Lesson 16
Solving Systems by Elimination (Part 3)
16.1: Multiplying Equations By a Number (10 minutes)
Warmup
In this warmup, experiment with the graphical effects of multiplying both sides of an equation in two variables by a factor.
Students are prompted to multiply one equation in a system by several factors to generate several equivalent equations. They then graph these equations on the same coordinate plane that shows the graphs of the original system. Students notice that no new graphs appear on the coordinate plane and reason about why this might be the case.
The work here reminds students that equations that are equivalent have all the same solutions, so their graphs are also identical. Later, students will rely on this insight to explain why we can multiply one equation in a system by a factor—which produces an equivalent equation—and solve a new system containing that equation instead.
Launch
Display the equation \(x+5 = 11\) for all to see. Ask students:
 "What is the solution to this equation?" (\(x=6\))
 "If we multiply both sides of the equation by a factor, say, 4, what equation would we have?" (\(4x + 20 = 44\)) "What is the solution to this equation?" (\(x=6\))
 "What if we multiply both sides by 100?" (\(100x + 500 = 1,\!100\)) "By 0.5?" (\(0.5x + 2.5 = 5.5\))
 "Is the solution to each of these equations still \(x=6\)?" (Yes)
Remind students that these equations are onevariable equations, and that multiplying both sides of a onevariable equation by the same factor produces an equivalent equation with the same solution.
Ask students: "What if we multiply both sides of a twovariable equation by the same factor? Would the resulting equation have the same solutions as the original equation?" Tell students that they will now investigate this question by graphing.
Arrange students in groups of 2–4 and provide access to graphing technology to each group. To save time, consider asking group members to divide up the tasks. (For example, one person could be in charge of graphing while the others write equivalent equations, and everyone analyze the graphs together.)
Student Facing
Consider two equations in a system:
\(\begin {cases}\begin {align} 4x + \hspace{2.2mm} y &= \hspace {2mm}1 &\quad&\text{Equation A}\\ x + 2y &= \hspace {2mm} 9&\quad&\text{Equation B} \end{align} \end{cases}\)
 Use graphing technology to graph the equations. Then, identify the coordinates of the solution.

