## Engage NY Eureka Math Precalculus Module 2 Lesson 15 Answer Key

### Eureka Math Precalculus Module 2 Lesson 15 Opening Exercise Answer Key

Opening Exercise:

Mariah was studying currents in two mountain streams. She determined that five times the current in stream A was 8 feet per second stronger than twice the current in stream B. Another day she found that double the current in stream A plus ten times the current in stream B was 3 feet per second. She estimated the current in stream A to be 1. 5 feet per second and stream B to be almost still (0 feet per second). Was her estimate reasonable? Explain your answer after completing parts (a) – (c).

a. Write a system of equations to model this problem.
5x = 8 + 2y
2x + 10y = 3

b. Solve the system using algebra.
Procedures may vary. An example of an appropriate algebraic procedure is shown.
Multiply the first equation by 5: 5(5x – 2y = 8) → 25x – 10y = 40
Add this equation to the second equation:

Back substitute the x-value into the first equation and isolate y.
5 . $$\frac{43}{27}$$ – 2y = 8
-2y = $$\frac{1}{27}$$
y = $$-\frac{1}{54}$$
Solution:
$$\left[\begin{array}{r} \frac{43}{27} \\ -\frac{1}{54} \end{array}\right]$$

c. Solve the system by representing it as a linear transformation of the point x and then applying the inverse of the transformation matrix L to the equation. Verify that the solution is the same as that found in part (b).

which is the same solution as in part (b).
Mariah ‘s estimate was reasonable because x = $$\frac{43}{27}$$ ≈ 1.59 ft/sec and y = –$$-\frac{1}{54}$$ ≈ -0.02 ft/sec. Since the signs are opposite, the currents are moving in opposite directions.

### Eureka Math Precalculus Module 2 Lesson 15 Example Answer Key

Example:

Dillon is designing a card game where different colored cards are assigned point values. Kryshna is trying to find the value of each colored card. Dillon gives him the following hints. If I have 3 green cards, 1 yellow card, and 2 blue cards in my hand, my total is 9. If I discard 1 blue card, my total changes to 7. If I have 1 card of each color (green, yellow, and blue), my cards total 1.

a. Write a system of equations for each hand of cards if x is the value of green cards, y is the value of yellow cards, and z is the value of blue cards.
3x + y + 2z = 9
3x + y + z = 7
x + y + z =1

b. Solve the system using any method you choose.
Answers will vary. An example of an appropriate response is shown.
Subtract the second equation from the first.

Subtract the third equation from the second.

Back substitute values of x and z into the first equation, and isolate y.
3(3) + y + 2(2) = 9
13 – y = 9
y = -4

Solution:
$$\left[\begin{array}{c} 3 \\ -4 \\ 2 \end{array}\right]$$ Green cards are worth 3 points, yellow cards -4 points, and blue cards 2 points.

c. Let x = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ and b = $$\left[\begin{array}{l} 9 \\ 7 \\ 1 \end{array}\right]$$. Find a matrix L so that the linear transformation equation Lx = b would produce image coordinates that are the same as the solution to the system of equations.
L = $$\left[\begin{array}{lll} 3 & 1 & 2 \\ 3 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]$$

d. Enter matrix L into a software program or app, and try to calculate its inverse. Does L have an inverse? If so, what is it?
L-1 = $$\left[\begin{array}{ccc} 0 & 0.50 & -0.50 \\ -1 & 0.50 & 1.50 \\ 1 & -1 & 0 \end{array}\right]$$

e. Calculate L-1$$\left[\begin{array}{l} 9 \\ 7 \\ 1 \end{array}\right]$$. Verify that the result is equivalent to the solution to the system you calculated in part (b). Why should the solutions be equivalent?

which is equivalent to the solution from part (b). This makes sense because the system of equations can be represented using the linear transformation equation Lx = b. From what we learned in Lesson 14, x = L-1b, so the solution found using inverse matrix operations should be the same as the solution set when solving the system of equations using algebra.

### Eureka Math Precalculus Module 2 Lesson 15 Exercise Answer Key

Exercise 1.
The system of equations is given:
2x – 4y + 6z = 14
9x – 3y + z = 10
5x + 9z = 1

a. Solve the system using algebra or matrix operations. If you use matrix operations, include the matrices you entered into the software and the calculations you performed to solve the system.
Answers will vary. An appropriate response is included.
Matrix method: A = $$\left[\begin{array}{ccc} 2 & -4 & 6 \\ 9 & -3 & 1 \\ 5 & 0 & 9 \end{array}\right]$$

b. Verify your solution is correct.
2($$-\frac{1}{85}$$) – 4($$-\frac{283}{85}$$) + 6($$\frac{2}{17}$$) = 14
9($$-\frac{1}{85}$$) – 3($$-\frac{283}{85}$$) + 1($$-\frac{2}{17}$$) = 10
5($$-\frac{1}{85}$$) + 9($$-\frac{2}{17}$$) = 1

c. Justify your decision to use the method you selected to solve the system.
Answers will vary. An example of an appropriate response is shown: The coefficients of the variables in the system did not contain corresponding variables that were identical or opposites, which might indicate that elimination is a slower method than using the matrix method.

