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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 9
Lesson 9: An Application of Linear Equations
99
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©2015 Great Minds. eureka-math.org This file derived from G8-M4-TE-1.3.0-09.2015
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 9: An Application of Linear Equations
Student Outcomes
Students know how to rewrite an exponential expression that represents a series as a linear equation.
Lesson Notes
The purpose of this lesson is to expose students to applications of linear equations. The Discussion revisits the Facebook
problem from Module 1 but this time in the context of a linear equation. This is an opportunity to highlight MP.1 (Make
sense of problems and persevere in solving them), as the Discussion requires students to work with equations in ways
they have not before. If the Discussion is too challenging for students, use Exercises 3β11, which is a series of more
accessible applications of linear equations.
Classwork
Discussion (30 minutes)
In Module 1, you saw the following problem:
You sent a photo of you and your family on vacation to seven Facebook friends. If each of them sends it to five
of their friends, and each of those friends sends it to five of their friends, and those friends send it to five
more, how many people (not counting yourself) will see your photo? Assume that no friend received the
photo twice.
In Module 1, you were asked to express your answer in exponential notation. The solution is given here:
(1) The number of friends you sent a photo to is 7.
(2) The number of friends 7 people sent the photo to is 7 β 5.
(3) The number of friends 7 β 5 people sent the photo to is (7 β 5) β 5.
(4) The number of friends (7 β 5) β 5 people sent the photo to is ((7 β 5) β 5) β 5.
Therefore, the total number of people who received the photo is
7 + 7 β 5 + 7 β 52 + 7 β 53.
Letβs refer to βyou sending the photoβ as the first step. Then, βyour friends sending the photo to their friendsβ
is the second step, and so on. In the original problem, there were four steps. Assuming the trend continues,
how would you find the sum after 10 steps?
We would continue the pattern until we got to the 10th step.
What if I asked you how many people received the photo after 100 steps?
It would take a long time to continue the pattern to the 100th
step.
We want to be able to answer the question for any number of steps. For that reason, we will work toward
expressing our answer as a linear equation.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 9
Lesson 9: An Application of Linear Equations
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For convenience, letβs introduce some symbols. Since we are talking about steps, we will refer to the sum after
step one as π1, the sum after step two as π2, the sum after step three as π3, and so on. Thus:
π1 = 7 (1)
π2 = 7 + 7 β 5 (2)
π3 = 7 + 7 β 5 + 7 β 52 (3)
π4 = 7 + 7 β 5 + 7 β 52 + 7 β 53 (4)
What patterns do you notice within each of the equations (1)β(4)?
They contain some of the same terms. For example, equation (2) is the same as (1), except equation (2)
has the term 7 β 5. Similarly, equation (3) is the same as (2), except equation (3) has the term 7 β 52.
What you noticed is true. However, we want to generalize in a way that does
not require us to know one step before getting to the next step. Letβs see what
other hidden patterns there are.
Letβs begin with equation (2):
π2 = 7 + 7 β 5
π2 β 7 = 7 β 5
π2 β 7 + 7 β 52 = 7 β 5 + 7 β 52
π2 β 7 + 7 β 52 = 5(7 + 7 β 5).
Notice that the grouping on the right side of the equation is exactly π2, so we have the following:
π2 β 7 + 7 β 52 = 5π2.
This equation is a linear equation in π2. It is an equation we know how to solve (pretend π2 is an π₯ if that
helps).
Letβs do something similar with equation (3):
π3 = 7 + 7 β 5 + 7 β 52
π3 β 7 = 7 β 5 + 7 β 52
π3 β 7 + 7 β 53 = 7 β 5 + 7 β 52 + 7 β 53
π3 β 7 + 7 β 53 = 5(7 + 7 β 5 + 7 β 52).
Again, the grouping on the right side of the equation is exactly the equation we began with, π3, so we have the
following:
π3 β 7 + 7 β 53 = 5π3.
This is a linear equation in π3.
Letβs work together to do something similar with equation (4):
π4 = 7 + 7 β 5 + 7 β 52 + 7 β 53.
What did we do first in each of the equations (2) and (3)?
We subtracted 7 from both sides of the equation.
Now we have the following:
π4 β 7 = 7 β 5 + 7 β 52 + 7 β 53.
Scaffolding:
Talk students through the manipulation of the equation. For example, βWe begin by subtracting 7 from both sides. Next, we will add the number 7 times 52 to both sides of the equation. Then, using the distributive property β¦.β
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 9
Lesson 9: An Application of Linear Equations
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What did we do next?
