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<h2> Activity (10 minutes) </h2>
<p>The activity allows students to practice solving systems of linear equations by substitution and reinforces the idea that there are multiple ways to perform substitution. Students are directed to find the solutions without graphing.</p>
<p>Monitor for the different ways that students use substitutions to solve the systems. Invite students with different approaches to share later.</p>
<h3> Launch </h3>
<p>Keep students in groups of 2. Give students a few minutes to work quietly and then time to discuss their work with a partner. If time is limited, ask each partner to choose two different systems to solve. Follow with a whole-class discussion.</p>
<br>
<div class="os-raise-extrasupport">
<div class="os-raise-extrasupport-header">
<p class="os-raise-extrasupport-title">Support for English Language Learners</p>
<p class="os-raise-extrasupport-name">MLR 7 Compare and Connect: Representing, Conversing</p>
</div>
<div class="os-raise-extrasupport-body">
<p>As students share their responses and reasoning with the class, call students’ attention to the different ways their classmates have chosen to make substitutions. Wherever possible, amplify student words and actions that involve the language of substitution:replace, substitute, solve for, and isolate.</p>
<p class="os-raise-text-italicize">Design Principle(s): Maximize meta-awareness; Support sense-making</p>
<p class="os-raise-extrasupport-title">Learn more about this routine</p>
<p>
<a href="https://www.youtube.com/watch?v=PF8fRA107OA;&rel=0" target="_blank">View the instructional video</a>
and
<a href="https://k12.openstax.org/contents/raise/resources/94a1159e7b81493c647515711f325771076d99b8" target="_blank">follow along with the materials</a>
to assist you with learning this routine.
</p>
<p class="os-raise-extrasupport-title">Provide support for students</p>
<p>
<a href="https://k12.openstax.org/contents/raise/resources/0b8a1a4ac3425e84a1d5452b3a5dffa38deb6b13" target="_blank">Distribute graphic organizers</a>
to the students to assist them with participating in this routine.
</p>
</div>
</div>
<br>
<br>
<div class="os-raise-extrasupport">
<div class="os-raise-extrasupport-header">
<p class="os-raise-extrasupport-title">Support for Students with Disabilities</p>
<p class="os-raise-extrasupport-name">Representation: Internalize Comprehension</p>
</div>
<div class="os-raise-extrasupport-body">
<p>
Differentiate the degree of difficulty or complexity by allowing students to complete problems in any order. Students may benefit from solving the system that uses the same letter variables \( ( x, y) \) first. Highlight connections between representations by inviting students use the same color to highlight a variable and the expression to be substituted for that variable. </p>
<p class="os-raise-text-italicize">Supports accessibility for: Conceptual processing</p>
</div>
</div>
<!--END ELL AND SWD GRAY BOX -->
<h4> Student Activity </h4>
<p>Here are the four systems of equations you saw in the previous exercise. Solve each system. Then, check your solutions by substituting them into the original equations to see if the equations are true.
</p>
<br>
<p>Use this system of equations to answer questions 1-3.</p>
<blockquote>\( \begin{cases} 5x - 2y = 26 \\ y + 4 = x \end{cases} \)</blockquote>
<ol class="os-raise-noindent">
<li>What is the \(x\) value?</li>
</ol>
<p><strong>Answer: </strong>6</p>
<ol class="os-raise-noindent" start="2">
<li>What is the \(y\) value?</li>
</ol>
<p><strong>Answer: </strong>2</p>
<ol class="os-raise-noindent" start="3">
<li>Explain your solution.</li>
</ol>
<p><strong>Answer: </strong> </p> </div>
<p>Your answer may vary, but here is a sample.</p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
5x-2y&=&26
\\ 5(y+4)-2y&=&26
\\ 5y+20-2y&=&26
\\ 3y+20&=&26
\\ 3y&=&6
\\ y&=&2
