o Section 4: Compound Inequalities

Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Section 1.4 Compound Inequalities 53 Version: Fall 2007 1.4 Compound Inequalities This section discusses a technique that is used to solve compound inequalities , which is a phrase that usually refers to a pair of inequalities connected either by the word “and” or the word “or.” Before we begin with the advanced work of solving these inequalities, let’s first spend a word or two (for purposes of review) discussing the solution of simple linear inequalities. Simple Linear Inequalities As in solving equations, you may add or subtract the same amount from both sides of an inequality. Property 1. Let a and b be real numbers with a < b . If c is any real number, then a + c < b + c and a c < b c. This utility is equally valid if you replace the “less than” symbol with > , , or . l⚏ Example 2. Solve the inequality x + 3 < 8 for x . Subtract 3 from both sides of the inequality and simplify. x + 3 < 8 x + 3 3 < 8 3 x < 5 Thus, all real numbers less than 5 are solutions of the inequality. It is traditional to sketch the solution set of inequalities on a number line. 5 We can describe the solution set using set-builder and interval notation. The solu- tion is ( −∞ , 5) = { x : x < 5 } . An important concept is the idea of equivalent inequalities . Copyrighted material. See: 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
54 Chapter 1 Preliminaries Version: Fall 2007 Equivalent Inequalities. Two inequalities are said to be equivalent if and only if they have the same solution set. Note that this definition is similar to the definition of equivalent equations. That is, two inequalities are equivalent if all of the solutions of the first inequality are also solutions of the second inequality, and vice-versa. Thus, in Example 2 , subtracting three from both sides of the original inequality produced an equivalent inequality. That is, the inequalities x +3 < 8 and x < 5 have the same solution set, namely, all real numbers that are less than 5. It is no coincidence that the tools in Property 1 produce equivalent inequalities. Whenver you add or subtract the same amount from both sides of an inequality, the resulting inequality is equivalent to the original (they have the same solution set). Let’s look at another example. l⚏ Example 3. Solve the inequality x 5 4 for x . Add 5 to both sides of the inequality and simplify. x 5 4 x 5 + 5 4 + 5 x 9 Shade the solution on a number line. 9 In set-builder and interval notation, the solution is [9 , ) = { x : x 9 } You can also multiply or divide both sides by the same positive number. Property 4. Let a and b be real numbers with a < b . If c is a real positive number, then ac < bc and a c < b c . Again, this utility is equally valid if you replace the “less than” symbol by > , , or . The tools in Property 4 always produce equivalent inequalities.
Image of page 2
Section 1.4 Compound Inequalities 55 Version: Fall 2007 l⚏ Example 5. Solve the inequality 3 x ≤ − 18 for x Divide both sides of the inequality by 3 and simplify.
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern