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<h3>Activity (15 minutes)</h3>
<p>This activity further develops the idea of equivalent equations and does so in the context of a situation. Students
pay attention to the moves that create equations with the same solution and those that don’t. They also make
sense of both the moves and the solutions in terms of the given situation. The work here gives students opportunities
to reason concretely and abstractly. In an upcoming lesson, they will think about why these moves lead to equations
with the same solution.</p>
<h3>Launch</h3>
<p>Arrange students in groups of 2 and provide access to calculators. Ask students to read the opening paragraph and
display the equation \( x - 0.1x + 2.70 = 56.70 \) for all to see. Ask students to explain
how the equation represents Noah’s purchase.</p>
<p>Then, give students a couple of minutes to discuss the first two questions with their partner and ask them to pause for a
whole-class discussion. Once students understand that the solution to the equation is the original price of a pair of
jeans and that substituting 60 for \( x \) makes the equation true, move on to
the rest of the activity.</p>
<p>Tell students that they will see a series of equations that are related to the original equation and that can be
interpreted in terms of the situation. For each equation, their job is to determine either what move was made to the
original equation, or what the equation means in context. Next, they should check if the equation has the same
solution as the original. (To help students understand the task and if needed, consider working on questions 3-5 as a
class.)</p>
<p>If time is limited, ask one partner in each group to take the equations from questions 3, 9, and 15 and the other partner to take the equations from questions 6, 12, and 18.
</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">Action and Expression: Internalize Executive Functions</p>
</div>
<div class="os-raise-extrasupport-body">
<p>Chunk this task into more manageable parts for
students who benefit from support with organizational skills in problem solving. Some students may benefit from a
graphic organizer that displays and the original equation before each of the related equations. This will help focus
students’ attention on the moves that may have been made.</p>
<p class="os-raise-text-italicize">Supports accessibility for: Organization;
Attention
</p>
</div>
</div>
<br>
<h4>Student Activity</h4>
<p>With a partner, discuss the following situations and equations. For each, consider:</p>
<ul>
<li>
What does the solution mean in the context of the situation?
</li>
<li>
Are the given values solutions?
</li>
</ul>
<p>If operations are applied correctly, the solution to an equation is also the solution to all equations equivalent to
it. If operations are incorrectly applied, the solution of each equation is different.</p>
<p>For numbers 1 and 2, use the following situation and equation:</p>
<p>Noah is buying a pair of jeans and using a coupon for 10% off. The total price is $56.70, which includes $2.70 in
sales tax. Noah’s purchase can be modeled by the equation:</p>
<p>\(x−0.1x+2.70=56.70\)</p>
<ol class="os-raise-noindent">
<li>
What does the solution to the equation mean in this situation?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.<br>
The solution is the price of a pair of jeans.</p>
<ol class="os-raise-noindent" start="2">
<li>
How can you verify that 70 is not a solution but 60 is the solution?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here are some samples.</p>
<p>70 can be substituted into the equation to make a false statement:<br>
\(70−0.1(70)+2.70=56.70\)<br>
\(65.70 ≠ 56.70\) ❌</p>
<p>60 can be substituted into the equation to make a true statement:<br>
\(
60−0.1(60)+2.70=56.70\)<br>
\(56.70=56.70\) \(\checkmark\) </p>
<p>Here are some equations that are related to \(x−0.1x+2.70=56.70\). Each equation is a result of performing one or
more moves on that original equation. Each can also be interpreted in terms of Noah’s purchase.</p>
<p>For numbers 3–5, use the equation \(100x−10x+270=5,670\).</p>
<ol class="os-raise-noindent" start="3">
<li>
What was done to Noah’s equation to make this one?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.<br>
Every term was multiplied by 100. The equation is equivalent to Noah’s equation.</p>
<ol class="os-raise-noindent" start="4">
<li>
What is the interpretation of this new equation?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.<br>
The price is now expressed in cents instead of dollars.</p>
<ol class="os-raise-noindent" start="5">
<li>
Is the solution the same?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.<br>
60 is a solution in both equations.</p>
<p>For numbers 6–8, use the equation \(x−0.1x=54\).</p>
<ol class="os-raise-noindent" start="6">
<li>
What was done to Noah’s equation to make this one?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.<br>
Subtract 2.70 from both sides of the equation. The equation is equivalent to Noah’s equation.</p>
<ol class="os-raise-noindent" start="7">
<li>
What is the interpretation of this new equation?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.<br>
The price is still in dollars.</p>
<ol class="os-raise-noindent" start="8">
<li>
Is the solution the same?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.<br>
60 is a solution in both equations.</p>
<p>For numbers 9–11, use the equation \(0.9x+2.70=56.70\).</p>
<ol class="os-raise-noindent" start="9">
<li>
What was done to Noah’s equation to make this one?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.<br>
The terms \(x\) and 0.1x were correctly combined to create an equivalent equation to Noah’s original equation.
