Eureka Math Algebra 1 Module 1 Lesson 16 Answer Key

Engage NY Eureka Math Algebra 1 Module 1 Lesson 16 Answer Key

Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key

Exercise 1.
Solve w2=121, for w. Graph the solution on a number line.
Answer:
w=±11
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 1

b. Solve w2<121, for w. Graph the solution on a number line, and write the solution set as a compound inequality.
Answer:
-11<w<11
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 2

c. Solve w2≥121, for w. Graph the solution on a number line, and write the solution set as a compound inequality.
Answer:
w≤-11 or w≥11
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 3

d. Quickly solve (x+7)2=121, for x. Graph the solution on a number line.
Answer:
x=-18 or x=4
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 4

e. Use your work from part (d) to quickly graph the solution on a number line to each inequality below.
i. (x+7)2<121
Answer:
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 5

ii. (x+7)2≥121
Answer:
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 6

Extension
Use the following to challenge students who finish early.
a. Poindexter says that (a+b)2 equals a2+2ab+b2. Is he correct?
b. Solve x2+14x+49<121, for x. Present the solution graphically on a number line.

Exercises 2–3
Give students four minutes to work on Exercises 2 and 3. Then, discuss the results as a class. Students are applying their knowledge from the previous lesson to solve an unfamiliar type of problem.

Exercise 2.
Consider the compound inequality -5<x<4. a. Rewrite the inequality as a compound statement of inequality. Answer: x>-5 and x<4

b. Write a sentence describing the possible values of x.
Answer:
x can be any number between -5 and 4.

c. Graph the solution set on the number line below.
Answer:
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 7

Exercise 3.
Consider the compound inequality -5<2x+1<4. a. Rewrite the inequality as a compound statement of inequality. Answer: 2x+1>-5 and 2x+1<4 b. Solve each inequality for x. Then, write the solution to the compound inequality. Answer: x>-3 and x<\(\frac{3}{2}\) OR -3<x<\(\frac{3}{2}\)

c. Write a sentence describing the possible values of x.
Answer:
x can be any number between -3 and \(\frac{3}{2}\).

d. Graph the solution set on the number line below.
Answer:
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 8

Review Exercise 3 with students to demonstrate how to solve it without rewriting it.
→ A friend of mine suggested I could solve the inequality as follows. Is she right?
-5<2x+1<4
-5-1<2x+1-1<4-1
-6<2x<3
-3<x<\(\frac{3}{2}\) Encourage students to articulate their thoughts and scrutinize each other’s reasoning. Point out to students that solving the two inequalities did not require any new skills. They are solved just as they learned in previous lessons. Have students verify their solutions by filling in a few test values. Remind students that the solution can be written two ways: x>-3 and x<\(\frac{3}{2}\) OR -3<x<\(\frac{3}{2}\)

Exercises 4–5
Give students four minutes to work on Exercises 4 and 5. Then, review the results as a class. Again, point out to students that solving the two inequalities did not require any new skills. They are solved just as they learned in previous lessons. Have students verify their solutions by filling in a few test values.

Exercise 4.
Given x<-3 or x>-1:
a. What must be true in order for the compound inequality to be a true statement?
Answer:
One of the statements must be true, so either x has to be less than -3, or it has to be greater than -1. (In this case, it is not possible that both are true.)

b. Write a sentence describing the possible values of x.
Answer:
x can be any number that is less than -3 or any number that is greater than -1.

c. Graph the solution set on the number line below.
Answer:
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 9

Exercise 5.
Given x+4<6 or x-1>3:
a. Solve each inequality for x. Then, write the solution to the compound inequality.
Answer:
x<2 or x>4

b. Write a sentence describing the possible values of x.
Answer:
x can be any number that is less than 2 or any number that is greater than 4.

c. Graph the solution set on the number line below.
Answer:
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 10

Exercise 6.
Solve each compound inequality for x, and graph the solution on a number line.
a. x+6<8 and x-1>-1
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 11
Answer:
x<2 and x>0 → 0<x<2

b. -1≤3-2x≤10
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 12
Answer:
x≥-\(\frac{7}{2}\) and x≤2 → –\(\frac{7}{2}\)≤x≤2

c. 5x+1<0 or 8≤x-5
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 13
Answer:
x<-\(\frac{1}{5}\) or x≥13 d. 10>3x-2 or x=4
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 14
Answer:
x<4 or x=4 → x≤4

e. x-2<4 or x-2>4
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 15
Answer:
x<6 or x>6 → x≠6

f. x-2≤4 and x-2≥4
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 16
Answer:
x=6

Debrief the exercise with the following questions:
→ Look at the solution to part (f) closely. Remind students that both statements must be true. Therefore, the solution is only x=6.
→ How would the solution to part (f) change if the “and” was an “or”? Let this discussion lead in to Exercise 7.

