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