</p>
<ol class="os-raise-noindent" start="10">
<li>
What is the interpretation of this new equation?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.</p>
<p>10% off means paying 90% of the original price. 90% of the original price plus sales tax is $56.70.</p>
<ol class="os-raise-noindent" start="11">
<li>
Is the solution the same?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.</p>
<p>60 is a solution in both equations.</p>
<p>For numbers 12–14, use the equation \(x−0.1x=56.70\).</p>
<ol class="os-raise-noindent" start="12">
<li>
What was done to Noah’s equation to make this one?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.</p>
<p>Sales tax was not included on the left but was included on the right. This equation is not equivalent to Noah’s
original equation.</p>
<ol class="os-raise-noindent" start="13">
<li>
What is the interpretation of this new equation?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.</p>
<p>The price after using the coupon for 10% off before sales tax is not the same as the price after using the coupon for
10% off including the sales tax, $56.70.</p>
<ol class="os-raise-noindent" start="14">
<li>
Is the solution the same?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.</p>
<p>60 is not a solution to this equation.</p>
<p>For numbers 15–17, use the equation \(x−0.1x=59.40\)</p>
<ol class="os-raise-noindent" start="15">
<li>
What was done to Noah’s equation to make this one?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.<br>
Subtract 2.70 from the left and add 2.70 to the right.</p>
<ol class="os-raise-noindent" start="16">
<li>
What is the interpretation of this new equation?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.</p>
<p>The same operation must be done to both sides of an equation to keep it balanced and to create an equivalent
equation. The price after using the coupon for 10% off is not equal to 56.70 plus tax.</p>
<ol class="os-raise-noindent" start="17">
<li>
Is the solution the same?
</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.<br>
60 is not a solution to this equation.</p>
<p>For numbers 18–20, use the equation \(2(x−0.1x+2.70)=56.70\).</p>
<ol class="os-raise-noindent" start="18">
<li>What was done to Noah’s equation to make this one?</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.<br>
The left side only was multiplied by 2.</p>
<ol class="os-raise-noindent" start="19">
<li>What is the interpretation of this new equation?</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.<br>
Two pairs of jeans after the 10% discount and sales tax cost 56.70.</p>
<ol class="os-raise-noindent" start="20">
<li>Is the solution the same?</li>
</ol>
<p><strong>Answer:</strong> Your answer may vary, but here is a sample.<br>
60 is not a solution to this equation.</p>
<ol class="os-raise-noindent" start="21">
<li>Based on your work above, which of the six equations are equivalent to the original equation, \(x−0.1x+2.70=56.70\)? Select <strong>three</strong> equations that are equivalent to the original.</li>
</ol>
<ul>
<li>
\(100x−10x+270=5670\)
</li>
<li>
\(x−0.1x=54
\)</li>
<li>
\(0.9x+2.70=56.70
\)</li>
<li>
\(x−0.1x=56.70
\)</li>
<li>
\(x−0.1x=59.40
\)</li>
<li>
\(2(x−0.1x+2.70)=56.70
\)</li>
</ul>
<p><strong>Answer:</strong></p>
<p>The correct answers are \(100x-10x+270=5670\), \(x-0.1x=54\), and \(0.9x+2.70=56.70\). </p>
<h4>Video: Looking at Equivalent Equations</h4>
<p>Watch the following video to learn more about why these are equivalent equations.</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/337b23511ca7ac0eb4d32eba29110835c40350e5">
<track default="true" kind="captions" label="On" src="https://k12.openstax.org/contents/raise/resources/4ee789e942e6571879491bca1b89dac5cb04a66a " srclang="en_us">
https://k12.openstax.org/contents/raise/resources/337b23511ca7ac0eb4d32eba29110835c40350e5
</video></div>
</div>
<br>
<br>
<h3>Activity Synthesis</h3>
<p>Display the original equation and the equations from questions 3, 6, 9. 12, 15, and 18 for all to see.<br>
Original Equation (Noah's) = x - 0.1x + 2.70 = 56.70<br>
Equation A (question 3) = 100x - 10x + 270 = 5,670<br>
Equation B (question 6) = x - 0.1x = 54<br>
Equation C (question 9) = 0.9x + 2.70 = 56.70<br>
Equation D (question 12) = x - 0.1x = 56.70<br>
Equation E (question 15) = x - 0.1x = 59.40<br>
Equation F (question 18) = 2(x - 0.1x + 2.70) = 56.70<br>
Invite students to share what was done to
the original equation to get each of those equations and whether they have the same solution. Along the way,
compile a list of moves that lead to equations with the same solution and those that lead to different
