Linear Inequalities, Absolute Value Equations, and Absolute Value Inequalities

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Section 1.2: Joint Inequalities and Domain Analysis

• Structure of Joint Inequalities:

  • A joint (or compound) inequality consists of an algebraic expression positioned between two boundary values separated by inequality symbols.
  • Example form: $a \text{ < } \text{Algebraic Expression} \text{ < } b$.

• Fundamental Operational Rule for Joint Inequalities:

  • Any operation performed to solve a joint inequality must be applied simultaneously to all three terms (left boundary, middle expression, and right boundary).

• Solving Joint Inequalities Step-by-Step Example:

  • Consider a joint inequality containing a fraction with a total denominator of $-3$:
  • Step 1: Multiply all three terms by $-3$ to eliminate the denominator.
  • Reversal Rule: Multiplying or dividing any inequality by a negative number requires reversing (flipping) the inequality signs.
  • Multiplying boundaries yields updated numerical limits (e.g., $-3$ on one side and $18$ on the other).
  • Step 2: Subtract $5$ from all three terms to isolate the linear term in the middle, yielding intermediate boundaries of $13$ and $-8$.
  • Step 3: Divide all three terms by $2$ to fully isolate $x$.
  • Final solution bounds: $x$ must be strictly greater than $-4$ and less than or equal to $\frac{13}{2}$.
  • Interval Notation:
  • Expressed as $\left(-4, \frac{13}{2}\right]$.
  • The soft parenthesis $\left(\right.$ indicates that $-4$ is excluded from the solution set.
  • The square bracket $\left]\right.$ indicates that $\frac{13}{2}$ is included in the solution set.

• Application of Inequalities: Finding the Domain of a Function:

  • Inequalities and algebraic restrictions are used to determine the set of real numbers that produce valid real outputs for a function.
  • Division by Zero Restriction: Division by zero is undefined in mathematics; denominators must never equal zero.
  • Function Example:
  • Given the function $f(x) = \frac{2}{x - 3}$.
  • To find non-permissible values, set the denominator equal to zero and solve: $x - 3 = 0 \implies x = 3$.
  • The domain consists of all real numbers except $x = 3$
  • Domain in Interval Notation:
  • $(-\infty, 3) \cup (3, \infty)$.
  • Soft parentheses around $3$ indicate that $3$ is strictly omitted.

Section 1.3: Absolute Value Equations

• Definition of Absolute Value:

  • Represented using vertical bars around a variable or expression: $|x|$.
  • Piecewise Definition:$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$
  • Geometric Meaning: Represents the distance of $x$ from $0$ on a real number line.
  • Numerical Demonstrations:
  • $|-3| = -(-3) = 3$
  • $|3| = 3$
  • $|0| = 0$
  • Both $-3$ and $3$ are exactly $3$ units away from $0$ on the number line.
  • Consequently, an absolute value equation like $|x| = 4$ has two distinct solutions: $x = 4$ and $x = -4$.

• Fundamental Property of Absolute Value Equations:

  • Let $X$ be an algebraic expression and $k$ be a positive real number ($k > 0$):     $|X| = k \implies X = k \quad \text{or} \quad X = -k$
  • Case when $k < 0$: If $k$ is a negative number, the equation $|X| = k$ has no solution, because absolute distance is strictly non-negative ($|X| \ge 0$).
  • Case when $k = 0$: If $|X| = 0$, then $X = 0$ is the single unique solution.

• Step-by-Step Procedure for Absolute Value Equations:

  • Isolation First: Absolute value terms must be isolated completely on one side of the equation before applying positive/negative splitting rules.
  • Example: Solve $-5|x - 7| + 2 = -13$
  • Step 1: Subtract $2$ from both sides:       $-5|x - 7| = -15$
  • Step 2: Divide both sides by $-5$:       $|x - 7| = 3$
  • Step 3: Apply absolute value property ($k = 3 > 0$):       $x - 7 = 3 \quad \text{or} \quad x - 7 = -3$
  • Step 4: Solve both individual linear equations:
    • $x - 7 = 3 \implies x = 10$
    • $x - 7 = -3 \implies x = 4$
  • Final Solutions: $x = 10$ or $x = 4$.

• Common Pitfalls and Non-Properties:

  • Distribution Error: You cannot distribute coefficients into absolute value bars.
  • $-5|x - 7| \neq |-5x + 35|$
  • Explanation: Distributing across absolute value bars scales and flips values non-uniformly. The expression $-5|x - 7|$ is strictly non-positive, whereas $|-5x + 35|$ is strictly non-negative; they are equal only when $x = 7$.
  • Premature Splitting Error: Do not split an equation into positive/negative cases prior to isolating the absolute value expression.