We added 7 β 5 raised to a power to both sides of the equation. When it was the second step, the
power of 5 was 2. When it was the third step, the power of 5 was 3. Now that it is the fourth step, the
power of 5 should be 4.
Now we have
π4 β 7 + 7 β 54 = 7 β 5 + 7 β 52 + 7 β 53 + 7 β 54.
What did we do after that?
We used the distributive property to rewrite the right side of the equation.
Now we have
π4 β 7 + 7 β 54 = 5(7 + 7 β 5 + 7 β 52 + 7 β 53).
What do we do now?
We substitute the grouping on the right side of the equation with π4.
Finally, we have a linear equation in π4:
π4 β 7 + 7 β 54 = 5π4.
Letβs look at the linear equations all together.
π2 β 7 + 7 β 52 = 5π2
π3 β 7 + 7 β 53 = 5π3
π4 β 7 + 7 β 54 = 5π4
What do you think the equation would be for π10?
According to the pattern, it would be π10 β 7 + 7 β 510 = 5π10.
This equation is solved differently from the equations that have been solved in order to show students the normal form
of an equation for the summation of a geometric series. In general, if π β 1, then
1 + π + π2 + β β β + ππβ1 + ππ =1 β ππ+1
1 β π.
Now letβs solve the equation. (Note, that we do not simplify (1 β 5) for the reason explained above.)
π10 β 7 + 7 β 510 = 5π10
π10 β 5π10 β 7 + 7 β 510 = 5π10 β 5π10
S10(1 β 5) β 7 + 7 β 510 = 0
π10(1 β 5) β 7 + 7 + 7 β 510 β 7 β 510 = 7 β 7 β 510
π10(1 β 5) = 7(1 β 510)
π10 =7(1 β 510)
(1 β 5)
π10 = 17β089β842
After 10 steps, 17,089,842 will see the photo!
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 9
Lesson 9: An Application of Linear Equations
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Exercises 1β2 (5 minutes)
Students complete Exercises 1β2 independently.
Exercises
1. Write the equation for the ππth step.
πΊππ β π + π β πππ = ππΊππ
2. How many people would see the photo after ππ steps? Use a calculator if needed.
πΊππ β π + π β πππ = ππΊππ
πΊππ β ππΊππ β π + +π β πππ = ππΊππ β ππΊππ
πΊππ(π β π) β π + π β πππ = π
πΊππ(π β π) β π + π + π β πππ β π β πππ = π β π β πππ
πΊππ(π β π) = π(π β πππ)
πΊππ =π(π β πππ)
(π β π)
πΊππ = ππβπππβπππβπππ
Exercises 3β11 as an Alternative to Discussion (30 minutes)
Students should be able to complete the following problems independently as they are an application of skills learned to
this point, namely, transcription and solving linear equations in one variable. Have students work on the problems one
at a time and share their work with the whole class, or assign the entire set and allow students to work at their own
pace. Provide correct solutions at the end of the lesson.
3. Marvin paid an entrance fee of $π plus an additional $π. ππ per game at a local arcade. Altogether, he spent
$ππ. ππ. Write and solve an equation to determine how many games Marvin played.
Let π represent the number of games he played.
π + π. πππ = ππ. ππ
π. πππ = ππ. ππ
π =ππ. ππ
π. ππ
π = ππ
Marvin played ππ games.
4. The sum of four consecutive integers is βππ. What are the integers?
Let π be the first integer.
π + (π + π) + (π + π) + (π + π) = βππ
ππ + π = βππ
ππ = βππ
π = βπ
The integers are βπ, βπ, βπ, and βπ.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 9
Lesson 9: An Application of Linear Equations
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5. A book has π pages. How many pages are in the book if Maria read ππ pages of a book on Monday, π
π the book on
Tuesday, and the remaining ππ pages on Wednesday?
Let π be the number of pages in the book.
π = ππ +π
ππ + ππ
π = πππ +π
ππ
π
ππ = πππ
π = πππ
The book has πππ pages.
6. A number increased by π and divided by π is equal to ππ. What is the number?
Let π be the number.
π + π
π= ππ
π + π = πππ
π = πππ
The number is πππ.
7. The sum of thirteen and twice a number is seven less than six times a number. What is the number?
Let π be the number.
ππ + ππ = ππ β π
ππ + ππ = ππ
ππ = ππ
π = π
The number is π.
8. The width of a rectangle is π less than twice the length. If the perimeter of the rectangle is ππ. π inches, what is the
area of the rectangle?
Let π represent the length of the rectangle.
