IM UNIT 2 Equations and Inequalities Teacher Guide

Activity Synthesis

Invite students to share their observations about the graphs they created. Ask
students why the graphs of equations A1, A2, and A3 all coincide with the graphs of the
original equation A. Discuss with students:

• "How can we explain the identical graphs?" (Equations A1, A2, and A3 are equivalent

to equation A, so they all have the same solutions as equation A, and their graphs are
the same line as the graph of A.)

• "What move was made to generate A1, A2, and A3? Why did it create equations that

are equivalent to A?" (The two sides of equation A was multiplied by the same factor,
which keeps the two sides equal.)

To further illustrate that equations A1, A2, and A3 are equivalent to equation A, and if time
permits, consider:

• Using tables to visualize the identical pairs. For example:

xy xy xy
01 01 01
1 -3 1 -3 1 -3
2 -7 2 -7 2 -7
3 -11 3 -11 3 -11

• Reminding students that, earlier in the unit, they saw that isolating one variable is a

way to see if two equations are equivalent. If we isolate in equations A, A1, A2, and

A3, the rearranged equation will be identical: .

16.2 Writing a New System to Solve a Given
System

15 minutes
In an earlier lesson, students learned that adding the two equations in a system creates a
new equation that shares a solution with the system. In the warm-up, they saw that

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multiplying an equation by a factor creates an equivalent equation that shares all the same
solutions as the original equation.

Here students learn that each time we perform a move that creates one or more new
equations, we are in fact creating a new system that is equivalent to the original system.
Equivalent systems are systems with the same solution set, and we can write a series of
them to help us get closer to finding the solution of an original system.

For instance, if and form a system, and is a multiple of

the second equation, the equations and form an equivalent

system that can help us eliminate and make progress toward finding the value of .

Students also learn to make an argument that explains why each new system is indeed
equivalent to the one that came before it (MP3), building on the work of justifying
equivalent equations in earlier lessons.

Building On

• HSA-REI.A.1

Addressing

• HSA-REI.C.5

Instructional Routines

• Think Pair Share

Launch

Arrange students in groups of 2. Give students 2–3 minutes of quiet work time, and then
1–2 minutes to discuss their thinking with their partner. Follow with a whole-class
discussion.

Anticipated Misconceptions

Some students may be confused by the subtraction symbol after the second equation,
wondering if some number is supposed to appear after the sign. Encourage them to
ignore the sign at first, find the relationship between the three equations, and then think
about what the sign might mean.

Student Task Statement
Here is a system you solved by graphing earlier.

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To start solving the system, Elena wrote:

And then she wrote:

1. What were Elena's first two moves? What might be possible reasons for those
moves?

2. Complete the solving process algebraically. Show that the solution is indeed
.

Student Response
Sample response:

1. First, she multiplied each side of the second equation by 4, which produces an
equivalent equation with in it. Next, she subtracted the new equation from the
first equation. Doing that creates another equation where the variable is
eliminated, leaving only one variable and making it possible to solve for .

2. can be rewritten as . Substituting 5 for in the first equation, we have

or , which means .

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Activity Synthesis

Invite students to share their analyses of Elena's moves. Highlight responses that point out

that Elena's moves enabled her to eliminate the -variable. (In other words, multiplying

equation B by 4 gives and subtracting this equation from equation A

removes the -variable.)

Display for all to see the two original equations in the system and the new equations Elena

wrote ( and ). Then, ask students to predict what the graphs of all

four equations might look like.

Next, use graphing technology to display all four graphs. Invite students to share their
observations about the graphs.

Students are likely to observe that the graphs all intersect at the same point, and

that there are only three lines, instead of 4. Discuss why only three lines are visible. Make

sure students understand that this is because the equations (equation B) and

are equivalent, so they share all the same solutions.

Then, focus students' attention on two things: the series of systems that came into play in
solving the original system, and the explanations that justify each step along the way.
Display the following systems and sequence the discussion as follows:

• "Here are the two equations in the original system. In

solving the system, what do we assume about the -
and -values in the equations?"

(We assume that there is a pair of - and -values that
make both equations true.)

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• "We didn't use the original two equations to solve.

Instead, we multiplied each side of equation B by 4 to
get equation B1. How do we know that the same
pair is also a solution to equation B1?"

(Multiplying each side of equation B by the same
number gives an equation that is equivalent to B. This
means it has all the same solutions as B, including the
pair that made the original system true.)

• "We couldn't yet solve the system with equations A and

B1, so we subtracted B1 from A and got equation C.

How do we know that the same pair from earlier

is also a solution to equation C?"

(When we subtracted from and

subtracted 36 from 1, we subtracted equal amounts

from each side of a true equation, which kept the two

sides equal. Even though the -value was eliminated in

the result, the -value that makes the original

equations true hasn't changed and is also a solution to

C.)

• Solving equation C gives us . How do we find the

-value?"

(Substituting this value into equation A or B and
solving it gives us the -value.)

• "If we substitute this pair of values for and in

equations A, B, B1, and C and evaluate the expressions,
can we expect to find true statements?"

(Yes. For A, it will be . For B, it will be . For

B1, it will be 36=36. For C, it will be .)

Explain that what we have done was to create equivalent systems—systems with the
exact same solution set—to help us get closer and closer to the solution of the original
system.

