Engage NY Eureka Math 8th Grade Module 4 Lesson 29 Answer Key
Eureka Math Grade 8 Module 4 Lesson 29 Example Answer Key
Example 1.
The sum of two numbers is 361, and the difference between the two numbers is 173. What are the two numbers?
Answer:
→ Together, we will read a word problem and work toward finding the solution.
→ The sum of two numbers is 361, and the difference between the two numbers is 173. What are the two numbers?
Provide students time to work independently or in pairs to solve this problem. Have students share their solutions and explain how they arrived at their answers. Then, show how the problem can be solved using a system of linear equations.
→ What do we need to do first?
We need to define our variables.
→ If we define our variables, we can better represent the situation we have been given. What should the variables be for this problem?
Let x represent one number, and let y represent the other number.
→ Now that we know the numbers are x and y, what do we need to do now?
We need to write equations to represent the information in the word problem.
→ Using x and y, write equations to represent the information we are provided.
The sum of two numbers is 361 can be written as x + y = 361. The difference between the two numbers is 173 can be written as x – y = 173.
→ We have two equations to represent this problem. What is it called when we have more than one linear equation for a problem, and how is it represented symbolically?
We have a system of linear equations.
x + y = 361
x – y = 173
→ We know several methods for solving systems of linear equations. Which method do you think will be the most efficient, and why?
We should add the equations together to eliminate the variable y because we can do that in one step.
Solve the system:
x + y = 361
x – y = 173
Sample student work:
x + y = 361
x – y = 173
x + x + y – y = 361 + 173
2x = 534
x = 267
267 + y = 361
y = 94
The solution is (267, 94).
→ Based on our work, we believe the two numbers are 267 and 94. Check to make sure your answer is correct by substituting the numbers into both equations. If it makes a true statement, then we know we are correct. If it makes a false statement, then we need to go back and check our work.
Sample student work:
267 + 94 = 361
361 = 361
267 – 94 = 173
173 = 173
→ Now we are sure that the numbers are 267 and 94. Does it matter which number is x and which number is y?
Not necessarily, but we need their difference to be positive, so x must be the larger of the two numbers to make sense of our equation x – y = 173.
Example 2.
There are 356 eighth – grade students at Euclid’s Middle School. Thirty – four more than four times the number of girls is equal to half the number of boys. How many boys are in eighth grade at Euclid’s Middle School? How many girls?
Answer:
→ Again, we will work together to solve the following word problem.
→ There are 356 eighth – grade students at Euclid’s Middle School. Thirty – four more than four times the number of girls is equal to half the number of boys. How many boys are in eighth grade at Euclid’s Middle School? How many girls? What do we need to do first?
We need to define our variables.
→ If we define our variables, we can better represent the situation we have been given. What should the variables be for this problem?
Let x represent the number of girls, and let y represent the number of boys.
Whichever way students define the variables, ask them if it could be done the opposite way. For example, if students respond as stated above, ask them if we could let x represent the number of boys and y represent the number of girls. They should say that at this stage it does not matter if x represents girls or boys, but once the variable is defined, it does matter.
→ Now that we know that x is the number of girls and y is the number of boys, what do we need to do now?
We need to write equations to represent the information in the word problem.
→ Using x and y, write equations to represent the information we are provided.
There are 356 eighth – grade students can be represented as x + y = 356. Thirty – four more than four times the number of girls is equal to half the number of boys can be represented as 4x + 34 = \(\frac{1}{2}\) y.
→ We have two equations to represent this problem. What is it called when we have more than one linear equation for a problem, and how is it represented symbolically?
We have a system of linear equations.
x + y = 356
4x + 34 = \(\frac{1}{2}\) y
→ We know several methods for solving systems of linear equations. Which method do you think will be the most efficient and why?
Answers will vary. There is no obvious “most efficient” method. Accept any reasonable responses as long as they are justified.
Solve the system:
x + y = 356
4x + 34 = \(\frac{1}{2}\) y
Sample student work:
x + y = 356
4x + 34 = \(\frac{1}{2}\) y
2(4x + 34 = \(\frac{1}{2}\) y)
8x + 68 = y
x + y = 356
8x + 68 = y
x + 8x + 68 = 356
9x + 68 = 356
9x = 288
x = 32
32 + y = 356
y = 324
The solution is (32, 324).
