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		<title>MathAdmin: Created page with &quot;&lt;div class=&quot;noautonum&quot;&gt;__TOC__&lt;/div&gt; ==Introduction== Inequalities allow us to talk about all of the numbers between two real numbers a and b. They are frequently used when ta...&quot;</title>
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		<updated>2015-10-03T01:20:27Z</updated>

		<summary type="html">&lt;p&gt;Created page with &amp;quot;&amp;lt;div class=&amp;quot;noautonum&amp;quot;&amp;gt;__TOC__&amp;lt;/div&amp;gt; ==Introduction== Inequalities allow us to talk about all of the numbers between two real numbers a and b. They are frequently used when ta...&amp;quot;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;&amp;lt;div class=&amp;quot;noautonum&amp;quot;&amp;gt;__TOC__&amp;lt;/div&amp;gt;&lt;br /&gt;
==Introduction==&lt;br /&gt;
Inequalities allow us to talk about all of the numbers between two real numbers a and b. They are frequently used when talking about safe temperature ranges for food. They can also be used when creating a modeling function since they tell&lt;br /&gt;
you for which input values the function makes sense.&lt;br /&gt;
&lt;br /&gt;
==Notation==&lt;br /&gt;
The collection of real numbers strictly between a and b is denoted (a, b), and is called an open interval. Another way to describe this collection of numbers is that it is all x such that &amp;lt;math&amp;gt; a &amp;lt; x &amp;lt; b&amp;lt;/math&amp;gt;&lt;br /&gt;
The collection  of numbers between a and b, inclusive, is denoted [a, b] and is called a closed interval. Another way to describe this collection of numbers is that it is all x such that &amp;lt;math&amp;gt; a \le x \le b&amp;lt;/math&amp;gt;&lt;br /&gt;
If we only want one of the endpoints we call the interval a half open interval. The notation corresponds as follows: [a, b) for &amp;lt;math&amp;gt;a \le x &amp;lt; b&amp;lt;/math&amp;gt; and (a, b] for &amp;lt;math&amp;gt; a &amp;lt; x \le b &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If we want to include &amp;lt;math&amp;gt; \infty \text{ or } -\infty &amp;lt;/math&amp;gt; we use parenthesis for any side of the inequality with an infinity. For example, if we wanted to denote &amp;lt;math&amp;gt; a \le \infty &amp;lt;/math&amp;gt; we would use &amp;lt;math&amp;gt; [a, \infty)&amp;lt;/math&amp;gt;&lt;br /&gt;
==Properties==&lt;br /&gt;
Before we start working on examples we need to discuss the properties. For the following properties a, b, and c are real number, unless otherwise stated.&lt;br /&gt;
&lt;br /&gt;
1) &amp;lt;math&amp;gt; a^2 \ge 0&amp;lt;/math&amp;gt; for any real number a.&lt;br /&gt;
&lt;br /&gt;
2) If &amp;lt;math&amp;gt; a &amp;lt; b \text{, then } a + c &amp;lt; b + c &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
3) If &amp;lt;math&amp;gt; a &amp;lt; b &amp;lt;/math&amp;gt; and if &amp;lt;math&amp;gt; c &amp;gt; 0 \text{, then } ac &amp;lt; bc &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4) If &amp;lt;math&amp;gt; a &amp;lt; b &amp;lt;/math&amp;gt; and if &amp;lt;math&amp;gt; c &amp;lt; 0 \text{, then } ac &amp;gt; bc &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
5) If &amp;lt;math&amp;gt; a &amp;gt; 0 \text{, then }\frac{1}{a} &amp;gt; 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
6) If &amp;lt;math&amp;gt; a &amp;lt; 0 \text{, then }\frac{1}{a} &amp;lt; 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Just like equations, there are some operations that leave the inequality direction unchanged.&lt;br /&gt;
&lt;br /&gt;
1) Simplifying either side of the inequality&lt;br /&gt;
&lt;br /&gt;
2) Adding or subtracting the same expression from both sides of the inequality&lt;br /&gt;
&lt;br /&gt;
3) Multiplying both sides of the inequality by the same positive expression&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
This section comprises some examples of problems involving inequalities that may arise&lt;br /&gt;
&lt;br /&gt;