One way to create an equivalent system is by multiplying one or both equations by a
factor. It helps to choose the factor strategically—one that would allow one variable to be

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eliminated when the two equations in the new system are added or subtracted. Elena
chose to multiply equation B by 4 so that the -variable could be eliminated.

Ask students:

• "Suppose we want to solve the system by first eliminating the -variable. What factor

should we choose? Which equation should we apply it to?" (We could multiply
equation A by 2, or multiply equation B by .)

• "Could we multiply equation A by 6 and multiply equation B by -3 as a way to

eliminate the -variable?" (Yes. Each new equation ( and )

would be equivalent to the original. Adding the two new equations would eliminate

the -variable.)

16.3 What Comes Next?

10 minutes
Earlier, students encountered the idea of equivalent systems. They described the moves
that lead to equivalent equations and explained why the solution to those equations was
the same as the solution to the original system. This activity reinforces that work.

Given an unordered set of equivalent systems, students work with a partner to arrange
the systems such that the order constitutes logical steps toward solving an original system.
Along the way, they take turns describing how each system came from the previous one
and explaining how they know it has the same solution as its predecessor. The work allows
students to practice constructing logical arguments and critiquing those of others (MP3).

Here is an image of the cards for reference and planning.

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Addressing

• HSA-REI.C.5

Instructional Routines

• MLR8: Discussion Supports
• Take Turns

Launch
Arrange students in groups of 2. Give one set of pre-cut slips or cards from the blackline
master to each group. Ask students to find the slip or card that shows a system of
equations and is labeled “Start here.”

Explain that all the other slips contain equivalent systems that represent steps in solving
the starting system. Ask students to arrange the slips in the order that would lead to its
solution, and to make sure they can explain what moves take each system to the next
system and why each system is equivalent to the one before it.

Partners should take turns finding the next step in the solving process and explaining their
reasoning. As one student explains, the partner's job is to listen and make sure they agree

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and the explanation makes sense. If they disagree, the partners should discuss until they
reach an agreement.

Give students 5 minutes to arrange the systems. Follow with a whole-class discussion.

Support for English Language Learners

Speaking, Representing: MLR8 Discussion Supports. Use this routine to support student
conversation as they take turns selecting cards and explaining how each system came
from the previous one. Display the following prompts for students to use as they
respond to reasoning shared by the partner who selected the card: “I agree
because….” or “I disagree because….” Encourage students not only to challenge each
other if they disagree with the card selected, but also to press for clear explanations
that use mathematical language.
Design Principle(s): Support sense-making

Support for English Language Learners

Action and Expression: Internalize Executive Functions. Chunk this task into more
manageable parts to support students who benefit from support with organization
and problem solving. For example, guide students in a strategy by prompting them to
begin by identifying the first and last cards. Next, ask students to identify the card
where the solution for first appears, and to explain how they know. Encourage
students to use these three cards as benchmarks for sorting the remaining cards.
Supports accessibility for: Organization; Attention

Anticipated Misconceptions
Students who are thinking algorithmically about solving systems of equations may think
that the first step should involve multiplying each side of the second equation by ,
becoming frustrated when no cards show that step. Encourage them to compare both the
first and the second equations in the starting card to the first and second equations on
other cards to gain some ideas about what steps might have been taken.

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Student Task Statement

Your teacher will give you some slips of paper with systems of equations written on
them. Each system represents a step in solving this system:

Arrange the slips in the order that would lead to a solution. Be prepared to:

• Describe what move takes one system to the next system.
• Explain why each system is equivalent to the one before it.

Student Response

Original: Step 1: Step 2: Step 3:

Step 4: Step 5: Step 6: Step 7:

Sample description of each move:

• From the original system to Step 1: Multiply the first equation by 5.

• From Step 1 to Step 2: Multiply the second equation by 4.

• From Step 2 to Step 3: Add the two equations in Step 2.

• From Step 3 to Step 4: Divide each side of by 102 to solve for .

• From Step 4 to Step 5: Substitute for in the first equation and evaluating the

expression.

• From Step 5 to Step 6: Subtract 35 from each side of the first equation.

• From Step 6 to Step 7: Divide each side of by 4 to solve for .

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Are You Ready for More? :
This system of equations has solution

Find the missing values and .

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Student Response
,

Activity Synthesis
Ask students to share the order in which the systems should be arranged to lead to the
solution. Display the ordered systems for all to see.

Point out to students that this particular solution path involves multiplying each of the two
equations by a factor in order to eliminate the -variable. Ask students if it's possible to
eliminate a variable by multiplying only one equation by a factor. (Yes, we could multiply
the first equation by to eliminate , or by 3 to eliminate . Or we could multiply the
second equation by to eliminate , or by to eliminate .)

See Lesson Synthesis for discussion questions and ways to help students connect the
ideas in the lesson.

16.4 Build Some Equivalent Systems

Optional: 15 minutes
This optional activity gives students another opportunity to identify moves that lead to
equivalent systems and to explain why the resulting systems have the same solution
(MP3). It also prompts students to build their own equivalent systems, which encourages
them to think strategically about what to do to reach the goal of solving the system, rather
than on a particular solution path.

As students work, prompt students to articulate the reasoning and assumptions. Ask
questions such as:

• "How do you know the factor to use when multiplying an equation so a variable can

be eliminated?"

• "Can the factor be a fraction? A negative number? Why or why not?"
• "Why doesn't adding one equation to another equation change the solution of the

system?"

