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<h4>Activity (15 minutes)</h4>
<p>In this activity, students see the same four pairs of equations as in the warm up. This time, their job is to find a way to solve the systems. Some students may choose to solve by graphing, but the systems lend themselves to being solved efficiently and precisely by substitution.</p>
<p>As students work, pay attention to the methods students use to solve the systems. Identify those who solve by substitution by replacing a variable or an expression in one equation with an equal value or equivalent expression from the other equation. Ask these students to share later.</p>
<h3>Launch</h3>
<p>Arrange students in groups of 2. Give students 6–8 minutes of quiet time to solve as many systems as they can and then a couple of minutes to share their responses and strategies with their partner.</p>
<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">Engagement: Internalize Self Regulation</p>
</div>
<div class="os-raise-extrasupport-body">
<p>
Chunk this task in half to differentiate the degree of difficulty or complexity. Present and discuss the first two problems before showing the last two. Support students in metacognitive development by having students articulate the steps they used to solve by substitution in these more straightforward examples. Record the steps on a chart or whiteboard to encode the process for later retrieval. Then, reveal the remaining problems. </p>
<p class="os-raise-text-italicize">Supports accessibility for: Organization; Attention</p>
</div>
</div>
<br>
<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>
<p>\( \begin{cases} x+2y=8 \\ x=-5 \end{cases} \)</p>
</blockquote>
<ol class="os-raise-noindent">
<li>What is the \(x\) value?</li>
</ol>
<p><strong>Answer: </strong> -5</p>
<!--TG QUESTION Answer FORMAT -->
<ol class="os-raise-noindent" start="2">
<li>What is the \(y\) value?</li>
</ol>
<p><strong>Answer: </strong>6.5 </p>
<ol class="os-raise-noindent" start="3">
<li>Explain your solution.</li>
</ol>
<p><strong>Answer: </strong> Compare your Answer: Your answer may vary, but here is a sample.</p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
-5+2y&=&8
\\ 2y&=&13
\\ y&=&6.5
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
x+2y&=&8
\\ x+2(6.5)&=&8
\\ x+13&=&8
\\ x&=&-5
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN SIDE BY SIDE EQUATIONSS-->
<p>\(\begin{array}{rcr}
-5+2(6.5) & &\\
-5+13\overset?=8 & & \\
8=8\checkmark& & -5=-5\checkmark
\end{array}\) </p>
<!--END SIDE BY SIDE EQUATIONS -->
<br>
<p>Use this system of equations to answer questions 4–6.</p>
<blockquote>
<p>\( \begin{cases} y=-7x+13 \\ y=-1 \end{cases} \)</p>
</blockquote>
<ol class="os-raise-noindent" start="4">
<li>What is the \(x\) value?</li>
</ol>
<p><strong>Answer: </strong>2 </p>
<br>
<ol class="os-raise-noindent" start="5">
<li>What is the \(y\) value?</li>
</ol>
<p><strong>Answer: </strong>-1</p>
<ol class="os-raise-noindent" start="6">
<li>Explain your solution.</li>
</ol>
<p><strong>Answer: </strong> </p>
<p>Compare your Answer: Your answer may vary, but here is a sample.</p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
-1&=&-7x+13
\\ -14&=&-7x
\\ 2&=&x
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
y&=&-7(2)+13
\\ y&=&-14+13
\\ y&=&-1
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN SIDE BY SIDE EQUATIONSS-->
<p>\(\begin{array}{rcr}
-1=-7(2)+13 & & \\
-1\overset?=-14+13 & & \\
-1=-1\checkmark& & -1=-1\checkmark
\end{array}\) </p>
<!--END SIDE BY SIDE EQUATIONS -->
<br>
<p>Use this system of equations to answer questions 7–9.</p>
<blockquote>
<p>\( \begin{cases} 3x=8 \\ 3x+y=15 \end{cases} \)</p>
</blockquote>
<br>
<!--Q7-->
<ol class="os-raise-noindent" start="7">
<li>What is the \(x\) value?</li>
</ol>
<p><strong>Answer:</strong> \(\frac83\)</p>