Write a few equations that are equivalent to equation A by multiplying both sides of it by the same number, for example, 2, 5, or \(\frac12\). Let’s call the resulting equations A1, A2, and A3. Record your equations here:
 Equation A1:
 Equation A2:
 Equation A3:
 Graph the equations you generated. Make a couple of observations about the graphs.
Student Response
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Activity Synthesis
Invite students to share their observations about the graphs they created. Ask students why the graphs of equations A1, A2, and A3 all coincide with the graphs of the original equation A. Discuss with students:
 "How can we explain the identical graphs?" (Equations A1, A2, and A3 are equivalent to equation A, so they all have the same solutions as equation A, and their graphs are the same line as the graph of A.)
 "What move was made to generate A1, A2, and A3? Why did it create equations that are equivalent to A?" (The two sides of equation A was multiplied by the same factor, which keeps the two sides equal.)
To further illustrate that equations A1, A2, and A3 are equivalent to equation A, and if time permits, consider:
 Using tables to visualize the identical \((x,y)\) pairs. For example:
\(4x+y=1\)
x y 0 1 1 3 2 7 3 11 \(20x+5y=5\)
x y 0 1 1 3 2 7 3 11 \(2x+\frac12y=\frac12\)
x y 0 1 1 3 2 7 3 11  Reminding students that, earlier in the unit, they saw that isolating one variable is a way to see if two equations are equivalent. If we isolate \(y\) in equations A, A1, A2, and A3, the rearranged equation will be identical: \(y=\text4x+1\).
16.2: Writing a New System to Solve a Given System (15 minutes)
Activity
In an earlier lesson, students learned that adding the two equations in a system creates a new equation that shares a solution with the system. In the warmup, they saw that multiplying an equation by a factor creates an equivalent equation that shares all the same solutions as the original equation.
Here students learn that each time we perform a move that creates one or more new equations, we are in fact creating a new system that is equivalent to the original system. Equivalent systems are systems with the same solution set, and we can write a series of them to help us get closer to finding the solution of an original system.
For instance, if \(4x+y=1\) and \(x+2y=9\) form a system, and \(4x+8y=36\) is a multiple of the second equation, the equations \(4x+y=1\) and \(4x+8y=36\) form an equivalent system that can help us eliminate \(x\) and make progress toward finding the value of \(y\).
Students also learn to make an argument that explains why each new system is indeed equivalent to the one that came before it (MP3), building on the work of justifying equivalent equations in earlier lessons.
Launch
Arrange students in groups of 2. Give students 2–3 minutes of quiet work time, and then 1–2 minutes to discuss their thinking with their partner. Follow with a wholeclass discussion.
Student Facing
Here is a system you solved by graphing earlier.
\(\begin {cases}\begin {align} 4x + \hspace{2.2mm} y &= \hspace {2mm}1 &\quad&\text{Equation A}\\ x + 2y &= \hspace {2mm} 9&\quad&\text{Equation B} \end{align} \end{cases}\)
To start solving the system, Elena wrote:
\(\begin {align} 4x + \hspace{2.2mm} y &= \hspace {2mm}1\\ 4x + 8y &= 36 \end{align}\)
And then she wrote:
\(\begin {align} 4x + \hspace{2.2mm} y &= \hspace {3mm}1\\ 4x + 8y &= \hspace{1mm}36 \hspace{1.5mm} \\ \overline {\hspace{8mm}\text7y} &\overline{\hspace{1mm}=\text35 \hspace{5mm}}\end{align}\)
 What were Elena's first two moves? What might be possible reasons for those moves?
 Complete the solving process algebraically. Show that the solution is indeed \(x=\text1, y=5\).
Student Response
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Anticipated Misconceptions
Some students may be confused by the subtraction symbol after the second equation, wondering if some number is supposed to appear after the sign. Encourage them to ignore the sign at first, find the relationship between the three equations, and then think about what the sign might mean.
Activity Synthesis
Invite students to share their analyses of Elena's moves. Highlight responses that point out that Elena's moves enabled her to eliminate the \(x\)variable. (In other words, multiplying equation B by 4 gives \(4x + 8y=36\) and subtracting this equation from equation A removes the \(x\)variable.)
Display for all to see the two original equations in the system and the new equations Elena wrote (\(4x + 8y=36\) and \(\text7y=\text35\)). Then, ask students to predict what the graphs of all four equations might look like.
Next, use graphing technology to display all four graphs. Invite students to share their observations about the graphs.
Students are likely to observe that the graphs all intersect at the same point, \((\text1,5)\) and that there are only three lines, instead of 4. Discuss why only three lines are visible. Make sure students understand that this is because the equations \(x+2y =9\) (equation B) and \(4x+8y=36\) are equivalent, so they share all the same solutions.
Then, focus students' attention on two things: the series of systems that came into play in solving the original system, and the explanations that justify each step along the way. Display the following systems and sequence the discussion as follows:
\(\begin {cases}\begin {align} 4x + \hspace{2.2mm} y &=1 &\quad&\text{A}\\ x + 2y &=9&\quad&\text{B} \end{align} \end{cases}\)

"Here are the two equations in the original system. In solving the system, what do we assume about the \(x\) and \(y\)values in the equations?"
(We assume that there is a pair of \(x\) and \(y\)values that make both equations true.)
\(\begin {cases}\begin {align} 4x + \hspace{2.2mm} y &=1 &\quad&\text{A}\\ 4x + 8y &= 36&\quad&\text{B1} \end{align} \end{cases}\)

"We didn't use the original two equations to solve. Instead, we multiplied each side of equation B by 4 to get equation B1. How do we know that the same \((x,y)\) pair is also a solution to equation B1?"
(Multiplying each side of equation B by the same number gives an equation that is equivalent to B. This means it has all the same solutions as B, including the pair that made the original system true.)
\(\begin {cases}\begin {align} 4x + \hspace{2.2mm} y &=1 &\quad&\text{A}\\ \text7y &= \text35&\quad&\text{C} \end{align} \end{cases}\)