Addressing

• HSA-REI.C.5
• HSA-REI.C.6

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Launch
Keep students in groups of 2, if desired.

Anticipated Misconceptions
If students struggle to create an equivalent system of their own, ask them to start by
deciding on a variable they'd like to eliminate. Then, ask them to think about a factor that,
when multiplied to one equation, would produce the same or opposite coefficients for that
variable. If they are uncomfortable using a fractional factor, ask if they could find a factor
to apply to each equation such that the resulting equations have the same or opposite
coefficients for the variable they wish to eliminate.

Student Task Statement
Here is a system of equations:

1. To solve this system, Diego wrote these equivalent systems for his first two
steps.

Step 1: Step 2:

Describe the move that Diego made to get each equivalent system. Be
prepared to explain how you know the systems in Step 1 and Step 2 have the
same solution as the original system.

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2. Write another set of equivalent systems (different than Diego's first two
steps) that will allow one variable to be eliminated and enable you to solve the
original system. Be prepared to describe the moves you make to create each
new system and to explain why each one has the same solution as the original
system.

3. Use your equivalent systems to solve the original system. Then, check your
solution by substituting the pair of values into the original system.

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Student Response
1. Sample response:

◦ Step 1: Multiply the second equation by -5. Multiplying each side of an equation

by the same number gives an equivalent equation with the same solution.

◦ Step 2: Add the resulting equation to the first equation to eliminate the variable

. Adding an equal amount to each side of an equation keeps the two sides
equal, so the solution for the first equation is also a solution for the sum.

2. Sample responses:

◦ Multiply the second equation by and then subtract the resulting equation

from the first equation to eliminate the variable .

◦ Multiply the first equation by -4 and the second equation by 6, and then add the

two resulting equations to eliminate the variable .

3.

Activity Synthesis
Invite students with different first steps to display their equivalent systems and solution
paths. Prompt them (or others students) to explain why each system generated share the
same solution as the original system or the system before it.

Verify that, regardless of the moves made, the different paths all led to the same pair of
values.

Lesson Synthesis

Refer to the systems that students have ordered in the "What Comes Next?" activity, Invite
students to explain what move takes one system to the next and how they know the new
system is equivalent to the one before it (even though one equation has been
transformed).

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As students share their responses, record the descriptions of moves and the justifications
for all to see. Consider using a graphic organizer, as shown here. (A completed organizer
could be preserved and used as a reference by students later.)

step system what move why is the new system equivalent to
was made? the previous one?

0

1

2

..
.

If time is limited, focus on discussing the justifications on why one system in the sequence
is equivalent to the before it (the last column in the graphic organizer).

Highlight statements such as: "Adding equal amounts to the two sides of an equation
keeps the two sides equal, so the same - and -values that make the first equation true
also makes this new equation true. This means that same pair of values is also a solution
to the new system." If no students explained in this way, demonstrate it.

16.5 Make Your Move

Cool Down: 5 minutes
Building On

• HSA-REI.C.6

Student Task Statement
Lin and Priya were working on solving this system of equations.

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Lin's first move is to multiply the first equation by 3.

Priya's first move is to multiply the second equation by 2.

1. Explain why either move creates a new equation with the same solutions as
the original equation.

2. Whose first move would you choose to do to solve the system? Explain your
reasoning.

Student Response

1. Sample responses:

◦ Multiplying the two sides of an equation by the same factor creates an

equivalent equation, which has the same solution as the original equation.

◦ Multiplying the two sides of an equation by the same number keeps the two

sides equal, so the solution of the first equation still works for the second one.

2. Sample responses:

◦ I would choose Priya's move. Multiplying the second equation by 2

gives , it can be subtracted from the first to eliminate the -variable.

◦ I would choose Lin's move. Multiplying the first equation by 3 gives .

Subtracting the second equation from this equation eliminates the -variable.

◦ Either person's move would work. Priya's move eliminates the -variable and

Lin's move eliminates the -variable.

Student Lesson Summary

We now have two algebraic strategies for solving systems of equations: by
substitution and by elimination. In some systems, the equations may give us a clue
as to which strategy to use. For example:

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In this system, is already isolated in one equation. We

can solve the system by substituting for in the

second equation and finding .

This system is set up nicely for elimination because of
the opposite coefficients of the -variable. Adding the
two equations eliminates so we can solve for .

In other systems, which strategy to use is less straightforward, either because no
variables are isolated, or because no variables have equal or opposite coefficients.
For example:

To solve this system by elimination, we first need to rewrite one or both equations
so that one variable can be eliminated. To do that, we can multiply both sides of an
equation by the same factor. Remember that doing this doesn't change the equality
of the two sides of the equation, so the - and -values that make the first equation
true also make the new equation true.

There are different ways to eliminate a variable with this approach. For instance, we
could:

• Multiply Equation A by 3 to get

. Adding this equation

to Equation B eliminates .

• Multiply Equation B by to get

. Subtracting this
equation from Equation A
eliminates .

• Multiply Equation A by to get

and multiply Equation

B by to get .

Subtracting one equation from the
other eliminates .

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Each multiple of an original equation is equivalent to the original equation. So each
new pair of equations is equivalent to the original system and has the same
solution.

Let’s solve the original system using the first equivalent
system we found earlier.

• Adding the two equations eliminates , leaving a

new equation , or .

• Putting together and the original

gives us another equivalent system.

• Substituting 7 for in the second equation allows

us to solve for .

