have just learned how to solve.
Therefore, we shall not solve any of the word problems in class.
Rather we will focus on constructing the equations. That is, we
will translate the word problems into equations.
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1.5.2 Projectile Problem
A stone is thrown downward from a height of 274.4 meters . The
stone will travel a distance of s meters in t seconds, where
s=4.9t2 +49t. How long will it take the stone to hit the ground?
Want: The time it will take the stone to hit the ground.
Variable: Let T seconds be the time it will take the stone
to hit the ground.
Equation: To find the equation, let’s try using a picture.
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1.5.2 Equation
T is the time it will take until the stone hits the ground.
Initial
Height
274.4
meters
Equation:
Distance traveled 4.9T2+49T
in T seconds
meters
initialheight
 distancetravelledin T seconds
274.4  4.9T  49T
2
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1.5.2 Summary
Want: The time it will take the stone to hit the ground.
Variables: Let T seconds be the time until the stone hits the ground.
4.9T  49T  274.4
Solutions to equation: T  4 or T  14
Equation:
2
Note that T=−14 cannot be the answer to the word problem, because
T denotes the time it will take the stone to hit the ground. So T>0.
Solution to word problem:
T 4
It will take 4 seconds for the stone to hit the ground.
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Extra Projectile Problem
A cannonball is launched from a cannon. The height, h, in feet of the
cannonball t seconds after it leaves the cannon is given by the equation
h=−16t2+63t+4. When will the cannon ball hit the ground?
Want: The time when the cannonball hits the ground.
Variable: Let T seconds be the time from when the
cannonball is launched until it hits the ground.
Equation: To find the equation, let’s try using a picture.
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Extra problem: Equation
The height, h, in feet of the cannonball t seconds after it leaves the
cannon is given by the equation h=−16t2+63t+4.
T is the time when the cannonball hits the ground.
After T seconds, the
cannonball is on the ground,
and has a height of 0 feet.
Equation:
0 feet  height after T seconds
0  16T  63T  4
2
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Extra problem: Summary
Want: The time when the cannonball hits the ground.
Variables: Let T seconds be the time from the cannonball is
launched until it hits the ground.
0  16T  63T  4
Solutions to equation:
T  0.0625 or T  4
Equation:
2
Note that T=−0.0625 cannot be the answer to the word problem,
because T denotes the time until the cannonball hits the ground. So T>0.
Solution to word problem:
T 4
The cannonball will hit the ground about 4 seconds
after it was launched.
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1.5.4 Geometry Problem
The area of a square is numerically 60 more than the perimeter.
Determine the length of the side of the square.
Want:
Length of the side of the square
Variables: Let S be the length of the side of the square.
Let A be the area of the square.
Let P be the perimeter of the square.
Equations: A  P  60 Information given in the problem.
A  S2
P  4S
Facts from geometry
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1.5.4 Equations
Want:
Length of the side of the square
Key Variable: Let S be the length of the side of the square.
Equations: A  P  60 A  S 2
P  4S
What method can we use to solve for S given these equations?
Substitution
A  P  60
P  4S
A  S2
S
2
 4 S  60
Now solve for S.
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1.5.4 Summary
Want:
Length of the side of the square
Key Variable: Let S be the length of the side of the square.
Key Equation:
S
2
 4 S  60
Solutions to equation:
S  10 or S  6
Note that S=−6 cannot be the answer to the word problem,
because S denotes the length of the side of the square. So S>0.
Solution to word problem:
S  10
The side of the square is 10 units long.
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1.5.5 Geometry Problem
The length of a rectangle is 7 centimeters longer than the width.
If the diagonal of the rectangle is 17 centimeters, determine the
length and width.
Want:
Length and Width of the rectangle
Variables: Let L cm be the length of the rectangle.
Let W cm be the width of the rectangle.
Let D cm be the perimeter of the square.
Equations: L  W  7
D  17
D
2
 L W
2
Information given in the problem.
2
Fact from geometry
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1.5.5 Equations
Want: Length and width of rectangle
Key Variables: Let L cm be the length of the rectangle.
Let W cm be the width of the rectangle.
Equations:
L  W  7 D  17
D  L W
2
2
2
What method can we use to solve for L and W given these equations?
Substitution
D
2
 L W
2
2
L W  7
D  17
17  W  7   W 2
2
2
Now solve for W.
Then solve for L.
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1.5.5 Summary
Want: Length and width of rectangle
Key Variable: Let L cm be the length of the rectangle.
Let W cm be the width of the rectangle.
Key Equations: 17 2  7  W 2  W 2 L  W  7
Solutions to first equation:
W  8 or W  15
Note that W=−15 cannot be the answer to the word problem,
because W denotes the widths of the rectangle. So W>0.
Solution to word problem:
W  8 and L  15
The width of the rectangle is 8 cm.
The length of the rectangle is 15 cm.
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Key Idea for many word problems
In the previous problems, the equations we are trying to find
are almost given, or else come from basic facts about geometry.
Next we will have to find the equations.
Amount 1  Amount 2  Total Amount
Quantity 1  Quantity 2  Total Quantity
Value 1  Value 2  Total Value
Cost 1  Cost 2  Total Cost
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Key Idea for many word problems
Once we have an equation of the form,
Amount 1  Amount 2  Total Amount
We may need to use formulas to replace the amount when
dealing with values and costs.
Value of coins  value of one coin  number of coins
Cost of jackets  cost of one jacket  number of jackets
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1.2.2 General problem
Tina has $6.30 in nickels and quarters in her coin purse. She has a
total of 54 coins. How many of each coin does she have?
Well, it depends on how many coins she
has that are not in her coin purse.
Want: The number of nickels and the number of quarters.
Variables: Let N be the number of nickels.
Let Q be the number of quarters.
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1.2.2 Equations
Amount 1  Amount 2  Total Amount
Number
of nickels

