# Lesson 17

Systems of Linear Equations and Their Solutions

## 17.1: A Curious System (10 minutes)

### Warm-up

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 \(y = 3 - x\). Substituting \(3-x\) for \(y\) in the second equation gives \(4x = 12 - 4(3-x)\) or \(4x = 12-12+4x\) or \(4x = 4x\). This equation is true no matter what \(x\) is.
- Solving by elimination: If we multiply the first equation by 4, rearrange the second equation to \(4x+4y=12\), and then subtract the second equation from the first, the result is \(4y=4y\). Subtracting \(4y\) from each side gives \(0=0\), which is true regardless of what \(x\) or \(y\) is.
- Reasoning about equivalent equations: If we rearrange the second equation so that the variables are on the same side \(4x+4y=12\), we can see that this equation is a multiple of the first and are equivalent. This means they have the exact same solution set, which contains infinite possible pairs of \(x\) and \(y\).

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).

### 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 Facing

Andre is trying to solve this system of equations: \(\begin {cases} x + y = 3\\ 4x = 12 - 4y \end{cases}\)

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."

- Choose any two numbers that add up to 3. Let the first one be the \(x\)-value and the second one be the \(y\)-value.
- 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.
- 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

For access, consult one of our IM Certified Partners.

### 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.

Make sure students see that the equations are equivalent, so all pairs of \(x\) and \(y\) 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)

### Activity

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.

### Launch

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

*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*

*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 Facing

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.

- 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.
- Find the price of a pool pass and the price of a gym membership by solving the system algebraically. Explain or show your reasoning.
- Use graphing technology to graph the equations in the system. Make 1-2 observations about your graphs.

### Student Response

For access, consult one of our IM Certified Partners.

### 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.
- 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)

### Activity

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:

### 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.

*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*

*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*

### Student Facing

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.

### Student Response

For access, consult one of our IM Certified Partners.

### Student Facing

#### Are you ready for more?

- In the cards, for each system with no solution, change a single constant term so that there are infinitely many solutions to the system.
- For each system with infinitely many solutions, change a single constant term so that there are no solutions to the system.
- Explain why in these situations it is impossible to change a single constant term so that there is exactly one solution to the system.

### Student Response

For access, consult one of our IM Certified Partners.

### 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.

### 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 \(y\)-intercept is different, so the lines must be parallel.
- Card 5: The second equation can be rearranged into \(5x - 20y = \text-60\) and multiplied by \(\frac15\) to give \(x-4y=\text-12\). Both equations now have \(x-4y\) on one side, but that expression is equal to 4 in the first equation and equal to -4 in the second. There are no pair of \(x\) and \(y\) 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: \(x -4y = \text-4\). We can then see that the second equation, \(4x-16y=\text-16\), is a multiple of the first equation, so they have all the same solutions.
- Card 8: The first equation can be rearranged to \(y=\text-25 x - \frac15\). We can then see that it is related to the second equation by a factor of 5, so they are equivalent equations.)

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 (10 minutes)

### Optional activity

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 \(5x-2y\) is equal to 10, it cannot also be equal to 4, so \(5x-2y=10\) and \(5x-2y=4\) 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 \(5x-2y =10\) in the form of \(y=mx+b\) gives \(y=\text-\frac52x + 10\). A second equation with a coefficient of \(\text-\frac52\) for \(x\) 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).

### 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 Facing

Here is an equation: \(5x-2y=10\).

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

- One solution
- No solutions
- Infinitely many solutions

### Student Response

For access, consult one of our IM Certified Partners.

### 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

### 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 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.

## 17.5: Cool-down - No Graphs, No Problem (5 minutes)

### Cool-Down

For access, consult one of our IM Certified Partners.

## Student Lesson Summary

### Student Facing

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.

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

System 1: \(\displaystyle \begin{cases} 3x+4y=8 \\ 3x-4y=8 \\ \end{cases} \)

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: \(\displaystyle \begin{cases} 3x+4y= 8\\ 6x+8y=16 \\ \end{cases} \)

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:\(\displaystyle \begin{cases} 3x+4y=8 \\ 3x+4y=\text-4 \\ \end{cases} \)

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 \(3x+4y =8\) and \(6x+8y=16\) are equivalent equations. Multiplying the first equation by 2 gives the second equation. Multiplying the second equation by \(\frac12\) gives the first equation. This means that any solution to the first equation is a solution to the second.
- Rearranging \(3x+4y=8\) into slope-intercept form gives \(y = \dfrac{8 - 3x}{4}\), or \(y=2 - \frac34x\). Rearranging \(6x+8y=16\) gives \(y=\dfrac {16-6x}{8}\), which is also \(y=2 - \frac34x\). Both lines have the same slope and the same \(y\)-value for the vertical intercept!

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

- Notice that in one equation \(3x+4y\) 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 \(x\)- and \(y\)-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 \(y=2 - \frac34x\) and \(y = \text-1 -\frac34x\). The two graphs have the same slope but the \(y\)-values of their vertical intercepts are different. This tells us that the lines are parallel and will never cross.