When we solve a system by elimination, we are essentially writing a series of
equivalent systems, or systems with the same solution. Each equivalent system
gets us closer and closer to the solution of the original system.

Glossary
• equivalent systems

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Lesson 16 Practice Problems

1. Problem 1
Statement

Solve each system of equations.
a.

b.

Solution

a.
b.

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2. Problem 2
Statement

Tyler is solving this system of equations:

He can think of two ways to eliminate a variable and solve the system:

◦ Multiply by 2, then subtract from the result.

◦ Multiply by 2, then add the result to .

Do both strategies work for solving the system? Explain or show your
reasoning.

Solution

Yes, both strategies work. The first strategy eliminates the -variable. The second

eliminates the -variable. The solution is regardless of the strategy used.

3. Problem 3

Statement

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Andre and Elena are solving this system of equations:

◦ Andre's first step is to write:
◦ Elena’s first step is to create a new system:

Do you agree with either first step? Explain your reasoning.

Solution

Sample responses:

◦ I agree with Andre's first step. The variable is equal to both and .

Writing makes it possible to solve for . Afterwards, we can use

the value of to find .

◦ I agree with Elena's first step. Multiplying the first equation by 3 gives two

equations with . Subtracting the two equations eliminates , making it

possible to solve for . Afterwards, we can use the value of to find .

◦ I agree with both first steps. Andre uses substitution and Elena uses elimination

to solve the system. Both ways give the same solution: .

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4. Problem 4
Statement

Select all systems that are equivalent to this system:
A.
B.
C.
D.
E.
F.

Solution

["B", "C", "D", "F"]

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5. Problem 5
Statement

Here is a system of equations with a solution:
a. Write a system of equations that is equivalent to this system. Describe
what you did to the original system to get the new system.

b. Explain how you know the new system has the same solution as the
original system.

Solution

a. Sample response:

I multiplied the second equation in the original system by 2 to get
. The first equation is unchanged.

b. Multiplying both sides of an equation by the same number keeps the two sides

equal, so the pair that makes the second equation true in the original

system also makes the new equation true. This means that the same pair

also makes the new system true.

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6. Problem 6

Statement

The cost to mail a package is $5.00. Noah has postcard stamps that are worth
$0.34 each and first-class stamps that are worth $0.49 each.

a. Write an equation that relates the number of postcard stamps , the
number of first-class stamps , and the cost of mailing the package.

b. Solve the equation for .

c. Solve the equation for .

d. If Noah puts 7 first-class stamps on the package, how many postcard
stamps will he need?

Solution (or equivalent)

a.

b.

c. (or equivalent)
d. . Noah will need 5 postcard stamps.

(From Unit 2, Lesson 8.)

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7. Problem 7

Statement

Here is a system of linear equations:

Find at least one way to solve the system by substitution and show your
reasoning. How many ways can you find? (Regardless of the substitution that
you do, the solution should be the same.)

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Solution . Substitute for the in

Sample responses:

◦ Rearrange the second equation to

the first equation:

Substituting 7.5 for in either equation and solving it gives .
for the in
◦ Rearrange the second equation to . Substitute

the first equation:

Substituting -1 for in either equation and solving it gives .
for the in the
◦ Rearrange the second equation to . Substitute

first equation:

Substituting -1 for in either equation and solving it gives .

(From Unit 2, Lesson 13.)

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8. Problem 8
Statement

Here is a system of equations:

Write an equation that results from subtracting the two equations.

Solution

or

(From Unit 2, Lesson 14.)

9. Problem 9

Statement

A grocery store sells bananas for dollars a pound and grapes for dollars a
pound. Priya buys 2.2 pounds of bananas and 3.6 pounds of grapes for $9.35.
Andre buys 1.6 pounds of bananas and 1.2 pounds of grapes for $3.68.

This situation is represented by the system of equations:

Explain why it makes sense in this situation that the solution of this system is

also a solution to .

Algebra1 Unit 2 Lesson 16 CC BY 2019 by Illustrative Mathematics 32

Solution

Sample response: represents the sum of Priya and Andre's

purchases. One side of the equation describes the combined cost based on the

weights of bananas and grapes. The other side shows the total cost in dollars. The

price of one pound of bananas and the price of one pound of grapes are unchanged,

so it makes sense that they are also a solution to this equation.

(From Unit 2, Lesson 15.)

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Lesson 17: Systems of Linear Equations and
Their Solutions

Goals
• Determine whether a system of equations will have 0, 1, or infinitely many solutions

by analyzing their structure or by graphing.

• Recognize that a system of linear equations can have 0, 1, or infinitely many

solutions.

• Use the structure of the equations in a linear system to make sense of the properties

of their graphs.

Learning Targets
• I can tell how many solutions a system has by graphing the equations or by analyzing

the parts of the equations and considering how they affect the features of the
graphs.

• I know the possibilities for the number of solutions a system of equations could have.

Lesson Narrative

This lesson serves two main goals. The first goal is to revisit the idea (first learned in
middle school) that not all systems of linear equations have a single solution. Some
systems have no solutions and others have infinitely many solutions. The second goal is to
investigate different ways to determine the number of solutions to a system of linear
equations.

Earlier in the unit, students learned that the solution to a system of equations is a pair of
values that meet both constraints in a situation, and that this condition is represented by a
point of intersection of two graphs. Here, students make sense of a system with no
solutions in a similar fashion. They interpret it to mean that there is no pair of values that
meet both constraints in a situation, and that there is no point at which the graph of each
each equation would intersect.