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
\\ 2+4&=&x
\\ 6&=&x
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN SIDE BY SIDE EQUATIONSS-->
<p>\(\begin{array}{rcr}
5(6)-2(2)=26 & &\\
30-4\overset?=26 & & 2+4\overset?=6 \\
26=26\checkmark & & 6=6\checkmark
\end{array}\) </p>
<!--END SIDE BY SIDE EQUATIONS -->
<br>
<p>Use this system of equations to answer questions 4-6.</p>
<blockquote><p>\( \begin{cases} 2m - 2p = -6\\ p = 2m + 10 \end{cases} \)</p></blockquote>
<ol class="os-raise-noindent" start="4">
<li>What is the \(m\) value?</li>
</ol>
<p><strong>Answer: </strong>-7</p>
<ol class="os-raise-noindent" start="5">
<li>What is the \(p\) value?</li>
</ol>
<p><strong>Answer: </strong>-4</p>
<ol class="os-raise-noindent" start="6">
<li>Explain your solution.</li>
</ol>
<p><strong>Answer: </strong> </p> </div>
<p>Your answer may vary, but here is a sample.</p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
2m-2p&=&-6
\\ 2m-2(2m+10)&=&-6
\\ 2m-4m-20&=&-6
\\ -2m-20&=&-6
\\ -2m&=&14
\\ m&=&-7
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
p&=&2m+10
\\ p&=&2(-7)+10
\\ p&=&-14+10
\\ p&=&-4
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN SIDE BY SIDE EQUATIONSS-->
<p>\(\begin{array}{rcr}
2(-7)-2(-4)=-6 & &-4=2(-7)+10\\
-14+8\overset?=-6 & & -4\overset?=(-14)+10 \\
-6=-6\checkmark & & -4=-4\checkmark
\end{array}\) </p>
<!--END SIDE BY SIDE EQUATIONS -->
<br>
<!-- END CLICK TO REVEAL NO TEXTBOX -->
<br>
<p>Use this system of equations to answer questions 7-9.</p>
<blockquote><p>\( \begin{cases} 2d = 8f \\ 18 - 4f = 2d \end{cases} \)</p></blockquote>
<ol class="os-raise-noindent" start="7">
<li>What is the \(d\) value?</li>
</ol>
<p><strong>Answer: </strong>6 </p>
<ol class="os-raise-noindent" start="8">
<li>What is the \(f\) value?</li>
</ol>
<p><strong>Answer: </strong>\(\frac32\) </p>
<ol class="os-raise-noindent" start="9">
<li>Explain your solution.</li>
</ol>
<p><strong>Answer: </strong> </p>
<p>Your answer may vary, but here is a sample.</p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
2d&=&8f
\\ d&=&4f
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
18-4f&=&2d
\\ 18-4f&=&2(4f)
\\ 18-4f&=&8f
\\ 18&=&12f
\\ \frac{18}{12}&=&f
\\ \frac32&=&f
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
2d&=&8(\frac32)
\\ 2d&=&12
\\ d&=&6
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN SIDE BY SIDE EQUATIONSS-->
<p>\(\begin{array}{rcr}
& &18-4f=2d \\
2d=8f & & 18-4(\frac32)=2(6) \\
2(6)=8(\frac32) & & 18-(2)(3)=12 \\
12\overset?=(4)(3) & & 18-6=12 \\
12=12\checkmark & & 12=12\checkmark
\end{array}\) </p>
<p>Use this system of equations to answer questions 10-12.</p>
<blockquote><p>\( \begin{cases} w + \frac17z = 4 \\ z = 3w -2 \end{cases} \)</p></blockquote>
<ol class="os-raise-noindent" start="10">
<li>What is the \(w\) value?</li>
</ol>
<p><strong>Answer: </strong>3 </p>
<ol class="os-raise-noindent" start="11">
<li>What is the \(z\) value?</li>
</ol>
<p><strong>Answer: </strong>7 </p>
<ol class="os-raise-noindent" start="12">
<li>Explain your solution.</li>
</ol>
<p><strong>Answer: </strong> </p>
<p>Your answer may vary, but here is a sample.</p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
w+\frac17z&=&4
\\ w+\frac17(3w-2)&=&4
\\ 7w+3w-2&=&28
\\ 10w-2&=&28
\\ 10w&=&30
\\ w&=&3
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
z&=&3w-2
\\ z&=&3(3)-2
\\ z&=&9-2
\\ z&=&7
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN SIDE BY SIDE EQUATIONSS-->
<p>\(\begin{array}{rcr}
w+\frac17z=4 & & z=3w-2 \\
3+\frac17(7)=4 & & 7=3(3)-2 \\
3+1\overset?=1=4 & & 7\overset?=1=9-2 \\
4=4 \checkmark & & 7=7 \checkmark \\
\end{array}\) </p>
<!--END SIDE BY SIDE EQUATIONS -->
<h3> Student Facing Extension </h3>
<h4> Are you ready for more? </h4>
<p>Solve this system with four equations.</p>
<p>Use this system of equations to answer questions 1–4.</p>
<blockquote>\( \begin{cases}3 x + 2y - z + 5w= 20 \\ y = 2z-3w\\ z=w+1 \\ 2w=8 \end{cases} \)</blockquote>
