One-Variable Compound Inequalities

INSTRUCTOR GUIDANCE EXAMPLE: Week Two Discussion One -Variable Compound Inequalities compound inequality and or intersection union This is my “and” compound inequality : -7 ≤ 5 + 3x ≤ 20 What that means is the inequality must fulfill two conditions at the same time. It means 5 + 3x must be equal to or less than 2 0 and also at the same time greater than or equal to - 7. I think of these as “between” inequalities because it turns out that the solution set for x will be between two numbers. Now I will find out what those two numbers are. -7 ≤ 5 + 3x ≤ 20 Subtract 5 from all three parts of the inequality. -7 – 5 ≤ 5 – 5 + 3x ≤ 20 – 5 -12 ≤ 3x ≤ 15 Divide all three parts by 3 -12 ≤ 3x ≤ 15 3 3 3 -4 ≤ x ≤ 5 So any value of x greater than or equal to -4 and less than or e qual to 5 will make this inequality true. This -4 ≤ x ≤ 5 is how this compound inequality is written algebraically. As an intersection of sets it would look like [ -4, ) (- , 5] which equals [ -4, 5] in interval notation. <---------------- [----------- |--------------- ]-------------- > Here is a number line graph of the -4 0 5 solution set . The square brackets mean that the end points are included in the solution set; notice the green highlighting extends through the square brackets as well. This is my “ or” compound inequality : 4 – x ≥ 1 or 6x – 3 > 27 What this means is that there are two conditions and one of them must be true with any given x from the solution set but both cannot be true at the same time. Since the solution will turn out to be two disjoint intervals, I am go ing to solve each part of the inequality separately. 4 – x ≥ 1 Subtract 4 from both sides. 4 – 4 – x ≥ 1 – 4 – x ≥ – 3 We must pay close attention to that negative in front of x. To remove it I must divide both sides of the inequality by -1 which also means I must flip the inequality symbol over so it points the other direction. – x ≤ – 3 Symbol is flipped. -1 -1 x ≤ 3 This is one part of my “or ” compound inequality . 6x – 3 > 27 Add 3 to both sides. 6x – 3 + 3 > 27 + 3 6x > 30 Divide bo th sides by 6 , but it is positive, so no flipping involved. 6x > 30 6 6 x > 5 This is the other part of my “or ” compound inequality . The complete solution set written algebraically is x ≤ 3 or x > 5 The solution set written in interval notation is the union of two intervals (- , 3] (5, ) Here is a number line graph of the solution set: <-------------- |------- ]------ (------------------------------ > 0 3 5 Notice that the 3 is included in the solution set b ut 5 is not.