Lesson 15

Solving Systems by Elimination (Part 2)

15.1: Is It Still True? (5 minutes)


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.

Student Facing

Here is an equation: \(50 + 1 = 51\).

  1. Perform each of the following operations and answer these questions: What does each resulting equation look like? Is it still a true equation?

    1. Add 12 to each side of the equation.
    2. Add \(10 + 2\) to the left side of the equation and 12 to the right side.
    3. Add the equation \(4 + 3 = 7\) to the equation \(50 + 1 = 51\).
  2. Write a new equation that, when added to \(50 +1 = 51\), gives a sum that is also a true equation.
  3. Write a new equation that, when added to \(50 +1 = 51\), gives a sum that is a false equation.

Student Response

For access, consult one of our IM Certified Partners.

Activity Synthesis

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.

15.2: Classroom Supplies (20 minutes)


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.

Reading, Listening, Conversing: MLR6 Three Reads. Ask students to keep their books or devices closed and display only the image with the opening task statement, without revealing the questions that follow. Use the first read to orient students to the situation by asking students to describe what the situation is about without using numbers (a teacher buys calculators and measuring tapes for her class). Use the second read to identify quantities and relationships; ask students: “What can be counted or measured?” Listen for, and amplify, the important quantities that vary in relation to each other in this situation: number of calculators purchased (for each order), number of measuring tapes purchased (for each order), and total cost of each order. After the third read, reveal the questions that follow and invite students to brainstorm possible strategies to find the solution to the system. This routine helps students interpret the language within a given situation needed to create an equation.
Design Principle(s): Support sense-making
Representation: Internalize Comprehension. Provide appropriate reading accommodations and supports to ensure student access to written directions, word problems and other text-based content. While reading the situation aloud, explicitly link the word problem to the corresponding algebraic expression. Point to both the sentence and its corresponding equation while reading. Pause in between sentences and reread or repeat gestures if necessary for increased comprehension. 
Supports accessibility for: Language; Conceptual processing

Student Facing

A teacher purchased 20 calculators and 10 measuring tapes for her class and paid \$495. Later, she realized that she didn’t order enough supplies. She placed another order of 8 of the same calculators and 1 more of the same measuring tape and paid \$178.50. 

This system represents the constraints in this situation:

\(\begin{cases} \begin {align}20c + 10m &= 495\\ 8c + \hspace{4.5mm} m &= 178.50 \end{align}\end{cases}\)

A calculator and a tape measure
  1. Discuss with a partner:
    1. In this situation, what do the solutions to the first equation mean? 
    2. What do the solutions to the second equation mean?
    3. For each equation, how many possible solutions are there? Explain how you know.
    4. In this situation, what does the solution to the system mean?
  2. Find the solution to the system. Explain or show your reasoning.
  3. To be reimbursed for the cost of the supplies, the teacher recorded: “Items purchased: 28 calculators and 11 measuring tapes. Amount: \$673.50.” 

    1. Write an equation to represent the relationship between the numbers of calculators and measuring tapes, the prices of those supplies, and the total amount spent.
    2. How is this equation related to the first two equations?
    3. In this situation, what do the solutions of this equation mean?
    4. How many possible solutions does this equation have? How many solutions make sense in this situation? Explain your reasoning.  

Student Response

For access, consult one of our IM Certified Partners.

Activity Synthesis

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:

Graph of a linear system. unit price of a measuring tape (dollars), unit price of a calculator (dollars).

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.)

15.3: A Bunch of Systems (10 minutes)


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.

Student Facing

Solve each system of equations without graphing and show your reasoning. Then, check your solutions.

A\(\begin {cases} \begin {align}2x + 3y &= \hspace {2mm}7\\ \text-2x +4y &= 14 \end {align} \end {cases} \)

B\(\begin {cases} \begin {align}2x + 3y &= \hspace {2mm}7\\ 3x -3y &= 3 \end {align} \end {cases} \)

C\(\begin {cases} \begin {align}2x + 3y &= 5\\ 2x +4y &= 9 \end {align} \end {cases}\)

D\(\begin {cases} \begin {align}2x + 3y &=16\\ 6x -5y &= 20 \end {align} \end {cases}\)


Student Response

For access, consult one of our IM Certified Partners.

Student Facing

Are you ready for more?

This system has three equations: \(\begin{cases}3 x + 2y - z = 7 \\ \text{-} 3x + y +2z =\text- 14 \\ 3x+y-z=10 \end{cases}\)

  1. Add the first two equations to get a new equation.
  2. Add the second two equations to get a new equation.
  3. Solve the system of your two new equations.
  4. What is the solution to the original system of equations?

Student Response

For access, consult one of our IM Certified Partners.

Activity Synthesis

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.