<ol class="os-raise-noindent" start="8">
<li>What is the \(y\) value?</li>
</ol>
<p><strong>Answer: </strong>7</p>
<p>Correct! 7</p>
<ol class="os-raise-noindent" start="9">
<li>Explain your solution.</li>
</ol>
<p>Compare your Answer: Your answer may vary, but here is a sample.</p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
3x&=&8
\\ x&=&\frac83
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
3(\frac83)+y&=&15
\\ 8+y&=&15
\\ y&=&=7
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN SIDE BY SIDE EQUATIONSS-->
<p>\(\begin{array}{rcr}
3(\frac83)+7=15 & &\\
-8+7\overset?=15 & & 3(\frac83)\overset?=8 \\
15=15\checkmark & & 8=8\checkmark
\end{array}\) </p>
<!--END SIDE BY SIDE EQUATIONS -->
<br>
<p>Use this system of equations to answer questions 10–12.</p>
<blockquote>
<p>\( \begin{cases} y = 2x - 7\\4 + y = 12 \end{cases} \)</p>
</blockquote>
<ol class="os-raise-noindent" start="10">
<li>What is the \(x\) value?</li>
</ol>
<p><strong>Answer: </strong>7.5 </p>
<ol class="os-raise-noindent" start="11">
<li>What is the \(y\) value?</li>
</ol>
<p><strong>Answer: </strong>8 </p>
<ol class="os-raise-noindent" start="12">
<li>Explain your solution.</li>
</ol>
<p><strong>Answer: </strong> </p>
<p>Compare your Answer: Your answer may vary, but here is a sample.</p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
4+y&=&12
\\ y&=&8
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}
8&=&2x-7
\\ 15&=&2x
\\ 7.5&=&x
\\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<br>
<!--BEGIN SIDE BY SIDE EQUATIONSS-->
<p>\(\begin{array}{rcr}
8=2(7.5)-7 & & \\
8=2(7.5)-7 & & 4+y=12\\
8\overset?=15-7 & & 4+8\overset?=12\\
8=8\checkmark & & 12=12\checkmark
\end{array}\) </p>
<h3>Anticipated Misconceptions</h3>
<p>Some students may not remember to find the value of the second variable after finding the first. They may need a reminder that the solution to a system of linear equations is a pair of values.</p>
<p>If some students struggle with the last system because the variable that is already isolated is equal to an expression rather than a number, ask what they would do if the first equation was \(y = a \) \(number\) instead of \( y = 2x − 7 \).</p>
<p>If students don’t know how to approach the last system, ask them to analyze both equations and see if the value of one of the variables could be found easily.</p>
<h3> Activity Synthesis</h3>
<p>Select previously identified students to share their responses and strategies. Display their work for all to see. Highlight the strategies that involve substitution and name them as such.</p>
<p>Make sure students see that the last two equations can be solved by substituting in different ways. Here are two ways for solving the third system, \(\left\{\begin{array}{l}3x\;=\;8\\3x\;+\;y\;=\;15\end{array}\right.\), by substitution:</p>
<p>Finding the value of \(x\) and substituting it into \(3x + 8 = 15\):</p>
<p>\(\begin{array}{rcl}3x\;&=&\;8\\x\;&=&\;\frac83\\&&\\3x\;+\;y\;&=&\;15\\3(\frac83)\;+\;y\;&=&\;15\\8\;+\;y\;&=&\;15\\y\;&=&\;7\end{array}\)</p>
<br>
<p>Substituting the value of 3x into \(3x + 8 = 15\):</p>
<p>\(\begin{array}{rcl}3x\;+\;y\;&=&\;15\\8\;+\;y\;&=&\;15\\y\;&=&\;7\\&&\end{array}\)</p>
<p>Here are two ways of solving the last system, \(\left\{\begin{array}{l}y\;=\;2x\;-7\\4\;+\;y\;=\;12\end{array}\right.\), by substitution:</p>
<p>Substituting \(2x − 7\) for \(y\) in the equation \(4 + y = 12\):</p>
<p>\(\begin{array}{rcl}4\;+\;y\;&=&\;12\\4\;+\;(2x\;-\;7)\;&=&\;12\\4\;+\;2x\;-\;7\;&=&\;12\\2x\;-7\;+\;4\;&=&\;12\\2x\;-\;3\;&=&\;12\\2x\;&=&\;15\\x\;&=&\;7.5\end{array}\)</p>
<p>\(\begin{array}{rcl}y\;&=&\;2x\;-\;7\\y\;&=&\;2(7.5)\;-\;7\\y\;&=&\;15\;-\;7\\y\;&=&\;8\end{array}\)</p>
<p>Rearranging or solving \( 4 + y = 12 \) to get \( y = 8 \) and then substituting 8 for \( y \) in the equation \( y = 2x − 7 \):</p>
