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<h2>Activity (20 minutes)</h2>
<p>In this activity, students write and graph equations to represent two constraints in the same situation and use tables and graphs to see possible values that satisfy the constraints. The work prompts them to think about a pair of values that simultaneously meets multiple constraints in the situation, which in turn helps them make sense of the phrase “a solution to both equations.”</p>
<p>Monitor for these likely strategies for completing the tables, from less precise to more precise:</p>
<ul>
<li>Guessing and checking</li>
<li>Using the graph (by identifying the point with a given \( x \)- or \( y \)-value and estimating the unknown value, or using technology to get an estimate)</li>
<li>Substituting the given value of one variable into the equation and solving for the other variable</li>
</ul>
<p>Identify students who use each strategy and ask them to share later.</p>
<h3>Launch</h3>
<p>Arrange students in groups of 2 and provide access to graphing technology. Explain to students that they will now examine the same situation involving two quantities (raisins and walnuts), but there are now two constraints (cost and weight).</p>
<p>Give students 3–4 minutes of quiet time to complete the first set of questions about the cost constraint, and then pause for a whole-class discussion. Select previously identified students to share their strategies, in the order listed in the Activity Narrative. If one of the strategies is not mentioned, bring it up. </p>
<p>Ask a student who used the graph to complete the table to explain how exactly they used the graph. For example: “How did you use the graph of \( 4x+8y=15 \) to find \( y \) when \( x \) is 2?” Make sure students recognize that this typically involves hovering over different points on the graph or tracing the graph to get the coordinates. </p>
<p>Ask students to proceed with the remainder of the activity. If time permits, pause again after the second set of questions (about the weight constraint) to discuss how the strategies for completing the second table are like or unlike those for completing the first table. Alternatively, give students a minute to confer with their partner before moving on to the last question.</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">Representation: Internalize Comprehension </p>
</div>
<div class="os-raise-extrasupport-body">
<p>Provide appropriate reading accommodations and supports to ensure student access to written directions, word problems, and other text-based content.</p>
<p class="os-raise-text-italicize">Supports accessibility for: Language; Conceptual processing</p>
</div>
</div>
<br>
<h4>Student Activity</h4>
<p>Use the situation you saw earlier to answer questions 1 -3.</p>
<blockquote>
<p>Diego bought some raisins and walnuts to make trail mix. Raisins cost $4 a pound, and walnuts cost $8 a pound. Diego spent $15 on both ingredients.</p>
</blockquote>
<ol class="os-raise-noindent">
<li> Write an equation to represent this constraint. Let \(x\) be the pounds of raisins and \(y\) be the pounds of walnuts. </li>
</ol>
<p><strong>Answer:</strong> </p>
<p>4x+8y=15</p>
<ol start="2">
<li> Use graphing technology to graph the equation. (Students were provided access to Desmos.) </li>
</ol>
<p><strong>Answer:</strong></p>
<img src="https://k12.openstax.org/contents/raise/resources/d965bc3596a4c65d4f18c5cca702881bf889798c" alt="Graph of a line. walnuts (pounds), raisin (pounds)."/> <br>
<br>
<ol start="3">
<li> Complete the table with the amount of one ingredient Diego could have bought given the other. Be prepared to show your reasoning. </li>
</ol>
<table class="os-raise-midsizetable">
<thead>
<tr>
<th scope="col">Raisins (lbs)</th>