Reading, Writing, Speaking: MLR3 Clarify, Critique, Correct. Before students share their solutions and strategies, present an incorrect method of solving a linear system. For example, “After the first step when solving the system \(2x + y = 9\) and \(x - y = 3\) by elimination, the resulting equation is \(x = 12\).” Ask students to identify the error, critique the reasoning, and write a correct explanation. As students discuss with a partner, listen for students who clarify the meaning of the solution to a system. Invite students to share their critiques and corrected explanations with the class. Listen for and amplify the language students use to describe what happens when adding two equations together and explain what the solution should be. This helps students evaluate and improve on the written mathematical arguments of others, as they clarify how to find the solution of a linear system by using the elimination method.
Design Principle(s): Optimize output (for explanation); Maximize meta-awareness

Lesson Synthesis

Lesson Synthesis

To highlight the idea that adding equal amounts to the two sides of a true equation keeps the equation true, display this system for all to see:

\(\begin{cases} \begin{align} \text-2x+8y&=20\\2x+ \hspace{2.2mm}y &=\hspace{2.2mm}7 \end{align} \end{cases}\)

Ask students if there is a pair of \(x\)- and \(y\)-values that make both equations true. Once students suggest \((2, 3)\) as the solution, use graphing technology to graph the equations and verify the solution. Display the graph for all to see.

Survey the class to see who have solved the system by adding the left side of each equation (\(2x + y\) and \(\text-2x+8y\) ) and then adding the right side (7 and 20). (Most or all students are likely to have done this.) Then, discuss with students:

  • "Why is it OK to add an expression, \(2x+y\), to one side of the first equation, but add a number, 7, to the other side?" (Because \(2x +y\) and 7 are equal)
  • "What is the new equation? (\(9y=27\)) Does the pair \((2,3)\) still make this equation true?" (Yes. Graph this equation to illustrate that it intersects the original graphs at the same point.)
  • "Suppose we add 7 to the left side of the first equation and \(2x+y\) to the right side. What is the new equation?" (\(\text-2x + 8y + 7 = 20 + 2x + y\)) "Would the same pair of \(x\) and \(y\) still work?" (Yes. Display the graph of the new equation so students could see that the new line intersects at the same point. Students could also verify by substitution.)
  • "Suppose that, instead of adding, we subtract 7 from the left side and subtract \(2x + y\) from the right side. What is the new equation?" (\(\text-2x+8-7 = 20 - 2x - y\)). "Would the same pair of \(x\) and \(y\) still work?" (Yes. Graph the new equation to show that this line, too, intersects at the same point. Or students could verify by substitution.)
  • "Why does the new equation still give the same solution even though we didn't add the numbers or expressions from the same side of each equation?" (If the two sides of an equation are equal, it doesn't matter which side of one equation is added to which side of another equation. The new equation will stay balanced.)

15.4: Cool-down - Putting New Equations to Work (5 minutes)


For access, consult one of our IM Certified Partners.

Student Lesson Summary

Student Facing

When solving a system with two equations, why is it acceptable to add the two equations, or to subtract one equation from the other?

Remember that an equation is a statement that says two things are equal. For example, the equation \(a = b\) says a number \(a\) has the same value as another number \(b\). The equation \(10 + 2 = 12\) says that \(10+2\) has the same value as 12.

If \(a = b\) and \(10 + 2 = 12\) are true statements, then adding \(10+2\) to \(a\) and adding \(12\) to \(b\) means adding the same amount to each side of \(a=b\). The result, \(a + 10 + 2 = b + 12\), is also a true statement.

As long as we add an equal amount to each side of a true equation, the two sides of the resulting equation will remain equal.

We can reason the same way about adding variable equations in a system like this:

\(\begin {cases} \begin {align} e + f = 17\\ \text-2e + f =\text-1 \end{align}\end {cases}\)

In each equation, if \((e,f)\) is a solution, the expression on the left of the equal sign and the number on the right are equal. Because \(\text-2e+f\) is equal to -1:

  • Adding \(\text-2e + f\) to \(e+f\) and adding -1 to 17 means adding an equal amount to each side of \(e+f=17\). The two sides of the new equation, \(\text-e + 2f = 16\), stay equal.

    The \(e\)- and \(f\)-values that make the original equations true also make this equation true.

\(\begin {align} e +\hspace{2mm} f &= 17\\ \text-2e +\hspace{2mm} f &=\hspace{0.8mm}\text-1 \quad+\\ \overline {\hspace{2mm}\text-e + 2f }& \overline{ \hspace{1mm}=16}  \end {align}\)

  • Subtracting \(\text-2e + f\) from \(e+f\) and subtracting -1 from 17 means subtracting an equal amount from each side of \(e+f=17\). The two sides of the new equation, \(3e = 18\), stay equal.

    The \(f\)-variable is eliminated, but the \(e\)-value that makes both the original equations true also makes this equation true.

\(\begin {align} e +\hspace{2mm} f &= 17\\ \text-2e +\hspace{2mm} f &=\hspace{0.8mm}\text-1 \quad-\\ \overline{\hspace{0.8mm}3e \hspace{9.5mm}} &\overline{\hspace{1mm}=18} \end {align}\)

From \(3e = 18\), we know that \(e=6\). Because 6 is also the \(e\)-value that makes the original equations true, we can substitute it into one of the equations and find the \(f\)-value.

The solution to the system is \(e=6, f=11\), or the point \((6,11)\) on the graphs representing the system. If we substitute 6 and 11 for \(e\) and \(f\) in any of the equations, we will find true equations. (Try it!)