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<h4>Activity (15 minutes)</h4>
<p>So far, students have seen only one-variable equations that have a solution. For these equations, performing acceptable moves always led to equivalent equations that have the same solution. In this activity, students encounter an example where the given equation has no solutions and performing the familiar moves leads to an untrue statement.</p>
<p>Prior to this point, students have added, subtracted, multiplied, and divided a number from both sides of an equation. They have also added a variable expression to (or subtracted a variable expression from) an equation. They recognize these moves as allowable for solving equations. Here, students also come across an equation that is divided by a variable expression and make sense of why it leads to a false statement.</p>
<h4>Launch</h4>
<p>Arrange students into groups of 3-4. Give groups 1-2 minutes of quiet time to analyze the first set of equations and then time to brainstorm why Noah's work results in a false statement. Follow with a class discussion.</p>
<p>Invite students to share their explanations as to why Noah ended up with \(6 = 1\). Students are likely to say that the moves Noah made were allowable and can be explained, but they may struggle to say why they end up with a false statement.</p>
<p>Draw students' attention to the third line of Noah's work. Ask them to interpret the expression on each side of the equal sign: \( x + 6\) can be interpreted as 6 more than a number and \(x + 1\) as 1 more than that same number. Then, ask them to try to find a value that could make that equation true. (There is none!)</p>
<p>Highlight that it's not possible for 6 more than some number, no matter what that number is, to be equal to 1 more than that number. If no value \(x\) could make the third equation true, the same could be said about the original equation—it had no solution. So even though Noah performed acceptable moves, the final equation is a false equation. Because all of these equations are equivalent, this means the original equation is also a false equation. It has no solutions because no value could make the equation true.</p>
<p>Repeat the process to analyze the second set of equations. See Activity Synthesis for discussion questions.</p>
<h4>Student Activity </h4>
<p>Noah is having trouble solving two equations. In each case, he took steps that he thought were acceptable but ended
up with statements that are clearly not true.</p>
<p>Analyze Noah’s work on each equation and the moves he made. Were they acceptable moves? Why do you think he
ended up with a false equation?</p>
<p>Discuss your observations with your group and be prepared to share your conclusions. If you get stuck, consider
solving each equation.</p>
<div class="os-raise-d-flex os-raise-justify-content-between"><p><strong>Example 1</strong><br>
<br>
<strong>Step 1 - </strong>Original equation.<br>
\( x+6 =4x+1-3x \)<br>
<br>
<strong>Step 2 - </strong>Apply the Commutative Property.<br>
\( x+6 =4x-3x+1 \)<br>
<br>
<strong>Step 3 - </strong>Combine like terms.<br>
\( x+6 =x+1 \)<br>
<br>
<strong>Step 4 - </strong>Subtract \(x\) from each side.<br>
\( 6 =1 \)</p>
<p><strong>Example 2</strong><br>
<br>
<strong>Step 1 - </strong>Original equation.<br>
\( 2(5+x)-1=3x+9 \)<br>
<br>
<strong>Step 2 - </strong>Apply the Distributive Property.<br>
\( 10+2x-1 =3x+9 \)<br>
<br>
<strong>Step 3 - </strong>Subtract 10 from each side.<br>
\( 2x-1 =3x-1 \)<br>
<br>
<strong>Step 4 - </strong>Add 1 to each side.<br>
\( 2x=3x \)<br>
</p></div>
<ol class="os-raise-noindent">
<li>Why did Noah arrive at no solution for the equation solved in Example 1?
</li>
</ol>
<p><strong>Answer:</strong> </p>
<p> Your answer may vary, but here is an example: <br>
All the moves that Noah made were acceptable, but because he ended up with a false statement, that means there is no value of \(x\) that makes the first equation true. We can see it in the third line: \(x+6=x+1\). There is no number that can make the equation true.</p>
<ol class="os-raise-noindent" start="2">
<li>Why did Noah arrive at no solution for the equation solved in Example 2?</li>
</ol>
<p><strong>Answer:</strong> </p>
<p>Your answer may vary, but here is an example:<br>
Noah's moves seem acceptable except for the last one. If he had subtracted \(2x\) from each side of the equation, he would have \(0=x\). Instead, he divided both sides by \(x\) which would give an undefined number if \(x\) is 0.
