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<h4>Activity (15 minutes)</h4>
<p>In this activity, students examine simple relationships that can each be expressed with many equations. Writing
multiple equations for the same relationship and doing so in context prepares students to later consider more
rigorously what makes equations equivalent.</p>
<p>As students write equations for the first relationship, identify students who use all numbers and those who use both
numbers and variables.</p>
<p>For the second question, students may explain Tyler’s claim concretely (using values of \( m \) and \( y \)) or
more abstractly (using the structure of the equations). For instance, they may reason that:</p>
<ul>
<li>When the middle child is 12, the youngest child is 7, and substituting \( m=12 \) and \( y=7 \) to \( 2m-2y \)
gives \( 2(12) - 2(7) \) or \( 24 - 14 \), which is 10. At all other ages of the two children, the expression \(
2m-2y \) always has a value of 10.</li>
<li>\( 2m-2y=10 \) is twice \( m-y=5 \). If the difference between \( m \) and \( y \) is 5, then twice the difference
between \( m \) and \( y \) must be twice 5, which is 10, so the two equations are still describing the same
relationship.
</li>
</ul>
<p>Identify students who reason in these ways so they can share later.</p>
<h4>Launch</h4>
<p>Give students a couple of minutes to answer the first question and ask them to pause for a discussion before moving
on.</p>
<p>Invite students to share their equations. Record and display the equations for all to see. Along the way, consider
organizing them into categories (numerical equations, equations in one variable, and equations in two variables).</p>
<p>Solicit from students some thoughts on the following questions. (It is not necessary to resolve the questions at the
moment.)</p>
<ul>
<li>“Consider the equations you wrote for each situation. Are they all equivalent? Why or why not?”</li>
<li>“What do you think it means for two equations to be equivalent?”</li>
</ul>
<p>Some students might say that each equation accurately represents the same relationship between the ages, so the
equations must be equivalent. Those who wrote variable equations might say that the equations are equivalent because
the same value for the variable makes each equation true. </p>
<p>After students have had a chance to consider these questions, prompt them to complete the remaining questions.</p>
<h4>Student Activity</h4>
<p>In questions 1-3:</p>
<blockquote><p>Write as many equations as possible that could represent the relationship between the ages of the
two children in each family described. Be prepared to explain what each part of your equation represents.
</p></blockquote>
<ol class="os-raise-noindent">
<li>In Family A, the youngest child is 7 years younger than the oldest, who is 18.
</li>
</ol>
<p><strong>Answer:</strong>
For \(y\), the age of the youngest child:
</p>
<ul>
<li>\( y=18-7\) </li>
<li>\(y+7=18\) </li>
<li>\(18-y=7\) </li>
</ul>
<p>Where 11 is the age of the youngest child:
</p>
<ul>
<li>\(18-7=11\) </li>
<li>\(11+7=18\) </li>
<li>\(18-11=7\) </li>
</ul>
<ol class="os-raise-noindent" start="2">
<li>In Family B, the middle child is 5 years older than the youngest child.
</li>
</ol>
<p><strong>Answer:</strong> For \(m\), the age of the middle child, and \(y\), the age of the youngest
</p>
<ul>
<li>\(m = y + 5\) </li>
<li>\(y = m - 5\) </li>
</ul>
<ol class="os-raise-noindent" start="3">
<li>Tyler thinks that the relationship between the ages of the children in Family B can be described with \(2m-2y=10\), where \(m\) is the age of the middle child and \(y\) is the age of the youngest. Describe how Tyler came up with this equation.
</li>
</ol>
<p><strong>Answer:</strong> </p>
<p>
The equation \(m=y+5\) represents the relationship of the ages in Family B, where \(m\) is the age of the middle child and \(y\) the age of the youngest. Subtracting \(y\) from both sides of the equation and multiplying both sides by 2 results in the equation: \(2m−2y=10\).
</p>
<ul>
<li>When the middle child is 12, the youngest child is 7, and that substituting \(m = 12\) and \(y = 7\) to \(2m - 2y\) gives \(2(12) - 2(7)\) or \(24-14\), which is 10. At all other ages of the two children, the expression \(2m - 2y\) always has a value of 10.</li>
<li>\(2m - 2y = 10\) is twice of \(m - y = 5\). If the difference between m and y is 5, then twice the difference between m and y must be twice of 5, which is 10, so the two equations are still describing the same relationship.</li>
</ul>
<ol class="os-raise-noindent" start="4">
<li><strong>Select three</strong> equations that are equivalent to \(3a + 6 = 15\).
