options: Represent a system of linear equations as a single matrix equation in a vector variable.

Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. For example, find the points of intersection between the line y = –3x and the circle x^(2 )+ y^(2) = 3.

Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.

Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions.

Find the inverse of a matrix if it exists and use it to solve systems of linear equations (using technology for matrices of dimension 3 × 3 or greater).
problem_activity: 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:
[IMG]
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
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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
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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.


correct_option_index: 2