</p>
<br>
<p>Analyze Noah’s work on each equation and the moves he made. Were they acceptable moves? Why do you think he ended up with a false equation?
</p>
<p>Discuss your observations with your group and be prepared to share your conclusions. If you get stuck, consider solving each equation.
</p>
<p><strong>Example 3</strong></p>
<p>Noah tried to solve a third equation and ended up with a true statement, without variables.</p>
<p><strong>Step 1 - </strong>Simplify each side of the equation as much as possible.
Use the Distributive Property to remove any parentheses.
Combine like terms.
<br>
\( 2 (4 + x) - 2 = 2x + 6 \)<br>\(2x+6=2x+6\)
</p>
<p><strong>Step 2 - </strong>Collect all the variable terms on one side of the equation.
Use the Addition or Subtraction Property of Equality.
Subtract \(2x\) from both sides.
<br>
\(2x-2x+6=6\)</p>
<p><strong>Step 3 - </strong>Collect all the constant terms on the other side of the equation.
Use the Addition or Subtraction Property of Equality.
Subtract 6 from each side.
<br>
\(0+6-6=6-6\)</p>
<p><strong>Step 4 - </strong>Make the coefficient of the variable term equal to 1.
This step is not needed.
<br>
\(0=0\)</p>
<p><strong>Step 5 - </strong> Check the solution.<br>
Since there is not a variable and the statement is true, any value of \(x\) would be a solution for the original equation. So, this equation has infinitely many solutions.
</p>
<h4>Student Facing Extension</h4>
<h4>Are you ready for more?</h4>
<ol class="os-raise-noindent">
<li>We cannot divide the number 100 by zero because dividing by zero is undefined. Instead, try dividing 100 by 10, then 1, then 0.1, then 0.01. What do you notice happens as you divide by smaller numbers?</li>
</ol>
<p><strong>Answer:</strong></p>
<p>Compare your answer: Your answer may vary, but here is a sample.</p>
<p>10, 100, 1000, 10,000. For example: The quotients are getting very large.</p>
<ol class="os-raise-noindent" start="2">
<li> Now try dividing the number –100 by 10, by 1, by 0.1, 0.01. What is the same and what is different when compared to question 1?</li>
</ol>
<p><strong>Answer:</strong></p>
<p>Compare your answer: Your answer may vary, but here is a sample.</p>
<p>–10, –100, –1000, –10,000. For example: The absolute values of the quotients are the same, but the sign is opposite.</p>
<ol class="os-raise-noindent" start="3">
<li>In middle school, you used tape diagrams to represent division. This tape diagram shows that \( 6 \div 2 = 3 \).</li>
</ol>
<p><img alt="Tape diagram, 3 parts, each marked 2." height="29" src="https://k12.openstax.org/contents/raise/resources/a2c2b4b0c73e105faf25cb06f344b74cd46bdd0d" width="171"></p>
<p>Draw a tape diagram that shows why \( 6 \div \frac12 = 12 \).</p>
<p><strong>Answer:</strong></p>
<p>Compare your answer: Your answer may vary, but here is a sample.</p>
<p><img alt="Tape diagram." height="29" src="https://k12.openstax.org/contents/raise/resources/ec9feea4e8cd7e31dc669c1194ac3b83e7863227" width="171"></p>
<br>
<ol class="os-raise-noindent" start="4">
<li>Try to draw a tape diagram that represents \( 6 \div 0 \). Explain why this is so difficult. </li>
</ol>
<p><strong>Answer:</strong></p>
<p>Compare your answer: Your answer may vary, but here is a sample.</p>
<p>For example: The zero takes up no space, so no matter how many copies we have it could never reach 6.</p>
<h4>Anticipated Misconceptions</h4>
<p>Some students may point to a step that is valid and mistakenly identify it as an error. For instance, in
the first set of steps, they may object to replacing \( +1 - 3x \) with
\( -3x+1 \), thinking that it should be rearranged to \( +3x-1 \). Push their reasoning with a simpler example. Ask, for
instance, if \( +7 - 4 \) is equivalent to
\( +4 - 7 \). Remind students that we can think of
\( +7 -4 \) as \( +7 + (-4) \) and
then apply the commutative property of addition to get \(
-4 + 7 \).</p>
<p>If students hypothesize about two equations being equivalent but are not sure how to check if it’s
actually the case, suggest that a good way to check is by finding the solution to one equation, then
checking whether that value is also a solution to the second equation.</p>
<h4>Activity Synthesis</h4>
<p>Invite students to share what they thought was the problem with Noah’s work. They are likely to say
that Noah seems to have performed allowable moves and did them correctly. Then, draw students’
attention to the second-to-last step: \( 2x = 3x \). Ask students:</p>
<ul>
<li>“Earlier, when looking at \( x + 6 = x + 1 \), we reasoned that
there’s no value of \( x \) that could make this equation true.
