## Engage NY Eureka Math Algebra 2 Module 1 Lesson 30 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 30 Example Answer Key

Example 1.

Determine the values for x, y, and z in the following system:

2x + 3y – z = 5

4x – y – z = – 1

x + 4y + z = 12

Answer:

2x + 3y – z = 5 …….. (1)

4x – y – z = – 1 …….. (2)

x + 4y + z = 12 …….. (3)

Suggest numbering the equations as shown above to help organize the process.

â†’ Eliminate z from equations (1) and (2) by subtraction. Replace equation (1) with the result.

â†’ Our goal is to find two equations in two unknowns. Thus, we will also eliminate z from equations (2) and (3) by adding as follows. Replace equation (3) with the result.

â†’ Our new system of three equations in three variables has two equations with only two variables in them:

– 2x + 4y = 6

4x – y – z = -1

5x + 3y = 11.

â†’ These two equations now give us a system of two equations in two variables, which we reviewed how to solve in Exercises 1 – 2.

– 2x + 4y = 6

5x + 3y = 11

At this point, let students solve this individually or with partners, or guide them through the process if necessary.

â†’ To get matching coefficients, we need to multiply both equations by a constant:

5(- 2x + 4y) = 5(6) Â Â â†’Â Â Â – 10x + 20y = 30

2(5x + 3y) = 2(11) Â Â â†’ Â Â 10x + 6y = 22.

â†’ Replacing the top equation with the sum of the top and bottom equations together gives the following:

26y = 52

10x + 6y = 22.

â†’ Replace y = 2 in one of the equations to find x:

5x + 3(2) = 11

5x + 6 = 11

5x = 5

x = 1.

â†’ Replace x = 1 and y = 2 in any of the original equations to find z:

2(1) + 3(2) – z = 5

2 + 6 – z = 5

8 – z = 5

z = 3.

â†’ The solution, x = 1, y = 2, and z = 3, can be written compactly as an ordered triple of numbers (1, 2, 3).

Consider pointing out to students that the point (1, 2, 3) can be thought of as a point in a three-dimensional coordinate plane, and that it is, like a two-by-two system of equations, the intersection point in three-space of the three planes given by the graphs of each equation. These concepts are not the point of this lesson, so addressing them is optional.

Point out that a linear system involving three variables requires three equations in order for the solution to possibly be a single point.

The following problems provide examples of situations that require solving systems of equations in three variables.

### Eureka Math Algebra 2 Module 1 Lesson 30 Exercise Answer Key

Determine the value of x and y in the following systems of equations.

Exercise 1.

2x + 3y = 7

2x + y = 3

Answer:

x = \(\frac{1}{2}\), y = 2

Exercise 2.

5x – 2y = 4

Answer:

– 2x + y = 2

x = 8, y = 18

Exercise 3.

A scientist wants to create 120 ml of a solution that is 30% acidic. To create this solution, she has access to a 20% solution and a 45% solution. How many milliliters of each solution should she combine to create the 30% solution?

Answer:

Milliliters of 20% solution: x ml

Milliliters of 45% solution: y ml

Write one equation to represent the total amounts of each solution needed:

x + y = 120.

Since 30% of 120 ml is 36, we can write one equation to model the acidic portion:

0.20x + 0.45y = 36.

Writing these two equations as a system:

x + y = 120

0.20x + 0.45y = 36

To solve, multiply both sides of the top equation by either 0.20 to eliminate x or 0.45 to eliminate y. The following steps will eliminate x:

0. 20(x + y) = 0.20(120)

0.20x + 0.45y = 40

which gives

0.20x + 0.20y = 24

0.20x + 0.45y = 36.

Replacing the top equation with the difference between the bottom equation and top equation results in a new system with the same solutions:

0.25y = 12

0.20x + 0.45y = 36.

The top equation can quickly be solved for y,

y = 48,

and substituting y = 48 back into the original first equation allows us to find x:

x + 48 = 120

x = 72.

Thus, we need 48 ml of the 45% solution and 72 ml of the 20% solution.

Exercise 4.

Given the system below, determine the values of r, s, and u that satisfy all three equations.

r + 2s – u = 8

s + u = 4

r – s – u = 2

Answer:

Adding the second and third equations together produces the equation r = 6. Substituting this into the first equation and adding it to the second gives 6 + 3s = 12, so that s = 2. Replacing s with 2 in the second equation gives u = 2. The solution to this system of equations is (6, 2, 2).

Exercise 5.

Find the equation of the form y = ax^{2}Â + bx + c whose graph passes through the points (1, 6), (3, 20), and (- 2, 15).

Answer:

We find a = 2, b = – 1, c = 5; therefore, the quadratic equation is y = 2x^{2} – x + 5.

â†’ Since we know three ordered pairs, we can create three equations.