Solve the following inequalities&lt;br /&gt;
1) &amp;lt;math&amp;gt; 3x + 5 \le 6x + 14&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
2) &amp;lt;math&amp;gt; -3 &amp;lt; 2x + 5 \le 10 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
3) &amp;lt;math&amp;gt; \frac{1}{3x - 5} &amp;gt; 0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Solutions:&lt;br /&gt;
&lt;br /&gt;
1) &amp;lt;center&amp;gt;&amp;lt;math&amp;gt; &lt;br /&gt;
\begin{array}{rcl}&lt;br /&gt;
3x + 5 &amp;amp; \le &amp;amp; 6x + 14\\&lt;br /&gt;
-9 &amp;amp; \le &amp;amp; 3x \\&lt;br /&gt;
-3 &amp;amp; \le &amp;amp; x&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&amp;lt;/center&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When solving inequalities, the final answer is sometimes required to be in interval notation.&lt;br /&gt;
For this problem that is &amp;lt;math&amp;gt; [-3, \infty)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
2) Here we can solve each inequality individually, and x has to satisfy both inequalities.&lt;br /&gt;
Thus, we have to solve &amp;lt;math&amp;gt; -3 &amp;lt; 2x + 5 \text{ and }2x + 5 \le 10 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For the first, we get -8 &amp;lt; 2x and -4 &amp;lt; x. For the last one we have &amp;lt;math&amp;gt; 2x \le 5 \text{ and } x \le \frac{5}{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
So &amp;lt;math&amp;gt; -4 &amp;lt; x \le \frac{5}{2}&amp;lt;/math&amp;gt; In interval notation, &amp;lt;math&amp;gt;(-4, \frac{5}{2}]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
3) For this problem we need the numerator and denominator to both be positive or both be negative. So we want to solve when 3x - 5 &amp;gt; 0. Notice that we do not include 3x - 5 = 0 since we cannot divide by 0.&lt;br /&gt;
Solving this inequality we find &amp;lt;math&amp;gt; x &amp;gt; \frac{5}{3}&amp;lt;/math&amp;gt;. In interval notation we have &amp;lt;math&amp;gt;(\frac{5}{3}, \infty)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Absolute Value and Inequalities==&lt;br /&gt;
&lt;br /&gt;
Inequalities involving absolute values are presented separately since the inequalities &amp;lt;math&amp;gt; \left| u\right| &amp;lt; a &amp;lt;/math&amp;gt; are equivalent to &amp;lt;math&amp;gt; -a &amp;lt; u &amp;lt; a \,&amp;lt;/math&amp;gt;, while &amp;lt;math&amp;gt;\, \left| u \right| &amp;gt; a &amp;lt;/math&amp;gt; is equivalent to u &amp;lt; -a or u &amp;gt; a.&lt;br /&gt;
In all of those inequalities &amp;lt;math&amp;gt; &amp;lt; \text{ or } &amp;gt; &amp;lt;/math&amp;gt; can be replaced with &amp;lt;math&amp;gt; \le \text{ or } \ge &amp;lt;/math&amp;gt;, respectively. Once we replace the inequality involving absolute values with an equivalent inequality&lt;br /&gt;
we can solve it the same way we solved the other inequalities before.&lt;br /&gt;
&lt;br /&gt;
Example:&lt;br /&gt;
Solve the inequality &amp;lt;math&amp;gt; \left| 4x + 6 \right| \le 5 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We use the first equivalence to create two connected inequalities: &amp;lt;math&amp;gt; -5 \le 4x + 6 \le 5&amp;lt;/math&amp;gt;.&lt;br /&gt;
We now separate the two inequalities and solve them individually reaching our final answer. Splitting the inequalities we get &amp;lt;math&amp;gt; -5 \le 4x + 6 \text{ and } 4x + 6 \le 5 &amp;lt;/math&amp;gt;.&lt;br /&gt;
Solving the first inequality we find &amp;lt;math&amp;gt; \frac{-11}{4} \le x&amp;lt;/math&amp;gt;. The second inequality yields the fact that &amp;lt;math&amp;gt; x \le \frac{-1}{4}&amp;lt;/math&amp;gt;.&lt;br /&gt;
Putting these together we can conclude, &amp;lt;math&amp;gt; \frac{-11}{6}\le x \le \frac{-1}{4}&amp;lt;/math&amp;gt; In interval notation the solution is &amp;lt;math&amp;gt; \left[\frac{-11}{4}, \frac{-1}{4}\right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Math_5|'''Return to Topics Page]]&lt;/div&gt;</summary>
		<author><name>MathAdmin</name></author>
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