4.5 Some Applications of Algebraic Equations
One of the primary uses of equations in algebra is to model and solve application problems. In
fact, much of the remainder of this book is based on the application of algebra to real-world
situations. The purpose of this section is to introduce the use of variables in equations as a
method of solving applications, and to familiarize you with some of the common applications in
algebra. Throughout this process, you should keep in mind the following general principles:
Steps for Solving Application Problems
1.
2.
3.
4.
5.
Familiarize yourself with the situation.
Understand how the quantities are related.
Draw a picture if appropriate.
Establish variable(s) needed for the problem.
Decide on the equation and eliminate unneeded variables.
Solve the equation for the unknown.
Answer the question and check your answer.
The examples in this section will consist of number problems, geometry problems, age problems,
and value problems.
Example 1
One number is three more than twice another. The sum of the two numbers is
–33. Find the two numbers.
Solution
There are two numbers (related to each other) whose sum is –33. Let n represent
one of the numbers. Since the other number is three more than twice this number,
the other number can be represented as 3 + 2n. Since their sum is –33, we have
the equation:
n + ( 3 + 2n ) = !33
Solving the equation:
n + ( 3 + 2n ) = !33
3n + 3 = !33
3n + 3 + (!3) = !33 + (!3)
3n = !36
1
1
• 3n = • (!36)
3
3
n = !12
One of the numbers is –12. Since the other number was represented as 3 + 2n, the
other number is 3 + 2(!12) = 3 + (!24) = !21 .Finally, as a check, note that
!12 + (!21) = !33 , as required.
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Example 2
A 5 foot board is cut so that one piece is 10 inches longer than the other piece.
Find the length of each piece.
Solution
First note that the units given in this problem are not the same. Converting feet to
inches, the board is 5 • 12 = 60 inches long. Since the board is being cut into two
pieces, these pieces must have a sum of 60 inches. Let l represent the length of
the shorter piece. Since the other piece is 10 inches longer than this, its length can
be represented as l + 10. Since their sum is 60 inches, we have the equation:
l + ( l + 10 ) = 60
Solving the equation:
l + ( l + 10 ) = 60
2l + 10 = 60
2l + 10 + (!10) = 60 + (!10)
2l = 50
1
1
• 2l = • 50
2
2
l = 25
The shorter piece is 25 inches long. Since the other piece was represented as
l + 10, the length of the other piece is 25 + 10 = 35 inches. Finally, as a check,
note that 25 + 35 = 60 inches, as required.
The next example is based on a geometry concept called perimeter of a rectangle. A rectangle
is a four-sided figure in which opposite sides are parallel and equal in length. The shorter side of
a rectangle is called the width (labeled w in the figure), and the longer side is called the length
(labeled l in the figure):
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The perimeter of the rectangle is the sum of the four sides:
Perimeter = w + w + l + l = 2w + 2l
Note that we combined like terms to produce a more simplified formula for the perimeter.
Example 3
The length of a rectangle is 6 inches more than twice the width. If the perimeter
of the rectangle is 102 inches, find the length and width.
Solution
We are given a rectangle with a perimeter of 102 inches. Let w represent the
width of the rectangle. Since the length is 6 inches more than twice the width, the
length can be represented as 2w + 6 . Since the perimeter is 102 inches, we have
the equation:
2w + 2 ( 2w + 6 ) = 102
Solving the equation:
2w + 2 ( 2w + 6 ) = 102
2w + 4w + 12 = 102
6w + 12 = 102
6w + 12 + (!12) = 102 + (!12)
6w = 90
1
1
• 6w = • 90
6
6
w = 15
The width of the rectangle is 15 inches. Since the length was represented as
2w + 6 , the length is given by 2 • 15 + 6 = 30 + 6 = 36 inches. As a check,
compute the perimeter:
Perimeter = 2•15 + 2•36 = 30 + 72 = 102 inches
The next example is also a geometry idea related to triangles. A triangle is any three-sided figure,
and the three interior angles of the triangle have measures which always add to 180°. This is
actually part of a more general result which will be discussed in Chapter 7.
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Example 4
The angles of a triangle are three consecutive even integers. Find each of their
measures.
Solution
We know the angles of a triangle add to 180°. Let x represent the smallest angle
of the triangle. Since even integers are always 2 apart, we can represent the next
(middle) angle as x + 2 (representing 2° more than x), and we can represent the
third (largest) angle as x + 4 (representing 2° more than the middle, or 4° more
than x). Since the sum of these angles is 180°, we have the equation:
x + ( x + 2 ) + ( x + 4 ) = 180
Solving the equation:
x + ( x + 2 ) + ( x + 4 ) = 180
3x + 6 = 180
3x + 6 + (!6) = 180 + (!6)
3x = 174
1
1
• 3x = • 174
3
3
x = 58
The smallest angle is 58°, the middle angle is 58° + 2° = 60°, and the largest
angle is 58° + 4° = 62°. As a check, add the angles: 58° + 60° + 62° = 180°.