Next, students use what they learned about the structure of equations and about
equivalent equations to reason about the number of solutions. For instance, students
recognize that equivalent equations have the same solution set. This means that if the two
equations in a system are equivalent, we can tell—without graphing—that the system has

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 1

infinitely many solutions. These exercises are opportunities to look for and make use of
structure (MP7).

Likewise, students are aware that the graphs of linear equations with the same slope but
different vertical intercepts are parallel lines. If the equations in a system can be
rearranged into slope-intercept form (where the slope and vertical intercept become
"visible"), it is possible to determine how many solutions a system has without graphing.

Alignments

Building On

• 8.EE.C.8: Analyze and solve pairs of simultaneous linear equations.

• HSA-CED.A.4: Rearrange formulas to highlight a quantity of interest, using the same

reasoning as in solving equations. For example, rearrange Ohm's law to

highlight resistance .

Addressing

• HSA-CED.A.3: Represent constraints by equations or inequalities, and by systems of

equations and/or inequalities, and interpret solutions as viable or nonviable options
in a modeling context. For example, represent inequalities describing nutritional and
cost constraints on combinations of different foods.

• HSA-REI.C.6: Solve systems of linear equations exactly and approximately (e.g., with

graphs), focusing on pairs of linear equations in two variables.

Instructional Routines

• Card Sort
• Graph It
• MLR6: Three Reads
• MLR8: Discussion Supports

Required Materials

Graphing technology
Examples of graphing technology are: a handheld graphing calculator, a computer with a
graphing calculator application installed, and an internet-enabled device with access to a
site like desmos.com/calculator or geogebra.org/graphing. For students using the digital
materials, a separate graphing calculator tool isn't necessary; interactive applets are

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 2

embedded throughout, and a graphing calculator tool is accessible on the student digital
toolkit page.

Pre-printed slips, cut from copies of the blackline master

Required Preparation
Acquire devices that can run Desmos (recommended) or other graphing technology. It is
ideal if each student has their own device. (Desmos is available under Math Tools.)

Student Learning Goals

• Let's find out how many solutions a system of equations could have.

17.1 A Curious System

Warm Up: 10 minutes
In grade 8, students learned that some systems of linear equations have many solutions.
This warm-up reminds students about this fact, while also prompting them to use what
they learned in this unit to better understand what it means for a system to have infinitely
many solutions.

The first equation shows two variables adding up to 3, so students choose a pair of values
whose sum is 3. They notice that all pairs chosen are solutions to the system. Next, they
try to find a strategy that can show that there are countless other pairs that also satisfy the
constraints in the system. Monitor for these likely strategies:

• Solving by graphing: The graphs of the two equations are the same line, so all the

points on the line are solutions to the system.

• Solving by substitution: The first equation can be rearranged to .
or
Substituting for in the second equation gives is.

or . This equation is true no matter what

• Solving by elimination: If we multiply the first equation by 4, rearrange the second

equation to , and then subtract the second equation from the first, the

result is . Subtracting from each side gives , which is true regardless

of what or is.

• Reasoning about equivalent equations: If we rearrange the second equation so that

the variables are on the same side , we can see that this equation is a

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multiple of the first and are equivalent. This means they have the exact same
solution set, which contains infinite possible pairs of and .

Identify students using different strategies and ask them to share their thinking with the
class later.

Making graphing technology available gives students an opportunity to choose
appropriate tools strategically (MP5).

Building On

• 8.EE.C.8

Addressing

• HSA-REI.C.6

Instructional Routines

• Graph It

Launch
Arrange students in groups of 3–4. Give students 1–2 minutes of quiet time to complete
the first two questions. Remind students that the numbers don't have be integers. Then,
give students a minute to share with their group what their two values are and whether
the pair is a solution to the second equation.

Pause for a brief whole-class discussion. Consider recording quickly all the pairs that
students have verified to be solutions to the second equation and displaying them for all
to see. Ask: "Which two numbers are the solution to the system?" (All of them)

Ask students to proceed to the last question. Students may conclude that seeing all the
pairs that satisfy both equations is enough to show that the system has many solutions. If
so, ask how many solutions they think there are, and ask: "Can you show that any
two numbers that add up to 3 is also a solution to the system, without having to do
calculations for each pair?"

Student Task Statement

Andre is trying to solve this system of equations:

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 4

Looking at the first equation, he thought, "The solution to the system is a pair of
numbers that add up to 3. I wonder which two numbers they are."

1. Choose any two numbers that add up to 3. Let the first one be the -value and
the second one be the -value.

2. The pair of values you chose is a solution to the first equation. Check if it is
also a solution to the second equation. Then, pause for a brief discussion with
your group.

3. How many solutions does the system have? Use what you know about
equations or about solving systems to show that you are right.

Student Response
1. Sample responses: 4 and -1, 0 and 3, and

2. Yes, it is also a solution to the second equation. Sample reasoning for 4 and -1:

Substituting 4 and -1 into the second equation gives or

, which is a true equation.

3. Infinitely many solutions. See Activity Narrative for some possible explanations.

Activity Synthesis

Ask the class how many solutions they think the system has. Select previously identified
students to explain how they know that the system has infinitely many solutions, or that
any pair of values that add up to 3 and make the first equation true also make the second
equation true.

