9e601938-e83d-455c-85b6-eca38972e218.html

<h3>Cool Down (5 minutes)</h3>
<h4>Student Activity</h4>

<p>Here is a system of equations:<br></p>
    <p>\( \left\{ \begin{array}{c l} 12a + 5b = -15\\ 8a + b = 11\end{array}\right. \)</p><br>

    <p>To solve this system, Diego wrote these equivalent systems for his first two steps.</p><br>

    <p><strong>Step 1</strong></p>
    <p>\( \left\{ \begin{array}{c l} 12a + 5b = -15\\ -40a + -5b = -55\end{array}\right. \)</p>
    <p><strong>Step 2</strong></p>
    <p>\( \left\{ \begin{array}{c l} 12a + 5b = -15\\ -28a = -70\end{array}\right. \)</p>


<!--Question 1 -->

    <ol class="os-raise-noindent">
      <li> Describe the move that Diego made between the original system and step 1. </li>
    </ol>  
    <p><strong>Answer: </strong></p>
    <p>Compare your answer: Your answer may vary, but here is a sample.</p>
    <p>Multiply the second equation by -5</p>
    <p>\(-5(8a+b=11)=-40a-5b-55\)</p>




<!--Question 2 -->

    <ol class="os-raise-noindent" start="2">
      <li> How do you know the system of equations in step 1 is equivalent to the original system? </li>
    </ol>     
    <p><strong>Answer: </strong></p>
    <p>Compare your answer:</p>
    <p>Multiplying each side of an equation by the same number gives an equivalent equation with the same solution.</p>  


<!--Question 3 -->

    <ol class="os-raise-noindent" start="3">
      <li> Describe the move that Diego made between step 1 and step 2. </li>
    </ol> 
    <p><strong>Answer: </strong></p>
    <p>Compare your answer: </p>
    <p>Add the resulting equation to the first equation to eliminate the variable \(b\).</p>
    <p>
    <!--BEGIN ALIGN TO EQUALS WITH LINE-->
     \(\begin{align*}12a+5b&=-15
     \\-40a-5b&=-55
     \\ \hline
     -28a&=-70\\
     \end{align*}\)<br><br>
     <!--BEGIN ALIGN TO EQUALS WITH LINE-->
    </p>


<!--Question 4 -->
    <ol class="os-raise-noindent" start="4">
      <li> How do you know the system of equations in step 2 is equivalent to the ones in step 1 (and to the original system) </li>
    </ol> 
    <p><strong>Answer: </strong></p>
    <p>Compare your answer:</p>
    <p>Adding an equal amount to each side of an equation keeps the two sides equal, so the solution for the first equation is also a solution for the sum.</p>


<!--Question 5 -->

    <ol class="os-raise-noindent" start="5">
      <li> Write another set of equivalent systems (different from Diego's first two steps) that will allow one variable to be eliminated and enable you to solve the original system. Be prepared to describe the moves you make to create each new system and to explain why each one has the same solution as the original system. </li>
    </ol> 
  <p><strong>Answer: </strong></p>
    <p>Compare your answer: Your answer may vary, but here is a sample.</p>
    <p>Original System of Equations</p>
    <p>\( \left\{ \begin{array}{c l} 12a + 5b = -15\\ 8a + b = 11\end{array}\right. \)</p><br>
    <p><strong>Step 1 - </strong>Multiply the second equation by&nbsp;\( \frac{3}{2} \)</p>
    <p>\(\frac32(8a+b=11)=12a+\frac32b=\frac{33}{2}\)</p>
      <br>
      <p><strong>Step 2 - </strong>Subtract the resulting equation from the first equation to eliminate the variable \( a \).</p>
      <p>
      <!--BEGIN ALIGN TO EQUALS WITH LINE-->
      \(\begin{align*}12a+5b&=-15
      \\-12a-\frac32b&=-\frac{33}{2}
      \\ \hline
      \frac72b&=-\frac{63}{2}\\
      \end{align*}\)<br><br>
      <!--BEGIN ALIGN TO EQUALS WITH LINE-->
      </p>
        <p><strong>Step 3 - </strong>Multiply the second equation by 2 to find the variable \(b\). <br>
      <p>
        <!--BEGIN ALIGN TO EQUALS WITH LINE-->
      \(\begin{align*}2(\frac72b&=-\frac{63}{2})
      \\7b&=-63
      \\b&=-9
      \end{align*}\)<br><br>
      <!--BEGIN ALIGN TO EQUALS WITH LINE-->    
    </p>


<!--Question 6 -->
    <ol class="os-raise-noindent" start="6">
      <li> Use your equivalent systems to solve the original system. Then, check your solution by substituting the pair of values into the original system. </li>
    </ol> 
    <p><strong>Answer: </strong></p>
    <p>Compare your answer:</p>
    <p>\(a=\frac52,\;b=-9\\\)</p>
    <p>Original System of Equations</p>
    <p>\( \left\{ \begin{array}{c l} 12a + 5b = -15\\ 8a + b = 11\end{array}\right. \)</p><br>
    <p>Substitute the values into <strong>both</strong> original systems to check the solution.</p>
    <br>
    <p>
       <!--BEGIN ALIGN TO EQUALS WITH LINE-->
       \(\begin{align*}12a+5b&=-15
       \\12(\frac52)+5(-9)&\overset?=-15
       \\30-45&\overset?=-15
       \\-15&=-15\checkmark
       \end{align*}\)<br><br>
       <!--BEGIN ALIGN TO EQUALS WITH LINE-->   
       <br> 
      <!--BEGIN ALIGN TO EQUALS WITH LINE-->
      \(\begin{align*}8a+b&=11
      \\8(\frac52)+(-9)&\overset?=11
      \\20-9&\overset?=11
      \\11&=11\checkmark
      \end{align*}\)<br><br>
      <!--BEGIN ALIGN TO EQUALS WITH LINE-->   
</p>