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<title>Solving Rational Inequalities</title>
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<h1 class="center">Solving Rational Inequalities</h1>
<h2>Rational</h2>
<p>A <a href="rational-expression.html">Rational Expression</a> looks like:</p>
<p class="center"><img src="images/rational-expression.svg" alt="Rational Expression"></p>
<h2>Inequalities</h2>
<p>Sometimes we need to solve rational <a href="inequality.html">inequalities</a> like these:</p>
<table align="center" cellpadding="5" border="0">
<tbody>
<tr>
<th> <div class="center"><b>Symbol</b></div></th>
<th> <div class="center"><b>Words</b></div></th>
<th> <div class="center"><b>Example</b></div></th>
</tr>
<tr height="6">
<td><br></td>
<td><br></td>
<td><br></td>
</tr>
<tr>
<td class="large">
<div class="center">&gt;</div></td>
<td>
<div class="center">greater than</div></td>
<td class="larger">
<div class="center">(x+1)/(3&minus;x) &gt; 2</div></td>
</tr>
<tr>
<td class="large">
<div class="center">&lt;</div></td>
<td>
<div class="center">less than</div></td>
<td class="larger">
<div class="center">x/(x+7) &lt; &minus;3</div></td>
</tr>
<tr>
<td class="large">
<div class="center">&ge;</div></td>
<td>
<div class="center">greater than or equal to</div></td>
<td class="larger">
<div class="center">(x&minus;1)/(5&minus;x) &ge;<b> </b>0</div></td>
</tr>
<tr>
<td class="large">
<div class="center">&le;</div></td>
<td>
<div class="center">less than or equal to </div></td>
<td class="larger">
<div class="center">(3&minus;2x)/(x&minus;1) &le; 2</div></td>
</tr>
<tr height="6">
<td><br></td>
<td><br></td>
<td><br></td>
</tr>
</tbody></table>
<h2>Solving</h2>
<p>Solving inequalities is very like <a href="equations-solving.html">solving equations</a> ... you do most of the same things.</p>
<table style="border: 0; margin:auto;">
<tbody>
<tr>
<td align="center"><img src="images/inequality-rational-graph-function.svg" alt="Graph of Rational Inequality"></td>
</tr>
<tr>
<td align="center"><span class="larger">When we solve <b>inequalities</b> <br>
we try to find <b>interval(s)</b>, <br>
</span>such as the ones marked "&lt;0" or "&gt;0"</td>
</tr>
</tbody></table>
<p>These are the steps:</p>
<div class="bigul">
<ul>
<li>find "points of interest":
<ul>
<li>the "=0" points (roots), and</li>
<li>"vertical asymptotes" (where the function is undefined)</li>
</ul></li>
<li>in between the "points of interest", the function is either <b>greater than zero</b> (&gt;0) or <b>less than zero</b> (&lt;0)</li>
<li>then pick a test value to find out which it is (&gt;0 or &lt;0)</li>
</ul>
<p>Here is an example:</p>
</div>
<div class="example">
<h3>Example: <span class="intbl"><em>3x&minus;10</em><strong>x&minus;4</strong></span> &gt; 2</h3>
<p><b>First</b>, let us simplify!</p>
<p class="larger">But You Cannot Multiply By (x&minus;4)</p>
<p>Because "x&minus;4" could be positive or negative ... we don't know if we should change the direction of the inequality or not. This is all explained on <a href="inequality-solving.html">Solving Inequalities</a>.</p>
<p>Instead, bring "2" to the left:</p>
<p class="center larger"><span class="intbl"><em>3x&minus;10</em><strong>x&minus;4</strong></span> &minus; 2 &gt; 0</p>
<p>Then multiply 2 by (x&minus;4)/(x&minus;4):</p>
<p class="center larger"><span class="intbl"><em>3x&minus;10</em><strong>x&minus;4</strong></span> &minus; 2<span class="intbl"><em>x&minus;4</em><strong>x&minus;4</strong></span> &gt; 0</p>
<p>Now we have a common denominator, let's bring it all together:</p>
<p class="center larger"><span class="intbl"><em>3x&minus;10 &minus; 2(x&minus;4)</em><strong>x&minus;4</strong></span> &gt; 0</p>
<p>Simplify:</p>
<p class="center larger"><span class="intbl"><em>x&minus;2</em><strong>x&minus;4</strong></span> &gt; 0</p>
<p>&nbsp;</p>
<p><b>Second</b>, let us find "points of interest".</p>
<p>At x=2 we have: <span class="larger">(0)/(x&minus;4) &gt; 0</span>, which is a "=0" point, or <b>root</b></p>
<p>At x=4 we have: <span class="larger">(x&minus;2)/(0) &gt; 0</span>, which is <b>undefined</b></p>
<p class="center larger"></p>
<p class="center larger"></p>
<p><b>Third</b>, do test points to see what it does in between:</p>
<p><b>At x=0:</b></p>
<ul>
<li>x&minus;2 = &minus;2, which is <b>negative</b></li>
<li>x&minus;4 = &minus;4, which is also <b>negative</b></li>
<li>So (x&minus;2)/(x&minus;4) must be <b>positive</b></li>
</ul>
<p>We can do the same for <b>x=3</b> and <b>x=5</b>, and end up with these results:</p>
<div class="simple">
<table style="border: 0; margin:auto;">
<tbody>
<tr align="center">
<th width="120">&nbsp;</th>
<th width="80">x=0</th>
<th width="80">x=2</th>
<th width="80">x=3</th>
<th width="80">x=4</th>
<th width="80">x=5</th>
</tr>
<tr align="center">
<td>&nbsp;</td>
<td>x&minus;2 &lt; 0</td>
<td>&nbsp;</td>
<td>x&minus;2 &gt; 0</td>
<td>&nbsp;</td>
<td>x&minus;2 &gt; 0</td>
</tr>
<tr align="center">
<td>&nbsp;</td>
<td>x&minus;4 &lt; 0 </td>
<td>&nbsp;</td>
<td>x&minus;4 &lt; 0 </td>
<td>&nbsp;</td>
<td>x&minus;4 &gt; 0</td>
</tr>
<tr align="center">
<td>(x&minus;2)/(x&minus;4) is</td>
<td>&gt; 0</td>
<td>0</td>
<td>&lt; 0</td>
<td>undefined</td>
<td>&gt; 0</td>
</tr>
</tbody></table>
</div>
<p>&nbsp;</p>
<p>That gives us a complete picture!</p>
<p class="larger">And where is it &gt; 0 ?</p>
<ul>
<li>Less than 2</li>
<li>More than 4</li>
</ul>
<p>So our result is: </p>
<p class="center"><span class="large">(&minus;&infin;, 2) <b>U</b> (4, +&infin;)</span></p>
</div>
<p>We did all that without drawing a plot!</p>
<p>But here is the <a href="../data/function-grapherd80d.html?func1=%28x-2%29/%28x-4%29&amp;xmax=10&amp;ymin=-6.17&amp;ymax=7.17">plot of (x&minus;2)/(x&minus;4)</a> so you can see:</p>
<p class="center"><img src="images/inequality-graph-xm2-xm4.gif" alt="Graph of Inequality" height="243" width="319"></p>
<p>&nbsp;</p>
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<a href="inequality.html">Inequalities</a>
<a href="../equal-less-greater.html">Less Than or Greater Than</a>
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