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2.8 Solving Linear Inequalities. Definition. An inequality is an algebraic expression related by < “is less than,” ≤ “is less than or equal to,” > “is greater than,” or ≥ “is greater than or equal to.”. Linear Inequality in One Variable
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Definition. An inequality is an algebraic expression related by < “is less than,” ≤ “is less than or equal to,” > “is greater than,” or ≥ “is greater than or equal to.” Linear Inequality in One Variable A linear inequality in one variable can be written in the form where A, B, and C represent real numbers, and A ≠ 0. We solve an inequality by finding all real number solutions of it. For example, the solution set {x | x ≤ 2} includes all real numbers that are less than or equal to 2, not just the integers less than or equal to 2. Slide 2.8-3
Objective 1 Graph intervals on a number line. Slide 2.8-4
Graph intervals on a number line. Graphing is a good way to show the solution set of an inequality. We graph all the real numbers belonging to the set {x | x ≤ 2} by placing a square bracket at 2 on a number line and drawing an arrow extending from the bracket to the left (to represent the fact that all numbers less than 2 are also part of the graph). Slide 2.8-5
Graph intervals on a number line. (cont’d) The set of numbers less than or equal to 2 is an example of an interval on the number line. To write intervals, we use interval notation. For example, the interval of all numbers less than or equal to 2 is written (−∞, 2]. The negative infinitysymbol −∞ does not indicate a number, but shows that the interval includes all real numbers less than 2. As on the number line, the square bracket indicates that 2 is part of the solution. A parentheses is always used next to the infinity symbol. The set of real numbers is written as (−∞, ∞). Slide 2.8-6
Graphing Intervals on a Number Line CLASSROOM EXAMPLE 1 Write each inequality in interval notation, and graph the interval. Solution: Solution: Slide 2.8-7
Some texts use a solid circle ● rather than a square bracket to indicate the endpoint is included in a number line graph. An open circle is used to indicate noninclusion, rather than a parentheses. Graph intervals on a number line. (cont’d) Keep the following important concepts regarding interval notation in mind: 1. A parenthesis indicates that an endpoint is not included in a solution set. 2. A bracket indicates that an endpoint is included in a solution set. 3. A parenthesis is always used next to an infinity symbol, −∞ or ∞. 4. The set of all real numbers is written in interval notation as (−∞,∞). Slide 2.8-8
Objective 2 Use the addition property of inequality. Slide 2.8-9
Use the addition property of inequality. Addition Property of Inequality If A, B, and C represent real numbers, then the inequalities and Have exactly the same solutions. That is, the same number may be added to each side of an inequality without changing the solutions. As with the addition property of equality, the same number may be subtracted from each side of an inequality. Slide 2.8-10
Use the addition property of inequality. (cont’d) Because an inequality has many solutions, we cannot check all of them by substitutions as we did with the single solution of an equation. Thus, to check the solutions of an inequality, first substitute into the equation the boundary point of the interval and another number from within the interval to test that they both result in true statements. Next, substitute any number outside the interval to be sure it gives a false statement. Slide 2.8-11
Using the Addition Property of Inequality CLASSROOM EXAMPLE 2 Solution: Solve the inequality, and graph the solution set. Slide 2.8-12
Objective 3 Use the multiplication property of inequality. Slide 2.8-13
Use the multiplication property of inequality. The addition property of inequality cannot be used to solve an inequality such as 4x ≥ 28. This inequality requires the multiplication. Multiply each side of the inequality 3 < 7 by the positive number 2. True Now multiply by each side of 3 < 7 by the negative number −5. False To get a true statement when multiplying each side by −5, we must reverse the direction of the inequality symbol. True Slide 2.8-14
Use the multiplication property of inequality. (cont’d) Multiplication Property of Inequality If A, B, andC, with C≠ 0, 1. if C is positive, then the inequalities and have exactly the same solutions; 2. if Cis negative, then the inequalities and have exactly the same solutions. That is, each of an inequality may be multiplied by the same positive number without changing the solutions. If the multiplier is negative, we must reverse the direction of the inequality symbol. As with the multiplication property of inequality, the same nonzero number may be divided into each side of an inequality. Slide 2.8-15
Using the Multiplication Property of Inequality CLASSROOM EXAMPLE 3 Solve the inequality, and graph the solution set. Solution: Slide 2.8-16
