## Engage NY Eureka Math 7th Grade Module 3 Lesson 13 Answer Key

### Eureka Math Grade 7 Module 3 Lesson 13 Example Answer Key

Example 1: Evaluating Inequalities—Finding a Solution

The sum of two consecutive odd integers is more than – 12. Write several true numerical inequality expressions.

Answer:

The sum of two consecutive odd integers is more than – 12. What is the smallest value that will make this true?

a. Write an inequality that can be used to find the smallest value that will make the statement true.

Answer:

x: an integer

2x + 1: odd integer

2x + 3: next consecutive odd integer

2x + 1 + 2x + 3 > – 12

b. Use if – then moves to solve the inequality written in part (a). Identify where the 0’s and 1’s were made using the if – then moves.

Answer:

4x + 4 > – 12

4x + 4 – 4 > – 12 – 4 If a > b, then a – 4 > b – 4.

4x + 0 > – 16 0 was the result.

(\(\frac{1}{4}\))(4x) > (\(\frac{1}{4}\))( – 16) If a > b, then a(\(\frac{1}{4}\)) > b(\(\frac{1}{4}\)).

x > – 4 1 was the result.

c. What is the smallest value that will make this true?

Answer:

To find the odd integer, substitute – 4 for x in 2x + 1.

2( – 4) + 1

– 8 + 1

– 7

The values that will solve the original inequality are all the odd integers greater than – 7. Therefore, the smallest values that will make this true are – 5 and – 3.

### Eureka Math Grade 7 Module 3 Lesson 13 Exercise Answer Key

Opening Exercise: Writing Inequality Statements

Tarik is trying to save $265.49 to buy a new tablet. Right now, he has $40 and can save $38 a week from his allowance.

Write and evaluate an expression to represent the amount of money saved after …

2 weeks

Answer:

40 + 38(2)

40 + 76

116

3 weeks

Answer:

40 + 38(3)

40 + 114

154

4 weeks

Answer:

40 + 38(4)

40 + 152

192

5 weeks

Answer:

40 + 38(5)

40 + 190

230

6 weeks

Answer:

40 + 38(6)

40 + 228

268

7 weeks

Answer:

40 + 38(7)

40 + 266

306

8 weeks

Answer:

40 + 38(8)

40 + 304

344

When will Tarik have enough money to buy the tablet?

Answer:

From 6 weeks and onward

Write an inequality that will generalize the problem.

Answer:

38w + 40 ≥ 265.49 Where w represents the number of weeks it will take to save the money.

Exercise 1.

Connor went to the county fair with $22.50 in his pocket. He bought a hot dog and drink for $3.75 and then wanted to spend the rest of his money on ride tickets, which cost $1.25 each.

a. Write an inequality to represent the total spent where r is the number of tickets purchased.

Answer:

1.25r + 3.75 ≤ 22.50

b. Connor wants to use this inequality to determine whether he can purchase 10 tickets. Use substitution to show whether he will have enough money.

Answer:

1.25r + 3.75 ≤ 22.50

1.25(10) + 3.75 ≤ 22.50

12.5 + 3.75 ≤ 22.50

16.25 ≤ 22.50

True

He will have enough money since a purchase of 10 tickets brings his total spending to $16.25.

c. What is the total maximum number of tickets he can buy based upon the given information?

Answer:

1.25r + 3.75 ≤ 22.50

1.25r + 3.75 – 3.75 ≤ 22.50 – 3.75

1.25r + 0 ≤ 18.75

(\(\frac{1}{1.25}\))(1.25r) ≤ (\(\frac{1}{1.25}\))(18.75)

r ≤ 15

The maximum number of tickets he can buy is 15.

Exercise 2.

Write and solve an inequality statement to represent the following problem:

On a particular airline, checked bags can weigh no more than 50 pounds. Sally packed 32 pounds of clothes and five identical gifts in a suitcase that weighs 8 pounds. Write an inequality to represent this situation.

Answer:

x: weight of one gift

5x + 8 + 32 ≤ 50

5x + 40 ≤ 50

5x + 40 – 40 ≤ 50 – 40

5x ≤ 10

(\(\frac{1}{5}\))(5x) ≤ (\(\frac{1}{5}\))(10)

x ≤ 2

Each of the 5 gifts can weigh 2 pounds or less.

### Eureka Math Grade 7 Module 3 Lesson 13 Problem Set Sample Answer Key

Question 1.

Match each problem to the inequality that models it. One choice will be used twice.