Is there a value of \( x \) that could make \( 2x = 3x \) true?” (Yes, 0.)</li>
<li>“If there is a value that can make \( 2x=3x \) true, what do we
know about the original equation? Is it also true?” (Yes, they are all equivalent equations.)
“What is its solution?” (0)</li>
<li>“So the equations are true up through the third step. The last step is where we have a false
statement, after Noah shows division by \( x \). If \( x \) is 0, what might be the problem with dividing by \( x \)?” (If \( x \) is 0, then
we are dividing both sides by 0, which gives an undefined result.)</li>
</ul>
<p>Explain that dividing by the variable in the equation is not done because if the solution happens to be
0, it could lead us to thinking that there is no solution while in fact there is (the solution is the
number 0).</p>
<p>Revisit the lists of acceptable and unacceptable moves compiled in earlier activities. Add
“dividing by the variable” and “dividing by 0” to the list of unacceptable moves.
</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 8 Discussion Supports: Speaking, Representing</p>
</div>
<div class="os-raise-extrasupport-body">
<p>Give students additional time to make sure
everyone in their group can explain their analysis of Noah’s moves. Invite groups to rehearse what
they will say when they share with the whole class. Rehearsing provides students with additional
opportunities to speak and clarify their thinking, and it will improve the quality of explanations shared
during the whole-class discussion.</p>
<p class="os-raise-text-italicize">Design Principle(s): Support sense-making; Cultivate
conversation
</p>
</div></div>
<br>
<h3>1.7.3: 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 equations has a solution?</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">Answers</th>
<th scope="col">Feedback</th>
</tr>
</thead>
<tbody>
<tr>
<td>\( 3x = x + 18 \)</td>
<td>That’s correct! When \(x\) is subtracted from each side and each side is divided by 2, \(
x \) will have a solution, \( x = 9 \).</td>
</tr>
<tr>
<td>\( 2x = 2x + 1 \)</td>
<td>Incorrect. Let’s try this another way: There are no values for \(x\) that will make this
equation true. The correct answer is \( 3x = x + 18 \). </td>
</tr>
<tr>
<td>\( x + 3 = x -2 \)</td>
<td>Incorrect. Let’s try this another way: Subtracting \(x\) from both sides results in a
false equation, \( 3 = -2 \). There are no values for \(x\) that will make this equation true. The
correct answer is \( 3x = x + 18 \). </td>
</tr>
<tr>
<td>\( 10x + 10 = 7x + 3x \)</td>
<td>Incorrect. Let’s try this another way: Combining the like terms on the right side and
subtracting \(10x\) from both sides results in a false statement, \( 10 = 0 \). There are no
values for \(x\) that will make this equation true. The correct answer is \( 3x = x + 18 \).
</td>
</tr>
</tbody>
</table>
<br>
<h3>1.7.3: 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>Equations Without Solutions </h4>
<p>What happens when we solve \( 5z = 5z - 1\)?<br></p>
<p><strong>Step 1 - </strong>Collect all the variable terms on one side of the equation.<br>Subtract \(5z\) to get the \(z\) terms together on one side.</p>
<p>\(\begin{aligned}5z-5z &=5z-5z-1\end{aligned}\)<br></p>
<p><strong>Step 2 - </strong>Simplify and the \(z\) terms are gone!.</p>
<p>\(\begin{aligned}0 &=1\end{aligned}\)</p>
<p>Since 0 does not equal \(-1\), this equation led to a false statement. There are no solutions to this equation.</p>
<br>
<p><strong>Example:</strong></p>
<p>Solve \( 5m+3(9+3m)=2(7m−11) \). </p>
<p><strong>Step 1 - </strong>Simplify each side of the equation as much as possible.