Number
of quarters

Total Number
of coins
What is the amount
being added?
Number
of coins
N  Q  54
Dollar Value
Dollar Value
of nickels

Dollar Value
of quarters

Dollar Value
of all coins
0.05N  0.25Q  6.30
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1.2.2 Summary
Tina has $6.30 in nickels and quarters in her coin purse. She has a
total of 54 coins. How many of each coin does she have?
Want: The number of nickels and the number of quarters.
Variables: Let N be the number of nickels.
Let Q be the number of quarters.
Equations:
N  Q  54
0.05 N  0.25Q  6.30
What method might we use to
solve this system of equations?
Elimination
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1.2.3 General problem
A company produces a pair of skates for $43.53 and sells a pair
for $89.95. If the fixed costs are $742.72, how may pairs must the
company produce and sell in order to break even?
Want:
The number of pairs of skates that must be
produced to break even.
Variables: Let S be the number of pairs of skates that
must be produced to break even.
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1.2.3 Equation
Fixed costs
Break even
Money Earned = Total Cost
Rent
Amount 1  Amount 2  Total Amount
Fixed Cost  Production Cost

What is the amount
being added?
Total Cost
742.72  43.53S  89.95S
Cost
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1.2.3 Summary
A company produces a pair of skates for $43.53 and sells a pair
for $89.95. If the fixed costs are $742.72, how may pairs must the
company produce and sell in order to break even?
Want:
The number of pairs of skates that must be
produced to break even.
Variables: Let S be the number of pairs of skates that
must be produced to break even.
Equation: 747.72  43.53S  89.95S
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Warm-up
 Page 9, How much pure salt is in 5 gallons of a 20% salt solution?
 Page 20 #3, An alloy contains 40% gold. Represent the number of
grams of gold present in G grams of the alloy.
 Page 9, What is the equation involving distance, rate, and time?
 Page 20 #1, A car is travelling M mph for H hours. Represent the
number of miles traveled.
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Key Idea for many word problems
In the previous problems, the equations we are trying to find
are almost given, or else come from basic facts about geometry.
Next we will have to find the equations.
Amount 1  Amount 2  Total Amount
Mass 1  Mass 2  Total Mass
Cost 1  Cost 2  Total Cost
Volume 1  Volume 2  Total Volume
Distance 1  Distance 2  Total Distance
Time 1  Time 2  Total Time
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Key Idea for many word problems
Once we have an equation of the form,
Amount 1  Amount 2  Total Amount
We may need to use formulas to replace the amount when
dealing with percents and rates.
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Key Idea for mixture problems
Below is the formula for percents.
volume of salt
Percent of salt 
volume of liquid
volume of salt
P% 
volume of liquid
P
volume of salt 
 volume of liquid
100
Page 9, How much pure salt is in 5 gallons of a 20% salt solution?
20
 5 gallons  0.20  5 gallons  1gallon
100
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Key Idea for mixture problems
Below is the formula for percents.
mass of gold
Percent of gold 
mass of alloy
mass of gold
P% 
mass of alloy
P
mass of gold 
 mass of alloy
100
Page 20 #3, An alloy contains 40% gold. Represent the number of
grams of gold present in G grams of the alloy.
40
 G grams  0.40  G grams  0.4G grams
100
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1.2.4 Mixture problem
A premium mix of nuts costs $12.99 per pound, while almonds
cost $6.99 per pound. A shop owner adds almonds into the
premium mix to get 90 pounds of nuts that cost $10.99 per pound.
How many pounds of almonds did she add?
Want:
The number of pounds of almonds she added.
Variables: Let A be the number of pounds of almonds she added.
Let P be the number of pounds of the premium mix of nuts.
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1.2.4 Equations
Amount 1  Amount 2  Total Amount
Pounds of  Pounds of
almonds
premium mix
A