<ol class="os-raise-noindent">
<li>What is the \(x\) value?</li>
</ol>
<p><strong>Answer: </strong>3 </p>
<ol class="os-raise-noindent" start="2">
<li>What is the \(y\) value?</li>
</ol>
<p><strong>Answer: </strong>-2 </p>
<ol class="os-raise-noindent" start="3">
<li>What is the \(z\) value?</li>
</ol>
<p><strong>Answer: </strong>5 </p>
<ol class="os-raise-noindent" start="4">
<li>What is the \(w\) value?</li>
</ol>
<p><strong>Answer: </strong>4 </p>
<h3> Anticipated Misconceptions </h3>
<p>When solving the second system, students are likely to substitute the expression \( 2m+10 \) for \( p \) in the first equation, \( 2m-2p=-6 \) . Done correctly, it should be written as \( 2m-2(2m+10)=-6 \) . Some students may neglect to write parentheses and write \( 2m-4m+10=-6 \) . Remind students that if \( p \) is equal to \( 2m+10 \) , then \( 2p \) is 2 times\( 2m+10 \) or \( 2(2m+10) \) . (Alternatively, use an example with a sum of two numbers for \( p \) : Suppose \( p=10 \) , which means \( 2p=2(10) \), or 20. If we express \( p \) as a sum of 3 and 7, or \( p=3+7 \) , then \( 2p=2(3+7) \) , not\( 2\cdot 3 + 7 \) . The latter has a value of 13, not 20.)</p>
<p>Some students who correctly write \( 2m-2(2m+10)=-6 \) may fail to distribute the subtraction and write the left side as \( 2m-4m+20 \) . Remind them that subtracting \( 2(2m+10) \) can be thought of as adding \( -2(2m+10) \) and ask how they would expand this expression.</p>
<h3> Activity Synthesis </h3>
<p>Select previously identified students to share their responses and reasoning. Display their work for all to see.</p>
<p>Highlight the different ways to perform substitution to solve the same system. For example:</p>
<ul>
<li>In the second system,\( \begin{cases} 2m - 2p = -6\\ p = 2m + 10 \end{cases} \) , we could substitute \( 2m+10 \) for \( p \) in the first equation, or we could substitute \( p-10 \) for \( 2m \) in the first equation.</li>
<li>In the third system,\( \begin{cases} 2d = 8f \\ 18 - 4f = 2d \end{cases} \) , we could substitute \( 8f \) for \( 2d \) in the second equation, or we could substitute \( \frac14 d \) for \( f \) in the second equation.</li>
</ul>
<!-- Begin Self Check Table -->
<h3>2.3.3: Self Check</h3>
<p><strong>Following the activity, students will answer the following question to check their understanding of the concepts explored in the activity. </strong></p>
<p class="os-raise-text-bold">QUESTION:</p>
<p>Solve the systems of equations by substitution.</p>
<P>\(\left\{\begin{array}{l}2x\;+\;y\;=\;-4\\3x\;+\;-2y\;=\;-6\end{array}\right.\)</P>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">Answers</th>
<th scope="col">Feedback</th>
</tr>
</thead>
<tbody>
<tr>
<td>\(x = -2\), \(y = 0\)</td>
<td>Correct! Check yourself: Substituting –2 for x and 0 for y makes each equation true.<br>
</td>
</tr>
<tr>
<td>\(x = 0\), \(y = 3\)</td>
<td>
Incorrect. Let’s try again in a different way: Did you solve the first equation for \(y\)? Did you substitute \(-2x-4\) into the second equation for \(y\)? The answer is \(x =-2 \) and \(y = 0\).<br>
</td>
</tr>
<tr>
<td>\(x = -4\), \(y = 0\)</td>
<td>
Incorrect. Let’s try again in a different way: Did you solve the first equation for \(y\)? Did you substitute \(-2x-4\) into the second equation for \(y\)? The answer is \(x =-2 \) and \(y = 0\).<br>
</td>
</tr>
<tr>
<td>\(x = -6\), \(y = 6\)</td>
<td>
Incorrect. Let’s try again in a different way: Did you solve the first equation for \(y\)? Did you substitute \(-2x-4\) into the second equation for \(y\)? The answer is \(x =-2 \) and \(y = 0\).<br>
</td>
</tr>
</tbody>
</table><br>
<!-- END SELF CHECK TABLE -->
<h3>2.3.3: Additional Resources</h3>
<p><em><strong>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.</strong></em></p>
<h3> Reinforcing the Substitution Method</h3>
<p >Solve the system by substitution:</p>
<!--BEGIN SYSTEM OF EQUATIONS-->
