Math 40
Prealgebra
Section 3.6 – Applications
3.6 Applications
REVIEW from Section 1.6
Five Step Word Problem Method
1)
2)
3)
4)
5)
Identify a variable.
Write an equation.
Solve the equation.
State your answer.
Check your answer.
Consecutive Integers
Word Expression
Algebraic Expression
Three consecutive
integers
Let x  first integer
Then x  1  second consecutive integer
x  2  third consecutive integer
Three consecutive
even integers
Let y  first even integer
Then y  2  second consecutive even integer
y  4  third consecutive even integer
Three consecutive
odd integers
Let z  first odd integer
Then z  2  second consecutive odd integer
z  4  third consecutive odd integer
Example
x
x 1
x2



4
 4  1  5
 4  2  6
y
y2
y4



8
 8   2  10  8   4  12
z
z2
z4



5
 5  2  7
5  4  9
Note: Notice the expressions used for odd and even integers are the same,
y  first even integer, y  2  second even integer, y  4  third even integer and
z  first odd integer, z  2  second odd integer, z  4  third odd integer .
This is because if you start with an even number to get to the next even number you will need to
add 2. Likewise, if you start with an odd number to get to the next odd number you will need to
add 2. So, what makes your consecutive integers odd or even, depends on which number you
start with, not the expression used.
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2015 Worrel
Math 40
Prealgebra
Section 3.6 – Applications
Example 1: The three sides of a triangle are consecutive integers and the perimeter is 72 inches. Find the
measure of each side of the triangle.
Solution:
k 1
k
k 2
1) Identify a variable. k = the length of the first side of the triangle
2) Write an equation. If the sides are consecutive integers and k = the length of the first side of the
triangle, then the lengths of the other two sides should be k  1 and k  2 . The perimeter of a
triangle is the sum of the lengths of the three sides. So we get,
 k    k  1   k  2   72
length of first side
length of second side
length of third side
Perimeter
k  k  1  k  2  72
3) Solve the equation.
k  k  1  k  2  72
3k  3  72
3 3
3k
69

3
3
k
 23
4) State your answer. We still need to plug the k  23 into k  1 and k  2 to find the lengths of the
other two sides.
k  23
k 1  
  1   23  1  24
k  2     2   23  2  25
So, the lengths of the sides of the triangle are 23 inches, 24 inches, and 25 inches.
5) Check your answer. Does the answer makes sense? Yes!
Did you answer what the problem is asking for? Yes!
You Try It 1: The three sides of a triangle are consecutive integers and the perimeter is 57 centimeters. Find
the measure of each side of the triangle.
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2015 Worrel
Math 40
Prealgebra
Section 3.6 – Applications
Example 2: The length and width of a rectangle are consecutive odd integers. The perimeter of the rectangle is
168 centimeters. Find the length and width of the rectangle.
Solution:
W
W 2
1) Identify a variable. W = the width of the rectangle
2) Write an equation. If the length and width of a rectangle are consecutive odd integers and
W is the width of the rectangle, then W  2 = L, the length of the rectangle. The perimeter of a
rectangle is given by the formula, P  2L  2W . So we get,
2 W  2   2 W   168
Perimeter
width
length
2 W  2   2 W   168
3) Solve the equation.
2 W  2   2 W   168
2W  4  2W
 168
4W  4  168
4
4W
4
4
164
4
W  41
4) State your answer. We still need to plug the W  41into W  2 to find the length of the
rectangle.
W  41
W 2

  2   41  2  43
So, width and length of the rectangle are 41 centimeters and 43 centimeters, respectively.
5) Check your answer. Does the answer makes sense? Yes!
Did you answer what the problem is asking for? Yes!
You Try It 2: The length and width of a rectangle are consecutive odd integers. The perimeter of the rectangle
is 120 meters. Find the length and width of the rectangle.
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2015 Worrel
Math 40
Prealgebra
Section 3.6 – Applications
Example 3: Ray inherits $15,000 and decides to invest in two different types of accounts, a savings account
paying 2% interest, and a certificate of deposit paying 4% interest. He decides to invest $3,000
more in the certificate of deposit than in savings. Find the amount invested in each account.
Solution: 1) Identify a variable. s = the amount invested in the savings account
From the information given, we know
the amount invested in the certificate of deposit is $3000 more than what is invested in the savings account
amount invested in the certificate of deposit