Exercise 2.
An athletic director at an all-boys high school is trying to find out how many coaches to hire for the football, basketball, and soccer teams. To do this, he needs to know the number of boys that play each sport. He does not have names or numbers but finds a note with the following information listed:

The total number of boys on all three teams is 86.
The number of boys that play football is 7 less than double the total number of boys playing the other two sports. The number of boys that play football is 5 times the number of boys playing basketball.

a. Write a system of equations representing the number of boys playing each sport where x is the number of boys playing football, y basketball, and z soccer.
x + y + z = 86
2(y + z) – 7 = x
x = 5y

b. Solve the system using algebra or matrix operations. If you use matrix operations, include the matrices you entered into the software and the calculations you performed to solve the system.
Answers will vary. An appropriate response is shown.
Matrix method:

c. Verify that your solution is correct.
1(55) + 1(11) + 1(20) = 86
2(11 + 20) — 7 = 55
55 = 5(11)

d. Justify your decision to use the method you selected to solve the system.
Answers will vary. An example of an appropriate response is shown: Using technology to solve the system using matrices generally takes less time than using algebra, especially when the solution set contains fractions.

Exercise 3.
Kyra had $20,000 to invest. She decided to put the money into three different accounts earning 3%, 5%, and 7% simple interest, respectively, and she earned a total of$920.00 in interest. She invested half as much money at 7% as at 3%. How much did she invest in each account?

a. Write a system of equations that models this situation.
x + y + z = 20000
0.03x+ 0.05y + 0.07z = 920
x – 2z = 0

b. Find the amount invested in each account.
She invested $8,000 at 3%,$8,000 at 5%, and $4,000 at 7%. ### Eureka Math Precalculus Module 2 Lesson 15 Problem Set Answer Key Question 1. A small town has received funding to design and open a small airport. The airport plans to operate flights from three airlines. The total number of flights scheduled is 100. The airline with the greatest number of flights is planned to have double the sum of the flights of the other two airlines. The plan also states that the airline with the greatest number of flights will have 40 more flights than the airline with the least number of flights. a. Represent the situation described with a system of equations. Define all variables. Answer: a + b + c = 100 a = 2(b + c) a = c +40 a: number of flights for the airline with the most flights b: number of flights for the airline with the second-greatest number of flights C: number of flights for the airline with the least number of flights Note: Variables used may differ. b. Represent the system as a linear transformation using the matrix equation Ax = b. Define matrices A, x, and b. Answer: Equation: $$\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -2 & -2 \\ 1 & 0 & -1 \end{array}\right]\left[\begin{array}{l} a \\ b \\ c \end{array}\right]=\left[\begin{array}{c} 100 \\ 0 \\ 40 \end{array}\right]$$ A:$$\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -2 & -2 \\ 1 & 0 & -1 \end{array}\right]$$ x: $$\left[\begin{array}{l} a \\ b \\ c \end{array}\right]$$ b:$$\left[\begin{array}{c} 100 \\ 0 \\ 40 \end{array}\right]$$ c. Explain how you can determine if the matrix equation has a solution without solving it. Answer: Answers will vary but should indicate that technology can be used to verify that matrix A has a nonzero determinant or that It has an inverse. d. Solve the matrix equation for x. Answer: x = A-1b = $$\left[\begin{array}{c} 66 \frac{2}{3} \\ 6 \frac{2}{3} \\ 26 \frac{2}{3} \end{array}\right]$$ e. Discuss the solution in context. Answer: The solution indicates that the airlines, from greatest to least number of flights, should have 66$$\frac{2}{3}$$, 26$$\frac{2}{3}$$ and 6$$\frac{2}{3}$$ flights, respectively. This does not make sense given that the number of flights must be a whole number. Therefore, the number of flights granted should be approximately 67, 27, and 7, which would satisfy the second and third conditions and would result in only 1 more than the total number of flights planned for the airport. Question 2. A new blockbuster movie opens tonight, and several groups are trying to buy tickets. Three types of tickets are sold: adult, senior (over 65), and youth (under 10). A group of 3 adults, 2 youths, and 1 senior pays$54.50 for their tickets. Another group of 6 adults and 12 youths pays $151.50. A final group of 1 adult, 4 youths, and 1 senior pays$49.00. What is the price for each type of ticket?

a. Represent the situation described with a system of equations. Define all variables.
3a + 2y + 1s = 54.50
6a + 12y = 151.50
1a + 4y + 1s = 49

a: price of an adult ticket
y: price of a youth ticket
s: price of a senior ticket
Note: Variables used may differ.

