Comprehensive Guide to Linear and Absolute Value Inequalities

Foundations of Linear and Absolute Value Inequalities

• Conceptual Overview: Linear inequalities, including linear and absolute value types, utilize inequality symbols such as greater than ($>$), less than ($<$), greater than or equal to ($\geq$), and less than or equal to ($\leq$). These operate similarly to linear equations but result in infinite sets of solutions rather than a single value.
• Comparison of Equations and Inequalities:     * Equation Example: $2x + 6 = 10$. Subtracting $6$ and dividing by $2$ yields exactly one answer: $x = 2$. Only this single number makes the equation true.     * Inequality Example: $2x + 6 \geq 10$. Subtracting $6$ and dividing by $2$ yields the result $x \geq 2$. This represents an infinite set of numbers (two or anything larger than two) that will satisfy the statement.
• The Critical Rule of Negative Multiplication and Division: In linear inequalities, the procedure is identical to equations with one major exception: if you multiply or divide by a negative number, you must reverse the inequality symbol to maintain a true statement.     * Numerical Proof: Consider the true statement $10 > 2$.         * Multiply by $-3$: The result is $-30$ and $-6$. Since $-30$ is further left on the number line than $-6$, the statement $-30 > -6$ is false. To correct it, the symbol must be reversed: $-30 < -6$.         * Divide by a negative: Similarly, dividing $10 > 2$ by $-2$ yields $-5$ and $-1$. Since $-5$ is not greater than $-1$, the symbol must reverse: $-5 < -1$.     * The Action vs. The Outcome: A common misconception (illustrated by an anecdote concerning an eighth-grade class) is that the symbol flips because the answer is negative. This is incorrect. The symbol flips solely because of the action of multiplying or dividing by a negative number. If the multiplier is positive, the symbol remains the same, regardless of whether the final solution is positive or negative.

Taxonomy of Compound Inequalities

• Definitions and Mechanics:     * Compound Word Analogy: Just as "cardboard" is a compound word formed by "card" and "board," a compound inequality is formed by joining two or more inequalities with the word and or or.     * Operations: There are eight distinct types of compound inequalities, though common textbooks (such as those by Bob Blitzer) often only focus on one.
• And (Intersection/Overlap): This represents the set of all numbers that occur in both groups simultaneously. When graphing an "and" statement, only the overlapping region is included in the final solution.
• Or (Union/Anything Graphed): This represents all numbers that are part of either set or both. If a number is shaded in at least one of the individual graphs, it is included in the final solution.
• Eight Specific Scenarios for Compound Inequalities:     * Group 1: Both going right:         * Type 1 (And): $x > 2$ and $x > 5$. Since values must satisfy both, the solution is restricted by the higher value: $x > 5$.         * Type 2 (Or): $x > 2$ or $x > 5$. Since anything shaded counts, any value starting from the lower number works: $x > 2$.     * Group 2: Both going left:         * Type 3 (And): $x < 2$ and $x < 5$. The overlap occurs only below the smaller number: $x < 2$.         * Type 4 (Or): $x < 2$ or $x < 5$. Everything shaded counts, starting from the higher number down: $x < 5$.     * Group 3: Opposing directions:         * Type 5 (And - Toward each other): $x > 2$ and $x < 5$. This creates a "sandwich problem" where $x$ is trapped between the two values: $2 < x < 5$.         * Type 6 (Or - Toward each other): $x > 2$ or $x < 5$. Because every part of the number line is covered by at least one ray, the solution is all real numbers ($-\infty < x < \infty$).         * Type 7 (And - Away from each other): $x < 2$ and $x > 5$. There is no overlap between a value being less than two and greater than five simultaneously. The result is no solution or the empty set ($\emptyset$).         * Type 8 (Or - Away from each other): $x < 2$ or $x > 5$. These are separate sets with no connection. There is no simpler way to write this; the statement remains $x < 2$ or $x > 5$.

Absolute Value Inequalities

• Recognition and Branching: To solve an absolute value inequality, first isolate the absolute value expression. Then, determine if it is an "And" or "Or" problem based on the symbol.     * Less Than ($<$ or $\leq$): This is an And problem (often resulting in a "sandwich").     * Greater Than ($>$ or $\geq$): This is an Or problem.
• Setting Up the Branches:     * First Branch: Write the inequality exactly as it appears without the absolute value bars.     * Second Branch: Write the inequality, but reverse the symbol and change the sign of the constant (e.g., if it was $< 7$, it becomes $> -7$).
• Example 8 Trace:     * Problem A: $|x - 4| < 3$. Since it uses "less than," it is an "And" problem. Branches: $x - 4 < 3$ AND $x - 4 > -3$. Solution: $x < 7$ and $x > 1$, resulting in the sandwich $1 < x < 7$.     * Problem B: $|5 - 2x| > 7$. Since it uses "greater than," it is an "Or" problem. Branches: $5 - 2x > 7$ OR $5 - 2x < -7$. After solving and reversing symbols for negative division ($-2x > 2 \rightarrow x < -1$ and $-2x < -12 \rightarrow x > 6$), the answer remains $x < -1$ or $x > 6$.
• Example 9: Multi-Step Isolation: $-2|3x + 5| + 7 \geq -13$     1. Subtract $7$: $-2|3x + 5| \geq -20$.     2. Divide by $-2$: This necessitates a symbol reversal, changing the problem from $\geq$ to $\leq$. The new statement is $|3x + 5| \leq 10$.     3. Because it is now a "less than or equal to" problem, it is an And problem.     4. Branches: $3x + 5 \leq 10$ AND $3x + 5 \geq -10$.     5. Solve: $3x \leq 5 \rightarrow x \leq \frac{5}{3}$; $3x \geq -15 \rightarrow x \geq -5$.     6. Combine into a sandwich: $-5 \leq x \leq \frac{5}{3}$.

Questions & Discussion

• Logical Operations: A student asked if there are other operations like "Not Or." The instructor clarified that while such operations exist in Boolean algebra and computer science applications, they are not used in this specific mathematical context.
• Sign Flipping Clarity: A student questioned why the sign didn't change when dividing $3x$ by $3$ in the final steps of Example 9. The instructor reiterated that the rule only applies when dividing/multiplying by a negative number. Since the $3$ in the denominator was positive, the sign was preserved.
• Graphing Rules: For graphing, a "bar under the symbol" ($\leq, \geq$) indicates a closed circle (inclusive). No bar ($<, >$) indicates an open circle (exclusive).
• Homework Procedures:     * Students scoring an A or B on an exam are automatically given a 100 on their homework because their performance implies completion.     * Students scoring C, D, or failing must email their homework.     * Submission requirement: Send ONLY the first page of each section (sections 1.2, 1.4, 1.5, 1.6, and 1.7).     * Format: All files must be sent as a PDF via email. Canvas messages and Google Drive links requiring passwords are not accepted.