→ What does the solution mean in context?
Since we let x represent the number of girls and y represent the number of boys, it means that there are 32 girls and 324 boys at Euclid’s Middle School in eighth grade.
→ Based on our work, we believe there are 32 girls and 324 boys. How can we be sure we are correct?
We need to substitute the values into both equations of the system to see if it makes a true statement.
32 + 324 = 356
356 = 356
4(32) + 34 = \(\frac{1}{2}\) (324)
128 + 34 = 162
162 = 162
Example 3.
A family member has some five – dollar bills and one – dollar bills in her wallet. Altogether she has 18 bills and a total of $62. How many of each bill does she have?
Answer:
→ Again, we will work together to solve the following word problem.
→ A family member has some five – dollar bills and one – dollar bills in her wallet. Altogether she has 18 bills and a total of $62. How many of each bill does she have? What do we do first?
We need to define our variables.
→ If we define our variables, we can better represent the situation we have been given. What should the variables be for this problem?
Let x represent the number of $5 bills, and let y represent the number of $1 bills.
Again, whichever way students define the variables, ask them if it could be done the opposite way.
→ Now that we know that x is the number of $5 bills and y is the number of $1 bills, what do we need to do now?
We need to write equations to represent the information in the word problem.
→ Using x and y, write equations to represent the information we are provided.
Altogether she has 18 bills and a total of $62 must be represented with two equations, the first being x + y = 18 to represent the total of 18 bills and the second being 5x + y = 62 to represent the total amount of money she has.
→ We have two equations to represent this problem. What is it called when we have more than one linear equation for a problem, and how is it represented symbolically?
We have a system of linear equations.
x + y = 18
5x + y = 62
→ We know several methods for solving systems of linear equations. Which method do you think will be the most efficient and why?
Answers will vary. Students might say they could multiply one of the equations by – 1, and then they would be able to eliminate the variable y when they add the equations together. Other students may say it would be easiest to solve for y in the first equation and then substitute the value of y into the second equation. After they have justified their methods, allow them to solve the system in any manner they choose.
→ Solve the system:
x + y = 18
5x + y = 62
Sample student work:
x + y = 18
5x + y = 62
x + y = 18
y = – x + 18
y = – x + 18
5x + y = 62
5x + ( – x) + 18 = 62
4x + 18 = 62
4x = 44
x = 11
11 + y = 18
y = 7
The solution is (11, 7).
→ What does the solution mean in context?
Since we let x represent the number of $5 bills and y represent the number of $1 bills, it means that the family member has 11 $5 bills, and 7 $1 bills.
→ The next step is to check our work.
It is obvious that 11 + 7 = 18, so we know the family member has 18 bills.
→ It makes more sense to check our work against the actual value of those 18 bills in this case. Now we check the second equation.
5(11) + 1(7) = 62
55 + 7 = 62
62 = 62
Example 4.
A friend bought 2 boxes of pencils and 8 notebooks for school, and it cost him $11. He went back to the store the same day to buy school supplies for his younger brother. He spent $11.25 on 3 boxes of pencils and 5 notebooks. How much would 7 notebooks cost?
Answer:
Solve the system:
2x + 8y = 11
3x + 5y = 11.25.
Sample student work:
2x + 8y = 11
3x + 5y = 11.25
3(2x + 8y = 11)
6x + 24y = 33
– 2(3x + 5y = 11.25)
– 6x – 10y = – 22.50
6x + 24y = 33
– 6x – 10y = – 22.50
6x + 24y – 6x – 10y = 33 – 22.50
24y – 10y = 10.50
14y = 10.50
y = \(\frac{10.50}{14}\)
y = 0.75
2x + 8(0.75) = 11
2x + 6 = 11
2x = 5
x = 2.50
The solution is (2.50, 0.75).
→ What does the solution mean in context?
It means that a box of pencils costs $2.50, and a notebook costs $0.75.
→ Before we answer the question that this word problem asked, check to make sure the solution is correct.