What is the amount
being added?
Total Pounds
in mixture
Mass
(Pounds)
P  90
Cost
Cost of
almonds

Cost of
premium mix

Total cost
of mixture
6.99 A  12.99P  10.99  90
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1.2.4 Summary
A premium mix of nuts costs $12.99 per pound, while almonds
cost $6.99 per pound. A shop owner adds almonds into the
premium mix to get 90 pounds of nuts that cost $10.99 per pound.
How many pounds of almonds did she add?
Want:
The number of pounds of almonds she added.
Key Variable: Let A be the number of pounds of almonds.
Equations:
A  P  90
6.99 A 12.99P  989.1
What method might we use to
solve this system of equations?
Elimination
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1.2.5 Mixture problem
A chemist wants to strengthen her 40L stock of 10% acid solution
to 20%. How much 24% solution does she have to add to the 40L
of 10% solution in order to obtain a mixture that is 20% acid?
Want: The volume of the 24% solution that the chemist
needs to add to obtain a mixture that is 20% acid.
Variable: Let V liters be the volume of the 24% solution
that needs to be added.
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1.2.5 Equation
(40+V) liters
of solution
40 liters
of solution
V liters
of solution
10% acid
24% acid
Beaker 1
20% acid
Beaker 3
Beaker 2
Amount 1  Amount 2  Total Amount
Volume of acid
in beaker 1

Volume of acid
in beaker 2

What is the amount
being added?
Total volume of
acid in beaker 3
Volume
of acid
0.1 40  0.24V  0.20  (40  V )
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1.2.5 Summary
A chemist wants to strengthen her 40L stock of 10% acid solution
to 20%. How much 24% solution does she have to add to the 40L
of 10% solution in order to obtain a mixture that is 20% acid?
Want: The volume of the 24% solution that the chemist
needs to add to obtain a mixture that is 20% acid.
Variable: Let V liters be the volume of the 24% solution
that needs to be added.
Equation:
4  0.24V  0.2(40  V )
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Key Idea for rate problems
Below is the formula for rates.
amount
rate 
time
distance
speed 
time
amount  rate  time
distance  speed  time
Page 20 #1, A car is travelling M mph for H hours. Represent the
number of miles traveled.
miles
 H hours  MH miles
# miles traveled  M
hour
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1.2.7 Distance-Rate-Time
Two motorcycles travel towards each other from Chicago and
Indianapolis (350km apart). One is travelling 110 km/hr, the other
90 km/hr. If they started at the same time, when will they meet?
Want: The time when they will meet.
Variable: Let T hours be the time from when they
started until they meet.
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1.2.7 Equation
110 km/hr
Distance
traveled by
Motorcycle 1
Distance
traveled by
Motorcycle 2
90 km/hr
Chicago
Motorcycle 1
Indianapolis
350 km Where they meet
Amount 1  Amount 2  Total Amount
Distance traveled Distance traveled