Exercise 7.
Solve each compound inequality for x, and graph the solution on a number line. Pay careful attention to the inequality symbols and the “and” or “or” statements as you work.
a. 1+x>-4 or 3x-6>-12
Answer:
x>-5
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 17

b. 1+x>-4 or 3x-6<-12
Answer:
x can be any real number. c. 1+x>4 and 3x-6<-12 Answer: x>3 and x<-2
No solution (empty set) since there are no numbers that satisfy both statements
Eureka Math Algebra 1 Module 1 Lesson 16 Exercise Answer Key 18

Eureka Math Algebra 1 Module 1 Lesson 16 Problem Set Answer Key

Solve each inequality for x, and graph the solution on a number line.

Question 1.
x-2<6 or \(\frac{x}{3}\)>4
Answer:
x<8 or x>12
Eureka Math Algebra 1 Module 1 Lesson 16 Problem Set Answer Key 30

Question 2.
-6<\($\frac{x+1}{4}$\)<3
Answer:
-25<x<11
Eureka Math Algebra 1 Module 1 Lesson 16 Problem Set Answer Key 31

Question 3.
5x≤21+2x or 3(x+1)≥24
Answer:
x≤7 or x≥7 → all real numbers
Eureka Math Algebra 1 Module 1 Lesson 16 Problem Set Answer Key 32

Question 4.
5x+2≥27 and 3x-1<29
Answer:
x≥5 and x<10 → 5≤x<10
Eureka Math Algebra 1 Module 1 Lesson 16 Problem Set Answer Key 33

Question 5.
0≤4x-3≤11
Answer:
\(\frac{3}{4}\)≤x≤\(\frac{7}{2}\)
Eureka Math Algebra 1 Module 1 Lesson 16 Problem Set Answer Key 34

Question 6.
2x>8 or -2x<4
Answer:
x>4 or x>-2 → x>-2
Eureka Math Algebra 1 Module 1 Lesson 16 Problem Set Answer Key 35

Question 7.
8≥-2(x-9)≥-8
Answer:
5≤x≤13
Eureka Math Algebra 1 Module 1 Lesson 16 Problem Set Answer Key 36

Question 8.
4x+8>2x-10 or \(\frac{1}{3}\) x-3<2
Answer:
x>-9 or x<15 → all real numbers
Eureka Math Algebra 1 Module 1 Lesson 16 Problem Set Answer Key 36.1

Question 9.
7-3x<16 and x+12<-8
Answer:
x>-3 and x<-20 → no solution
Eureka Math Algebra 1 Module 1 Lesson 16 Problem Set Answer Key 37

Question 10.
If the inequalities in Problem 8 were joined by “and” instead of “or,” what would the solution set become?
Answer:
-9 < x < 15
Eureka Math Algebra 1 Module 1 Lesson 16 Problem Set Answer Key 38

Question 11.
If the inequalities in Problem 9 were joined by “or” instead of “and,” what would the solution set become?
Answer:
x > -3 or x < -20
Eureka Math Algebra 1 Module 1 Lesson 16 Problem Set Answer Key 39

Eureka Math Algebra 1 Module 1 Lesson 16 Exit Ticket Answer Key

Question 1.
Solve each compound inequality for x, and graph the solution on a number line.
a. 9+2x<17 or 7-4x<-9
Answer:
x<4 or x>4 or x≠4
Eureka Math Algebra 1 Module 1 Lesson 16 Exit Ticket Answer Key 20

b. 6≤\(\frac{x}{2}\)≤11
Answer:
12 ≤ x ≤ 22
Eureka Math Algebra 1 Module 1 Lesson 16 Exit Ticket Answer Key 21

Question 2.
a. Give an example of a compound inequality separated by “or” that has a solution of all real numbers.
Answer:
Sample response: x>0 or x<2 b. Take the example from part (a), and change the “or” to an “and.” Explain why the solution set is no longer all real numbers. Use a graph on a number line as part of your explanation. Answer: Sample response: x>0 and x<2
In the first example, only one of the inequalities needs to be true to make the compound statement true. Any number selected is either greater than 0 or less than 2 or both. In the second example, both inequalities must be true to make the compound statement true. This restricts the solution set to only numbers between 0 and 2.

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