b. Represent the system as a linear transformation using the matrix equation Ax = b.
Ax = b$$\left[\begin{array}{ccc} 3 & 2 & 1 \\ 6 & 12 & 0 \\ 1 & 4 & 1 \end{array}\right]\left[\begin{array}{l} a \\ y \\ s \end{array}\right]=\left[\begin{array}{c} 54.50 \\ 151.50 \\ 49 \end{array}\right]$$

c. Explain how you can determine if the matrix equation has a solution without solving it.
Answers will vary but should Indicate that technology can be used to verify that matrix A has a nonzero determinant or that it has an inverse.

d. Solve the matrix equation for x.
x = A-1b = $$\left[\begin{array}{c} 10.25 \\ 7.50 \\ 8.75 \end{array}\right]$$

e. Discuss the solution in context.
An adult ticket costs $10.25, a youth ticket costs$7.50, and a senior ticket costs $8.75. f. How much would it cost your family to attend the movie? Answer: Answers will vary. Question 3. The system of equations is given: 5w – 2x + y + 3z = 2 4w – x + 6y + 2z = 10 w – x – y – z = 3 2w + 7x – 3y + 5z = 12 a. Write the system using a matrix equation in the form Ax = b. Answer: $$\left[\begin{array}{cccc} 5 & -2 & 1 & 3 \\ 4 & -1 & 6 & 2 \\ 1 & -1 & -1 & -1 \\ 2 & 7 & -3 & 5 \end{array}\right]\left[\begin{array}{c} w \\ x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 2 \\ 0 \\ 3 \\ 12 \end{array}\right]$$ b. Write the matrix equation that could be used to solve for x. Then use technology to solve for x. Answer: c. Verify your solution using back substitution. Answer: d. Based on your experience solving this problem and others like it in this lesson, what conclusions can you draw about the efficiency of using technology to solve systems of equations compared to using algebraic methods? Answer: Answers will vary. An example of an appropriate response would be that, in general, as the number of equations and variables in a system increases, the more efficient it is to use matrices and technology to solve systems when compared to using algebraic methods. Question 4. In three-dimensional space, a point x is reflected over the xz plane resulting in an image point of $$\left[\begin{array}{c} -3 \\ 1 \\ -2 \end{array}\right]$$. a. Write the transformation as an equation in the form Ax = b, where A represents the transformation of point x resulting in image point b. Answer: $$\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} -3 \\ 1 \\ -2 \end{array}\right]$$ b. Use technology t calculate A-1. Answer: A-1 = $$\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$ c. Calculate A-1b to solve the equation for x. Answer: x = $$\left[\begin{array}{l} -3 \\ -1 \\ -2 \end{array}\right]$$ d. Verify that this solution makes sense geometrically. Answer: Reflecting a point over the xz plane effectively reflects it over the plane y = 0, which will change the sign of the y-coordinate and will leave the x- and z-coordinates unchanged. 5. Jamie needed money and decided it was time to open her piggy bank. She had only nickels, dimes, and quarters. The total value of the coins was$85. 50. The number of quarters was 39 less than the number of dimes. The total value of the nickels and dimes was equal to the value of the quarters. How many of each type of coin did Jamie have? Write a system of equations and solve.
Let x represent the number of quarters, y represent the number of dimes, and z represent the number of nickels.
0.25x + 0.10y + 0.05z = 85.50
X = y – 39
0.05z + 0.10y = 0.25x
x = 171, y = 210, z = 435
Jamie has 171 quarters, 210 dimes, and 435 nickels.

### Eureka Math Precalculus Module 2 Lesson 15 Exit Ticket Answer Key

Question 1.
The lemonade sales at a baseball game are described as follows:

The number of small lemonades purchased is the number of mediums sold plus double the number of larges sold. The total number of all sizes sold is 70. One-and-a-half times the number of smalls purchased plus twice the number of mediums sold is 100.

Use a system of equations and its matrix representation to determine the number of small, medium, and large lemonades sold.
$$\left[\begin{array}{ccc} -1 & 1 & 2 \\ 1 & 1 & 1 \\ 1.5 & 2 & 0 \end{array}\right]\left[\begin{array}{c} s \\ m \\ l \end{array}\right]=\left[\begin{array}{c} 0 \\ 70 \\ 100 \end{array}\right]$$
A-1: $$\left[\begin{array}{ccc} \frac{-4}{9} & \frac{8}{9} & \frac{-2}{9} \\ \frac{1}{3} & \frac{-2}{3} & \frac{2}{3} \\ \frac{1}{9} & \frac{7}{9} & \frac{-4}{9} \end{array}\right]$$
A-1b:$$\left[\begin{array}{l} 40 \\ 20 \\ 10 \end{array}\right]$$