<th scope="col">Walnuts (lbs)</th>
</tr>
<tr>
<td>0</td>
<td></td>
</tr>
</thead>
<tbody>
<tr>
<td>0.25</td>
<td></td>
</tr>
<tr>
<td></td>
<td>1.375</td>
</tr>
<tr>
<td></td>
<td>1.25</td>
</tr>
<tr>
<td>1.75</td>
<td></td>
</tr>
<tr>
<td>3</td>
<td></td>
</tr>
</tbody>
</table>
<br>
<p><strong>Answer:</strong></p>
<table class="os-raise-midsizetable">
<thead>
<tr>
<th scope="col">Raisins (lbs)</th>
<th scope="col">Walnuts (lbs)</th>
</tr>
<tr>
<th scope="col">0</th>
<th scope="col">1.875</th>
</tr>
</thead>
<tbody>
<tr>
<td>0.25</td>
<td>1.75</td>
</tr>
<tr>
<td>1</td>
<td>1.375</td>
</tr>
<tr>
<td>1.25</td>
<td>1.25</td>
</tr>
<tr>
<td>1.75</td>
<td>1</td>
</tr>
<tr>
<td>3</td>
<td>0.375</td>
</tr>
</tbody>
</table>
<br>
<p>Use the following new piece of information to answer questions 4 - 6.</p>
<p>Diego bought a total of 2 pounds of raisins and walnuts combined.</p>
<ol class="os-raise-noindent" start="4">
<li> Write an equation to represent this constraint. Let \(x\) be the pounds of raisins and \(y\) be the pounds of walnuts. </li>
</ol>
<p><strong>Answer:</strong> \(x+y=2\)</p>
<ol class="os-raise-noindent" start="5">
<li> Use graphing technology to graph the equation. (Students were provided access to Desmos.) </li>
</ol>
<p><strong>Answer:</strong></p>
<img src="https://k12.openstax.org/contents/raise/resources/a126efd0b1465dbec019f08b2165a251e7afb50d" alt="Graph of a line. walnuts (pounds), raisin (pounds)."/> <br>
<br>
<ol class="os-raise-noindent" start="6">
<li> Complete the table with the amount of one ingredient Diego could have bought given the other. Be prepared to show your reasoning. </li>
</ol>
<table class="os-raise-midsizetable">
<thead>
<tr>
<th scope="col">Raisins (lbs)</th>
<th scope="col">Walnuts (lbs)</th>
</tr>
<tr>
<td>0</td>
<td></td>
</tr>
</thead>
<tbody>
<tr>
<td>0.25</td>
<td></td>
</tr>
<tr>
<td></td>
<td>1.375</td>
</tr>
<tr>
<td></td>
<td>1.25</td>
</tr>
<tr>
<td>1.75</td>
<td></td>
</tr>
<tr>
<td>3</td>
<td></td>
</tr>
</tbody>
</table>
<br>
<p><strong>Answer:</strong></p>
<table class="os-raise-midsizetable">
<thead>
<tr>
<th scope="col">Raisins (lbs)</th>
<th scope="col">Walnuts (lbs)</th>
</tr>
<tr>
<th scope="col">0</th>
<th scope="col">2</th>
</tr>
</thead>
<tbody>
<tr>
<td>0.25</td>
<td>1.75</td>
</tr>
<tr>
<td>0.625</td>
<td>1.375</td>
</tr>
<tr>
<td>0.75</td>
<td>1.25</td>
</tr>
<tr>
<td>1.75</td>
<td>0.25</td>
</tr>
<tr>
<td>3</td>
<td>-1 (not possible)</td>
</tr>
</tbody>
</table>
<br>
<p>Use the following information to answer questions 7-9.</p>
<p>Diego spent $15 and bought exactly 2 pounds of raisins and walnuts.</p>
<ol class="os-raise-noindent" start="7">
<li> How many pounds of raisins did he buy? </li>
</ol>
<p><strong>Answer:</strong> 0.25</p>
<ol class="os-raise-noindent" start="8">
<li> How many pounds of walnuts did he buy? </li>
</ol>
<p><strong>Answer:</strong> 1.75</p>
<ol class="os-raise-noindent" start="9">
<li> Explain or show how you know how many pounds he purchased of each type. </li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here are some samples.</p>
<ul>
<li> It’s the only pair of values that appears in both tables. </li>
<li> If the two equations are graphed on the same coordinate plane, it’s where the two graphs intersect, so the pair of values meets both requirements. </li>
</ul>
<img src="https://k12.openstax.org/contents/raise/resources/ae6240fe3890762ae7950155690f4d3f8186d0b9" alt="Graph of a linear system. walnuts (pounds), raisin (pounds)."/> <br>
<br>
<h3>Anticipated Misconceptions</h3>
<p>Some students might think that the values in the second table need to reflect a total cost of $15. Clarify that the table represents only the constraint that Diego bought a total of 2 pounds of raisins and walnuts.</p>