A favorite problem of many algebra books is age problems. The next examples illustrate a
technique for organizing information in age problems. This technique can be extended to many
other application problems.
Example 5
Fred is 12 years older than Barney. In four years Fred will be twice as old as
Barney. How old are each of them now?
Solution
If b represents Barney’s age, then b + 12 will represent Fred’s age. In four years
each person will be exactly four years older, so we add 4 to each of their current
ages. This can be organized in the following chart:
age now age in 4 years
Fred
b + 12
b + 16
Barney
b
b+4
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We still need an equation in this problem. Since Fred will be twice as old as
Barney in four years, the equation is:
b + 16 = 2 ( b + 4 )
Solving the equation:
b + 16 = 2 ( b + 4 )
b + 16 = 2b + 8
b + ( !2b ) + 16 = 2b + ( !2b ) + 8
!b + 16 = 8
!b + 16 + (!16) = 8 + (!16)
!b = !8
!1 • ( !b ) = !1 • ( !8 )
b=8
Barney is 8 years old, and Fred is 8 + 12 = 20 years old. As a check, in four years
Barney will be 12 and Fred will be 24, which is twice Barney’s age.
Example 6
In seven years Jon will be twice as old as he was nine years ago. How old is Jon
now?
Solution
Let j represent Jon’s age now. Again organizing the ages in a chart:
age now age in 7 years age 9 years ago
Jon
j
j+7
j!9
Jon’s age in 7 years will be twice his age 9 years ago, so our equation is:
j + 7 = 2 ( j ! 9)
Solving the equation:
j + 7 = 2 ( j ! 9)
j + 7 = 2 j ! 18
j + ( !2 j ) + 7 = 2 j + ( !2 j ) ! 18
! j + 7 = !18
! j + 7 + (!7) = !18 + (!7)
! j = !25
!1 • ( ! j ) = !1 • ( !25 )
j = 25
Jon’s current age is 25 years old. As a check, in 7 years Jon will be 25 + 7 = 32,
and 9 years ago Jon was 25 – 9 = 16, so his age in 7 years is double his age
9 years ago.
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The next type of problem we will look at is called a value problem. Our approach for solving it
will also involve a chart to organize information.
Example 7
A total of 100 tickets to a school play are sold consisting of child seats and adult
seats. The child seats sell for $4 and the adult seats sell for $6. If a total of $520
is collected for the tickets, how many adult and how many child seats are sold?
Solution
Let c represent the number of child seats sold. Since there are a total of
100 tickets sold among child seats and adult seats, there must be 100 – c adult
seats sold. Now organize the information in the following chart:
price number of tickets total amount collected
child
4
c
4c
adult
6
100 ! c
6 (100 ! c )
Note that the total amount collected is the price multiplied by the number of
tickets. These total amounts collected must add up to $520 (the total amount
collected for the tickets), so the equation is:
4c + 6 (100 ! c ) = 520
Solving the equation:
4c + 6 (100 ! c ) = 520
4c + 600 ! 6c = 520
!2c + 600 = 520
!2c + 600 + (!600) = 520 + (!600)
!2c = !80
1
1
! • ( !2c ) = ! • ( !80 )
2
2
c = 40
There were 40 child seats sold and 100 ! 40 = 60 adult seats sold. As a check,
we compute the total amount collected:
40 child @ $4: $160
60 adult @ $6: $360
total: $520
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Example 8
Bernice has $1, $5, and $10 bills in her purse. She has twice as many $1 bills as
$5 bills, and ten less $10 bills as $5 bills. If she has $308 in her purse, how many
of each type of bill does she have in her purse?
Solution
Let f represent the number of $5 bills in her purse. Since she has twice as many
$1 bills, then 2f represents the number of $1 bills. Since she has ten less $10 bills,
then f – 10 represents the number of $10 bills. Now organize the information in
the following chart:
bill number of bills total value
ones $1
2f
1• 2 f
fives $5
f
5• f
tens $10
f ! 10
10 • ( f ! 10 )
Note that the total value is the value of the bill multiplied by the number of bills.
Since these total values must add up to $308 (the amount in her purse), we have
the equation:
1 • 2 f + 5 • f + 10 • ( f ! 10 ) = 308
Solving the equation:
1 • 2 f + 5 • f + 10 • ( f ! 10 ) = 308
2 f + 5 f + 10 f ! 100 = 308
17 f ! 100 = 308
17 f ! 100 + 100 = 308 + 100
17 f = 408
1
1
• 17 f =
• 408
17
17
f = 24
Bernice has 24 $5 bills, 48 $1 bills (twice as many), and 14 $10 bills (ten less).