Use the Distributive Property to remove any parentheses.
Combine like terms.<br>
\( 5m+27+9m=14m−22 \)<br>
\(14m+27=14m−22\)</p>
<p><strong>Step 2 - </strong>Collect all the variable terms on one side of the equation.<br>
Use the Addition or Subtraction Property of Equality.<br>
\(14m−14m+27=14m−14m−22\)<br>
\(27=-22\)
</p>
<p><strong>Step 3 - </strong> Check the solution.<br>
The equation \(27=−22\) is NOT a true statement. The equation 5m+3(9+3m)=2(7m−11) has no solutions.
</p>
<h4>Infinitely Many Solutions</h4>
<p>What if an equation results in a true statement without variables?</p>
<p>The equations \( -4 = -4 \), \( 5 = 5 \), and \( 0 = 0 \) are all true statements because one side is equal to the other. This means that any value of the variable in the original equation would be a solution. When this happens in the final step of an equation, there are infinitely many solutions.</p>
<p><strong>Example</strong><br>
Solve \( 2y +6 =2(y+3) \)</p>
<p><strong>Step 1 - </strong>Simplify each side of the equation as much as possible.
Use the Distributive Property to remove any parentheses.
Combine like terms.
</p>
<p>\( 2y +6 =2(y+3) \)</p>
<p><strong>Step 2 - </strong> Collect all the variable terms on one side of the equation.
Use the Addition or Subtraction Property of Equality.<br>
\( 2y−2y+6=2y−2y+6\)<br>
\(6=6\)
</p>
<p>\(2y+6=2y+6\)</p>
<p><strong>Step 3 - </strong> Check the solution.
We say the solution to the equation is all of the real numbers.An equation that is true for any value of the variable is called an <span class="os-raise-ib-tooltip" data-store="glossary-tooltip" data-schema-version="1.0">identity</span>.
</p>
<p>The solution of an identity is all real numbers.</p>
<br>
<h4>Try It: Equations with No Solutions and Infinitely Many Solutions </h4>
<p>1. Solve \( 10x - 4 = 2(5x -2) \)and determine the number of solutions.</p>
<p><strong>Answer:</strong></p>
<p>Your answer may vary, but here is a sample.</p>
<p>Solve this linear equation using a general strategy:</p>
<p><strong>Step 1 - </strong>Original equation.<br></p>
<p>\( 10x - 4 = 2(5x -2) \)</p>
<p><strong>Step 2 - </strong>Distribute<br></p>
<p>\( 10x - 4 =10x - 4 \)</p>
<p><strong>Step 3 - </strong>Collect all the variable terms on one side of the equation. Use the Addition or Subtraction Property of Equality. <br></p>
<p>\(\begin{aligned}10x - 10x - 4 &=10x - 10x - 4 \\-4 &=-4 \end{aligned}\)<br></p>
<p><strong>Step 4 - </strong>Check the solution.</p>
<p>This is a true statement so there are infinitely many solutions.</p>
<br>
<p>2. Solve \( 3x - 4 = 3(x -2) \) and determine the number of solutions.</p>
<p><strong>Answer:</strong></p>
<p> Your answer may vary, but here is a sample.</p>
<p>Solve this linear equation using a general strategy:</p>
<p><strong>Step 1 - </strong>Original equation.<br></p>
<p>\( 3x - 4 = 3(x -2) \)</p>
<p><strong>Step 2 - </strong>Distribute<br></p>
<p>\( 3x - 4 = 3x -6 \)</p>
<p><strong>Step 3 - </strong>Collect all the variable terms on one side of the equation. Use the Addition or Subtraction Property of Equality. <br></p>
<p>\(\begin{aligned}3x - 3x - 4 &=3x - 3x - 6 \\-4 &=-6 \end{aligned}\)<br></p>
<p><strong>Step 4 - </strong>Check the solution</p>
<p>The equation \( -4=-6\) is NOT a true statement. The equation \( ( 3x - 4 = 3(x -2) \) has no solutions. </p>