Lesson 14
Solving Systems by Elimination (Part 1)
14.1: Notice and Wonder: Hanger Diagrams (5 minutes)
Warmup
The purpose of this warmup is to give students an intuitive and concrete way to think about combining two equations that are each true.
Students are presented with diagrams of three balanced hangers, which suggest that the weights on the two sides of each hanger are equal. Each side of the last hanger shows the combined objects from the corresponding side of the first two hangers. Students can reason that if 2 circles weigh the same as 1 square, and 1 circle and 1 triangle weigh the same as 1 pentagon, then the combined weight of 3 circles and 1 triangle should also be equal to the combined weight of 1 square and 1 pentagon.
Launch
Display the hanger diagrams for all to see. Ask students to think of at least one thing they notice and at least one thing they wonder. Give students 1 minute of quiet think time, and then 1 minute to discuss the things they notice and wonder with their partner, followed by a wholeclass discussion.
Student Facing
What do you notice? What do you wonder?
Student Response
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Activity Synthesis
Ask students to share the things they noticed and wondered. Record and display their responses for all to see. If possible, record the relevant reasoning on or near the image. After all responses have been recorded without commentary or editing, ask students, “Is there anything on this list that you are wondering about now?” Encourage students to respectfully disagree, ask for clarification, or point out contradicting information.
The idea to emphasize is that the weights on each side of the third hanger come from combining the weights on the corresponding sides of the first two hangers. If no one points this out, raise it as a point for discussion. Ask students:
 “What do you notice about the left side of the last hanger? What about the right side?"
 "If we only saw the first two hangers but knew that the third hanger has the combined weights from the corresponding side of the first two hangers, could we predict whether the weights on the third hanger would balance? "Why or why not?” (Yes. We can think of it as adding the weights from the second hanger to the first one, or vice versa. If the same weight is added to each side of a balanced hanger, the hanger would still be balanced.)
14.2: Adding Equations (15 minutes)
Activity
In the warmup of this lesson, students saw a visual representation of two equations being added to form a third equation. Because the first two equations are balanced, the third is also balanced. In this activity, students continue to develop the idea of adding two equations to form a third equation and use it to help them solve systems of linear equations.
Along the way, students examine the work of others and practice explaining their reasoning and critiquing that of others (MP3). They also see that sometimes adding equations is a productive way to solve systems, but other times it isn't.
Launch
Arrange students in groups of 2. Give students 2 minutes of quiet time to think about the first set of questions and then time to share their thinking with their partner. Follow with a wholeclass discussion before students proceed to the second set of questions.
Invite students to share their analysis of Diego's work—what Diego has done to solve the system and why he might have done it that way. Discuss questions such as:
 "In this case, what happens when the equations are added? Why might it be helpful to do so?" (The expressions with \(x\) add up to 0, so it's removed from the equation, making it possible to solve for \(y\).)
 "How does finding the value of \(y\) help with solving the system?" (Once we know the value of one variable, we can use it to find the value of the other, by substituting it back into one of the equations and solving that equation.)
 "How can we be sure that \(x=1\) and \(y=2\) simultaneously make both equations true and is a solution to the system?" (We can substitute those values into the equations and see if the equations are true. We can also graph the system and see if it intersects at \((1,2)\).)
Next, ask students to complete the remainder of the activity.
Supports accessibility for: Conceptual processing; Memory
Student Facing
Diego is solving this system of equations:
\(\begin{cases} \begin {align}4x + 3y &= 10\\ \text4x + 5y &= 6 \end{align} \end{cases}\)
Here is his work:
\(\begin {align}4x + 3y &= 10\\ \text4x + 5y &= \hspace{2mm}6 \quad+\\ \overline {\quad 0 + 8y} &\overline{ \hspace{1mm}= 16 \qquad}\\ y &= 2 \end{align} \)
\(\begin {align}4x + 3(2) &= 10\\ 4x + 6 &= 10\\ 4x &= 4\\ x &= 1 \end{align} \)

Make sense of Diego’s work and discuss with a partner:
 What did Diego do to solve the system?
 Is the pair of \(x\) and \(y\) values that Diego found actually a solution to the system? How do you know?

Does Diego’s method work for solving these systems? Be prepared to explain or show your reasoning.
a. \(\begin {cases} \begin {align}2x + y &= 4\\ x  y &= 11 \end {align} \end {cases} \)
b. \(\begin {cases} \begin{align} 8x + 11y &= 37\\ 8x + \hspace{4.5mm} y &= \hspace{2mm} 7 \end{align} \end{cases}\)
Student Response
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Activity Synthesis
Invite students to share their responses to the last set of questions and discuss whether Diego's method works for solving the two systems. Ask students:
 “Why does adding the equations work for solving the system with \(2x + y =4\) and \(xy=11\), but doesn't work for \(8x + 11y =37\) and \(8x + y =7\)?” (In the resulting equation, there are still two variables whose values we don't know.)
