1. Word problems with Linear and Quadratic Equations
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Chapter 1 Section 1.2 & 1.5
Word problems with Linear and Quadratic Equations

2. Warm-up Write each description as a mathematical statement.
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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 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?

3. Page 9 #1 Variable: Let x be the number. Expression:
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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:

4. Page 9 #2 Variable: Let x be the number. Equation:
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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:

5. 1.2.1 Want: The smallest of the three integers. Variable:
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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
17, 18 ,19
25, 26, 27
x, x+1, x+2
Equation:

6. 1.5.1 Want: The number. Variable: Let x be the number. Equation:
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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:

7. Approach to word problems
Prepared by Doron Shahar
Approach to word problems
Solving word problems involves two key steps:
Constructing the equations you are to solve.
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.

8. 1.5.2 Projectile Problem Want:
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1.5.2 Projectile Problem
A stone is thrown downward from a height of 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.

9. Prepared by Doron Shahar
1.5.2 Equation
T is the time it will take until the stone hits the ground.
Initial Height
Distance traveled
in T seconds
4.9T2+49T meters
274.4
meters
Equation:

10. 1.5.2 Summary It will take 4 seconds for the stone to hit the ground.
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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.
Equation:
Solutions to equation:
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:
It will take 4 seconds for the stone to hit the ground.

11. Extra Projectile Problem
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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.

12. Extra problem: Equation
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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:

13. Extra problem: Summary
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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.
Equation:
Solutions to equation:
Note that T=− 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:
The cannonball will hit the ground about 4 seconds
after it was launched.

14. 1.5.4 Geometry Problem Want: Length of the side of the square
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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:
Information given in the problem.
Facts from geometry

15. 1.5.4 Equations Want: Length of the side of the square Key Variable:
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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:
What method can we use to solve for S given these equations?
Substitution
Now solve for S.

16. 1.5.4 Summary The side of the square is 10 units long. Want:
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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:
Solutions to equation:
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:
The side of the square is 10 units long.

17. 1.5.5 Geometry Problem Want: Length and Width of the rectangle
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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:
Information given in the problem.
Fact from geometry

18. 1.5.5 Equations Want: Length and width of rectangle
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1.5.5 Equations
Want:
Length and width of rectangle
Let L cm be the length of the rectangle.
Key Variables:
Let W cm be the width of the rectangle.
Equations:
What method can we use to solve for L and W given these equations?
Substitution
Now solve for W. Then solve for L.

19. 1.5.5 Summary The width of the rectangle is 8 cm.
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1.5.5 Summary
Want:
Length and width of rectangle
Let L cm be the length of the rectangle.
Key Variable:
Let W cm be the width of the rectangle.
Key Equations:
Solutions to first equation:
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:
The width of the rectangle is 8 cm.
The length of the rectangle is 15 cm.

20. Key Idea for many word problems
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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.

21. Key Idea for many word problems
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Key Idea for many word problems
Once we have an equation of the form,
We may need to use formulas to replace the amount when dealing with values and costs.

22. 1.2.2 General problem Well, it depends on how many coins she
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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.

23. 1.2.2 Equations What is the amount being added? Number of coins Number
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1.2.2 Equations
What is the amount being added?
Number
of coins
Number
of nickels
Number
of quarters
Total Number
of coins
Dollar Value
Dollar Value
of nickels
Dollar Value
of quarters
Dollar Value
of all coins

24. 1.2.2 Summary Want: The number of nickels and the number of quarters.
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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:
What method might we use to solve this system of equations?
Elimination

25. Prepared by Doron Shahar
1.2.3 General problem
A company produces a pair of skates for $43.53 and sells a pair for $ 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.

26. 1.2.3 Equation Rent What is the amount being added? Fixed costs
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1.2.3 Equation
Fixed costs
Break even
Rent
Money Earned = Total Cost
What is the amount being added?
Cost
Fixed Cost
Production Cost
Total Cost

27. Prepared by Doron Shahar
1.2.3 Summary
A company produces a pair of skates for $43.53 and sells a pair for $ 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:

28. Prepared by Doron Shahar
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.

29. Key Idea for many word problems
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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.

30. Key Idea for many word problems
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Key Idea for many word problems
Once we have an equation of the form,
We may need to use formulas to replace the amount when dealing with percents and rates.

31. Key Idea for mixture problems
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Key Idea for mixture problems
Below is the formula for percents.
Page 9, How much pure salt is in 5 gallons of a 20% salt solution?

32. Key Idea for mixture problems
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Key Idea for mixture problems
Below is the formula for percents.
Page 20 #3, An alloy contains 40% gold. Represent the number of
grams of gold present in G grams of the alloy.

33. 1.2.4 Mixture problem Want: The number of pounds of almonds she added.
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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.