ππ + π(ππ β π) = ππ. π
ππ + ππ β ππ = ππ. π
ππ β ππ = ππ. π
ππ = ππ. π
π =ππ. π
π
π = π. π
The length of the rectangle is π. π inches, and the width is ππ. π inches, so the area is πππ. ππ π’π§π.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 9
Lesson 9: An Application of Linear Equations
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9. Two hundred and fifty tickets for the school dance were sold. On Monday, ππ tickets were sold. An equal number
of tickets were sold each day for the next five days. How many tickets were sold on one of those days?
Let π be the number of tickets sold on one of those days.
πππ = ππ + ππ
πππ = ππ
ππ = π
ππ tickets were sold on each of the five days.
10. Shonna skateboarded for some number of minutes on Monday. On Tuesday, she skateboarded for twice as many
minutes as she did on Monday, and on Wednesday, she skateboarded for half the sum of minutes from Monday and
Tuesday. Altogether, she skateboarded for a total of three hours. How many minutes did she skateboard each day?
Let π be the number of minutes she skateboarded on Monday.
π + ππ +ππ + π
π= πππ
ππ
π+
ππ
π+
ππ + π
π= πππ
ππ
π= πππ
ππ = πππ
π = ππ
Shonna skateboarded ππ minutes on Monday, ππ minutes on Tuesday, and ππ minutes on Wednesday.
11. In the diagram below, β³ π¨π©πͺ ~ β³ π¨β²π©β²πͺβ². Determine the length of π¨πͺ Μ Μ Μ Μ Μ and π©πͺΜ Μ Μ Μ .
ππ
ππ. π β π=
ππ
ππ β π. π
ππ(ππ β π. π) = ππ(ππ. π β π)
πππ β π. π = πππ. π β πππ
ππππ β π. π = πππ. π
ππππ = πππ. π
π =πππ. π
πππ
π = π. π
The length of π¨πͺΜ Μ Μ Μ is π. π π¦π¦, and the length of π©πͺΜ Μ Μ Μ is ππ. π π¦π¦.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 9
Lesson 9: An Application of Linear Equations
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Closing (5 minutes)
Summarize, or ask students to summarize, the main points from the lesson:
We can rewrite equations to develop a pattern and make predictions.
We know that for problems like these, we can generalize equations so that we do not have to do each step to
get our answer.
We learned how equations can be used to solve problems.
Exit Ticket (5 minutes)
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 9
Lesson 9: An Application of Linear Equations
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Name Date
Lesson 9: An Application of Linear Equations
Exit Ticket
1. Rewrite the equation that would represent the sum in the fifth step of the Facebook problem:
π5 = 7 + 7 β 5 + 7 β 52 + 7 β 53 + 7 β 54.
2. The sum of four consecutive integers is 74. Write an equation, and solve to find the numbers.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 9
Lesson 9: An Application of Linear Equations
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Exit Ticket Sample Solutions
1. Rewrite the equation that would represent the sum in the fifth step of the Facebook problem:
πΊπ = π + π β π + π β ππ + π β ππ + π β ππ.
πΊπ = π + π β π + π β ππ + π β ππ + π β ππ
πΊπ β π = π β π + π β ππ + π β ππ + π β ππ
πΊπ β π + π β ππ = π β π + π β ππ + π β ππ + π β ππ + π β ππ
πΊπ β π + π β ππ = π(π + π β π + π β ππ + π β ππ + π β ππ)
πΊπ β π + π β ππ = π(πΊπ)
πΊπ β ππΊπ β π + π β ππ = π
πΊπ β ππΊπ = π β (π β ππ)
(π β π)πΊπ = π β (π β ππ)
(π β π)πΊπ = π(π β ππ)
πΊπ =π(π β ππ)
π β π
2. The sum of four consecutive integers is ππ. Write an equation, and solve to find the numbers.
Let π be the first number.
π + (π + π) + (π + π) + (π + π) = ππ
ππ + π = ππ
ππ = ππ
π = ππ
The numbers are ππ, ππ, ππ, and ππ.
Problem Set Sample Solutions
Assign the problems that relate to the elements of the lesson that were used with students.
1. You forward an e-card that you found online to three of your friends. They liked it so much that they forwarded it
on to four of their friends, who then forwarded it on to four of their friends, and so on. The number of people who
saw the e-card is shown below. Let πΊπ represent the number of people who saw the e-card after one step, let πΊπ
represent the number of people who saw the e-card after two steps, and so on.