<p>\(\left\{\begin{array}{l}4x\;+\;2y\;=\;4\\6x\;+\;-y\;=\;8\end{array}\right.\)</p>
<!--END SYSTEM OF EQUATIONS -->
<p>We need to solve one equation for one variable. We will solve the first equation for \(y\).</p>
<!--BEGIN STEPS-->
<div>
<p><strong>Step 1 - </strong>Solve the first equation for \(y\).<br></p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
4x+2y&=&4
\\2y&=&-4x+4
\\y&=&-2x+2
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<p><strong>Step 2 - </strong>Substitute \(−2x+2\) for \(y\) in the second equation. <br></p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
6x-y&=&8
\\6x-(-2x+2)&=&8
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<p><strong>Step 3 - </strong>Solve the equation for \(x\).</p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
6x-(-2x+2)&=&8
\\6x+2x-2&=&8
\\8x-2&=&8
\\8x&=&10
\\x&=&\frac{10}{8}
\\x&=&\frac54
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<p><strong>Step 4 - </strong>Substitute \(x=\frac54\) into \(4x+2y=4\) to find \(y\).<br></p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
4x+2y&=&4
\\ 4(\frac54)+2y&=&4
\\5+2y&=&4
\\2y&=&-1
\\y&=&\frac{-1}{2}
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<p><strong>Step 5 - </strong>Check the ordered pair. \((\frac54,\frac{-1}{2})\) in both equations.<br></p>
<!--BEGIN SIDE BY SIDE EQUATIONS-->
<p>\(\begin{array}{rcr}
& & 6x-y=8\\
& & 6(\frac54)-(-\frac12)\overset?=8\\
4x+2y=4 & & \frac{30}{4}-(-\frac12)\overset?=8\\
4(-\frac12)+2(\frac54)\overset?=4 & & \frac{15}{2}-(-\frac12)\overset?=8\\
5-1\overset?=4 & & \frac{15}{2}+\frac12\overset?=8 \\
4=4\checkmark & & 8=8\checkmark\\
\end{array}\) </p>
<!--END SIDE BY SIDE EQUATIONS -->
</div>
<!-- END STEPS -->
<h4>Try It: Reinforcing the Substitution Method</h4>
<br>
<p>Solve the system by substitution:</p>
<!--BEGIN SYSTEM OF EQUATIONS-->
<p>\(\left\{\begin{array}{l}x\;-\;4y\;=\;-4\\-3x\;+\;4y\;=\;0\end{array}\right.\)</p>
<!--END SYSTEM OF EQUATIONS -->
<p><strong>Answer: </strong></p>
<p>Your answer may vary, but here is a sample.</p>
<p><strong>Step 1 - </strong>Solve the first equation for \( x \) to isolate a variable.<br></p>
<p>\(\begin{array}{rcl}\;\;x\;-\;4y\;&=&\;-4\\x\;-\;4y\;{\style{color:red}+}\;{\style{color:red}4}{\style{color:red}y}\;&=&\;-4\;{\style{color:red}+}\;{\style{color:red}4}{\style{color:red}y}\\x\;&=&\;-4\;+\;4y\end{array}\)</p>
<p><strong>Step 2 - </strong> Replace the \(x\) in the second equation with the expression \( -4 + 4y \).<br></p>
<p>\(\begin{array}{rcl}-3x\;+\;4y\;&=&\;0\\-3{\style{color:red}(}{\style{color:red}-}{\style{color:red}4}\;{\style{color:red}+}\;{\style{color:red}4}{\style{color:red}y}{\style{color:red})}\;+\;4y\;&=&\;0\end{array}\)</p>
<p><strong>Step 3 - </strong> Solve the equation with just one variable.</p>
<p>\(\begin{array}{rcl}-3(-4\;+\;4y)\;+\;4y\;&=&\;0\\12 - 12y\;+\;4y\;&=&\;0\\-8y\;&=&\;-12\\y\;&=&\;\;\frac32\end{array}\)</p>
<p><strong>Step 4 - </strong> Use the first equation and replace \(y\) with \(\frac32\).</p>
<p>\(\begin{array}{rcl}x\;-\;4y\;&=&\;-4\\x\;-\;4(\frac32)\;&=&\;-4\\x\;-\;6\;&=&\;-4\\x\;&=&\;2\end{array}\)</p>
<p><strong>Step 5 - </strong> The ordered pair is \((x, y)\).</p>
<p> \((2,\frac32)\)</p>
<p><strong>Step 6 - </strong> Substitute \( x=6 \), \( y=1 \) to check the solution in both equations.</p>
<p>\(\begin{array}{r}x\;-\;4y\;=\;-4\\\;{\style{color:red}(}{\style{color:red}2}{\style{color:red})}\;-\;4{\style{color:red}(}\frac{\style{color:red}3}{\style{color:red}2}{\style{color:red})}\stackrel{?}{=}-4\\\;2\;-\;6\stackrel{?}{=}-4\\-4\;=\;-4\;{\style{color:red}✔}\end{array}\)<br></p>
<p><br></p>
<p>\(\begin{array}{r}-3x\;+\;4y\;=\;0\\\;-3{\style{color:red}(}{\style{color:red}2}{\style{color:red})}\;-\;4{\style{color:red}(}\frac{\style{color:red}3}{\style{color:red}2}{\style{color:red})}\stackrel{?}{=}0\\\;-6\;+\;6\stackrel{?}{=}0\\0\;=\;0\;{\style{color:red}✔}\end{array}\)
<!-- END TRY IT -->