_________ The sum of three times a number and – 4 is greater than 17. a. 3x + – 4 ≥ 17

_________ The sum of three times a number and – 4 is less than 17. b. 3x + – 4 < 17

_________ The sum of three times a number and – 4 is at most 17. c. 3x + – 4 > 17

_________ The sum of three times a number and – 4 is no more than 17. d. 3x + – 4 ≤ 17

_________ The sum of three times a number and – 4 is at least 17.

Answer:

c The sum of three times a number and – 4 is greater than 17. a. 3x + – 4 ≥ 17

b The sum of three times a number and – 4 is less than 17. b. 3x + – 4 < 17

d The sum of three times a number and – 4 is at most 17. c. 3x + – 4 > 17

d The sum of three times a number and – 4 is no more than 17. d. 3x + – 4 ≤ 17

a The sum of three times a number and – 4 is at least 17.

Question 2.

If x represents a positive integer, find the solutions to the following inequalities.

a. x < 7

Answer:

x < 7 or 1, 2, 3, 4, 5, 6

b. x – 15 < 20

Answer:

x < 35

c. x + 3 ≤ 15

Answer:

x ≤ 12

d. – x > 2

Answer:

There are no positive integer solutions.

e. 10 – x > 2

Answer:

x < 8

f. – x ≥ 2

Answer:

There are no positive integer solutions.

g. \(\frac{x}{3}\) < 2

x < 6 Answer: h. – \(\frac{x}{3}\) > 2

Answer:

There are no positive integer solutions.

i. 3 – \(\frac{x}{4}\) > 2

Answer:

x < 4

Question 3.

Recall that the symbol ≠ means not equal to. If x represents a positive integer, state whether each of the following statements is always true, sometimes true, or false.

a. x > 0

Answer:

Always true

b. x < 0

Answer:

False

c. x > – 5

Answer:

Always true

d. x > 1

Answer:

Sometimes true

e. x ≥ 1

Answer:

Always true

f. x ≠ 0

Answer:

Always true

g. x ≠ – 1

Answer:

Always true

h. x ≠ 5

Answer:

Sometimes true

Question 4.

Twice the smaller of two consecutive integers increased by the larger integer is at least 25.

Model the problem with an inequality, and determine which of the given values 7, 8, and/or 9 are solutions. Then, find the smallest number that will make the inequality true.

Answer:

2x + x + 1 ≥ 25

The smallest integer would be 8.

For x = 7:

2x + x + 1 ≥ 25

2(7) + 7 + 1 ≥ 25

14 + 7 + 1 ≥ 25

22 ≥ 25

False

For x = 8:

2x + x + 1 ≥ 25

2(8) + 8 + 1 ≥ 25

16 + 8 + 1 ≥ 25

25 ≥ 25

True

For x = 9:

2x + x + 1 ≥ 25

2(9) + 9 + 1 ≥ 25

18 + 9 + 1 ≥ 25

28 ≥ 25

True

The smallest integer would be 8.

Question 5.

a. The length of a rectangular fenced enclosure is 12 feet more than the width. If Farmer Dan has 100 feet of fencing, write an inequality to find the dimensions of the rectangle with the largest perimeter that can be created using 100 feet of fencing.

Answer:

Let w represent the width of the fenced enclosure.

w + 12: length of the fenced enclosure

w + w + w + 12 + w + 12 ≤ 100

4w + 24 ≤ 100

b. What are the dimensions of the rectangle with the largest perimeter? What is the area enclosed by this rectangle?

Answer:

4w + 24 ≤ 100

4w + 24 – 24 ≤ 100 – 24

4w + 0 ≤ 76

(\(\frac{1}{4}\))(4w) ≤ (\(\frac{1}{4}\))(76)

w ≤ 19

Maximum width is 19 feet.

Maximum length is 31 feet.

Maximum area: A = lw

A = (19)(31)

A = 589

The area is 589 ft < sup > 2 < /sup > .

Question 6.

At most, Kyle can spend $50 on sandwiches and chips for a picnic. He already bought chips for $6 and will buy sandwiches that cost $4.50 each. Write and solve an inequality to show how many sandwiches he can buy. Show your work, and interpret your solution.

Answer:

Let s represent the number of sandwiches.

4.50s + 6 ≤ 50

4.50s + 6 – 6 ≤ 50 – 6

4.50s ≤ 44

(\(\frac{1}{4.50}\))(4.50s) ≤ (\(\frac{1}{4.50}\))(44)

s ≤ 9 \(\frac{7}{9}\)

At most, Kyle can buy 9 sandwiches with $50.

### Eureka Math Grade 7 Module 3 Lesson 13 Exit Ticket Answer Key

Question 1.

Shaggy earned $7.55 per hour plus an additional $100 in tips waiting tables on Saturday. He earned at least $160 in all. Write an inequality and find the minimum number of hours, to the nearest hour, that Shaggy worked on Saturday.

Answer:

Let h represent the number of hours worked.

7.55h + 100 ≥ 160

7.55h + 100 – 100 ≥ 160 – 100

7.55h ≥ 60

(\(\frac{1}{7.55}\))(7.55h) ≥ (\(\frac{1}{7.55}\))(60)

h ≥ 7.9

If Shaggy earned at least $160, he would have worked at least 8 hours.