solutions. </p>
<p>Ask students to observe the list and see what kinds of moves produce Equations A–C and D–F. Record
the moves that create equations with the same solutions, such as:</p>
<ul>
<li>adding or multiplying both sides of the equal sign by the same number.</li>
<li>applying properties of operations (commutative, associative, or distributive).</li>
<li>combining like terms.</li>
</ul>
<p>Also record the moves that create equations with different solutions, such as:</p>
<ul>
<li>adding or multiplying a different number to the two sides, or performing an operation to only one side.</li>
<li>adding different expressions to the two sides.</li>
<li>performing different operations on each side.</li>
</ul>
<p>(The lists don’t need to be comprehensive because students will examine these moves more closely later.)
</p>
<p>Next, give students a couple of examples of how the equations and their solutions could be interpreted in
context. For example:</p>
<ul>
<li>Equation A shows the cost calculation in cents, so the original price of a pair of jeans is unchanged.</li>
<li>Equation D shows the discounted price, excluding tax, to be $56.70, but in the original
equation, the price $56.70 included tax. This means that, in Equation D, the original price of
one pair of jeans is different from the first equation. </li>
</ul>
<p>Prompt students to interpret 1–2 other equations and to explain why the solution in each equation (or the
price of one pair of jeans) is equal or unequal to that in the initial equation. Here are possible
interpretations:</p>
<ul>
<li>Equation B shows the discounted price before sales tax to be $54, which is $2.70
less than $56.70. This is also the case in the initial equation, so the original price of the
jeans is still the same. </li>
<li>Equation C shows the discounted price as 90% of the original price, which is the same as 10% less than the
original price (100%). The original price is unchanged.</li>
<li>Equation E shows the discounted price, before tax, to be $59.40. This is more than the pre-tax
price in the initial equation, so the original price of the jeans here must be higher.</li>
<li>Equation F shows the price of 2 pairs of jeans, including the discount and tax, to be $56.70.
This means the price of a pair of jeans must be much less than in the initial equation.</li>
</ul>
<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"><!--Extra Support Name-->MLR8 Discussion Supports: Speaking </p>
</div>
<div class="os-raise-extrasupport-body">
<!--Support Content--><p>Use this routine to support whole-class discussion. For each observation that is shared, ask students to restate what they heard using precise mathematical language. Consider providing students time to restate what they hear to a partner before selecting one or two students to share with the class. Ask the original speaker if their peer was accurately able to restate their thinking. Call students' attention to any words or phrases that helped clarify the original statement. This provides more students with an opportunity to produce language as they interpret the reasoning of others.
</p>
<p class="os-raise-text-italicize"><!--Support Italics-->Design Principle(s): Support sense-making</p>
</div>
</div>
<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">
<!--Extra Support Name-->
Representation: Develop Language and Symbols </p>
</div>
<div class="os-raise-extrasupport-body">
<!--Support Content-->
<p>Create a display of important terms and vocabulary. Keep this display visible throughout the remainder of the unit. Invite students to suggest language or diagrams to include that will support their understanding of combining like terms, and the commutative, associative, and distributive properties. </p>
<p class="os-raise-text-italicize">
<!--Support Italics-->
Supports accessibility for: Conceptual processing; Language</p>
</div>
</div>
<br>
<h3>1.6.3: 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 of the following is a correct next step to create an equation equivalent to \( x - 0.2x - 4.2 =13.4 \)?
</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">Answers</th>
<th scope="col">Feedback</th>
</tr>
</thead>
<tbody>
<tr>
<td>\( 0.8x - 4.2 = 13.4 \)</td>
<td>That’s correct! Check yourself: Combine like terms so that \( x - 0.2x=0.8x \).</td>
</tr>
<tr>
<td>\( 10(x - 0.2x-4.2) = 13.4 \)</td>
<td>Incorrect. Let’s try again a different way: Multiplying by 10 can help get rid of the decimal,
however, you must multiply both sides by the same quantity. The correct answer is \( 0.8x - 4.2 =13.4 \).