πΊπ = π
πΊπ = π + π β π
πΊπ = π + π β π + π β ππ
πΊπ = π + π β π + π β ππ + π β ππ
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 9
Lesson 9: An Application of Linear Equations
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a. Find the pattern in the equations.
πΊπ = π + π β π
πΊπ β π = π β π
πΊπ β π + π β ππ = π β π + π β ππ
πΊπ β π + π β ππ = π(π + π β π)
πΊπ β π + π β ππ = ππΊπ
πΊπ = π + π β π + π β ππ
πΊπ β π = π β π + π β ππ
πΊπ β π + π β ππ = π β π + π β ππ + π β ππ
πΊπ β π + π β ππ = π(π + π β π + π β ππ)
πΊπ β π + π β ππ = ππΊπ
πΊπ = π + π β π + π β ππ + π β ππ
πΊπ β π = π β π + π β ππ + π β ππ
πΊπ β π + π β ππ = π β π + π β ππ + π β ππ + π β ππ
πΊπ β π + π β ππ = π(π + π β π + π β ππ + π β ππ)
πΊπ β π + π β ππ = ππΊπ
b. Assuming the trend continues, how many people will have seen the e-card after ππ steps?
πΊππ β π + π β πππ = ππΊππ
πΊππ β ππΊππ β π + π β πππ = π
πΊππ(π β π) = π β π β πππ
πΊππ(π β π) = π(π β πππ)
πΊππ =π(π β πππ)
(π β π)
πΊππ = πβπππβπππ
After ππ steps, π, πππ, πππ people will have seen the e-card.
c. How many people will have seen the e-card after π steps?
πΊπ =π(π β ππ)
(π β π)
For each of the following questions, write an equation, and solve to find each answer.
2. Lisa has a certain amount of money. She spent $ππ and has π
π of the original amount left. How much money did she
have originally?
Let π be the amount of money Lisa had originally.
π β ππ =π
ππ
βππ = βπ
ππ
πππ = π
Lisa had $πππ originally.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 9
Lesson 9: An Application of Linear Equations
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3. The length of a rectangle is π more than π times the width. If the perimeter of the rectangle is ππ. π ππ¦, what is the
area of the rectangle?
Let π represent the width of the rectangle.
π(π + ππ) + ππ = ππ. π
π + ππ + ππ = ππ. π
π + ππ = ππ. π
ππ = ππ. π
π =ππ. π
π
π = π. π
The width of the rectangle is π. π ππ¦, and the length is π. π ππ¦, so the area is ππ. ππ ππ¦π.
4. Eight times the result of subtracting π from a number is equal to the number increased by ππ. What is the number?
Let π be the number.
π(π β π) = π + ππ
ππ β ππ = π + ππ
ππ β ππ = ππ
ππ = ππ
π = π
The number is π.
5. Three consecutive odd integers have a sum of π. What are the numbers?
Let π be the first odd number.
π + (π + π) + (π + π) = π
ππ + π = π
ππ = βπ
π = βπ
The three consecutive odd integers are βπ, π, and π.
6. Each month, Liz pays $ππ to her phone company just to use the phone. Each text she sends costs her an additional
$π. ππ. In March, her phone bill was $ππ. ππ. In April, her phone bill was $ππ. ππ. How many texts did she send
each month?
Let π be the number of texts she sent in March.
ππ + π. πππ = ππ. ππ
π. πππ = ππ. π
π =ππ. π
π. ππ
π = πππ
She sent πππ texts in March.
Let π be the number of texts she sent in April.
ππ + π. πππ = ππ. ππ
π. πππ = ππ. ππ
π =ππ. ππ
π. ππ
π = πππ
She sent πππ texts in April.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 9
Lesson 9: An Application of Linear Equations
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7. Claudia is reading a book that has πππ pages. She read some of the book last week. She plans to read ππ pages
today. When she does, she will be π
π of the way through the book. How many pages did she read last week?
Let π be the number of pages she read last week.
π + ππ =π
π(πππ)
π + ππ = πππ
π = πππ
Claudia read πππ pages last week.
8. In the diagram below, β³ π¨π©πͺ ~ β³ π¨β²π©β²πͺβ². Determine the measure of β π¨.
ππ β π = π + ππ
ππ β π = ππ
ππ = ππ
π = π
The measure of β π¨ is ππΒ°.
9. In the diagram below, β³ π¨π©πͺ ~ β³ π¨β²π©β²πͺβ². Determine the measure of β π¨.
πππ β π = ππ + ππ
ππ β π = ππ
ππ = ππ
π = ππ
The measure of β π¨ is πππΒ°.