Students are likely to think of using graphs. It is not essential to elicit all strategies shown
in the Activity Narrative, but if no one thinks of an algebraic explanation, be sure to bring
one up. The last strategy mentioned in the Narrative (reasoning about equivalent
equations) is likely to be intuitive to students.

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 5

Make sure students see that the equations are equivalent, so all pairs of and values
that make one equation true also make the other equation true, giving an infinite number
of solutions.

17.2 What's the Deal?

10 minutes
In the warm-up, students saw that some systems have infinitely many solutions. In this
activity, they encounter a situation that can be represented with a system of equations but
the system has no solutions. Students write equations to represent the two constraints in
the situation and then solve the system algebraically and graphically.

As students work, notice the different ways they reach the conclusion that the systems
have no solutions. Identify students with varying strategies and ask them to share later.

Building On

• 8.EE.C.8

Addressing

• HSA-CED.A.3
• HSA-REI.C.6

Instructional Routines

• Graph It
• MLR6: Three Reads

Launch
Keep students in groups of 3–4 and provide access to graphing technology.

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 6

Support for English Language Learners

Reading: MLR6 Three Reads. Use this routine to support reading comprehension of this
problem. Ask students to keep their books or devices closed and display only the
image and the task statement without revealing the questions that follow. Use the
first read to orient students to the situation. After a shared reading, ask students:
“What is this situation about?” (a recreation center sells pool passes and gym
memberships). After the second read, students list any quantities that can be counted
or measured, without focusing on specific values (total amount of money paid by the
family and by the individual, the number of pool passes and number of gym
memberships purchased by each). During the third read, the questions are revealed.
Invite students to discuss possible strategies, referencing the relevant quantities
named after the second read.
Design Principle: Support sense-making

Support for Students with Disabilities

Representation: Access for Perception. Read the “What’s the Deal?” scenario aloud.
Students who both listen to and read the information will benefit from extra
processing time. Encourage students to annotate their document as they follow
along.
Supports accessibility for: Language; Conceptual processing

Student Task Statement
A recreation center is offering special prices on its pool passes and gym
memberships for the summer. On the first day of the offering, a family paid $96 for
4 pool passes and 2 gym memberships. Later that day, an individual bought a pool
pass for herself, a pool pass for a friend, and 1 gym membership. She paid $72.

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 7

1. Write a system of equations that represents the relationships between pool
passes, gym memberships, and the costs. Be sure to state what each variable
represents.

2. Find the price of a pool pass and the price of a gym membership by solving
the system algebraically. Explain or show your reasoning.

3. Use graphing technology to graph the equations in the system. Make 1-2
observations about your graphs.

Student Response

1. , where is the price of each pool pass and is the price of gym
membership for 1 person.

2. The system has no solutions. Sample reasoning: Solving by elimination or

substitution leads to a false equation (such as ).

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 8

Sample observations: The graphs are two
parallel lines that don’t intersect. The lines
have the same slope but different vertical
intercepts.

3.

Activity Synthesis
Invite previously identified students to share their response to the second question.
Record or display their reasoning for all to see. After each student shares, ask if anyone
else reasoned the same way.

Next, select other students to share their observations about the graphs. Ask students:

• “How can we tell from the graphs that there are no solutions?” (The lines are parallel.)

• “How can we tell for sure that the lines are parallel and never intersect?” (The slope of

both graphs is -2 but they have different intercepts.)

• “Why do parallel lines mean no solutions?” (A solution is a pair of values that satisfy

both equations and are on both graphs. There are no points that are on both lines
simultaneously.)

• “What does ‘no solutions’ mean in this situation, in terms of price of pool passes and

gym memberships?” (The prices for a pool pass and for a gym membership are
different for the two purchases.)

Here are some ways to think about the situation:

• The family purchased twice the number of of pool passes and gym memberships as

the individual did, but they did not pay twice as much, so the prices of passes and
memberships must have been different for the two purchases.

• The person who bought half as many passes and memberships did not pay half as

much, which meant that different prices applied to the two transactions.

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 9

• The special rates for a family of 4 did not apply to individuals, hence the different

prices.

17.3 Card Sort: Sorting Systems

15 minutes
In earlier activities, students gained some insights on the structure of equations in systems
that have infinitely many solutions and those that have no solutions. In this activity, they
apply those insights to sort systems of equations based on the number of solutions (one
solution, many solutions, or no solutions).

Students could solve each system algebraically or graphically and sort afterwards, but
given the number of systems to be solved, they will likely find this process to be time
consuming. A more productive way would be to look for and make use of the structure in
the equations in the systems (MP7), for example, by looking out for equivalent equations,
equations with the same slope but different vertical intercepts, variable expressions with
the same or opposite coefficients, and so on.

To effectively make use of the structure of the systems, students need to attend closely to
all parts of each equation—the signs, variables, coefficients and constants—and to
rearrange equations with care (MP6).

As students discuss their thinking in groups, make note of the different ways they use
structure to complete the task. Encourage students who are solving individual systems to
analyze the features of the equations and see if they could reason about the solutions or
gain information about the graphs that way.

In this activity, students are analyzing the structure of equations in the systems, so
technology is not an appropriate tool.

Here is an image of the cards for reference and planning:

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 10

Building On

• HSA-CED.A.4

Addressing

• HSA-REI.C.6

Instructional Routines

• Card Sort
• MLR8: Discussion Supports

Launch
Arrange students in groups of 2. Give one set of pre-cut slips or cards from the blackline
master to each group.