</td>
</tr>
<tr>
<td>\( x - 0.2x = 9.2 \)</td>
<td>Incorrect. Let’s try again a different way: Since the 4.2 was subtracted in the original equation,
it should have been added to both sides. The correct answer is \( 0.8x -4.2 =13.4 \).</td>
</tr>
<tr>
<td>\( 2(x - 0.2x) = 17.6 \)</td>
<td>Incorrect. Let’s try again a different way: If one side is multiplied by 2, the other side must
also be multiplied by 2. The correct answer is \( 0.8x -4.2 =13.4 \).</td>
</tr>
</tbody>
</table>
<br>
<h2>1.6.3: Additional Resources</h2>
<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>
<h3>Properties of Equality</h3>
<br>
<div class="os-raise-graybox">
<h4>Properties of Equality</h4>
<hr>
<p>When you add, subtract, multiply, or divide the same quantity from both sides of an equation, you
still have equality.</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col"></th>
<th scope="col"></th>
</tr>
</thead>
<tbody>
<tr>
<td>
<p>Subtraction Property of Equality</p>
<p>For any real numbers \( a \), \( b \), and \( c \),<br>if \( a = b \),<br>then \( a−c = b−c
\).</p>
</td>
<td>
<p>Addition Property of Equality</p>
<p>For any real numbers \( a \), \( b \), and \( c \),<br>if \( a = b \),<br>then \( a+c = b+c \).</p>
</td>
</tr>
<tr>
<td>
<p>Division Property of Equality</p>
<p>For any real numbers \( a \), \( b \), and \( c \), and \( c \neq 0 \),<br>if \( a = b \),<br>then \(
\frac{a}{c} = \frac{b}{c} \).</p>
</td>
<td>
<p>Multiplication Property of Equality</p>
<p>For any real numbers \( a \), \( b \), and \( c \),<br>if \( a = b \),<br>then \( ac = bc \).</p>
</td>
</tr>
</tbody>
</table>
</div>
<br>
<br>
<p>For equations to be equivalent, inverse operations are used and must be applied to both sides of an equation so
it remains balanced.</p>
<p>Inverse operations are operations that “undo” other operations.</p>
<p><strong>Example 1 </strong></p>
<p>Solve \( 3x =15 \).</p>
<p>Since 3 is being multiplied by \( x \), to solve for \( x \), divide both sides by 3.</p>
<p>\( \frac{3x}{3}=\frac{15}{3} \)</p>
<p>\( x=5 \)</p>
<p><strong>Example 2</strong></p>
<p>Solve \( \frac{x}{2}+4=12 \).</p>
<p><strong>Step 1 - </strong>Subtract 4 from both sides.<br>
\(\frac{x}{2}+4−4=12−4\)</p>
<p><strong>Step 2 - </strong>Simplify.<br>
\(\frac{x}{2}=8\)</p>
<p><strong>Step 3 - </strong>Multiply both sides by 2.<br>
\(2 \times \frac{x}{2}=8 \times 2\)</p>
<p><strong>Step 4 -</strong> Simplify.<br>
\(x =16\)
</p>
<h3>Try It: Properties of Equality</h3>
<p>For questions 1 - 2, use the scenario:</p>
<p>Denae bought 6 pounds of grapes for $10.74.</p>
<ol class="os-raise-noindent">
<li>
Write an equation for the situation.
</li>
</ol>
<p><strong>Answer:</strong> Let \(g\)=one pound of grapes.</p>
<p>6g=10.74</p>
<ol class="os-raise-noindent" start="2">
<li>
Solve the equation.
</li>
</ol>
<p><strong>Answer:</strong></p>
<p>
<strong>STEP 1:</strong> Identify the variable. <br>
\(g\): cost for one pound of grapes<br>
<br>
<strong>Step 2: </strong>Write an equation.<br>
\(6g=10.74\)<br>
<br>
<strong>STEP 3:</strong> Solve for g.<br>
Divide BOTH sides by 6.<br>
\(\frac{6g}{6}=\frac{10.74}{6}\)<br>
<br>
<strong>Step 4: </strong>Simplify to solve.<br>
\(g=1.79\)<br>
</p>
<p>The grapes cost $1.79 per pound.</p>