Lesson 15

In this warm-up, students reason about whether and when the sums of equations are true. The work here prepares students for the next activity, where they begin to think about why the values that simultaneously satisfy two equations in a system also satisfy the the equation that is a sum of those two equations.

Ask students to share their responses to the first three questions. Discuss questions such as:

• "Why do you think the resulting equations remain true even after we add numbers or expressions that look different to each side?" (The numbers being added to the two sides are always equal amounts, even though they are written in a different form.)
• "Suppose we subtract 12 from the left side of \(50+1=51\) and subtract \(5+7\) from the right side. Would the resulting equation still be true?" (Yes. \(50+1-12\) is 39 and \(51 - (5+7)\) is also 39. The same amount, 12, is subtracted from both sides.)
• "Here is another equation: \(50 + 1 = 60\). Suppose we subtract 12 from the left side of that equation and \(5+7\) from the right side. Would the resulting equation still be true? Why or why not?" (No. The given equation is a false statement. Subtracting the same amount from both sides will keep it false.)

Invite students to share their equations for the last two questions. Display the equations for all to see. If no one shares an equation that uses a variable, give an example or two (as shown in the Student Responses).

Make sure students understand that adding (or subtracting) the same amount to each side of a true equation keeps the two sides equal, resulting in an equation that is also true. Adding (or subtracting) different amounts to each side of a true equation, however, makes the two sides unequal and thus produces a false equation.

This activity serves two goals. The first goal is to further build students' intuition about the new variable equation that comes from adding two variable equations in a system, and about the solution to that new equation. This is done by grounding the addition and the sum in a familiar context.

The second goal is to support students in reasoning about why the new equation shares a solution with the original system. The context gives students a concrete mental reference, which can be helpful for interpreting the intersection of the graphs of all three equations and for thinking about the solution that all three equations share.

In this activity, students also encounter a system that they can solve by graphing and by substitution, but cannot be easily solved simply by adding or subtracting the equations. This observation may pique students' curiosity and make them wonder if it is possible to solve all systems by elimination.

As students work, identify students who solve the system by graphing and those who solve by substitution. Ask them to share their work later.

Arrange students in groups of 2 and provide access to graphing technology.

Read the opening paragraphs with the class. Invite students to explain what each equation represents in this situation. Then, give students 2 minutes of quiet time to think about the first set of questions, and a few minutes to discuss their thinking with their partner. Follow with a whole-class discussion.

Ask students to share their interpretation of the solutions to each equation and how many solutions are possible for each equation. Make sure students recognize that the solutions are pairs of \(c\)- and \(m\)-values that make each equation true, and that each equation can have many solutions because there are many possible prices of calculators and measuring tapes that can make each equation true.

Before students proceed to the rest of the activity, ask: “If we are solving the system, what are we really looking for?” Be sure students see that to solve the system is to find a pair of unit prices that make the equations for both purchases true.

Select previously identified students to share their solution and strategy for solving the system. Display their work (especially student-generated graphs) for all to see, or consider diplaying this graph:

Emphasize that 21.50 and 6.50 are the unit prices of the two supplies that make both equations true.

Next, discuss students' responses to the last set of questions. Ask questions such as:

• "The first equation was \(20c + 10m = 495\). We added \(8c + m\) to the left side and 178.50 to the right side to get \(28c + 11m = 673.50\). Are the two sides of the resulting equation still equal? Why or why not?" (Yes. We’re adding equal amounts to each side, so the two sides are still equal. The cost of buying 8 calculators and 1 measuring tape is $178.50.)
• "Predict what the graphs of all three equations would look like." (They would intersect at a common point.) Consider graphing all three equations for all to see.
• "In this situation, why does it make sense for the graphs intersect at the same point? What does that point mean?" (The prices of a calculator and of a measuring tape in the first two orders are also the prices in the combined order. The unit prices of the items didn’t change, so it makes sense that they work for all three equations.)
• "In general, why does it make sense for a third equation that is the sum of the two equations in a system to share a common solution with the other two?" (Adding the two equations means adding equal values to each side of one equation, which keeps the two sides of the sum equal. If there is a pair of \(c\) and \(m\) values that make the first two equations true, that pair also makes the third equation true. )
• "Were you able to solve the system by adding or subtracting the equations, without graphing? Why or why not?" (No, because no variables get eliminated when we add or subtract.)

In this activity, students practice using algebra to solve systems of linear equations in two variables and checking their solutions. Students do not have to use elimination, but the equations in the first three systems conveniently have opposites for the coefficients of one variable, so one variable can be easily eliminated.

In the last system, none of the coefficients of the variables are opposites. Some students may choose to solve the system by substitution, but the process would be pretty cumbersome. The complication students encounter here motivates the need for another move, which students will explore in the next lesson.

Students who opt to use technology to check their solutions practice choosing tools strategically (MP5).

Keep students in groups of 2 and provide continued access to graphing technology, in case needed for checking solutions.

If time is limited, ask one partner in each group to solve the first two of the systems and the other partner to solve the last two, and then check each other's solutions.

Invite students to share their solutions and strategies for the first three systems and how they check their solutions. Then, focus the discussion on the last system. Solicit the strategies students used for approaching that system. If someone solved it by substitution, display the work for all to see. If no one did, ask if it is possible to do. (If time permits, consider asking students to attempt to do so, or demonstrating that strategy to illustrate that it is not exactly efficient.)

Discuss questions such as:

• “What happens if we add or subtract the equations?” (No variables are eliminated, so it doesn’t help with solving.)
• “Why were you able to solve the first three systems by elimination but not the last one?” “What features were in the equations in those systems but not in the last one?” (In the first three systems, at least one variable in each pair of equations have the same or opposite coefficients, so when the terms were added or subtracted, the result is 0.)

We need new moves! Tell students that they will explore another way to solve a system by elimination in an upcoming lesson.