<p>The idea of finding an \( (x,y) \) pair that satisfies multiple constraints should be familiar from middle school. If students struggle to answer the last question, ask them to study the values in the table. Ask questions such as: “If Diego bought 0.25 pound of raisins, would he meet both the cost and weight requirements?” and “Which combinations of raisins and walnuts would allow him to meet both requirements? How many combinations are there?”</p>
<h3>Activity Synthesis</h3>
<p>Invite students to share their response and reasoning for the last question. Display the graphs representing the system (either created by a student or as shown in the Student Response). Discuss with students:</p>
<ul>
<li>“Can you find other combinations of raisins and walnuts, besides 0.25 pound and 1.75 pounds, that meet both cost and weight constraints?” (No)</li>
<li>“How many possible combinations of raisins and walnuts meet both constraints? How do we know?” (One combination, because the graphs intersect only at one point.)</li>
</ul>
<p>Explain to students that the two equations written to represent the constraints form a<em> </em><strong>system of equations</strong>. We use a curly bracket to indicate a system, like this: </p>
<p>\( \begin{cases}\begin {align} 4x + 8y &= 15\\ x+ \hspace{2mm}y&=2 \end{align} \end{cases} \)</p>
<p>Highlight that the solution to the system is a pair of values (in this case, pounds of raisins and walnuts) that meet both constraints. This means the pair of values is a solution to both equations. Graphing is an effective way to see the solution to both equations, if one exists.</p>
<h2>2.1.2: Self Check</h2>
<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>A school group bought 12 tickets to visit a history museum.</p>
<ul>
<li>Adult tickets are $5 each.</li>
<li>Student tickets are $8 each.</li>
</ul>
<p>The school paid a total of $90. How many adult tickets, \( x \), and student tickets, \( y \), did the school buy?</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">Answers</th>
<th scope="col">Feedback</th>
</tr>
</thead>
<tbody>
<tr>
<td> 5 student tickets and 10 adult tickets </td>
<td> Incorrect. Let’s try again in a different way: Does the number of tickets bought equal 12? \( 5 + 10 = 15 \), not 12. The answer is 10 student tickets and 2 adult tickets. </td>
</tr>
<tr>
<td> 9 student tickets and 3 adult tickets </td>
<td> Incorrect. Let’s try again in a different way: Does the total spent equal $90? \( 9(8) + 3(5) = 87 \), not 90. The answer is 10 student tickets and 2 adult tickets. </td>
</tr>
<tr>
<td> 10 student tickets and 2 adult tickets </td>
<td> Correct! Check yourself: 10 student tickets and 2 adult tickets totals 12 tickets and costs $90. </td>
</tr>
<tr>
<td> 11 student tickets and 1 adult ticket </td>
<td> Incorrect. Let’s try again in a different way: Does the total spent equal $90? \( 11(8) + 1(5) = 93 \), not 90. The answer is 10 student tickets and 2 adult tickets. </td>
</tr>
</tbody>
</table>
<br>
<h3>2.1.2: 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>Solve Systems of Linear Equations </h4>
<p>In Unit 1, you learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations.</p>
<p>An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.</p>
<p>\(\left\{\begin{array}{l}2x\;+\;y\;=\;7\\x\;-\;2y\;=\;6\end{array}\right.\)</p>
<p>A linear equation in two variables, such as \( 2x + y = 7 \), has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation, and every solution to the equation is a point on the line.</p>
<p>To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.</p>
<p>To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs \((x, y)\) that make both equations true. These are called the solutions to a system of equations.</p>
<div class="os-raise-graybox">
