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<h4>Activity (15 minutes)</h4>
<p>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 warm up, they saw that multiplying an equation by a factor creates an equivalent equation that shares all the same solutions as the original equation.</p>
<p>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.</p>
<p>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 \( y \).</p>
<p>Students also learn to make an argument that explains why each new system is indeed equivalent to the one that came before it, building on the work of justifying equivalent equations in earlier lessons.</p>
<h4>Launch</h4>
<p>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 whole-class discussion.</p>
<h4>Student Activity</h4>
<p>Here is a system you solved by graphing earlier.</p><br>
<p>\( \begin {cases}\begin {align} 4x + \hspace{2.2mm} y &= \hspace {2mm}1 &\quad&\text{Equation A}\\ x + 2y &= \hspace {2mm} 9&\quad&\text{Equation B} \end{align} \end{cases} \)</p>
<p>To start solving the system, Elena wrote:</p>
<p>\( \begin {align} 4x + \hspace{2.2mm} y &= \hspace {2mm}1\\ 4x + 8y &= 36 \end{align} \)</p>
<p>And then she wrote:</p>
<p>\( \begin {align} 4x + \hspace{2.2mm} y &= \hspace {3mm}1\\ 4x + 8y &= \hspace{1mm}36 \hspace{1.5mm}- \\ \overline {\hspace{8mm}\text-7y} &\overline{\hspace{1mm}=\text-35 \hspace{5mm}}\end{align} \)</p>
<ol class="os-raise-noindent">
<li>What was Elena’s first move?</li>
</ol>
<p><strong>Answer: </strong>First, she multiplied each side of the second equation by 4, which produces an equivalent equation with \(4x\) in it.</p>
<ol class="os-raise-noindent" start="2">
<li>What was Elena’s second move?</li>
</ol>
<p><strong>Answer: </strong>Next, she subtracted the new equation from the first equation. Doing that creates another equation where the variable \(x\) is eliminated, leaving only one variable and making it possible to solve for \(y\).</p>
<ol class="os-raise-noindent" start="3">
<li>What might be possible reasons for those moves?</li>
</ol>
<p><strong>Answer: </strong>Doing these steps creates another equation where the \(x\) variable is eliminated, leaving only one variable making it possible to solve for \(y\). </p>
<ol class="os-raise-noindent" start="4">
<li>Complete the solving process algebraically. Show that the solution is indeed \(x=-1\), \(y=5\).</li>
</ol>
<p><strong>Answer: </strong>\( -7y=-35 \) can be rewritten as \(y=5\). Substituting 5 for \(y\) in the first equation, we have \(4x+5=1\) or \(4x=-4\), which means \(x=-1\).</p>
<h4>Video: Writing a New System to Solve a Given System</h4>
<p>Watch the following video to learn more about writing a new system of equations to solve a situation of constraints.</p>
<div class="os-raise-d-flex-nowrap os-raise-justify-content-center">
<div class="os-raise-video-container"><video controls="true" crossorigin="anonymous">
<source src="https://k12.openstax.org/contents/raise/resources/5d5e3a9d49372cd8561211fb5b459e2977617763">
<track default="true" kind="captions" label="On" src="https://k12.openstax.org/contents/raise/resources/58f9c94ed72c794d910edc1862583e7906b54eb9" srclang="en_us">https://k12.openstax.org/contents/raise/resources/5d5e3a9d49372cd8561211fb5b459e2977617763
</video></div>
</div>
<br><br>
<h4>Anticipated Misconceptions</h4>
<p>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.</p>
<h4>Activity Synthesis</h4>
<p>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.)<br></p>
<div>
<p>Display the two original equations in the system alongside the new equations Elena wrote (\( 4x + 8y=36 \) and \( \text-7y=\text-35 \)). Then, ask students to predict what the graphs of all four equations might look like.</p>
<p>Next, use graphing technology to display all four graphs. Invite students to share their observations about the graphs.</p>
<div><img alt="Graph of a linear system." height="182" src="https://k12.openstax.org/contents/raise/resources/4b726eb4b446c7635aea8198af9f4b43868bfa92" width="201"></div>
<p>Students are likely to observe that the graphs all intersect at the same point, \( (\text-1,5) \) 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.</p>
<p>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:</p>
<div>
<div>
<div>
<p>\( \begin {cases}\begin {align} 4x + \hspace{2.2mm} y &=1 &\quad&\text{A}\\ x + 2y &=9&\quad&\text{B} \end{align} \end{cases} \)</p>
</div>
<div>
<ul>
<li>
<p>"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?"</p>
<p>(We assume that there is a pair of \( x \)- and \( y \)-values that make both equations true.)</p>
</li>
</ul>
</div>
</div>
<div>
<div>
<p>\( \begin {cases}\begin {align} 4x + \hspace{2.2mm} y &=1 &\quad&\text{A}\\ 4x + 8y &= 36&\quad&\text{B1} \end{align} \end{cases} \)</p>
</div>
<div>
<ul>
<li>
<p>"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?"</p>
<p>(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.)</p>