</li>
</ol>
<ul>
<li>\(a + 3 =12\)</li>
<li>\(3a=9\)</li>
<li>\(a+2=5\)</li>
<li>\(\frac{1}{3} a =1\)</li>
</ul>
<p><strong>Answer:</strong> </p>
<ul>
<li>\(3a=9\)</li>
<li>\(a+2=5\)</li>
<li>\(\frac{1}{3} a =1\)</li>
</ul>
<ol class="os-raise-noindent" start="5">
<li>Explain your reasoning for the equivalent equations in number 4.
</li>
</ol>
<p><strong>Answer:</strong> </p>
<ul>
<li>The equation \(3a + 6=15\) can be divided by 3 on both sides and becomes \(a +2 = 5\).</li>
<li>When 6 is subtracted from both sides of the equation \(3a +6 = 15\), the equation becomes 3a = 9.</li>
<li>When the equation \(3a + 6 = 15\) has 6 subtracted from both sides and becomes \(3a = 9\), then both sides are divided by 9, the equivalent equation is \(\frac{1}{3} a = 1\).</li>
</ul>
<h4>Student Facing Extension</h4>
<h4>Are you ready for more?</h4>
<p>Here is a puzzle:</p>
<p>Write two expressions that are equivalent to \(p + Q\).</p>
<ul>
<li>\(m+m=N \)</li>
<li>\(N+N=p \)</li>
<li>\(m+p=Q\)</li>
<li>\(p+Q=? \)</li>
</ul>
<p><strong>Answer:</strong> </p>
<p>Your answer may vary, but here is a sample.</p>
<li>\(2p+m\)</li>
<li>\(9m\)</li>
</ul>
<br>
<h4>Anticipated Misconceptions</h4>
<p>Some students may say that \( 2m-2y=10 \) doesn’t describe the relationship between the ages because it
doesn’t involve the number 5. Encourage students to think of some values for \( m \) and \( y \) that
<em>do</em> make the equation true. After they find some pairs of \( m \) and \( y \), ask them what they notice about
the pairs.</p>
<h4>Activity Synthesis</h4>
<p>Select previously identified students to share their explanation of why Tyler's claim is true. Use the order shown in the activity description at the beginning of the teacher guide (description prior to launch). If no students mention the last approach, introduce it to the class.
</p>
<ul>
<li>
Equivalent equations have identical solutions, or identical values that make each of the equations true. All of the
equations in the last question are equivalent because they have 3 as the solution and they all have no other
solutions.
</li>
<li>Suppose we start with a true equation, where the two sides are equal. If we perform the same operation to both
sides of the equation and get a new equation where the two sides are also equal, we can say that the two equations
are equivalent. For instance:
<ul>
<li>Subtracting 6 from both sides of \( 3a + 6 = 15 \) gives \( 3a=9 \). If \( 3a+6 \) and 15 are equal, then the
expressions or numbers we get by subtracting 6 from each one are also equal. We can conclude that \( 3a+6=15 \)
and \( 3a=9 \) are equivalent.</li>
<li>Dividing both sides of \( 3a + 6 = 15 \) by 3 gives \( a + 2 = 5 \). If \( 3a+6 \) is equal to 15, the result
of dividing \( 3a+6 \) by 3 is equal to dividing 15 by 3. We can conclude that \( 3a+6=15 \) and \( a + 2 = 5 \)
are equivalent.</li>
</ul>
</li>
</ul>
<p>Ask students:</p>
<ul>
<li>“How can we show that \( \frac13a = 1 \) is equivalent to \( 3a = 9 \)?” (Multiplying both sides of \(
\frac13a = 1 \) by 9 gives \( 3a=9 \).)</li>
<li>“How can we show that \( a + 2 = 5 \) is equivalent to \( \frac13a = 1 \)?” (Subtracting 2 from both
sides of \( a + 2 = 5 \) and then multiplying both sides of the resulting equation, \( a=3 \), by \( \frac13 \)
gives \( \frac13a = 1 \).)</li>
</ul>
<p>If no students notice that we have made these moves when solving equations, bring it to their attention. Highlight
that solving an equation essentially involves writing a series of equivalent equations that eventually isolates the
variable on one side.</p>
<h3>1.6.2 Self-Check</h3>
<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>Which equation is equivalent to \( 3x-6=18 \)?</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">Answers</th>
