Section 3.2 Chapter 3 Graphing Fundamentals Solving and Graphing Compound Inequalities of a Single Variable TERMINOLOGY 3.2 Prerequisite Terms: Absolute Value Inequality Your definition New Terms to Learn: Compound Inequality Your definition Formal definition Example READING ASSIGNMENT 3.2 Sections 8.3 and 8. READING AND SELF-DISCOVERY QUESTIONS 3.2 1. What are the conjunctions that are used presenting the solution set to a compound inequality? 2. How can you identify a compound inequality? 118
3. What are two ways of presenting a compound inequality? KEY CONCEPTS 3.2 A compound inequality is two or more inequalities linked together with a conjunction (and, or). For example, [x < 4 or x > ] as well as [x > 4 and x < ]. Such inequalities occur in situations where conditions are put on the variables. Limitation/Caution: In the and situation both inequalities must be true, whereas in the or situation only one needs to be true (including when both are true). A three-part inequality has two inequality signs in the same direction, for example, 2 < 3 < which is the compound inequality [2 < 3 and 3 < ]. Limitation/Caution: The inequality signs need to be in the same direction and the compound inequality 2 < 3 and 4 < cannot be written as a three-part inequality. TECHNIQUE 3.2 CHANGING AN ABSOLUTE VALUE INEQUALITY INTO A COMPOUND INEQUALITY Absolute value inequalities occur in scientific measurements which are not exact. A measured length, x, is off by no more than 2 cm either way from the exact length, R, that is to say, R cm x cm < 2 cm or R x < 2 Limitation/Caution: Absolute value inequalities with more than one absolute value are difficult to understand and to solve. x < 2 implies 2 < x < 2 x > 2 implies x < 2 or x > 2 2 0 2 2 0 2 : : Change into a compound inequality: y 2 < 3 3 < y 2 < 3 Change into a compound inequality: 3 2x > 3 2x < or 3 2x > 119
Chapter 3 Graphing Fundamentals METHODOLOGY 3.2 SOLVING COMPOUND INEQUALITIES The methodology highlights, in its steps, how to change compound linear inequalities into simple linear inequalities and then solve these linear inequalities and then present the solution set symbolically and graphically including the appropriate notation for boundaries of the solution set. Limitation/Caution: The methodology does not work for non-linear compound inequalities. Solve for x: 1 ( x 3) < 1 (4x + 2) 10 Solve for x: 1 2(3x 1) < x 3 Steps Discussion 1 Set up the problem ( x 3) < (4x + 2) 10 2 Separate Compound Inequality If you are presented with a compound inequality, you must solve each inequality separately and then move to step 4 Problem 1: ( x 3) < (4x + 2) Problem 2: 1 (4 x + 2) 10 3 3 Solve the inequalities Use the Methodology for Solving Inequalities in One Variable. ( x) (3) < (4 x) + (2) 2 3 x 3 4x 2 6 < 6 + 2 2 3 3 3x 9 < 8x + 4 9 < x + 4 13 < x 13 < x (4 x) + (2) 10 3 3 4x 2 3 + 3 10 3 3 4x + 2 30 4x + 2 30 4x 28 x 7 120
Steps 3 Solve the inequalities Discussion Use the Methodology for Solving Inequalities in One Variable. 4 Present the compound solution set Both solutions must be true. If appropriate, present as a three-part inequality. 13 < x and x 7 13 < x 7 Describe the solution set Identify the end of the line segments, determine if open or closed, and identify direction of the solution sets Point is 13 is open and point is 7 is closed. 6 Present the compound solution set graphically Determine the end point(s), open or closed and the interval(s) 13 2 0 7 121
Chapter 3 Graphing Fundamentals Steps 7 Test and Validate solution set Discussion Pick two or more points, some in the solution set and some not in solution set (note: pick easy calculation points if possible) We choose 0, 7 (in the solution set) and 10, 20 (not in the solution set) ( x 3) < (4x + 2) 10 (0 3) < (4 0 + 2) 10 3 2 < 10 (yes) 1 (4 7 + 2) 10 3 1 28 (28) 10, 10 (yes) 3 3 ( 10 3) < (4( 10) + 2) 10 13 38 < 10 (no) (19 3) < (4 19 + 2) 10 (16) < (78) 10 8 < 26 10 (no) TIP FOR SUCCESS 3.2 You can solve a compound inequality by applying the properties to the whole inequality rather than splitting it into two parts if there are constants on both ends. 122 Example: 7 < 4x < 11 7 + < 4x + < 11+ 12 < 4x < 16 3 < x < 4 Validate with x = 0, 6 (not in the solution set) and x = 3. (in the solution set) 7 < 4 (0) < 11 7 < < 11 (no) 7 < 4(6) < 11 7 < 24 < 11 7 < 19 < 11 (no) 7 < 4 (3.) < 11 7 < 14 < 11 7 < 9 < 11 (yes)
CRITICAL THINKING QUESTIONS 3.2 1. What three things do you have to know to be able to represent a solution set graphically and in interval notation? 2. How do you represent an end point not in the solution set symbolically, graphically, and in interval notation? 3. How is the methodology for solving a compound inequality different from the methodology for solving a simple inequality? 4. What is the most common error that occurs when multiplying by the multiplicative inverse of a negative coefficient?. Conceptually, how is the meaning of and involved in finding the common solution set of a compound inequality? 6. What are the issues you can identify for why solving linear inequalities are difficult for some students? 123
Chapter 3 Graphing Fundamentals DEMONSTRATE YOUR UNDERSTANDING 3.2 Solve each of the following inequalities. 1. 1 2 < ( 2) 3 x < x 2. 2 < 2x 7 3. 7 3x + 3 4. 3( x 1) ( x) < 3 4 124
. 2x > 7 6. On the number line, represent the set of numbers for which 9 x > or x < 2. 2 IDENTIFY AND CORRECT THE ERRORS 3.2 In the second column, identify the error you find in each of the following worked solutions and describe the error made. Solve the problem correctly in the third column. Problem Describe Error Correct Process 1. Solve: 2 2 x < 4 Wrong process: Failure to change the direction of the 2 2 x < 4 2 2 x < 4 2 0 x < 2 0 x < 2 inequality when multiplying by a negative number. 2. Solve: 7 < x 4 < Wrong process: Misinterpreting the and conjunction. 7 < x 4 < 7 < x 4 and x 4 < 3 < x and x < 9 So, x < 9 12
Chapter 3 Graphing Fundamentals Problem Describe Error Correct Process 3. Solve: 2x < ( x 3) < 7 Wrong process: Failure to change the direction of the inequality when multiplying by a negative number. 2x < ( x 3) < 7 2x < ( x 3) and ( x 3) < 7 2x < x 1 and x 1 < 7 7x < 1 and x < 22 1 22 x < and x < 7 1 x < 7 4. Solve: 2x 3 > 4 Wrong process: Misinterpreting an absolute value inequality with an and rather than an or. 2x 3 > 4 ( ) 2x 3 < 4 and (2x 3) > 4 2x < 1 and 2x > 7 1 7 x > and x > 2 2 So, 7 x > 2. Solve: 8 < 2x 4 < 2 Wrong process: Making an incorrect three-part 8 < 2x 4 < 2 8 < 2x 4 and 2x 4 < 2 4 < 2x and 2x > 6 2 > x and x < 3 2 > x < 3 inequality for the solution set. 126