</li>
</ul>
</div>
</div>
<div>
<div>
<p>\( \begin {cases}\begin {align} 4x + \hspace{2.2mm} y &=1 &\quad&\text{A}\\ \text-7y &= \text-35&\quad&\text{C} \end{align} \end{cases} \)</p>
</div>
<div>
<ul>
<li>
<p>"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?"</p>
<p>(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.)</p>
</li>
</ul>
</div>
</div>
<div>
<div>
<p>\( \begin {cases}\begin {align} 4x + \hspace{2.2mm} y &=1 &\quad&\text{A}\\ y &= 5&\quad&\text{Solution} \end{align} \end{cases} \)</p>
</div>
<div>
<ul>
<li>
<p>Solving equation C gives us \( y=5 \). How do we find the \( x \)-value?"</p>
<p>(Substituting this value into equation A or B and solving it gives us the \( x \)-value.)</p>
</li>
</ul>
</div>
</div>
<div>
<div>
<p>\( \begin {cases}\begin {align} x&= \text-1 &\quad&\text{Solution}\\ y &= \hspace {1.5mm}5 &\quad&\text{Solution}\end{align} \end{cases} \)</p>
</div>
<div>
<ul>
<li>
<p>"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?"</p>
<p>(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 \( \text-35=\text-35 \).)</p>
</li>
</ul>
</div>
</div>
</div>
<p>Explain that what we have done was to create <strong>equivalent systems</strong>, or systems with the exact same solution set, to help us get closer and closer to the solution of the original system.<br></p>
<p>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.</p>
</div>
<p>Ask students:</p>
<ul>
<li>"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 12.)</li>
<li>"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 \( -3x−6y=-27) \) would be equivalent to the original. Adding the two new equations would eliminate the \( y \)-variable.)</li>
</ul>
<h4>2.6.2 Self Check</h4>
<p><strong><em>Following the activity, students will answer the following question to check their understanding of the concepts explored in the activity.</em></strong></p>
<!-- Begin Self Check Table -->
<p class="os-raise-text-bold">QUESTION:</p>
<p>Solve the system of equations using elimination.<br></p>
<p>\( \left\{ \begin{array}{c l} 6x – 5y = –34 \\ 2x + 6y = 4 \end{array}\right. \)</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">Answers</th>
<th scope="col">Feedback</th>
</tr>
</thead>
<tbody>
<tr>
<td>\( (5, –1) \)</td>
<td>Incorrect. Let’s try again a different way: Does the ordered pair make both equations true? Did you multiply the second equation by 3? The answer is \( (–4, 2) \).<br></td>
</tr>
<tr>
<td>\( (–10, 4) \)</td>
<td>Incorrect. Let’s try again a different way: Does the ordered pair make both equations true? Did you multiply the second equation by 3? The answer is \( (–4, 2) \).<br></td>
</tr>
<tr>
<td>\( (–4, 2) \)</td>
<td>That’s correct! Check yourself: The ordered pair makes both equations true.<br></td>
</tr>
<tr>
<td>\( (2, 0) \)</td>
<td>Incorrect. Let’s try again a different way: Does the ordered pair make both equations true? Did you multiply the second equation by 3? The answer is \( (–4, 2) \).<br></td>
</tr>
</tbody>
</table><br>
<!-- END SELF CHECK TABLE -->
<br>
<h4>2.6.2: Additional Resources</h4>
<p><strong><em>The following content is available to students who would like more support based on their experience with the self check. Students will not automatically have access to this content, so you may wish to share it with those who could benefit from it. </em></strong></p>
<h4>Solving a System of Equations by Elimination Using Multiplication</h4>
<p>Let’s look
at an example where we need to multiply both equations by constants in order to
make the coefficients of one variable opposites.</p>
<p>Solve the
system by elimination: \(\left\{\begin{array}{l}\;4x-3y=9\\7x+2y=-6\end{array}\right.\)</p>
<p>In this
example, we cannot multiply just one equation by any constant to get opposite
coefficients. So we will strategically multiply both equations by different
constants to get the opposites.</p>
<p><strong>Step 1 - </strong>Write both equations in standard form. If any coefficients are fractions, clear them.<br><br>
Both equations are in standard form.<br>
\(4x-3y=9\)<br>
\(7x+2y=-6\)<br><br>
<strong>Step 2 - </strong>Check to see if the coefficients of one variable are opposites or equivalent. <br><br>
To get opposite coefficients of \(y\), we will need to multiply the first equation by 2 and the second equation by 3.<br>
Multiply the first equation by 2 and the second equation by 3.<br><br>
\(2(4x-3y) = 2(9)\)<br>
\(3(7x+2y) = 3(-6)\)<br><br>
<strong>Step 3 - </strong>Simplify<br><br>
\(8x−6y=18\)<br>
\(21x+6y=−18\)<br><br>
<strong>Step 4 - </strong> Add the equations to eliminate one variable. <br>
\(\begin{array}{r}8x-6y=18\\21x+6y=-18\\ \hline 29x=0\end{array}\)<br><br>
<strong>Step 5 - </strong> Solve for the remaining variable.<br><br>
\(x=0\)<br><br>
<strong>Step 6 - </strong>Substitute the solution into one of the original equations. Then solve for the other variable.
