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<h4>Solve Application Problems Using Direct Variation</h4>
<p>In applications using direct variation, generally we will know values of one pair of the variables and will be asked to find the equation that relates \( x \) and \( y \). Then we can use that equation to find values of \( y \) for other values of \( x \).</p>
<p>Here&rsquo;s how to solve applications of direct variation problems.</p>
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  <p><strong>Step 1</strong> - Write the equation for direct variation. Identify the variables.</p>
  <p><strong>Step 2</strong> - Substitute the given values for the variables.</p>
  <p><strong>Step 3</strong> - Solve for the constant of variation.</p>
  <p><strong>Step 4</strong> - Write the equation that relates \(x\) and \(y\) using the constant of variation.</p>
  <p><strong>Step 5</strong> - Substitute the value for the variable.</p>
  <p><strong>Step 6</strong> - Solve.</p>
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<p>Let&rsquo;s look at an application problem:</p>
<p>If the cost of a pizza varies directly with its diameter, and an 8-inch pizza costs $12, how much would a 6-inch pizza cost?</p>
<p><strong>Step 1</strong> - Write the equation for direct variation. Identify the variables.<br>
  \( y = kx \)</p>
<p> \(y\): cost of pizza, \(x\): diameter</p>
<p><strong>Step 2</strong> - Substitute the given values for the variables.<br>
  \(12 = k(8)\)</p>
<p><strong>Step 3</strong> - Solve for the constant of variation.<br>
  \(k= \frac32 \)</p>
<p><strong>Step 4</strong> - Write the equation that relates x and y using the constant of variation.<br>
  \( y= \frac{3}{2}x \)</p>
<p><strong>Step 5</strong> - Substitute the value for the variable.<br>
  \( y= \frac{3}{2}(6) \)</p>
<p><strong>Step 6</strong> - Solve.<br>
  \( y = 9 \)</p>
<p>What does this mean in the context of the problem?</p>
<p>A pizza that has a diameter of 6 inches costs $9.<br>
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<h4>
  Try It: Solve Application Problems Using Direct Variation</h4>
<p>The number of buckets of water, \(b\), filled by a spring varies directly with the number of minutes, \(m\), passed. The spring fills 3 buckets of water in 14 minutes.</p>

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    <p>How many buckets would be filled in 42 minutes?</p>
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  <div class="os-raise-ib-cta-prompt">
    <p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
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  <p>Compare your answer: Here is how to model this application problem of direct variation.</p>
  <p><strong>Step 1</strong> - Write the equation for direct variation. Identify the variables.<br>
    \( y = kx \)<br>
    \( b=km \)</p>
  <p> \(b\): number of buckets, \(m\): number of minutes</p>
  <p><strong>Step 2</strong> - Substitute the given values for the variables.<br>
    \( 3=k(14) \)</p>
  <p><strong>Step 3</strong> - Solve for the constant of variation.<br>
    \( k= 3/14 \)</p>
  <p><strong>Step 4</strong> - Write the equation that relates \(m\) and \(b\) using the constant of variation.<br>
    \( b= \frac{3}{14}m \)</p>
  <p><strong>Step 5</strong> - Substitute the value for the variable.<br>
    \( b= \frac{3}{14}(42) \)</p>
  <p><strong>Step 6</strong> - Solve.<br>
    \(b= 9\) buckets</p>
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