<th scope="col">Feedback</th>
</tr>
</thead>
<tbody>
<tr>
<td>\( 3x = 24 \)</td>
<td>That’s correct! Check yourself: Use inverse operations and add 6 to each side of the equation. </td>
</tr>
<tr>
<td>\( 3x = 12 \)</td>
<td>Incorrect. Let’s try again a different way: Use inverse operations to add 6 to each side of the
equation. The answer is \(3x = 24\).</td>
</tr>
<tr>
<td>\( x - 2 =18 \)</td>
<td>Incorrect. Let’s try again a different way: If you divide terms by 3, you must divide ALL terms in the
equation by 3. The answer is \(3x = 24\).</td>
</tr>
<tr>
<td>\( 3x + 12 = 0 \)</td>
<td>Incorrect. Let’s try again a different way: If you use inverse operations to bring 18 over to the other
side of the equation, you must subtract 18 from both sides of the equation. The answer is \(3x = 24\).</td>
</tr>
</tbody>
</table>
<br>
<h3>1.6.2: Additional Resources</h3>
<p class="os-raise-text-bold"><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></p>
<h4>Finding Equivalent Equations</h4>
<p>Equivalent equations are equations that have the same solutions. They are often found by using inverse
operations or by multiplying each term in the equation by the same value.</p>
<br>
<p><strong>Example</strong></p>
<p>Write three equivalent equations to \(4x - 2 = 8\) using one of three strategies.</p>
<ul>
<li>
Using an inverse operation
</li>
<li>
Using division
</li>
<li>
Using multiplication
</li>
</ul>
<p><strong>Using an inverse operation:</strong><br>
An equivalent equation can be determined by adding 2 to both sides.</p>
<p>\(4x - 2 = 8\)<br>
\(4x - 2 + 2 = 8 + 2\)<br>
\(4x = 10\)<br>
\((4x - 2 = 8)\) and \((4x = 10)\) are equivalent.</p>
<p><strong>Using division:</strong><br>
An equivalent equation can be determined by dividing by 2 (or multiplying by \(\frac12\)).</p>
<p>\(4x - 2 = 8\)<br>
\((\frac{1}{2})(4x - 2) = (\frac{1}{2}) (8)\)<br>
\(2x - 1 = 4\)</p>
<p>\((4x - 2 = 8)\) and \((2x - 1 = 4)\) are equivalent.</p>
<p><strong>Using multiplication:</strong><br>
An equivalent equation can be determined by multiplying the same number to both sides.</p>
<p>What is the equivalent equation if you multiply each term in the equation \(4x - 2 = 8\) by 3?</p>
<p><strong>Answer: </strong>\(12x - 6 = 24\), because:</p>
<p> \(4x - 2 = 8\)<br>
\((3) (4x - 2) = (3) (8)\)<br>
\(12x - 6 = 24\)</p>
<h4>Try It: Finding Equivalent Equations </h4>
<ol class="os-raise-noindent">
<li>
Determine the equivalent equation if you subtract 5 from each side of the equation, \(3x + 5 = 15\).
</li>
</ol>
<p><strong>Answer:</strong> \(3x = 10\)</p>
<ol class="os-raise-noindent" start="2">
<li>
Explain how \((3x = 10)\) is related to \((x = \frac{10}{3})\).
</li>
</ol>
<p><strong>Answer:</strong> Divide both sides by 3.<br>
\(\frac{3x}{3}=\frac{10}{3}\) so \(x=\frac{10}{3}\)</p>
<ol class="os-raise-noindent" start="3">
<li>
Multiply every term by the same number to determine an equivalent equation to \(3x + 5 = 15\). (Answers vary response.)
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here are some samples.</p>
<ul>
<li>\(
2(3x + 5 = 15)\) becomes \(6x +10 =30\)
</li>
<li>
\(4(3x + 5 = 15)\) becomes \(12x +20 - 60\)
</li>
<li>
\(-2(3x + 5 = 15)\) becomes \(-6x -10 = 30\)
</li>
</ul>
<ol class="os-raise-noindent" start="4">
<li>
Is \((3x + 5 = 15)\) equivalent to \((6x + 15 = 60)\)?
</li>
</ol>
<p><strong>Answer:</strong> No</p>
<ol class="os-raise-noindent" start="5">
<li>
Explain your reasoning for your answer to number 4 and whether the equations \((3x + 5 = 15)\) and \((6x + 15 = 60)\) are equivalent.
</li>
</ol>
<p><strong>Answer:</strong> No. Each term in the first equation was multiplied by a different number to get equation two.</p>