6 = a + b + c

20 = 9a + 3b + c

15 = 4a – 2b + c

Ask students to explain where the three equations came from. Then have them use the technique from Example 1 to solve this system.

Have students use a graphing utility to plot y = 2x^{2} – x + 5 along with the original three points to confirm their answer.

### Eureka Math Algebra 2 Module 1 Lesson 30 Problem Set Answer Key

Solve the following systems of equations.

Question 1.

x + y = 3

y + z = 6

x + z = 5

Answer:

x = 1, y = 2, z = 4 or (1, 2, 4)

Question 2.

r = 2(s – t)

2t = 3(s – r)

r + t = 2s – 3

Answer:

r = 2, s = 4, t = 3, or (2, 4, 3)

Question 3.

2a + 4b + c = 5

a – 4b = – 6

2b + c = 7

Answer:

a = – 2, b = 1, c = 5 or (- 2, 1, 5)

Question 4.

2x + y – z = – 5

4x – 2y + z = 10

2x + 3y + 2z = 3

x = \(\frac{1}{2}\), y = – 2, z = 4 or (\(\frac{1}{2}\), – 2, 4)

Question 5.

r + 3s + t = 3

2r – 3s + 2t = 3

– r + 3s – 3t = 1

r = 3, s = \(\frac{1}{3}\), t = -1 or (3, \(\frac{1}{3}\), -1)

Question 6.

x – y = 1

2y + z = – 4

x – 2z = – 6

Answer:

x = – 2, y = -3, z = 2 or (- 2, – 3,2)

Question 7.

x = 3(y – z)

y = 5(z – x)

x + y = z + 4

Answer:

x = 3, y = 5, z = 4 or (3, 5, 4)

Question 8.

p + q + 3r = 4

2q + 3r = 7

p – q – r = – 2

Answer:

p = 2, q = 5, r = – 1 or (2, 5, – 1)

Question 9.

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) = 5

\(\frac{1}{x}+\frac{1}{y}\) = 2

\(\frac{1}{x}-\frac{1}{y}\) = – 2

Answer:

x = 1, y = 1, z = \(\frac{1}{3}\) or (1, 1, \(\frac{1}{3}\))

Question 10.

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) = 6

\(\frac{1}{b}+\frac{1}{c}\) = 5

\(\frac{1}{a}-\frac{1}{b}\) = – 1

Answer:

a = -1, b = \(\frac{1}{2}\), c = \(\frac{1}{3}\) or (1, \(\frac{1}{2}\), \(\frac{1}{3}\))

Question 11.

Find the equation of the form y = ax^{2} + bx + c whose graph passes through the points (1, – 1), (3, 23), and (- 1, 7).

Answer:

y = 4x^{2} – 4x – 1

Question 12.

Show that for any number t, the values x = t + 2, y = 1 – t, and z = t + 1 are solutions to the system of equations below.

x + y = 3

y + z = 2

(In this situation, we say that t parameterizes the solution set of the system.)

Answer:

x + y = (t + 2) + (1 – t) = 3

y + z = (1 – t) + (t + 1) = 2

Question 13.

Some rational expressions can be written as the sum of two or more rational expressions whose denominators are the factors of its denominator (called a partial fraction decomposition). Find the partial fraction decomposition for \(\frac{1}{n(n+1)}\) by finding the value of A that makes the equation below true for all n except 0 and – 1.

Answer:

Adding \(\frac{1}{n+1}\) to both sides of the equations, we have

so A = 1 and thus \(\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\).

Question 14.

A chemist needs to make 40 ml of a 15% acid solution. He has a 5% acid solution and a 30% acid solution on hand. If he uses the 5% and 30% solutions to create the 15% solution, how many ml of each does he need?

Answer:

He needs 24 ml of the 5% solution and 16 ml of the 30% solution.

Question 15.

An airplane makes a 400-mile trip against a head wind in 4 hours. The return trip takes 2. 5 hours, the wind now being a tail wind. If the plane maintains a constant speed with respect to still air, and the speed of the wind is also constant and does not vary, find the still-air speed of the plane and the speed of the wind.

Answer:

The speed of the plane in still wind is 130 mph, and the speed of the wind is 30 mph.

Question 16.

A restaurant owner estimates that she needs the same number of pennies as nickels and the same number of dimes as pennies and nickels together. How should she divide $26 between pennies, nickels, and dimes?

Answer:

She will need 200 dimes ($20 worth), 100 nickels ($5 worth), and 100 pennies ($1 worth) for a total of $26.

### Eureka Math Algebra 2 Module 1 Lesson 30 Exit Ticket Answer Key

For the following system, determine the values of p, q, and r that satisfy all three equations:

2p + q – r = 8

q + r = 4

p – q = 2.

Answer:

p = 4, q = 2, r = 2, or equivalently (4, 2, 2)