"We couldn't yet solve the system with equations A and B1, so we subtracted B1 from A and got equation C. How do we know that the same \((x,y)\) pair from earlier is also a solution to equation C?"
(When we subtracted \(4x+8y\) from \(4x+1\) and subtracted 36 from 1, we subtracted equal amounts from each side of a true equation, which kept the two sides equal. Even though the \(x\)value was eliminated in the result, the \(y\)value that makes the original equations true hasn't changed and is also a solution to C.)
\(\begin {cases}\begin {align} 4x + \hspace{2.2mm} y &=1 &\quad&\text{A}\\ y &= 5&\quad&\text{Solution} \end{align} \end{cases}\)

Solving equation C gives us \(y=5\). How do we find the \(x\)value?"
(Substituting this value into equation A or B and solving it gives us the \(x\)value.)
\(\begin {cases}\begin {align} x&= \text1 &\quad&\text{Solution}\\ y &= \hspace {1.5mm}5 &\quad&\text{Solution}\end{align} \end{cases}\)

"If we substitute this pair of values for \(x\) and \(y\) in equations A, B, B1, and C and evaluate the expressions, can we expect to find true statements?"
(Yes. For A, it will be \(1=1\). For B, it will be \(9=9\). For B1, it will be 36=36. For C, it will be \(\text35=\text35\).)
Explain that what we have done was to create equivalent systems—systems with the exact same solution set—to help us get closer and closer to the solution of the original system.
One way to create an equivalent system is by multiplying one or both equations by a factor. It helps to choose the factor strategically—one that would allow one variable to be eliminated when the two equations in the new system are added or subtracted. Elena chose to multiply equation B by 4 so that the \(x\)variable could be eliminated.
Ask students:
 "Suppose we want to solve the system by first eliminating the \(y\)variable. What factor should we choose? Which equation should we apply it to?" (We could multiply equation A by 2, or multiply equation B by \(\frac12\).)
 "Could we multiply equation A by 6 and multiply equation B by 3 as a way to eliminate the \(y\)variable?" (Yes. Each new equation (\(24x+6y=6\) and \(\text3x 6y = \text27\)) would be equivalent to the original. Adding the two new equations would eliminate the \(y\)variable.)
16.3: What Comes Next? (10 minutes)
Activity
Earlier, students encountered the idea of equivalent systems. They described the moves that lead to equivalent equations and explained why the solution to those equations was the same as the solution to the original system. This activity reinforces that work.
Given an unordered set of equivalent systems, students work with a partner to arrange the systems such that the order constitutes logical steps toward solving an original system. Along the way, they take turns describing how each system came from the previous one and explaining how they know it has the same solution as its predecessor. The work allows students to practice constructing logical arguments and critiquing those of others (MP3).
Here is an image of the cards for reference and planning.
Launch
Arrange students in groups of 2. Give one set of precut slips or cards from the blackline master to each group. Ask students to find the slip or card that shows a system of equations and is labeled “Start here.”
Explain that all the other slips contain equivalent systems that represent steps in solving the starting system. Ask students to arrange the slips in the order that would lead to its solution, and to make sure they can explain what moves take each system to the next system and why each system is equivalent to the one before it.
Partners should take turns finding the next step in the solving process and explaining their reasoning. As one student explains, the partner's job is to listen and make sure they agree and the explanation makes sense. If they disagree, the partners should discuss until they reach an agreement.
Give students 5 minutes to arrange the systems. Follow with a wholeclass discussion.
Design Principle(s): Support sensemaking
Supports accessibility for: Organization; Attention
Student Facing
Your teacher will give you some slips of paper with systems of equations written on them. Each system represents a step in solving this system:
\(\begin {cases}\frac45 x + 6y = 15\\ \textx + 18y = 11 \end{cases}\)
Arrange the slips in the order that would lead to a solution. Be prepared to:
 Describe what move takes one system to the next system.
 Explain why each system is equivalent to the one before it.
Student Response
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Student Facing
Are you ready for more?
This system of equations has solution \((5,\text2)\): \(\begin {cases}Ax  By = 24\\ Bx + Ay = 31 \end{cases}\)
Find the missing values \(A\) and \(B\).
Student Response
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Anticipated Misconceptions
Students who are thinking algorithmically about solving systems of equations may think that the first step should involve multiplying each side of the second equation by \(\frac45\), becoming frustrated when no cards show that step. Encourage them to compare both the first and the second equations in the starting card to the first and second equations on other cards to gain some ideas about what steps might have been taken.
Activity Synthesis
Ask students to share the order in which the systems should be arranged to lead to the solution. Display the ordered systems for all to see.
Point out to students that this particular solution path involves multiplying each of the two equations by a factor in order to eliminate the \(x\)variable. Ask students if it's possible to eliminate a variable by multiplying only one equation by a factor. (Yes, we could multiply the first equation by \(\frac54\) to eliminate \(x\), or by 3 to eliminate \(y\). Or we could multiply the second equation by \(\frac45\) to eliminate \(x\), or by \(\frac13\) to eliminate \(y\).)
See Lesson Synthesis for discussion questions and ways to help students connect the ideas in the lesson.
16.4: Build Some Equivalent Systems (15 minutes)
Optional activity
This optional activity gives students another opportunity to identify moves that lead to equivalent systems and to explain why the resulting systems have the same solution (MP3). It also prompts students to build their own equivalent systems, which encourages them to think strategically about what to do to reach the goal of solving the system, rather than on a particular solution path.
As students work, prompt students to articulate the reasoning and assumptions. Ask questions such as:
 "How do you know the factor to use when multiplying an equation so a variable can be eliminated?"
 "Can the factor be a fraction? A negative number? Why or why not?"
 "Why doesn't adding one equation to another equation change the solution of the system?"
Launch
Keep students in groups of 2, if desired.
Student Facing
Here is a system of equations:
\(\begin {cases} \begin {align}12a + 5b &= \text15\\8a + \hspace{2mm}b &= \hspace{1.5mm}11 \end{align} \end {cases} \)

