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 in which they reach the conclusion that the systems have no solutions. Identify students with varying strategies, and ask them to share later.

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 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.

In earlier activities, students gained some insights into 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 afterward, 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.

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 students to look for similarities in the structure of the equations and to see how the structure might be related to the number of solutions. 

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 multiplied by to give . Both equations now have on one side, but that expression is equal to -12 in the first equation and equal to 12 in the second. There is 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.)

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.