by motorcycle 1
by motorcycle 2
Motorcycle 2
What is the amount
being added?
distance
 Total
traveled by both
110 T  90 T  350
Distance
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1.2.7 Summary
Two motorcycles travel towards each other from Chicago and
Indianapolis (about 350km apart). One is travelling 110 km/hr, the
other 90 km/hr. If they started at the same time, when will they meet?
Want: The time when they will meet.
Variable: Let T hours be the time from when they
started until they meet.
Equation:
110T  90T  350
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1.5.3 Distance-Rate-Time
Amy travels 450 miles in her car at a certain speed. If the car had
gone 15 mph faster, the trip would have taken 1 hour less.
Determine the speed of Amy’s car.
Want: The speed of Amy’s car.
Variables: Let A mph be the speed of Amy’s car.
Let T hours by the time it takes Amy to drive 450 miles.
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1.5.3 Equations
A mph
Amy’s car
(A+15) mph
Amy’s car
for T hours
450 miles
distance
speed 
time
450
A
T
for (T-1) hours
450 miles
450
A 15 
T 1
Prepared by Doron Shahar
1.5.3 Summary
Amy travels 450 miles in her car at a certain speed. If the car had
gone 15 mph faster, the trip would have taken 1 hour less.
Determine the speed of Amy’s car.
Want: The speed of Amy’s car.
Key Variable: Let A mph be the speed of Amy’s car.
Equations:
A  450 T
A  15  450 (T 1)
What method might we use to
solve this system of equations?
Substitution
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1.2.9 Shared Work Problem
Suppose a journeyman and apprentice are working on making
cabinets. The journeyman is twice as fast as his apprentice. If they
complete one cabinet in 14 hours, how many hours does it take
for the journeyman working alone to make one cabinet?
Want: The time it takes the journeyman working alone
to make one cabinet.
Variables: Let J hours be the time it takes the journeyman
to make one cabinet working alone.
Let A hours be the time it takes the apprentice
to make one cabinet working alone.
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1.2.9 Equations
amount
rate 
time
The journeyman is twice as fast as his apprentice.
Journeyman’s rate

2 × Apprentice’s rate
1
1
 2
A
J
Multiply both sides of
the equation by AJ
A  2J
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1.2.9 Equations
Amount 1  Amount 2  Total Amount
Amount of the cabinet built
What is the amount being added? The answer is NOT time!!
Amount of the cabinet
built by the journeyman
in 14 hours

Amount of the cabinet
built by the apprentice
in 14 hours

Amount of the
cabinet built by
both in 14 hours
Journeyman ' s rate 14  Apprentice ' s rate 14  1
1 J 14  1 A 14  1
Prepared by Doron Shahar
1.2.9 Summary
Want: The time it takes the journeyman working alone
to make one cabinet.
Key Variable: Let J be the time it takes the journeyman to
make one cabinet working along.
Equations:
A  2J
What method might we use to
solve this system of equations?
14 14
 1
J A
Substitution
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1.5.6 Shared Work Problem
It take Julia 16 minutes longer to chop vegetables than it takes Bob.
Working together, they are able to chop the vegetables in 15 minutes.
How long will it take each of them if they work by themselves?
Want: The time it takes each of them working alone to
chop the vegetables.
Variables: Let J minutes be the time it takes Julia to chop
the vegetables working alone.
Let B minutes be the time it takes Bob to chop
the vegetables working alone.
Prepared by Doron Shahar
1.5.6 Equations
It take Julia 16 minutes longer to chop the vegetables than it takes Bob.
J  B  16
Amount 1  Amount 2  Total Amount
Amount of vegetables chopped
What is the amount being added? The answer is NOT time!!
Amount of the
Amount of the
Amount of the
vegetables chopped  vegetables chopped  vegetables chopped
by Julia in 15 minutes by Bob in 15 minutes by both in 15 minutes
Julia' s rate 15  Bob' s rate 15  1
1 J 15  1 B 15  1
Prepared by Doron Shahar
1.5.6 Summary
Want: The time it takes each of them working alone to
chop the vegetables.
Variables: Let J minutes be the time it takes Julia to chop
the vegetables working alone.
Let B minutes be the time it takes Bob to chop
the vegetables working alone.
Equations:
J  B  16
15 15
 1
J B
What method might we use to
solve this system of equations?
Substitution
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1.2.8 Shared Work Problem
Suppose it takes Mike 3 hours to grade one set of homework and it
takes Jenny 2 hours to grade one set of homework. If they grade
together, how long will it take to grade one set of homework?
Want: The time it will take them to grade one set of
homework if they work together.
Variables: Let T hours be the time it will take them to grade
one set of homework if they work together.
Prepared by Doron Shahar
1.2.8 Equations
Amount 1  Amount 2  Total Amount
What is the amount being added?
Amount of the
homework set 
graded by Mike
in T hours
Amount of the
homework set
graded by Jenny
in T hours