<p>\(\begin{array}{rcl}y\;&=&\;2x\;-\;7\\8\;&=&\;2x\;-\;7\\15\;&=&\;2x\\7.5\;&=&\;x\end{array}\)</p>
<p>In each of these two systems, students are likely to notice that one way of substituting is much quicker than the other. Emphasize that when one of the variables is already isolated or can easily be isolated, substituting the value of that variable (or the expression that is equal to that variable) into the other equation in the system can be an efficient way to solve the system.</p>
<h3>2.3.2: Self Check</h3>
<!-- Begin Self Check Table -->
<p class="os-raise-text-bold"><em>After the activity, students will answer the following question to check their understanding of the concepts explored in the activity.</em></p>
<p class="os-raise-text-bold">QUESTION:</p>
<p>Solve the following system of equations. Check to see if your solution makes each equation true.</p>
<p>\(\left\{\begin{array}{l}3x\;+\;y\;=\;6\\x\;+\;2y\;=\;7\end{array}\right.\)</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">Answers</th>
<th scope="col">Feedback</th>
</tr>
</thead>
<tbody>
<tr>
<td>\( x = 3 \), \( y = 1 \)</td>
<td>Incorrect. Let’s try again in a different way: Did you switch the values for each variable? The answer is \(x = 1\) and \(y = 3\). <br></td>
</tr>
<tr>
<td>\( x = -3 \), \( y = 5 \)</td>
<td>Incorrect. Let’s try again a different way: This point is located above the line that is graphed. When a
point is a solution to an equation, it must be located on the graph of the equation. The answer is \( (3, 1) \). </td>
</tr>
<tr>
<td>\( x = 3 \), \( y = -3 \)</td>
<td>Incorrect.
Let’s try again a different way: To get to \((1, 3)\), you move 1 right from the origin and 3 up, so this
point is not on the line. When a point is a solution to an equation, it must be located on the graph of the
equation. The answer is \( (3, 1) \).</td>
</tr>
<tr>
<td>\( x = 1 \), \( y = 3 \)</td>
<td>Incorrect. Let’s
try again a different way: To get to \((-3, 3)\), move 3 left and then 3 up. This point is not a solution to the
equation of the line graphed. When a point is a solution to an equation, it must be located on the graph of the
equation. The answer is \( (3, 1) \).</td>
</tr>
</tbody>
</table>
<br>
<!-- END SELF CHECK TABLE -->
<h3>2.3.2: Additional Resources</h3>
<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 Systems of Linear Equations</h4>
<p>Let’s solve this system of linear equations using the substitution method.</p>
<p>\(\left\{\begin{array}{l}2x\;+\;y\;=\;7\\x\;-\;2y\;=\;6\end{array}\right.\)</p>
<p>We will first solve one of the equations for either \(x\) or \(y\). We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.</p>
<p>Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!</p>
<p>After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.</p>
<br>
<div class="os-raise-graybox">
<h4>How to Solve a System of Equations by Substitution</h4>
<p><strong>Step 1 - </strong>Solve one of the equations for either variable.</p>
<p><strong>Step 2 - </strong>Substitute the expression from Step 1 into the other equation.</p>
<p><strong>Step 3 - </strong>Solve the resulting equation.</p>
<p><strong>Step 4 - </strong>Substitute the solution from Step 3 into one of the original equations to find the other variable.</p>
<p><strong>Step 5 - </strong>Write the solution as an ordered pair.</p>
<p><strong>Step 6 - </strong>Check that the ordered pair is a solution to both original equations.</p>
</div>
<br>
<p>Solve the system by substitution: \(\left\{\begin{array}{l}2x\;+\;y\;=\;7\\x\;-\;2y\;=\;6\end{array}\right.\)</p>
<h5>Solution</h5>
<p><strong>Step 1 - </strong>Solve one of the equations for either variable.</p>
<p>We’ll solve the first equation for y.</p>
<p>\(\begin{array}{c}\left\{\begin{array}{l}2x+y=7\\x-2y=6\end{array}\right.\end{array}\)</p>