Give students 7–8 minutes to sort the cards into groups. Emphasize to students that
they should be prepared to explain how they place each system. Follow with a whole-class
discussion.

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 11

Support for English Language Learners

Conversing: MLR8 Discussion Supports. Use this routine to support small-group
discussion during the card sort. Encourage students to take turns selecting a card and
to explain to their partner whether the systems has no solutions, one solution, or
infinitely many solutions. Display the following sentence frames for all to see: “This
system has _____ solutions because….” Encourage students not only to challenge each
other if they disagree, but also to press for clear explanations that use mathematical
language.
Design Principle(s): Support sense-making; Maximize meta-awareness

Support for Students with Disabilities

Representation: Internalize Comprehension. Chunk this task into more manageable parts
to differentiate the degree of difficulty or complexity by beginning with fewer cards.
For example, give students cards a subset of the cards to start with and introduce the
remaining cards once students have completed their initial set of matches. Be sure to
include at least one card from each category in the initial set.
Supports accessibility for: Conceptual processing; Organization

Anticipated Misconceptions
Some students may not know how to begin sorting the cards. Suggest that they
try solving 2–3 systems. Ask them to notice if there's a point in the solving process when
they realize how many solutions the system has or what the graphs of the two equations
would look like. Encourage to look for similarities in the structure of the equations and see
how the structure might be related to the number of solutions.

Student Task Statement
Your teacher will give you a set of cards. Each card contains a system of equations.

Sort the systems into three groups based on the number of solutions each system
has. Be prepared to explain how you know where each system belongs.

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 12

Student Response

• No solutions: Cards 2 and 5
• One solution: Cards 1, 4, 6, and 9
• Infinitely many solutions: Cards 3, 7, and 8

Are You Ready for More?
1. In the cards, for each system with no solution, change a single constant term
so that there are infinitely many solutions to the system.

2. For each system with infinitely many solutions, change a single constant term
so that there are no solutions to the system.

3. Explain why in these situations it is impossible to change a single constant
term so that there is exactly one solution to the system.

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 13

Student Response
Sample responses:

1. Card 2: change 3 to 13, Card 5: change -12 to 12

2. Any change to a constant term would work.

3. Changing a constant term won’t change the slope of the lines, so the graphs will
either be distinct parallel lines or the same line.

Activity Synthesis

Invite groups to share their sorting results and record them. Ask the class if they agree or
disagree. If there are disagreements, ask students who disagree to share their reasoning.

Display all the systems—sorted into groups—for all to see and discuss the characteristics
of the equations in each group. Ask students questions such as:

• “How can we tell from looking at the equations in cards 2 and 5 that the systems have

no solution?” (Possible reasoning:

◦ Card 2: The equations are in slope-intercept form. The slope is 2 for both

graphs, but the -intercept is different, so the lines must be parallel.

◦ Card 5: The second equation can be rearranged into and
on one
multiplied by to give . Both equations now have

side, but that expression is equal to 4 in the first equation and equal to -4 in the
second. There are no pair of and that can make both equations true.)

• “What about the equations in cards 3, 7, and 8? What features might give us a clue

that the systems have many solutions?” (Possible reasoning:

◦ Card 3: The coefficients and constants in the second equation are 3 times those

in the first, so they are equivalent equations.

◦ Card 7: The first equation can be rearranged into the same form as the first:

. We can then see that the second equation, , is a

multiple of the first equation, so they have all the same solutions.

◦ Card 8: The first equation can be rearranged to . We can then see

that it is related to the second equation by a factor of 5, so they are equivalent
equations.)

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 14

We can reason that all the other systems have one solution by a process of
elimination—by noticing that they don’t have the features of systems with many solutions
or systems with no solutions.

17.4 One, Zero, Infinitely Many

Optional: 10 minutes
This optional activity gives students another opportunity to apply what they learned about
the features of systems of linear equations with one solution, zero solutions, and many
solutions. In earlier activities, students were given systems of equations and were asked to
determine the number of solutions. Here, they are given one equation and are to write a
second one such that the two equations form a system with one solution, zero solutions,
and many solutions.

To answer the first question (a system with one solution), students could write a second
equation with randomly chosen parameters. Answering the second and third questions,
however, relies on an understanding of what "zero solutions" and "infinitely many
solutions" mean and how these conditions are visible in the pair of equations and
graphically.

For example, students could reason that in a system with no solutions:

• The two equations have the same variable expressions on one side but different

numbers on the other side, and then write a second equation accordingly. For

instance, if is equal to 10, it cannot also be equal to 4, so and

would form a system with no solutions.

• The graphs of the equations have the same slope, but they cross the vertical axis at

different points. Rewriting in the form of gives

. A second equation with a coefficient of for but a different

constant would have a graph that is parallel to the graph of the first equation,
forming a system with no solutions.

Regardless of the form students use to write the second equation, they need to choose the
parameters strategically to achieve a system with the desired number of solutions. The
work here prompts students to look for and make use of structure (MP7).

Addressing

• HSA-REI.C.6

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 15

Launch

Consider keeping students in groups of 2. Encourage students to think quietly about the
first question and then discuss their response with their partner, and to do the same with
the remaining two questions.