To solve this system, Diego wrote these equivalent systems for his first two steps.
Step 1:
\(\begin {cases} \begin {align}12a + \hspace{1.5mm}5b &= \text15\\\text40a + \text5b &= \text55 \end{align} \end {cases} \)
Step 2:
\(\begin {cases} \begin {align}12a + 5b &= \text15\\\text28a \hspace{8.5mm}&= \text70 \end{align} \end {cases} \)
 Write another set of equivalent systems (different than Diego's first two steps) that will allow one variable to be eliminated and enable you to solve the original system. Be prepared to describe the moves you make to create each new system and to explain why each one has the same solution as the original system.
 Use your equivalent systems to solve the original system. Then, check your solution by substituting the pair of values into the original system.
Student Response
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Anticipated Misconceptions
If students struggle to create an equivalent system of their own, ask them to start by deciding on a variable they'd like to eliminate. Then, ask them to think about a factor that, when multiplied to one equation, would produce the same or opposite coefficients for that variable. If they are uncomfortable using a fractional factor, ask if they could find a factor to apply to each equation such that the resulting equations have the same or opposite coefficients for the variable they wish to eliminate.
Activity Synthesis
Invite students with different first steps to display their equivalent systems and solution paths. Prompt them (or others students) to explain why each system generated share the same solution as the original system or the system before it.
Verify that, regardless of the moves made, the different paths all led to the same pair of values.
Lesson Synthesis
Lesson Synthesis
Refer to the systems that students have ordered in the "What Comes Next?" activity, Invite students to explain what move takes one system to the next and how they know the new system is equivalent to the one before it (even though one equation has been transformed).
As students share their responses, record the descriptions of moves and the justifications for all to see. Consider using a graphic organizer, as shown here. (A completed organizer could be preserved and used as a reference by students later.)
step  system  what move was made?  why is the new system equivalent to the previous one? 