<br><br>
\(\begin{array}{rcl}7x+2y&=&-6
\\ 7(0)+2y&=&-6
\\2y&=&-6
\\y&=&-3
\\ \end{array}\)<br><br>
<strong>Step 7 - </strong>Check that the ordered pair is a solution to both original equations.<br><br>
<!--BEGIN SIDE BY SIDE EQUATIONS-->
\(\begin{array}{rr}
4x-3y=9 & 7x+2y=-6\\
4(0)-3(-3) \overset?=9 & 7(0)+2(-3)\overset?=-6 \\
9=9\checkmark & -6=-6\checkmark
\end{array}\) <br><br>
<!--END SIDE BY SIDE EQUATIONS -->
The solution is \((0,-3)\).
</p>
<br>
<h4>Try It: Solving a System of Equations by Elimination Using Multiplication</h4>
<p>Solve the following system of equations using elimination:</p>
<p>\( \left\{ \begin{array}{c l} 7x+8y=4 \\ 3x-5y= 27 \end{array}\right. \)</p>
<p><strong>Answer: </strong></p>
<p><strong>Step 1 - </strong>Write both equations in standard form. If any coefficients are fractions, clear them.<br><br>
\(7x+8y=4\)<br>
\(3x–5y=27\)<br><br>
<strong>Step 2 - </strong>Check to see if the coefficients of one variable are opposites or equivalent. To get opposite coefficients of \(y\), we will need to multiply the first equation by 5 and the second equation by 8.<br><br>
\(5(7x+8y)=5(4)\)
\(8(3x–5y)=8(27)\)<br><br>
<strong>Step 3 - </strong>Simplify. <br><br>
\(35x+40y=20\)<br>
\(24x–40y=216\)<br><br>
<strong>Step 4 - </strong>Add the equations to eliminate one variable. <br><br>
\(\begin{array}{r}35x+40y=20\\24x–40y=216\\ \hline 59x=236\\\end{array}\)<br><br>
<strong>Step 5 - </strong>Solve for the remaining variable.<br><br>
\(59x=236\)<br>
\(x=4\)<br><br>
<strong>Step 6 - </strong>Substitute the solution into one of the original equations. Then solve for the other variable.<br><br>
\(59x=236\)<br>
\(x=4\)<br><br>
<!--BEGIN ALIGN TO EQUALS -->
\(\begin{array}{rcl}7x+8y&=&4
\\ 7(4)+8y&=&4
\\28+8y&=&4
\\8y&=&-24
\\y&=&-3
\\ \end{array}\)<br><br>
<!--END ALIGN TO EQUALS -->
<strong>Step 7 - </strong>Check that the ordered pair is a solution to both original equations.<br><br>
<!--BEGIN SIDE BY SIDE EQUATIONS-->
\(\begin{array}{rr}
7x+8y=4 & 3x–5y=277\\
7(4)+8(−3) \overset?=4 & 3(4)–5(−3)\overset?=277 \\
20−24 \overset?=4 & 3(4)–5(−3)\overset?=277 \\
4=4\checkmark & 27=27\checkmark
\end{array}\) <br><br>
<!--END SIDE BY SIDE EQUATIONS -->
The solution is \((0,-3)\).
</p>