## Engage NY Eureka Math Algebra 1 Module 1 Lesson 22 Answer Key

### Eureka Math Algebra 1 Module 1 Lesson 22 Example Answer Key

Example 1.

Solve the following system of equations.

y=2x+1

x-y=7

Graphically:

Answer:

Algebraically:

Answer:

x – (2x + 1) = 7

x = -8

y = 2(-8) + 1

y = -15

Solution:

(-8, -15)

Reinforce that even though the “and” is not stated explicitly, it is implied when given a system of equations. Problems written using this notation are asking one to find the solution(s) where y=2x+1 and x-y=17. Work the problem using substitution. The elimination method is reviewed in the next lesson.

Example 2.

Now suppose the system of equations from Exercise 1 part (c) was instead a system of inequalities:

3x+y≥5

3x+y≤8

Graph the solution set.

Answer:

→ How did the solution set change from Exercise 1 part (c) to Example 2? What if we changed the problem to 3x+y≤5 and 3x+y≥8?

→ There would be no solution.

The solution to a system of inequalities is where their shaded regions intersect. Let this idea lead into Example 3.

Example 3.

Graph the solution set to the system of inequalities.

2x-y<3 and 4x+3y≥0

Answer:

### Eureka Math Algebra 1 Module 1 Lesson 22 Opening Exercise Answer Key

Consider the following compound sentence: x+y>10 and y=2x+1.

a. Circle all the ordered pairs (x,y) that are solutions to the inequality x+y>10.

b. Underline all the ordered pairs (x,y) that are solutions to the equation y=2x+1.

Answer:

c. List the ordered pair(s) (x,y) from above that are solutions to the compound sentence x+y>10 and y=2x+1.

Answer:

(5,11) and (12,25)

d. List three additional ordered pairs that are solutions to the compound sentence x+y>10 and y=2x+1.

Answer:

(4,9), (6,13), and (7,15)

Ask:

→ How many possible answers are there to part (d)?

→ Can anyone come up with a non-integer solution?

Discuss that just as they saw with compound equations in one variable, solving pairs of equations in two variables linked by AND is given by common solution points.

→ How does the solution set change if the inequality is changed to x+y≥10?

→ The point (3,7) would be added to the solution set.

Have students complete parts (e) and (f) in pairs and discuss responses.

e. Sketch the solution set to the inequality x+y>10 and the solution set to y=2x+1 on the same set of coordinate axes. Highlight the points that lie in BOTH solution sets.

Answer:

f. Describe the solution set to x+y>10 and y=2x+1.

Answer:

All points that lie on the line y=2x+1 and above the line y=-x+10.

→ Which gives a more clear idea of the solution set: the graph or the verbal description?

→ Answers could vary. The verbal description is pretty clear, but later in the lesson we will see systems with solution sets that would be difficult to describe adequately without a graph.

### Eureka Math Algebra 1 Module 1 Lesson 22 Exercise Answer Key

Have students complete Exercise 1 individually.

Exercise 1.

Solve each system first by graphing and then algebraically.

a. y=4x-1

y=-\(\frac{1}{2}\) x+8

Answer:

y=4x-1

y=-\(\frac{1}{2}\) x+8

(2,7)

b. 2x+y=4

2x+3y=9

Answer:

2x+y=4

2x+3y=9

(\(\frac{3}{4}\), \(\frac{5}{2}\))

c. 3x+y=5

3x+y=8

Answer:

3x+y=5

3x+y=8

No solution

As students finish, have them put both the graphical and algebraic approaches on the board for parts (a)–(c) or display student work using a document camera. Discuss as a class.

→ Were you able to find the exact solution from the graph?

→ Not for part (b).

→ Solving by graphing sometimes only yields an approximate solution.

→ How can you tell when a system of equations will have no solution from the graph?

→ The graphs do not intersect. For linear systems, this occurs when the lines have the same slope but have different y-intercepts, which means the lines will be parallel.

→ What if a system of linear equations had the same slope and the same y-intercept?

→ There would be an infinite number of solutions (all points that lie on the line).

Exercise 2.

Graph the solution set to each system of inequalities.

a. x-y>5

x>-1

Answer:

b. y≤x+4

y≤4-x

y≥0

Answer:

→ Where does the solution to the system of inequalities lie?

→ Where the shaded regions overlap.

→ What is true about all of the points in this region?

→ These points are the only ones that satisfy both inequalities. Verify this by testing a couple of points from the shaded region and a couple of points that are not in the shaded region to confirm this idea to students. Exercise

→ Could you express the solution set of a system of inequalities without using a graph?

→ Yes, using set notation, but a graph makes it easier to visualize and conceptualize which points are in the solution set.

→ How can you check your solution graph?

→ Test a few points to confirm that the points in the shaded region satisfy all the inequalities.

### Eureka Math Algebra 1 Module 1 Lesson 22 Problem Set Answer Key

Question 1.

Estimate the solution to the system of equations by graphing and then find the exact solution to the system algebraically.

4x+y=-5

x+4y=12

Answer:

Estimated solution: (-2.1,3.5)

Exact solution: (-\(\frac{32}{15}\),\(\frac{53}{15}\))

Question 2.

a. Without graphing, construct a system of two linear equations where (0,5) is a solution to the first equation but is not a solution to the second equation, and (3,8) is a solution to the system.

Answer:

The first equation must be y=x+5; the second equation could be any equation that is different from y=x+5, and whose graph passes through (3,8); for example, y=2x+2 will work.

b. Graph the system and label the graph to show that the system you created in part (a) satisfies the given conditions.

Answer:

Question 3.

Consider two linear equations. The graph of the first equation is shown. A table of values satisfying the second equation is given. What is the solution to the system of the two equations?

Answer:

The form of the second equation can be determined exactly to be y=4x-10. Since the first equation is only given graphically, one can only estimate the solution graphically. The intersection of the two graphs appears to occur at (2,-2). This may not be exactly right. Solving a system of equations graphically is always subject to inaccuracies associated with reading graphs.

Question 4.

Graph the solution to the following system of inequalities:

x≥0

y<1 x+3y>0

Answer:

Question 5.

Write a system of inequalities that represents the shaded region of the graph shown.

Answer:

y≥-x+6

y<1

Question 6.

For each question below, provide an explanation or an example to support your claim.

a. Is it possible to have a system of equations that has no solution?

Answer:

Yes, for example, if the equations’ graphs are parallel lines.

b. Is it possible to have a system of equations that has more than one solution?

Answer:

Yes, for example, if the equations have the same graph, or in general, if the graphs intersect more than once.

c. Is it possible to have a system of inequalities that has no solution?

Answer:

Yes, for example, if the solution sets of individual inequalities, represented by shaded regions on the coordinate plane, do not overlap.

### Eureka Math Algebra 1 Module 1 Lesson 22 Exit Ticket Answer Key

Question 1.

Estimate the solution to the system of equations whose graph is shown.

Answer:

(5.2,0.9)

Question 2.

Write the two equations for the system of equations, and find the exact solution to the system algebraically.

Answer:

y=-x+6

y=\(\frac{3}{4}\) x-3

\(\frac{3}{4}\) x-3=-x+6

\(\frac{7}{4}\) x=9

x=\(\frac{36}{7}\)

y=-\(\frac{36}{7}\)+6=\(\frac{6}{7}\)

(\(\frac{36}{7}\),\(\frac{6}{7}\))

Question 3.

Write a system of inequalities that represents the shaded region on the graph shown to the right.

Answer:

y≥-x+6

y≥\(\frac{3}{4}\) x-3