<p><strong>Determining if Ordered Pairs Solve a System </strong><br>
<br>
<strong>Step 1</strong> - Substitute the values of both variables into the first equation and simplify. Verify if the equation is true.<br>
<strong>Step 2</strong> - Substitute the values of both variables into the second equation and simplify. Verify if the equation is true. <br>
<strong>Step 3</strong> - If the ordered pair makes both equations true, then it is a solution to the system. </p>
</div>
<br>
<h5>Example 1</h5>
<p>Determine whether the ordered pair \( (-2, -1) \) is a solution to the system\( \left\{\begin{array}{l}x\;-\;y\;=\;-1\\2x\;-\;y\;=\;-5\end{array}\right. \).</p>
<br>
<p>Here is how to do this:</p>
<p>\(\left\{\begin{array}{c}x-y=-1\\2x-y=-5\end{array}\right.\)</p>
<p>We substitute \(x=-2\) and \(y=-1\) into both equations. </p>
<p>\(\begin{array}{rr}
x-y=-1 & 2x-y=-5 \\
-2-(-1) \stackrel{?}{=} -1 & 2(-2)-(-1)\stackrel{?}{=} -5 \\
-1=-1 \checkmark & -3 \neq -5
\end{array}\) </p>
<p>\((-2,-1)\) does not make both equations true. </p>
<p> \((-2,-1)\) is not a solution. </p>
<br>
<h5>Example 2</h5>
<p>Determine whether the ordered pair \( (-4, -3) \) is a solution to the system\( \left\{\begin{array}{l}x\;-\;y\;=\;-1\\2x\;-\;y\;=\;-5\end{array}\right. \).</p>
<br>
<p>Here is how to do this:</p>
<p>We substitute \(x=-4\) and \(y=-3\) into both equations.</p>
<p> \(\begin{array}{rr}x-y=-1 & 2x-y=-5 \\
-4-(-3) \stackrel{?}{=} -1 & 2(-4)-(-3) \stackrel{?}{=} -5 \\
-1=-1 \checkmark & -5=-5 \checkmark \end{array}\) </p>
<p>\((-4,-3)\) is true for both equations. It is a solution.</p>
<br>
<h3>Try It: Solve Systems of Linear Equations</h3>
<p>Determine whether the ordered pair is a solution to the system</p>
<p>\(3x+y=0\)</p>
<p>\(x+2y=−5\)</p>
<ol class="os-raise-noindent">
<li> Is \((1, -3)\) a solution to the system of equations? </li>
</ol>
<p><strong>Answer:</strong> Yes</p>
<ol class="os-raise-noindent" start="2">
<li> Explain how you know if it is or is not a solution of the system. </li>
</ol>
<p><strong>Answer:</strong> When \((1, -3)\) is substituted into the equations of the system, it makes both of them true.</p>
<p>\(\begin{array}{ll}
3(1)+-3=0 & 1+2(-3)=-5 \\
3−3=0 & 1−6=−5 \\
0=0 & −5=−5
\end{array}\)</p>
<ol class="os-raise-noindent" start="3">
<li> Is \((0, 0)\) a solution to the system of equations? </li>
</ol>
<p><strong>Answer:</strong> No</p>
<ol class="os-raise-noindent" start="4">
<li> Explain how you know if it is or is not a solution of the system. </li>
</ol>
<p><strong>Answer:</strong> When \((0, 0)\) is substituted into the equations of the system, it only makes one of the equations true, not both of them.</p>
<p>\(\begin{array}{ll}
3(0)+0=0 & 0+2(0)=−5 \\
0+0=0 & 0+0=−5 \\
0=0 & 0 \neq −5
\end{array}\)</p>
<br>
<h3>Solve a System of Linear Equations by Graphing</h3>
<p>The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And by finding what the lines have in common, we’ll find the solution to the system.</p>
<p><strong>Example</strong><br>
Solve the system by graphing \(\left\{\begin{array}{l}2x\;+\;y\;=\;7\\x\;-\;2y\;=\;6\end{array}\right.\)</p>
<p><b>Step 1 -</b> Graph the first equation.</p>
<p>To graph the
first line, write the equation in slope-intercept form.
\(2x+y=7\) <br>
\(y=-2x+7\) <br>
So, the slope is -2 and the \(y\)-intercept is 7. </p>
<img alt class="img-fluid atto_image_button_text-bottom" height="204" role="presentation" src="https://k12.openstax.org/contents/raise/resources/6c0b6d4112ab008727d6aab970f577aed3940ecd" width="200">
</p>
<p><b>Step 2 -</b> Graph the second equation on the same rectangular coordinate system. </p>
<p>This line may be easier to graph using the intercepts. <br>
<br>
To find the \(x\)-intercept, substitute \(y = 0\) into the equation.