Objective 4 Solve linear inequalities by using both properties of inequality. Slide 2.8-17
Solve linear inequalities by using both properties of inequality. Solving a Linear Inequality Step 1:Simplify each side separately. Use the distributive property to clear parentheses and combine like terms on each side as needed. Step 2:Isolate the variable terms on one side. Use the addition property of inequality to get all terms with variables on one side of the inequality and all numbers on the other side. Step 3:Isolate the variable.Use the multiplication property of inequality to change the inequality to the form “variable < k” or “variable > k,” where k is a number. Remember: Reverse the direction of the inequality symbol only when multiplying or dividing each side of an inequality by a negative number.. Slide 2.8-18
Solving a Linear Inequality CLASSROOM EXAMPLE 4 Solve the inequality, and then graph the solution set. Solution: Slide 2.8-19
Solving a Linear Inequality CLASSROOM EXAMPLE 5 Solve the inequality, and graph the solution set. Solution: Slide 2.8-20
CLASSROOM EXAMPLE 6 Solving a Linear Inequality with Fractions Solve and graph the solution set. Solution: Multiply by 4. Distributive property. Multiply. Distributive property. Subtract 11. Slide 2.8- 20
CLASSROOM EXAMPLE 6 Solving a Linear Inequality with Fractions (cont’d) Subtract 11. Subtract 3m. Divide 2. Reverse the inequality symbol when dividing by a negative number. The solution set is the interval [−13/2, ). [ Slide 2.8- 21
Objective 5 Solve applied problems by using inequalities. Slide 2.8-23
Solve applied problems by using inequalities. Inequalities can be used to solve applied problems involving phrases that suggest inequality. The table gives some of the more common such phrases, along with examples and translations. In general, to find the average of n numbers, add the numbers and divide by n. We use the same six problem-solving steps from Section 2.4, changing Step 3 to “Write an inequality.”, instead of “Write an equation.” Do not confuse statements such as “5 is more than a number” with phrases like “5 more than a number.” The first of these is expressed as 5 > x, while the second is expressed as x + 5 or 5 + x. Slide 2.8-24
CLASSROOM EXAMPLE 7 Using a Linear Inequality to Solve a Rental Problem A rental company charges $5 to rent a leaf blower, plus $1.75 per hr. Marge Ruhberg can spend no more than $26 to blow leaves from her driveway and pool deck. What is the maximumamount of time she can use the rented leaf blower? Step 1 Readthe problem again. What is to be found? The maximum time Marge can afford to rent the blower. What is given? The flat rate to rent the leaf blower, the additional hourly charge to rent the leaf blower, and the maximum amount that Marge can spend. Step 2 Assign a variable. Let h = the number of hours she can rent the blower. Solution: Slide 2.5- 24
CLASSROOM EXAMPLE 7 Using a Linear Inequality to Solve a Rental Problem (cont’d) Step 3 Write an inequality. She must pay $5, plus $1.75 per hour for h hours and no more than $26. Cost of is no renting more than 26 5 + 1.75h≤ 26 Step 4 Solve.1.75h≤ 21 h≤ 12 Step 5 State the answer. She can use the leaf blower from a maximum of 12 hours. Step 6 Check. If she uses the leaf blower for 12 hr, she will spend 5 + 1.75(12) = 26 dollars, the maximum. Subtract 5. Divide by 1.75. Slide 2.5- 25
Finding an Average Test Score CLASSROOM EXAMPLE 8 Maggie has scores of 98, 86, and 88 on her first three tests in algebra. If she wants an average of at least 90 after her fourth test, what score must she make on that test? Solution: Let x = Maggie’s fourth test score. Maggie must get greater than or equal to an 88. Slide 2.8-27
Objective 6 Solve linear inequalities with three parts. Slide 2.8-28
For some applications, it is necessary to work with a three-part inequality such as where x +2 is between 3 and 8. To solve this inequality, we subtract 2 from each of the three parts of the inequality. Solve linear inequalities with three parts. Inequalities that say the one number is between two other numbers are three-part inequalities. For example, says that 5 is between −3 and 7. Slide 2.8-29
Solve linear inequalities with three parts. (cont’d) The idea is to get the inequality in the form a number< x <another number, using “is less than.” The solution set can then easily be graphed. When inequalities have three parts, the order of the parts is important. It would be wrong to write an inequality as 8 < x + 2 < 3, since this would imply 8 < 3, a false statement. In general, three-part inequalities are written so that the symbols point in the same direction and both point toward the lesser number. Slide 2.8-30
Solving Three-Part Inequalities CLASSROOM EXAMPLE 9 Write the inequality in interval notation, and graph the interval. Solution: Slide 2.8-31
Solving Three-Part Inequalities CLASSROOM EXAMPLE 10 Solve the inequality, and graph the solution set. Solution: Remember to work with all three parts of the inequality. Slide 2.8-32