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<h4>Solve a System of Linear Inequalities by Graphing</h4><br>
<p>The solution to a single linear inequality is the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To find this region, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph.</p>
<p><strong>Example</strong></p>
<p>Solve the system by graphing:&nbsp;</p>
<p>\(\left\{\begin{array}{l}y\geq2x-1\\y&lt;x+1\end{array}\right.\)</p>
<br>
<p><strong>Answer:</strong></p>
  <!--STEPS -->   
<p><b>Step 1 - </b>Graph the first inequality.</p>
<ul>
    <li>Graph the boundary line.</li>
    <li>Shade in the side of the boundary line where the inequality is true.</li>
</ul>
<p>We will graph \(y\geq2x-1\)</p>
<p>We graph the line \(y=2x-1.\) It is a solid line because the inequality sign is \(\geq\).</p>
<p>We choose \((0,0)\) as a test point. It is a solution to \(y\geq2x-1\), so we shade in above the boundary line.</p>
<p><img alt class="img-fluid atto_image_button_text-bottom" height="196" role="presentation" src="https://k12.openstax.org/contents/raise/resources/2c192eef3f8f3211f2f50bc847ff83edee22d4f6" width="200"></p>

<p><b>Step 2 - </b>On the same grid, graph the second inequality.</p>
<ul>
  <li>Graph the boundary line.</li>
  <li>Shade the side of that boundary line where the inequality is true.</li>
</ul>
<p>We will graph \(y&lt;x+1\) on the same grid. </p>
<p>We graph the line \(y=x+1\).  It is a dotted line because the inequality sign is \(&lt;\).</p>
<p>Again, we use \((0,0)\) as a test point. It is a solution so we shade in that side of the line \(y=x+1\).</p>
<p><img alt class="img-fluid atto_image_button_text-bottom" height="207" role="presentation" src="https://k12.openstax.org/contents/raise/resources/68df81a6e50ed422c52b13c8aa624bc3ae30fbe5" width="200"></p>

<p><b>Step 3 - </b>The solution is the region where the shading overlaps.</p>
<p>The point where the boundary lines intersect is not a solution because it is not a solution to \(y&lt;x+1\).</p>
<p>The solution is all points in the area shaded twice, which appears as the darkest shaded region.</p>

<p><b>Step 4 - </b>Check by choosing a test point.</p>
<p>We&rsquo;ll use \((-1, -1)\) as a test point.</p>
<p>Is \((-1,1)\) a solution to \(y\geq2x-1\)?</p>
<p>\(-1\overset?\geq2(-1)-1\)</p>
<p>\(-1\geq-3\) \(\checkmark\)</p>
<p>Is \((-1,-1)\) a solution to \(y&lt;x+1\)?</p>
<p>\(-1\overset?\geq-1+1\)</p>
<p>\(-1&lt;0\) \(\checkmark\)</p>
<p>The region containing \((-1,-1)\) is the solution to this system.</p>  

<h4>Try It: Solve a System of Linear Inequalities by Graphing <br></h4>
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<div class="os-raise-ib-cta" data-button-text="Solution" data-fire-event="Reveal1" data-schema-version="1.0">
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    <ol class="os-raise-noindent">
      <li>
       Solve the system by graphing
      </li>
    </ol>
    <p>\(\left\{\begin{array}{l}y&lt;3x+2\\y&gt;-x-1\end{array}\right.\)</p>

  </div>
  <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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<div class="os-raise-ib-content" data-wait-for-event="Reveal1" data-schema-version="1.0">
<p>Compare your answer: Here is how to solve the system by graphing. </p>
   <p role="presentation">Where the shaded regions overlap.</p>

  <p><img alt="The figure shows a graph plotted for the inequalities y less than three times x plus two and y greater than minus x minus one. Two lines intersect each other on the graph. An area to the right of both the lines is colored in grey. It is the solution." class="img-fluid atto_image_button_text-bottom" height="296" src="https://k12.openstax.org/contents/raise/resources/a9a0b1afe58d20ccfba1413e0cd2b02cf38b33ed" width="300"></p>

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