Instructor Guidance Example Week Two Discussion One Variable

Instructor Guidance Example Week Two Discussionone Variable Compound

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 20 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. -12/3 ≤ 3x/3 ≤ 15/3 => -4 ≤ x ≤ 5 So any value of x greater than or equal to -4 and less than or equal 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] in interval notation. Here is a number line graph of the -4 to 5 solution set. The square brackets mean that the end points are included in the solution set; the green highlighting extends through the square brackets as well. This is my “or” compound inequality: 4 – x ≥ 1 or 6x – 3 > 27. This means there are two conditions and one of them must be true with any given x, but both cannot be true at the same time. Since the solution will turn out to be two disjoint intervals, I am going to solve each part separately.

First part: 4 – x ≥ 1. Subtract 4 from both sides: 4 – 4 – x ≥ 1 – 4 => – x ≥ – 3. To remove the negative sign in front of x, divide both sides by -1 and flip the inequality: x ≤ 3. Second part: 6x – 3 > 27. Add 3 to both sides: 6x > 30. Divide both sides by 6: x > 5. The complete solution set written algebraically is x ≤ 3 or x > 5. The solution set in interval notation is (-∞, 3] ∪ (5, ∞). Here is a number line graph: from negative infinity to 3, inclusive, and then from greater than 5 to infinity. Notice that 3 is included in the solution set but 5 is not.

Paper For Above instruction

The study of inequalities, particularly compound inequalities involving "and" and "or" relations, is fundamental in algebra and its applications. These inequalities help define ranges of solutions that satisfy multiple conditions simultaneously or alternative conditions separately. Understanding how to solve and interpret these inequalities is essential for advanced mathematics, science, and engineering.

Firstly, an "and" compound inequality, such as -7 ≤ 5 + 3x ≤ 20, indicates that a value of x must satisfy both inequalities simultaneously. To solve this, one isolates x by manipulating the inequality while maintaining its compound structure. Subtracting 5 from all parts gives -12 ≤ 3x ≤ 15, and dividing through by 3 yields -4 ≤ x ≤ 5. Graphically, this solution set is represented as a closed interval [-4, 5], indicating that x can take any value within and including these endpoints. Numerically, the solution set includes all real numbers between -4 and 5.

In contrast, an "or" compound inequality, such as 4 – x ≥ 1 or 6x – 3 > 27, defines two separate solution sets. Solving each independently: for the first inequality, subtracting 4 yields –x ≥ –3, and dividing by -1 and flipping yields x ≤ 3; for the second, adding 3 yields 6x > 30, and dividing by 6 gives x > 5. The overall solution encompasses all x satisfying either condition, represented as the union of two intervals: (-∞, 3] and (5, ∞). On a number line, these intervals are visually distinguished, with the boundary points highlighted accordingly. This illustrates the disjoint nature of solutions for "or" inequalities and emphasizes the importance of understanding the logical connection between multiple conditions.

Practically, these concepts are vital when defining permissible ranges of variables in various contexts, such as engineering tolerances, economic thresholds, and scientific measurements. Mathematical proficiency in solving inequalities also underpins calculus, optimization, and statistical modeling, making it a critical skill for students pursuing STEM fields.

Moreover, visual representations like number line graphs enhance comprehension by providing intuitive insight into the solution sets’ structure. Algebraic solutions combined with graphical interpretations form a comprehensive approach to understanding inequalities.

In conclusion, mastering the solving and interpretation of compound inequalities is crucial for mathematical literacy and practical problem-solving. Recognizing the difference between "and" (intersection) and "or" (union) solutions aids in approaching complex real-world scenarios systematically and effectively, reinforcing the importance of algebraic reasoning in various disciplines.

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