 “What if we subtract the equations? Would that help us solve the last system?” (Yes, subtracting the second equation from the first gives \(10y=30\) or \(y=3\), which we can then use to find the \(x\)value.)
 “How is adding the two equations here like and unlike combining the shapes in the two hanger diagrams earlier? How is it different?” (It is alike in that the result is another equation with the combined parts on the left side staying on the left and combined parts on the right stay on the right. It is unlike the hanger diagrams in that these equations use numbers and variables, and one of the parts on the left add up to 0 and disappears from the third equations.)
Make sure students see that if we choose to add or subtract strategically, in each of the new equations, one variable is eliminated, making it possible to solve for the other variable. When the value of that variable is substituted to either of the original equations, we can solve for the variable that was eliminated. Tell students that this method of solving a system is called solving by elimination.
Point out that there is nothing wrong about adding the equations in the last system. It simply doesn't get us anywhere closer to the solution and is therefore unproductive.
Design Principle(s): Support sensemaking; Optimize output (for explanation)
14.3: Adding and Subtracting Equations to Solve Systems (15 minutes)
Activity
Earlier, students saw that adding or subtracting the equations in a system creates a third equation that can help us solve the system. In this activity, students reconnect the idea of the solution to a system to the intersection of the graphs of the equations. They graph each original pair of equations and the equation that results from adding or subtracting them. They then observe that the graph of the third equation intersects the other two graphs at the exact same point—at the intersection of the first two.
At this point, students simply get a graphical confirmation that adding or subtracting equations can help them find the solution to a system. They are not yet expected to be able to articulate why this is the case. That understanding will be developed over a few upcoming lessons.
Launch
Remind students that earlier they added or subtracted pairs of equations to form new equations. Explain that they will now graph each pair of equations in the systems given earlier, as well as the third equation that came from adding or subtracting those equations, and then make some observations about them.
Arrange students into groups of 3 and provide access to graphing technology. Assign one system for each group member to graph. Ask students to discuss their observations after graphing.
If possible, make available graphing technology that allows users to enter linear equations in standard form, such as Desmos (available under Math Tools). Otherwise, give students time to rearrange the equations into a form that can be used with the technology and to check their equivalent equations. If time is limited, provide these equivalent equations:
System A
\(\begin {cases}\begin {align}y &= \text\frac43x+\frac{10}{3}\\ y&=\hspace{2mm}\frac45x + \frac65 \end{align} \end{cases}\)
System B
\(\begin {cases}\begin {align}y &= \text2x+4\\ y&=\hspace{3mm}x11 \end{align} \end{cases}\)
System C
\(\begin {cases}\begin {align}y &= \text\frac{8}{11}x+\frac{37}{11}\\ y&=\hspace{3.7mm}\text8x + 7 \end{align} \end{cases}\)
Supports accessibility for: Memory; Organization
Student Facing
Here are three systems of equations you saw earlier.
System A
\(\begin {cases}\begin {align}4x + 3y &= 10\\ \text4x + 5y &= \hspace{2mm}6 \end{align} \end{cases}\)
System B
\(\begin {cases} \begin {align}2x + y &= 4\\ x  y &= 11 \end {align} \end {cases} \)
System C
\(\begin {cases} \begin{align} 8x + 11y &= 37\\ 8x + \hspace{4mm} y &= \hspace{2mm} 7 \end{align} \end{cases}\)
For each system:
 Use graphing technology to graph the original two equations in the system. Then, identify the coordinates of the solution.
 Find the sum or difference of the two original equations that would enable the system to be solved.
 Graph the third equation on the same coordinate plane. Make an observation about the graph.
Student Response
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Student Facing
Are you ready for more?
Mai wonders what would happen if we multiply equations. That is, we multiply the expressions on the left side of the two equations and set them equal to the expressions on the right side of the two equations.
 In system B write out an equation that you would get if you multiply the two equations in this manner.
 Does your original solution still work in this new equation?
 Use graphing technology to graph this new equation on the same coordinate plane. Why is this approach not particularly helpful?
Student Response
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Anticipated Misconceptions
When solving system B, some students may not notice that the \(y\)variable in one equation has a positive coefficient and the other has a negative coefficient, and consequently decide to subtract the second equation from the first, rather than to add the two equations. They may struggle to figure out why the solution pair they find doesn't match what is on the graph. Suggest that they express the second equation in terms of addition, \(x+ (\texty) = 11\), and try eliminating one variable again.