Student Task Statement .
Here is an equation:

Create a second equation that would make a system of equations with:

1. One solution

2. No solutions

3. Infinitely many solutions

Student Response
Sample response:

1.

2.

3.

Activity Synthesis
Invite students to share their equations. If possible, use graphing technology to graph each
equation students share and to verify that the resulting system indeed has the specified
number of solutions. Display the graphs for all to see.

Focus the discussion on how students wrote an equation that would produce a system
with no solutions and a system with infinitely many solutions. Highlight strategies that
show an understanding of equivalent equations, of the meaning of solutions to equations
in two variables, and of the graphical features of these systems.

Lesson Synthesis

To help students summarize and organize the insights they gained in the lesson, consider
asking them to complete (collaboratively in small groups or as a class) a graphic organizer

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 16

with the following components for each type of system (no solutions, many solutions, and
one solution).

If time is limited, focus on describing some characteristics of the equations in a system
with each number of solutions (one, many, none) and sketching their graphs. Consider
using a graphic organizer such as this one.

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 17

17.5 No Graphs, No Problem

Cool Down: 5 minutes
Graphing technology should not be used in this cool-down.

If time is limited, ask students to choose one system and explain how they could tell that it
has no solutions or infinitely many solutions.

Addressing

• HSA-REI.C.6

Student Task Statement
Mai is given these two systems of linear equations to solve:

System 1: System 2:

She analyzed them for a moment, and then—without graphing the equations—said,
"I got it! One of the systems has no solution and the other has infinitely many
solutions!" Mai is right!

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 18

Which system has no solution and which one has many solutions? Explain or show
how you know (without graphing the equations).

Student Response

The first system has no solutions and the second system has infinitely many solutions.
Sample explanations:

• Multiplying the first equation by 4 gives . Subtracting the second
, which is a false equation and tells us
equation from the first equation gives

that the system has no solutions.

• (in the second equation) is 4 times (in the first equation), but 64 is not

4 times 13. This means there is no pair of - and -values that could make both

equations simultaneously true.

• If we isolate in the first equation, we have . If we do the same with the

second equation, we have or . The graphs of the two

equations have the same slope (-5) but they intersect the -axis at different points (13

and 16).

The second system has infinitely many solutions. Sample explanations:

• Multiplying the first equation by 4 gives . Rearranging the second

equation so that the variables are on the left side also gives . The two

equations are identical, so they have the same solutions.

• The second equation can be rearranged to . We can see that it is 4

times the first equation, which means the two equations, and

, are equivalent and have all the same solutions.

• If we isolate in each equation, we'd have the exact same equation, , so

they have all the same solutions.

Student Lesson Summary

We have seen many examples of a system where one pair of values satisfies both
equations. Not all systems, however, have one solution. Some systems have many
solutions, and others have no solutions.

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 19

Let's look at three systems of equations and their graphs.

System 1:

The graphs of the equations in System 1 intersect at
one point. The coordinates of the point are the one pair
of values that are simultaneously true for both
equations. When we solve the equations, we get exactly
one solution.

System 2:

The graphs of the equations in System 2 appear to be
the same line. This suggests that every point on the line
is a solution to both equations, or that the system has
infinitely many solutions.

System 3:

The graphs of the equations in System 3 appear to be
parallel. If the lines never intersect, then there is no
common point that is a solution to both equations and
the system has no solutions.

How can we tell, without graphing, that System 2 indeed has many solutions?

• Notice that and are equivalent equations.

Multiplying the first equation by 2 gives the second equation. Multiplying the

second equation by gives the first equation. This means that any solution to

the first equation is a solution to the second.

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 20

• Rearranging into slope-intercept form gives , or

. Rearranging gives , which is

also . Both lines have the same slope and the same -value for the
vertical intercept!

How can we tell, without graphing, that System 3 has no solutions?

• Notice that in one equation equals 8, but in the other equation

it equals -4. Because it is impossible for the same expression to equal 8 and -4,

there must not be a pair of - and -values that are simultaneously true for

both equations. This tells us that the system has no solutions.

• Rearranging each equation into slope-intercept form gives and

. The two graphs have the same slope but the -values of their

vertical intercepts are different. This tells us that the lines are parallel and will
never cross.

Lesson 17 Practice Problems

1. Problem 1

Statement

Here is a system of equations:

a. Solve the system by graphing the equations (by hand or using
technology).

b. Explain how you could tell, without graphing, that there is only one
solution to the system.

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 21

Solution .

a. The solution is

b. Sample explanations:
i. If we solve the system algebraically (by substitution or by elimination), we
would find one value of and one value of that make both equations
true.

ii. If we rewrite both equations into slope-intercept form, we would have
and . The graphs of the two equations have

different slopes (3 and ), and lines with different slopes can only
intersect at one point.

2. Problem 2

Statement

Consider this system of linear equations:

a. Without graphing, determine how many solutions you would expect this
system of equations to have. Explain your reasoning.

b. Try solving the system of equations algebraically and describe the result
that you get. Does it match your prediction?

Solution

a. No solutions. Sample reasoning: The two equations have the same coefficient

for , so their graphs have the same slope or the lines are parallel. This means

the lines never cross and there is no pair that is true for both equations.

b. Sample response: Solving by substitution, I end up with the equation (or

some multiple of this equation), which is a false statement. This means that the

system has no solutions, which matches my prediction.

Algebra1 Unit 2 Lesson 17 CC BY 2019 by Illustrative Mathematics 22