Sample student work:
2(2.50) + 8(0.75) = 11
5 + 6 = 11
11 = 11
3(2.50) + 5(0.75) = 11.25
7.50 + 3.75 = 11.25
11.25 = 11.25
→ Now that we know we have the correct costs for the box of pencils and notebooks, we can answer the original question: How much would 7 notebooks cost?
The cost of 7 notebooks is 7(0.75) = 5.25. Therefore, 7 notebooks cost $7.25.
→ Keep in mind that some word problems require us to solve the system in order to answer a specific question, like this example about the cost of 7 notebooks. Other problems may just require the solution to the system to answer the word problem, like the first example about the two numbers and their sum and difference. It is always a good practice to reread the word problem to make sure you know what you are being asked to do.
Eureka Math Grade 8 Module 4 Lesson 29 Exercise Answer Key
Exercises
Exercise 1.
A farm raises cows and chickens. The farmer has a total of 42 animals. One day he counts the legs of all of his animals and realizes he has a total of 114. How many cows does the farmer have? How many chickens?
Answer:
Let x represent the number of cows and y represent the number of chickens. Then:
x + y = 42
4x + 2y = 114
– 2(x + y = 42)
– 2x – 2y = – 84
– 2x – 2y = – 84
4x + 2y = 114
– 2x – 2y + 4x + 2y = – 84 + 114
– 2x + 4x = 30
2x = 30
x = 15
15 + y = 42
y = 27
The solution is (15, 27).
4(15) + 2(27) = 114
60 + 54 = 114
114 = 114
The farmer has 15 cows and 27 chickens.
Exercise 2.
The length of a rectangle is 4 times the width. The perimeter of the rectangle is 45 inches. What is the area of the rectangle?
Answer:
Let x represent the length and y represent the width. Then:
x = 4y
2x + 2y = 45
2(4y) + 2y = 45
8y + 2y = 45
10y = 45
y = 4.5
x = 4(4.5)
x = 18
The solution is (18, 4.5).
2(18) + 2(4.5) = 45
36 + 9 = 45
45 = 45
Since 18×4.5 = 81, the area of the rectangle is 81 in^2.
Exercise 3.
The sum of the measures of angles x and y is 127°. If the measure of ∠x is 34° more than half the measure of ∠y, what is the measure of each angle?
Answer:
Let x represent the measure of ∠x and y represent the measure of ∠y. Then:
x + y = 127
x = 34 + \(\frac{1}{2}\) y
34 + \(\frac{1}{2}\) y + y = 127
34 + \(\frac{3}{2}\) y = 127
\(\frac{3}{2}\) y = 93
y = 62
x + 62 = 127
x = 65
The solution is (65,62).
65 = 34 + \(\frac{1}{2}\) (62)
65 = 34 + 31
65 = 65
The measure of ∠x is 65°, and the measure of ∠y is 62°.
Eureka Math Grade 8 Module 4 Lesson 29 Problem Set Answer Key
Question 1.
Two numbers have a sum of 1,212 and a difference of 518. What are the two numbers?
Answer:
Let x represent one number and y represent the other number.
x + y = 1212
x – y = 518
x + y + x – y = 1212 + 518
2x = 1730
x = 865
865 + y = 1212
y = 347
The solution is (865, 347).
865 – 347 = 518
518 = 518
The two numbers are 347 and 865.
Question 2.
The sum of the ages of two brothers is 46. The younger brother is 10 more than a third of the older brother’s age. How old is the younger brother?
Answer:
Let x represent the age of the younger brother and y represent the age of the older brother.
x + y = 46
x = 10 + \(\frac{1}{3}\) y
10 + \(\frac{1}{3}\) y + y = 46
10 + \(\frac{4}{3}\) y = 46
\(\frac{4}{3}\) y = 36
y = 27
x + 27 = 46
x = 19
The solution is (19,27).
19 = 10 + \(\frac{1}{3}\) (27)
19 = 10 + 9
19 = 19
The younger brother is 19 years old.
Question 3.
One angle measures 54 more degrees than 3 times another angle. The angles are supplementary. What are their measures?
Answer:
Let x represent the measure of one angle and y represent the measure of the other angle.
x = 3y + 54
x + y = 180
3y + 54 + y = 180
4y + 54 = 180
4y = 126
y = 31.5
x = 3(31.5) + 54
x = 94.5 + 54
x = 148.5
The solution is ( 148.5, 31.5).