As a check, we compute the total money in her purse:
48 @ $1:
$48
24 @ $5:
$120
14 @ $10:
$140
total:
$308
Terminology
Steps for solving application problems
age problem
perimeter of a rectangle
value problem
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Exercise Set 4.5
Solve each of the following number problems. Use the steps for solving application problems.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
One number is two more than three times another. The sum of the two numbers is 34.
Find the two numbers.
One number is five more than three times another. The sum of the two numbers is 49.
Find the two numbers.
One number is four less than three times another. The sum of the two numbers is 52.
Find the two numbers.
One number is six less then three times another. The sum of the two numbers is 54.
Find the two numbers.
One number is three more than six times another. The sum of the two numbers is –46.
Find the two numbers.
One number is six less than five times another. The sum of the two numbers is –78.
Find the two numbers.
One number is twelve less than six times another. The sum of the two numbers is
–187. Find the two numbers.
One number is ten less than five times another. The sum of the two numbers is –100.
Find the two numbers.
The sum of twice a number and 4 is –12. Find the number.
The sum of three times a number and 7 is –23. Find the number.
The difference of three times a number and 6 is –24. Find the number.
The difference of four times a number and 5 is –53. Find the number.
A 4 foot board is cut so that one piece is 6 inches longer than the other piece. Find the
length of each piece.
A 6 foot board is cut so that one piece is 8 inches shorter than the other piece. Find the
length of each piece.
A 5 foot board is cut so that one piece is twice as long as the other piece. Find the
length of each piece.
A 5 foot board is cut so that one piece is three times as long as the other piece. Find
the length of each piece.
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Solve each of the following perimeter problems. Use the steps for solving application problems.
17. The length of a rectangle is 3 inches more than twice the width. If the perimeter of the
rectangle is 42 inches, find the length and width.
18. The length of a rectangle is 1 inch more than twice the width. If the perimeter of the
rectangle is 44 inches, find the length and width.
19. The length of a rectangle is 2 meters less than twice the width. If the perimeter of the
rectangle is 26 meters, find the length and width.
20. The length of a rectangle is 2 meters less than twice the width. If the perimeter of the
rectangle is 38 meters, find the length and width.
21. The length of a rectangle is 3 feet more than twice the width. If the perimeter of the
rectangle is 42 feet, find the length and width.
22. The length of a rectangle is 3 feet more than twice the width. If the perimeter of the
rectangle is 60 feet, find the length and width.
23. A rectangular room is being designed such that the length is 2.5 times the width. If the
perimeter of the room is 56 feet, find the length and width.
24. A rectangular room is being designed such that the length is 3.5 times the width. If the
perimeter of the room is 72 feet, find the length and width.
25. A rectangular garden is being planted such that the length is 1 meter less than three
times the width. If the perimeter of the garden is 54 meters, find the length and width.
26. A rectangular garden is being planted such that the length is 3 meters less than three
times the width. If the perimeter of the garden is 58 meters, find the length and width.
27. The length and width of a rectangle are two consecutive even integers. If the perimeter
of the rectangle is 44 feet, find the length and width.
28. The length and width of a rectangle are two consecutive odd integers. If the perimeter
of the rectangle is 88 inches, find the length and width.
29. The length and width of a rectangle are two consecutive integers. If the perimeter of
the rectangle is 86 meters, find the length and width.
30. The length and width of a rectangle are two consecutive integers. If the perimeter of
the rectangle is 198 feet, find the length and width.
Solve each of the following triangle problems. Use the steps for solving application problems.
31. The smallest angle in a triangle is 5° less than the middle angle, and the largest angle
is 5° more than the middle angle. Find the three angles of the triangle.
32. The smallest angle in a triangle is 8° less than the middle angle, and the largest angle
is 2° more than the middle angle. Find the three angles of the triangle.
33. The middle angle in a triangle is twice the smallest angle, and the largest angle is triple
the smallest angle. Find the three angles of the triangle.
34. The middle angle in a triangle is three times the smallest angle, and the largest angle is
six times the smallest angle. Find the three angles of the triangle.
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35. One angle in a triangle is 40°, and the other two angles are equal in measure. Find the
measure of the other two angles.
36. One angle in a triangle is 90°, and the other two angles are equal in measure. Find the
measure of the other two angles.
37. The angles in a triangle are three consecutive integers. Find the measure of the three
angles.
38. The angles in a triangle are integers which are 10° apart. Find the measure of the three
angles.
Solve each of the following age problems. Use the steps for solving application problems.
39. Kyle is four years older than Travis. Eight years ago Kyle was twice as old as Travis.
How old are each of them now?
1
40. Brett is five years older than Lia. Three years ago Brett was 1 times as old as Lia.
2
How old are each of them now?
41. Marble is ten years older than Frank. Five years ago Marble was twice as old as Frank.
How old are each of them now?