<p>\( \begin{array}{rcl}2x+y&=&7\\y&=&7-2x\\&&\end{array}\)</p>
<p><strong>Step 2 - </strong>Substitute the expression from Step 1 into the other equation.</p>
<p>\( \begin{array}{rcl}x-2y&=&6\\x-2(7-2x)&=&6\\&&\end{array}\)</p>
<p><strong>Step 3 - </strong>Solve the resulting equation.</p>
<p>Now we have an equation with just 1 variable. We know how to solve this!</p>
<p>\( \begin{array}{rcl}x-2(7-2x)&=&6\\x-14+4x&=&6\\5x&=&20\\x&=&4\\&&\end{array}\)</p>
<p><strong>Step 4 - </strong>Substitute the solution from Step 3 into one of the original equations to find the other variable.</p>
<p>We’ll use the first equation and replace \(x\) with 4.</p>
<p>\(\begin{array}{rcl}2x+y&=&7\\2(4)+y&=&7\\8+y&=&7\\y&=&-1\\&&\end{array}\)</p>
<p><strong>Step 5 - </strong>Write the solution as an ordered pair.</p>
<p>The ordered pair is \((x,y)\).</p>
<p>\((4,-1)\)</p>
<p><strong>Step 6 - </strong>Check that the ordered pair is a solution to both original equations.</p>
<p>Substitute \(x=4\), \(y=-1\) into both equations and make sure they are both true.</p>
<!--BEGIN SIDE BY SIDE EQUATIONSS-->
<p>\(\begin{array}{rcr}
2x+y=7 & &x-2y=6 \\
2(4)+(-1)\overset?=7 & & 4-2(-1)\overset?=6 \\
7=7\checkmark & & 6=6\checkmark
\end{array}\) </p>
<!--END SIDE BY SIDE EQUATIONS -->
<p>Both equations are true.</p>
<p>\((4,-1)\) is the solution to the system.</p>
<br>
<h3>Try It: Solving Systems of Linear Equations</h3>
<br>
<!--Text Entry Interaction Start -->
<p>Solve the system by substitution, then choose the correct ordered pair. \(\left\{\begin{array}{l}-2x\;+\;y\;=\;-11\\x\;+\;3y\;=\;9\end{array}\right.\)</p>
<ul>
<li>\((9,-11)\) </li>
<li>\((1,6) \)</li>
<li>\((6,1)\) </li>
<li>\((-11,9)\) </li>
</ul>
<p><strong>Answer: </strong>\((6,1)\) </p>
<p>Compare your Answer: </p>
<p><strong>Step 1 - </strong>Solve the first equation for \( y \) to isolate a variable.</p>
<!--BEGIN ALIGN TO EQUALS-->
<p>\(\begin{array}{rcl}-2x + y &= &-11
\\ -2x\; {\style{color:red} + }\;{\style{color:red} 2}{\style{color:red} x } + y &=&-11\;{\style{color:red} + }{\style{color:red} 2}\;{\style{color:red}x }
\\ y&=&-11 + 2x \\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<p><strong>Step 2 - </strong>Replace the \(y\) in the second equation with the expression \( -11 + 2x \).</p>
<!--BEGIN ALIGN TO EQUALS-->
<p>\(\begin{array}{rcl}x + 3y&= &9
\\ x + 3\;{\style{color:red}(}{\style{color:red}–}{\style{color:red}1}{\style{color:red}1}\;{\style{color:red} +}\;{\style{color:red} 2}{\style{color:red}x}{\style{color:red})} &=&9
\\ x - 33 + 6x&=&9 \\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<p><strong>Step 3 - </strong>Solve the equation with just one variable.</p>
<p>Now we have an equation with just 1 variable. We know how to solve this!</p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}7x &= &42
\\ x&=&6 \\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<p><strong>Step 4 - </strong>Use the first equation and replace \( x \) with 6.</p>
<!--BEGIN ALIGN TO EQUALS -->
<p>\(\begin{array}{rcl}-2x + y &= &-11
\\ -2{\style{color:red}(}{\style{color:red}6}{\style{color:red})} + y &=&-11
\\ -12 + y&= &-11
\\ y&=&1 \\ \end{array}\)</p>
<!--END ALIGN TO EQUALS -->
<p><strong>Step 5 - </strong>The ordered pair is \((x, y)\).</p>
<p> \((6, 1)\)</p>
<p><strong>Step 6 - </strong>Substitute \( x=6 \), \( y=1 \) to check the solution in both equations.</p>
<!--BEGIN SIDE BY SIDE EQUATIONSS-->
<p>\(\begin{array}{rcr}
-2x + y = -11 & \\
-2{\style{color:red}(}{\style{color:red}6}{\style{color:red})} + {\style{color:red}(}{\style{color:red}1}{\style{color:red})}\overset?=-11 & & x + 3y = 9\\
-12 + 1\overset?=-11 & &{\style{color:red}(}{\style{color:red}6}{\style{color:red})} + 3{\style{color:red}(}{\style{color:red}1}{\style{color:red})}\overset?=9\\
-11 = -11{\style{color:red}\checkmark } & & 9 = 9{\style{color:red}\checkmark }
\end{array}\) </p>