Amount of the
homework set graded
The answer is NOT time!!
Amount of the
homework set
graded by both
in T hours
Mike' s rate T  Jenny' s rate T  1
1 3 T  1 2 T  1
Prepared by Doron Shahar
1.2.8 Summary
Suppose it takes Mike 3 hours to grade one set of homework and it
takes Jenny 2 hours to grade one set of homework. If they grade
together, how long will it take to grade one set of homework?
Want: The time it will take them to grade one set of
homework if they work together.
Variables: Let T be the time it will take them to grade
one set of homework if they work together.
1
1
Equation:
T  T 1
3
2
Prepared by Doron Shahar
Key Idea for relative motion problems
Downstream
Upstream
Directions of Boats
B mph
B mph
Direction of Current
C mph
The boat’s speed
going downstream is
( B  C ) mph
going upstream is
( B  C ) mph
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1.2.6 Relative Motion Problem
A boat travels down a river with a current. Travelling with the current,
a trip of 66 miles takes 3 hours while the return trip travelling against
the current takes 4 hours. How fast is the current?
Want: The speed of the current.
Variables: Let C mph be the speed of the current.
Let B mph be the speed of the boat in still water.
Prepared by Doron Shahar
1.2.6 Equations
B mph
For 3 hours
C mph
66 miles
distance
speed 
time
with current
with current
66
BC 
3
distance
speed 
against current time
against current
For 4 hours
B mph
C mph
66
B C  4
Prepared by Doron Shahar
1.2.6 Summary
A boat travels down a river with a current. Travelling with the current,
a trip of 66 miles takes 3 hours while the return trip travelling against
the current takes 4 hours. How fast is the current?
Want: The speed of the current.
Key Variable: Let C mph be the speed of the current.
Equations:
66
BC 
3
66
B C 
4
What method might we use to
solve this system of equations?
Elimination
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1.5.7 Relative Motion
Maria traveled upstream along a river in a boat a distance of 39
miles and the came right back. If the speed of the current was 1.3
mph and the total trip took 16 hours, determine the speed of the
boat relative to the water.
Want: The speed of the boat relative to the water.
Variables:
Let B mph be the speed of the boat relative to the water.
Let T hours be the time of the trip upstream.
Prepared by Doron Shahar
1.2.6 Equations
For T hours
B mph
distance
speed 
against current time
1.3 mph
against current
39
B 1.3  T
39 miles
distance
speed 
time
with current
with current
B mph
For 16−T hours
1.3 mph
39
B  1.3 
16  T
Prepared by Doron Shahar
1.5.7 Summary
Maria traveled upstream along a river in a boat a distance of 39
miles and the came right back. If the speed of the current was 1.3
mph and the total trip took 16 hours, determine the speed of the
boat relative to the water.
Want: The speed of the boat relative to the water.
Key Variable: Let B mph be the relative speed of the boat.
Equations:
39
B  1.3 
T
39
B  1.3 
16  T
What method might we use to
Substitution or Elimination
solve this system of equations?
Prepared by Doron Shahar
Review of Word Problems
 Don’t forget units
 Sanity checks: Check that your answer makes sense
Examples:
 Lengths, times, and speeds should not be negative.
 If Michael and Rachel can each build a bookcase working
alone in under a day, it should not take them more than a
day to build a bookcase working together.
If people do not believe that mathematics is simple, it is only because
they do not realize how complicated life is.—John Louis von Neumann