148.5 + 31.5 = 180
180 = 180
One angle measures 148.5°, and the other measures 31.5°.
Question 4.
Some friends went to the local movie theater and bought four large buckets of popcorn and six boxes of candy. The total for the snacks was $46.50. The last time you were at the theater, you bought a large bucket of popcorn and a box of candy, and the total was $9.75. How much would 2 large buckets of popcorn and 3 boxes of candy cost?
Answer:
Let x represent the cost of a large bucket of popcorn and y represent the cost of a box of candy.
4x + 6y = 46.50
x + y = 9.75
– 4(x + y = 9.75)
– 4x – 4y = – 39
4x + 6y = 46.50
– 4x – 4y = – 39
4x + 6y – 4x – 4y = 46.50 – 39
6y – 4y = 7.50
2y = 7.50
y = 3.75
x + 3.75 = 9.75
x = 6
The solution is (6, 3.75).
4(6) + 6(3.75) = 46.50
24 + 22.50 = 46.50
46.50 = 46.50
Since one large bucket of popcorn costs $6 and one box of candy costs $3.75, then
2(6) + 3(3.75) = 12 + 11.25 = 23.25, and two large buckets of popcorn and three boxes of candy will cost $23.25.
Question 5.
You have 59 total coins for a total of $12.05. You only have quarters and dimes. How many of each coin do you have?
Answer:
Let x represent the number of quarters and y represent the number of dimes.
x + y = 59
0.25x + 0.1y = 12.05
– 4(0.25x + 0.1y = 12.05)
– x – 0.4y = – 48.20
x + y = 59
– x – 0.4y = – 48.20
x + y – x – 0.4y = 59 – 48.20
y – 0.4y = 10.80
0.6y = 10.80
y = \(\frac{10.80}{0.6}\)
y = 18
x + 18 = 59
x = 41
The solution is (41,18).
0.25(41) + 0.1(18) = 12.05
10.25 + 1.80 = 12.05
12.05 = 12.05
I have 41 quarters and 18 dimes.
Question 6.
A piece of string is 112 inches long. Isabel wants to cut it into 2 pieces so that one piece is three times as long as the other. How long is each piece?
Answer:
Let x represent one piece and y represent the other.
x + y = 112
3y = x
3y + y = 112
4y = 112
y = 28
x + 28 = 112
x = 84
The solution is (84, 28).
3(28) = 84
84 = 84
One piece should be 84 inches long, and the other should be 28 inches long.
Eureka Math Grade 8 Module 4 Lesson 29 Exit Ticket Answer Key
Question 1.
Small boxes contain DVDs, and large boxes contain one gaming machine. Three boxes of gaming machines and a box of DVDs weigh 48 pounds. Three boxes of gaming machines and five boxes of DVDs weigh 72 pounds. How much does each box weigh?
Answer:
Let x represent the weight of the gaming machine box, and let y represent the weight of the DVD box. Then:
3x + y = 48
3x + 5y = 72
– 1(3x + y = 48)
– 3x – y = – 48
– 3x – y = – 48
3x + 5y = 72
3x – 3x + 5y – y = 72 – 48
4y = 24
y = 6
3x + 6 = 48
3x = 42
x = 14
The solution is (14, 6).
3(14) + 5(6) = 72
72 = 72
The box with one gaming machine weighs 14 pounds, and the box containing DVDs weighs 6 pounds.
Question 2.
A language arts test is worth 100 points. There is a total of 26 questions. There are spelling word questions that are worth 2 points each and vocabulary word questions worth 5 points each. How many of each type of question are there?
Answer:
Let x represent the number of spelling word questions, and let y represent the number of vocabulary word questions.
x + y = 26
2x + 5y = 100
– 2(x + y = 26)
– 2x – 2y = – 52
– 2x – 2y = – 52
2x + 5y = 100
2x – 2x + 5y – 2y = 100 – 52
3y = 48
y = 16
x + 16 = 26
x = 10
The solution is (10, 16).
2(10) + 5(16) = 100
100 = 100
There are 10 spelling word questions and 16 vocabulary word questions.