1
42. Jack is 1 times as old as Jill. Eight years ago Jack was twice as old as Jill. How old
2
are each of them now?
43. Jeremy is 18 years older than Julia. In six years Jeremy will be twice as old as Julia.
How old are each of them now?
44. Janice is 18 years older than Frank. Three years ago Janice was three times as old as
Frank. How old are each of them now?
45. Don is 25 years older than Darlene. In ten years Don will be twice as old as Darlene.
How old are each of them now?
46. Fran is six years older than Martha. In two years Fran will be twice as old as Martha
was two years ago. How old are each of them now?
47. In five years Lia will be twice as old as she was five years ago. How old is she now?
2
48. In five years Brett will be 1 as old as he was five years ago. How old is he now?
3
49. In three years Louis will be twice as old as he was seven years ago. How old is he
now?
50. In three years Daniel will be twice as old as he was five years ago. How old is he now?
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Solve each of the following value problems. Use the steps for solving application problems.
51. A total of 600 tickets to a school play are sold consisting of child seats and adult seats.
The child seats sell for $5 and the adult seats sell for $8. If a total of $4200 is collected
for the tickets, how many child and how many adult tickets are sold?
52. A total of 400 tickets to a high school basketball game are sold consisting of student
seats and general public seats. The student seats sell for $4 and the general public seats
sell for $5. If a total of $1850 is collected for the tickets, how many student and how
many general public tickets are sold?
53. Tickets for a small arena (2000 seat) performance of Bob Dylan are sold for $25 and
$40. If the concert is sold out and a total of $62,000 is collected, how many of each
type of ticket are sold?
54. Tickets for a small arena (2500 seat) performance of Neil Young are sold for $40 and
$75. If the concert is sold out and a total of $138,500 is collected, how many of each
type of ticket are sold?
55. Tickets for a 16,000 seat Rush concert are sold for $30 and $40. If the concert is sold
out and a total of $520,000 is collected, how many of each type of ticket are sold?
56. Tickets for a 15,000 seat Rolling Stones concert are sold for $75 and $100. If the
concert is sold out and a total of $1,250,000 is collected, how many of each type of
ticket are sold?
57. Tickets for a recent Metallica concert were sold for $50, $75, and $100. They sold
twice as many $75 tickets as $100 tickets, and three times as many $50 tickets as $100
tickets. If the total concert receipts for tickets was $4 million, how many people
attended the concert?
58. Tickets for the most recent Pink Floyd concert were sold for $75, $100, and $120.
They sold twice as many $100 tickets as $120 tickets, and twice as many $75 tickets
as $100 tickets. If the total concert receipts for tickets was $6.2 million, how many
people attended the concert?
59. Sandy has $82 in her purse consisting of $1 and $5 bills. If she has 30 bills in her
purse, how many of each type of bill does she have?
60. Martha has $52 in her purse consisting of $1 and $5 bills. If she has 20 bills in her
purse, how many of each type of bill does she have?
61. Frank has nickels and dimes in his pocket. He has 25 coins in his pocket with a total
value of $1.85. How many of each type of coin does he have?
62. Jerry has nickels and dimes in his pocket. He has 30 coins in his pocket with a total
value of $2.15. How many of each type of coin does he have?
63. Pete has nickels and quarters in his pocket. He has 30 coins in his pocket with a total
value of $3.10. How many of each type of coin does he have?
64. Frank finds some nickels and quarters in Todd’s pocket. He finds 40 coins with a total
value of $4.60. How many of each type of coin does he find?
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65. Mary has $92 in $1, $5, and $10 bills. She has twice as many $5 bills as $10 bills, and
three times as many $1 bills as $10 bills. How many of each type of bill does she
have?
66. George has $134 in $1, $5, and $10 gambling chips. He has the same amount of $5
and $10 chips, and 30 total chips. How many of each type of chip does he have?
67. Frank finds $2.25 in pennies, nickels, and quarters on his floor. He finds 25 total coins,
and one less quarter than nickel. How many of each type of coin does he find?
68. Frank finds $2.66 in pennies, nickels, and quarters in his washing machine (after doing
Todd’s laundry). He finds 30 total coins, and the same amount of pennies as nickels.
How many of each type of coin does he find?
69. Karla finds a sale on peeled tomatoes in 11 oz, 15 oz, and 25 oz cans. She buys 30
cans total, and the same amount of 11 oz as 15 oz cans. If she buys a total of 462 oz
of peeled tomatoes, how many of each size can does she buy?
70. After buying tomatoes, Karla (from Exercise #69) hears about a tennis shoe sale at her
favorite discount store. Shoes are priced at $2.50, $4, and $6. She buys two more pairs
of $4 shoes than $6 shoes, and four more pairs of $2.50 shoes than $6 shoes. If she
spends $118 total, how many of each price of shoe does she buy?
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