0  \(\begin {cases}\frac45 x + 6y = 15\\ \textx + 18y = 11 \end{cases}\)  
1  \(\begin {cases} \begin{align}4x + 30y = 75\\ \textx + 18y = 11 \end{align} \end{cases}\)  
2  
. . . 
If time is limited, focus on discussing the justifications on why one system in the sequence is equivalent to the before it (the last column in the graphic organizer).
Highlight statements such as: "Adding equal amounts to the two sides of an equation keeps the two sides equal, so the same \(x\) and \(y\)values that make the first equation true also makes this new equation true. This means that same pair of values is also a solution to the new system." If no students explained in this way, demonstrate it.
16.5: Cooldown  Make Your Move (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
We now have two algebraic strategies for solving systems of equations: by substitution and by elimination. In some systems, the equations may give us a clue as to which strategy to use. For example:
\(\begin{cases} y=2x11 \\ 3x+2y=18 \\ \end{cases}\)
In this system, \(y\) is already isolated in one equation. We can solve the system by substituting \(2x11\) for \(y\) in the second equation and finding \(x\).
\(\begin{cases} \begin {align} 3xy&=\text17 \\ \text3x+4y&=23 \\ \end {align} \end{cases}\)
This system is set up nicely for elimination because of the opposite coefficients of the \(x\)variable. Adding the two equations eliminates \(x\) so we can solve for \(y\).
In other systems, which strategy to use is less straightforward, either because no variables are isolated, or because no variables have equal or opposite coefficients. For example:
\(\begin{cases} \begin {align} 2x+3y&=15 \quad&\text{Equation A}\\ 3x9y&=18 \quad&\text{Equation B} \\ \end{align}\end{cases}\)
To solve this system by elimination, we first need to rewrite one or both equations so that one variable can be eliminated. To do that, we can multiply both sides of an equation by the same factor. Remember that doing this doesn't change the equality of the two sides of the equation, so the \(x\) and \(y\)values that make the first equation true also make the new equation true.
There are different ways to eliminate a variable with this approach. For instance, we could:
 Multiply Equation A by 3 to get \(6x +9y = 45\). Adding this equation to Equation B eliminates \(y\).
\(\displaystyle \begin{cases} \begin {align} 6x+9y&=45 &\quad&\text{Equation A1} \\ 3x9y&=18 &\quad&\text{Equation B}\end{align}\end{cases}\)

Multiply Equation B by \(\frac23\) to get \(2x  6y = 12\). Subtracting this equation from Equation A eliminates \(x\).
\(\begin{cases} \begin {align} 2x+3y&=15 &\quad&\text{Equation A}\\ 2x  6y &=12 &\quad&\text{Equation B1} \\ \end{align}\end{cases}\)
 Multiply Equation A by \(\frac12\) to get \(x+\frac32y = 7\frac12\) and multiply Equation B by \(\frac13\) to get \(x  3y = 6\). Subtracting one equation from the other eliminates \(x\).
\(\begin{cases} \begin {align} x+\frac32y&= 7\frac12 &\quad&\text{Equation A2}\\ x  3y &= 6 &\quad&\text{Equation B2} \\ \end{align}\end{cases}\)
Each multiple of an original equation is equivalent to the original equation. So each new pair of equations is equivalent to the original system and has the same solution.
Let’s solve the original system using the first equivalent system we found earlier.
\(\displaystyle \begin{cases} \begin {align} 6x+9y&=45 &\quad&\text{Equation A1} \\ 3x9y&=18 &\quad&\text{Equation B}\end{align}\end{cases}\)
 Adding the two equations eliminates \(y\), leaving a new equation \(9x=63\), or \(x=7\).
\(\displaystyle \begin {align} 6x+9y&=45 \\ 3x9y&=18 \quad+\\ \overline {9x + 0\hspace{2mm}}&\overline{= 63}\\x &=7 \end{align}\)
 Putting together \(x=7\) and the original \(3x9y=18\) gives us another equivalent system.
\(\displaystyle \begin{cases} \begin {align} x&=7 \\ 3x9y&=18 \end{align}\end{cases}\)
 Substituting 7 for \(x\) in the second equation allows us to solve for \(y\).
\(\displaystyle \begin{align} 3(7)  9y &=18\\ 21 9y&=18\\ \text9y &= \text3\\ y&=\frac13 \end{align}\)
When we solve a system by elimination, we are essentially writing a series of equivalent systems, or systems with the same solution. Each equivalent system gets us closer and closer to the solution of the original system.
\(\begin{cases} \begin {align} 2x+3y&=15\\ 3x9y&=18\\ \end{align}\end{cases}\)
\(\displaystyle \begin{cases} \begin {align} 6x+9y&=45\\ 3x9y&=18 \end{align}\end{cases}\)
\(\displaystyle \begin{cases} \begin {align} x&=7 \\ 3x9y&=18 \end{align}\end{cases}\)
\(\displaystyle \begin{cases} \begin {align} x&=7 \\ y&=\frac13\end{align}\end{cases}\)