3000

So, the amount invested in the certificate of deposit = 3000  s
Account Type
Amount Invested
s
Savings Account (2% interest)
3000  s
Certificate of Deposit (4% interest)
15000
Total
s
A table can be useful to
organize the information given.
2) Write an equation. Using the table above, we can see that if we add the amounts invested in
each account, we should get the total amount invested, $15,000. Hence we get,
 s    3000  s   15000
3) Solve the equation. Since there is no coefficient in front of the parentheses (other than the
invisible 1), we can remove them.
s  3000  s  15000
2 s  3000  15000
 3000  3000
2s
2
12000
2
s  6000
4) State your answer. We still need to plug the s  6000 into 3000  s to find the amount invested
in the certificate of deposit.
s  6000

3000  s  3000  
  3000   6000   9000
So, Ray invested $6,000 in the savings account and $9,000 in the certificate of deposit.
5) Check your answer. Does the answer makes sense? Yes!
Did you answer what the problem is asking for? Yes!
You Try It 3: Dylan invests a total of $2,750 in two accounts, a savings account paying 3% interest, and a
mutual fund paying 5% interest. He invests $250 less in the mutual fund than in savings. Find
the amount invested in each account.
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2015 Worrel
Math 40
Prealgebra
Section 3.6 – Applications
Example 4: Jose cracks open his piggy bank and finds that he has $3.45, all in nickels and dimes. He has 12
more dimes than nickels. How many dimes and nickels does Jose have?
Solution: 1) Identify a variable. n = the number of nickels
From the information given, we know
the number of dimes is 12 more than the number of nickels
the number of dimes

12

So, the number of dimes = 12  n
Type of
Value Number
Coin
of Coin of Coins
n
Nickel
5
12  n
Dime
10
n
Total Value
5n
10 12  n 
Note: We will write the values using
cents so we do not need to
worry about decimals.
345
Total Amount
To complete the table, we multiply the value of the coin with the number of coins to get the total
value of each coin.
2) Write an equation. Using the table above, we can see that if we add the total values of each
coin, we should get the total amount, 345. Hence we get,
5n  10 12  n   345
3) Solve the equation.
5n  10 12  n   345
5n  120  10n  345
15n  120  345
 120  120
15n
15
225
15
n  15
4) State your answer. We still need to plug the n  15 into 12  n to find the number of dimes.
n  15
12  n  12  

  12  15  27
So, Jose has 15 nickels and 27 dimes.
5) Check your answer. Does the answer makes sense? Yes!
Did you answer what the problem is asking for? Yes!
You Try It 4: David keeps his change in a bowl made by his granddaughter. There is $1.95 in change in the
bowl, all in dimes and quarters. There are two fewer quarters than dimes. How many dimes and
quarters does he have in the bowl?
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2015 Worrel
Math 40
Prealgebra
Section 3.6 – Applications
Example 5: A large children’s organization purchases tickets to the circus. The organization has a strict rule
that every five children must be accompanied by one adult. Hence, the organization orders five
times as many child tickets as it does adult tickets. Child tickets are $3 each and adult tickets are
$6 each. If the total costs of tickets is $420, how many child and adult tickets were purchased?
Solution: 1) Identify a variable. A = the number of Adult tickets purchased
From the information given, we know
the number of Child tickets is 5 times the number of Adult tickets
the number of Child tickets

5

So, the Child tickets purchased = 5A
Type of Price per Number of Tickets
Ticket
Ticket
Purchased
Adult
6
A
5A
Child
3
A
Cost
6A
3  5 A   15 A
420
Total Cost
To complete the table, we multiply the price per ticket with the number of tickets purchased to
get the cost.
2) Write an equation. Using the table above, we can see that if we add the cost of each
type of ticket, we should get the total cost, 420. Hence we get,
6 A  15 A  420
3) Solve the equation.
6 A  15 A  420
21A 420

21
21
A  20
4) State your answer. We still need to plug the A  20 into 5A to find the number of Child tickets
purchased.
A  20
5A  5
  5  20   100
So, the organization purchased 20 Adult tickets and 100 Child tickets.
5) Check your answer. Does the answer makes sense? Yes!
Did you answer what the problem is asking for? Yes!
You Try It 5: Emily purchased tickets to the IMAX theater for her family. An adult ticket costs $12 and a child
ticket costs $4. She buys two more child tickets than adult tickets and the total cost is $136.
How many adult and child tickets did she buy?
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