Activity Synthesis
Display the graphs that students generated for all to see and ask students to share their observations. Highlight that the graph of the new equation intersects the graphs of the equations in the original system at the same point.
Time permitting, ask students to subtract the equations they previously added (or add the equations they previously subtracted) and then to graph the resulting equation on the same coordinate plane. Ask them to comment on the graphs. Students are likely to see that the graphs of the new equations are no longer horizontal or vertical lines, but they still intersect at the same point as the original graphs.
Invite students to make some conjectures as to why the graph of the new equation intersects the other two graphs at the same point. Without confirming or correcting their conjectures, tell students that they will investigate this question in the coming activities.
Lesson Synthesis
Lesson Synthesis
Now that students have an additional strategy for solving systems in their toolkit, invite students to reflect on three systems seen in the synthesis of a previous lesson, in which they made a case for solving one by substitution.
System 1
\(\begin {cases} 3m + n = 71\\2mn =30 \end {cases}\)
System 2
\(\begin {cases} 4x + y = 1\\y = \text2x+9 \end {cases}\)
System 3
\(\displaystyle \begin{cases} 5x+4y=15 \\ 5x+11y=22 \end{cases}\)
Ask students to discuss the following questions with a partner and be prepared to report their partner's responses:
 "Look at a system that you would have chosen to solve by substitution. Would you still choose to solve it by substitution now? Why or why not?"
 "Look at a system that you would not have chosen to solve by substitution. Would it help to solve by elimination? Why or why not?"
With a little bit of rearranging (of the equations), all of these systems could be solved by substitution or elimination. Students should at least recognize that systems 1 and 3 can be efficiently solved by elimination, while system 2 can be efficiently solved by substitution.
14.4: Cooldown  What to Do with This System? (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
Another way to solve systems of equations algebraically is by elimination. Just like in substitution, the idea is to eliminate one variable so that we can solve for the other. This is done by adding or subtracting equations in the system. Let’s look at an example.
\(\begin {cases} \begin {align}5x+7y=64\\ 0.5x  7y = \text9 \end{align} \end {cases}\)
Notice that one equation has \(7y\) and the other has \(\text7y\).
If we add the second equation to the first, the \(7y\) and \(\text7y\) add up to 0, which eliminates the \(y\)variable, allowing us to solve for \(x\).
\(\begin {align} 5x+7y&=64\\ 0.5 x  7y &= \text9 \quad+\\ \overline {5.5 x + 0} &\overline {\hspace{1mm}= 55}\\ 5.5x &= 55 \\x &=10 \end {align}\)
Now that we know \(x = 10\), we can substitute 10 for \(x\) in either of the equations and find \(y\):
\(\begin {align} 5x+7y&=64\\ 5(10)+7y &= 64\\ 50 + 7y &= 64\\ 7y &=14 \\y &=2 \end {align}\)
\(\begin {align} 0.5 x  7y &= \text9\\ 0.5(10)  7y &= \text9\\ 5  7y &= \text9\\ \text7y &= \text14\\ y &= 2 \end {align}\)
In this system, the coefficient of \(y\) in the first equation happens to be the opposite of the coefficient of \(y\) in the second equation. The sum of the terms with \(y\)variables is 0.
What if the equations don't have opposite coefficients for the same variable, like in the following system?
\(\begin {cases}\begin {align}8r + 4s &= 12 \\ 8r + \hspace{2.3mm}s &= \hspace{1mm}\text3 \end{align} \end {cases}\)
Notice that both equations have \(8r\) and if we subtract the second equation from the first, the variable \(r\) will be eliminated because \(8r8r\) is 0.
\(\begin {align} 8r + 4s &= 12\\ 8r + \hspace{2.3mm}s &= \hspace{1mm}\text3 \quad\\ \overline {\hspace{2mm}0 + 3s }& \overline{ \hspace{1mm}=15} \\ 3s &= 15 \\s &=5 \end {align}\)
Substituting 5 for \(s\) in one of the equations gives us \(r\):
\(\begin {align} 8r + 4s &= 12\\ 8r + 4(5) &= 12\\ 8r + 20 &=12 \\ 8r &= \text8 \\ r &= \text1 \end {align}\)
Adding or subtracting the equations in a system creates a new equation. How do we know the new equation shares a solution with the original system?
If we graph the original equations in the system and the new equation, we can see that all three lines intersect at the same point, but why do they?
In future lessons, we will investigate why this strategy works.