\(x-2y=6\)<br>
\(x-2(0)=6\)<br>
\(x=6\)<br>
<br>
To find the \(y\)-intercept, substitute \(x = 0\) into the equation. <br>
\(x - 2y = 6\)<br>
\(0 - 2y = 6\)<br>
\(-2y = 6\)<br>
\(y = -3\)<br>
So, the intercepts are \((6,0)\) and \((0,-3)\).<br>
<img alt class="img-fluid atto_image_button_text-bottom" height="204" role="presentation" src="https://k12.openstax.org/contents/raise/resources/c4e74281a1dfc70805384671b85e95fc5437bd07" width="200"> </p>
<p><b>Step 3 -</b> Determine whether the lines intersect, are parallel, or are the same line. </p>
<p>Look at the graph of the lines.
In this case, the lines intersect. </p>
<p><b>Step 4</b>-
Estimate the solution to the system. </p>
<ul>
<li>If the lines intersect, identify the point of
intersection. Check to make sure it is a solution to both equations. This is
the solution to the system.<br>
</li>
<li>If the lines are parallel, the system has no
solution.<br>
</li>
<li>If the lines are the same, the system has an infinite
number of solutions. </li>
</ul>
<p>The lines intersect at \((4,-1)\).</p>
<p><strong>Step 5 </strong>- Check the solution in both equations.</p>
<div class="os-raise-d-flex os-raise-justify-content-evenly">
<p> \(\begin{array}{l}2x+y=7\\2(4)+(-1)\overset?=\;7\\8-1\overset?=7\\7=7✔\end{array}\)</p>
<p> \(\begin{array}{l}x-2y=6\\4-2(-1)\overset?=\;6\\6=6✔\end{array}\)</p>
</div>
<br>
<p>The steps to use to solve a system of linear equations by graphing are shown here.</p>
<div class="os-raise-graybox">
<p><strong>Using a Graph to Solve a System</strong></p>
<p><strong>Step 1</strong> - Graph the first equation. </p>
<p><strong>Step 2</strong> - Graph the second equation on the same rectangular coordinate system. </p>
<p><strong>Step 3</strong> - Determine whether the lines intersect. </p>
<p><strong>Step 4</strong> - Estimate the solution to the system. </p>
<p> If the lines intersect, identify the point of intersection. This is the solution to the system. </p>
<p><strong>Step 5</strong> - Check the solution in both equations.</p>
</div>
<br>
<h3>Try It: Solve a System of Linear Equations by Graphing</h3>
<p> Solve the system by graphing<br>
\(\left\{\begin{array}{l}x\;-\;3y\;=\;-3\\x\;+\;y\;=\;5\end{array}\right.\)</p>
<p><strong>Answer:</strong></p>
<p>Here is how to graph the system:</p>
<p><strong>Step 1</strong> - Graph the first equation. <br>
<img alt="GRAPH OF A LINE WITH \(y\)-intercepts OF 1 AND \(x\)-intercepts OF −3.
" class="img-fluid atto_image_button_text-bottom" height="337" src="https://k12.openstax.org/contents/raise/resources/73a9cf97fb40f2375b1af2f999af5131a6545627" width="250"><br>
<br>
<strong>Step 2 -</strong> Graph the second equation on the same rectangular coordinate system.<br>
<img alt="GRAPH OF A LINE WITH \(y\)-intercepts OF 1 AND \(x\)-intercepts OF −3. GRAPH OF A SECOND LINE WITH \(y\)-intercepts OF 5 AND \(x\)-intercepts OF " class="img-fluid atto_image_button_text-bottom" height="241" src="https://k12.openstax.org/contents/raise/resources/0f21b31e1111536a59875dda19c5712005405efd" width="250"></p>
<p><strong>Step 3 - </strong> Determine whether the lines intersect. <br>
The lines intersect at one point. </p>
<p><strong>Step 4 - </strong> Estimate the solution to the system. </p>
<p> If the lines intersect, identify the point of intersection. This is the solution to the system. </p>
<p>The lines intersect at \((3, 2)\). </p>
<p><strong>Step 5 -</strong> Check the solution in both equations.</p>
<div class="os-raise-d-flex os-raise-justify-content-evenly">
<p>\(\begin{array}{l}x-3y=-3\\3-3(2)\;\overset?=-3\\3-6\overset?=-3\\-3=-3{\style{color:red}✔}\end{array}\)</p>
<p> \(\begin{array}{l}x+y=5\\3+2\;\overset?=5\\5=5{\style{color:red}✔}\end{array}\)</p>
</div>