34. 1.2.4 Equations What is the amount being added? Mass (Pounds)
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1.2.4 Equations
What is the amount being added?
Mass (Pounds)
Pounds of almonds
Pounds of premium mix
Total Pounds
in mixture
Cost
Cost of almonds
Cost of premium mix
Total cost
of mixture

35. 1.2.4 Summary Want: The number of pounds of almonds she added.
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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:
What method might we use to solve this system of equations?
Elimination

36. Prepared by Doron Shahar
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.

37. 1.2.5 Equation What is the amount being added? (40+V) liters
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1.2.5 Equation
(40+V) liters
of solution
40 liters
of solution
V liters
of solution
20% acid
10% acid
24% acid
Beaker 3
Beaker 1
Beaker 2
What is the amount being added?
Volume of acid
Volume of acid in beaker 1
Volume of acid in beaker 2
Total volume of
acid in beaker 3

38. Prepared by Doron Shahar
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:

39. Key Idea for rate problems
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Key Idea for rate problems
Below is the formula for rates.
Page 20 #1, A car is travelling M mph for H hours. Represent the
number of miles traveled.

40. 1.2.7 Distance-Rate-Time Want: The time when they will meet. Variable:
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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.

41. 1.2.7 Equation What is the amount being added? 90 km/hr 110 km/hr
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1.2.7 Equation
90 km/hr
110 km/hr
Distance
traveled by Motorcycle 1
Distance
traveled by Motorcycle 2
Chicago
Indianapolis
Motorcycle 1
Motorcycle 2
350 km
Where they meet
What is the amount being added?
Distance
Distance traveled by motorcycle 1
Distance traveled by motorcycle 2
Total distance traveled by both

42. 1.2.7 Summary Want: The time when they will meet. Variable:
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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:

43. 1.5.3 Distance-Rate-Time Want: The speed of Amy’s car. Variables:
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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.

44. 1.5.3 Equations A mph for T hours Amy’s car 450 miles (A+15) mph
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1.5.3 Equations
A mph
for T hours
Amy’s car
450 miles
(A+15) mph
for (T-1) hours
Amy’s car
450 miles

45. 1.5.3 Summary Want: The speed of Amy’s car. Key Variable:
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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:
What method might we use to solve this system of equations?
Substitution

46. 1.2.9 Shared Work Problem Want:
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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.

47. 1.2.9 Equations The journeyman is twice as fast as his apprentice.
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1.2.9 Equations
The journeyman is twice as fast as his apprentice.
Journeyman’s rate
2 × Apprentice’s rate
Multiply both sides of the equation by AJ

48. 1.2.9 Equations What is the amount being added?
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1.2.9 Equations
Amount of the cabinet built
What is the amount being added?
The answer is NOT time!!
Amount of the cabinet built by the apprentice
in 14 hours
Amount of the cabinet built by both in 14 hours
Amount of the cabinet built by the journeyman in 14 hours

49. 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:
What method might we use to solve this system of equations?
Substitution

50. 1.5.6 Shared Work Problem Want:
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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.

51. 1.5.6 Equations What is the amount being added?
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1.5.6 Equations
It take Julia 16 minutes longer to chop the vegetables than it takes Bob.
Amount of vegetables chopped
What is the amount being added?
The answer is NOT time!!
Amount of the vegetables chopped
by Julia in 15 minutes
Amount of the vegetables chopped
by Bob in 15 minutes
Amount of the vegetables chopped
by both in 15 minutes

52. 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:
What method might we use to solve this system of equations?
Substitution

53. 1.2.8 Shared Work Problem Want:
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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.

54. 1.2.8 Equations What is the amount being added?
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1.2.8 Equations
Amount of the homework set graded
What is the amount being added?
The answer is NOT time!!
Amount of the homework set graded by both in T hours
Amount of the homework set graded by Mike
in T hours
Amount of the homework set graded by Jenny
in T hours

55. 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.
Equation:

56. Key Idea for relative motion problems
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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
going upstream is

57. 1.2.6 Relative Motion Problem
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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.

58. 1.2.6 Equations For 3 hours B mph C mph 66 miles For 4 hours B mph
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1.2.6 Equations
For 3 hours
B mph
C mph
66 miles
For 4 hours
B mph
C mph

59. 1.2.6 Summary Want: The speed of the current. Key Variable:
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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:
What method might we use to solve this system of equations?
Elimination

60. Prepared by Doron Shahar
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.

61. 1.2.6 Equations For T hours B mph 1.3 mph 39 miles B mph
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1.2.6 Equations
For T hours
B mph
1.3 mph
39 miles
B mph
For 16−T hours
1.3 mph

62. 1.5.7 Summary Want: The speed of the boat relative to the water.
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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:
What method might we use to solve this system of equations?
Substitution or Elimination

63. Review of Word Problems
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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