• Multiplicative Property of Absolute Value:

  • For any algebraic expressions $A$ and $B$:     $|A \times B| = |A| \times |B|$
  • Directionality: Can be used to split a product inside an absolute value into separate absolute values, or combine two absolute values into one.
  • Proof of $|-b| = |b|$:
  • Let $A = -1$ and $B = b$
  • $|-b| = |(-1) \times b| = |-1| \times |b| = 1 \times |b| = |b|$
  • Simplification Example Using Multiplicative Property:
  • Given: $|-2x| + 5 = 13$
  • Step 1: Isolate absolute value: $|-2x| = 8$
  • Step 2: Apply multiplicative property: $|-2| \times |x| = 8 \implies 2|x| = 8$
  • Step 3: Divide by $2$: $|x| = 4$
  • Solutions: $x = \pm 4$

• Equations with Two Absolute Value Expressions:

  • General Form: $|A| = |B|$
  • Four logical scenarios exist:
  1. $A = B$
  2. $A = -B$
  3. $-A = B$ (algebraically equivalent to Case 2)
  4. $-A = -B$ (algebraically equivalent to Case 1)
  • Simplified Operational Rule: To solve $|A| = |B|$, evaluate only two linear equations:     $A = B \quad \text{or} \quad A = -B$
  • Worked Example: Solve $|2x + 7| = |x - 1|$
  • Case 1: $2x + 7 = x - 1$$2x - x = -1 - 7 \implies x = -8$
  • Case 2: $2x + 7 = -(x - 1)$$2x + 7 = -x + 1 \implies 3x = -6 \implies x = -2$
  • Solutions: $x = -8$ or $x = -2$.
  • Verification tests: Substituting $x = 4$ yields $15 = 3$ (false); substituting $x = -2$ yields $3 = 3$ (true); substituting $x = -8$ yields $9 = 9$ (true).

Section 1.3: Absolute Value Inequalities

• Less Than Absolute Value Inequalities ($<$ or $\le$):

  • Conceptual Meaning: The inequality $|x| < k$ seeks all points $x$ whose distance from $0$ on the number line is strictly less than $k$ units.
  • Property 1 ("Less Than" Rule):
  • If $X$ is an algebraic expression and $k > 0$:       $|X| < k \implies -k < X < k$$|X| \le k \implies -k \le X \le k$
  • Converts the absolute value inequality directly into a single joint inequality.
  • Geometric Graph: Represents a single bounded line segment centered relative to the origin.
  • Special Case ($k < 0$):
  • $|X| < k$ where $k < 0$ has no solution (the solution set is the empty set, $\emptyset$).
  • Terminology Distinction: "Undefined" refers to invalid operations like division by zero. An inequality with no solution is well-defined, possessing a solution set containing zero elements (an empty bag).
  • Worked Example: Solve $\left| \frac{3x + 2}{4} \right| \le 1$
  • Step 1: Clear denominator by multiplying by $4$ ($4 > 0$ preserves inequality direction):       $|3x + 2| \le 4$
  • Step 2: Convert to joint inequality via Property 1:       $-4 \le 3x + 2 \le 4$
  • Step 3: Subtract $2$ from all three parts:       $-6 \le 3x \le 2$
  • Step 4: Divide all three parts by $3$:       $-2 \le x \le \frac{2}{3}$
  • Solution set in interval notation: $\left[-2, \frac{2}{3}\right]$.

• Greater Than Absolute Value Inequalities ($>$ or $\ge$):

  • Conceptual Meaning: The inequality $|x| > k$ seeks all points $x$ whose distance from $0$ on the number line exceeds $k$ units.
  • Property 2 ("Greater Than" Rule):
  • If $X$ is an algebraic expression and $k > 0$:       $|X| > k \implies X < -k \quad \text{or} \quad X > k$$|X| \ge k \implies X \le -k \quad \text{or} \quad X \ge k$
  • Converts the absolute value inequality into two separate, disjoint linear inequalities.
  • Geometric Graph: Represents two unbounded rays pointing outward toward $-\infty$ and $+\infty$
  • Interval Notation Formatting:
  • Represented as a union of disjoint intervals: $(-\infty, -k) \cup (k, \infty)$.
  • The left boundary of each interval block must always contain the smaller value or $-\infty$; the right boundary contains the larger value or $+\infty$
  • Worked Example: Solve $-\frac{1}{3} \left| \frac{x}{2} + 3 \right| \le -2$
  • Step 1: Multiply both sides by $-3$ to isolate absolute value (flip inequality sign):       $\left| \frac{x}{2} + 3 \right| \ge 6$
  • Step 2: Apply Property 2 to split into two disjoint linear inequalities:       $\frac{x}{2} + 3 \le -6 \quad \text{or} \quad \frac{x}{2} + 3 \ge 6$
  • Step 3: Solve Inequality 1:       $\frac{x}{2} + 3 \le -6 \implies \frac{x}{2} \le -9 \implies x \le -18$
  • Step 4: Solve Inequality 2:       $\frac{x}{2} + 3 \ge 6 \implies \frac{x}{2} \ge 3 \implies x \ge 6$
  • Solution set in interval notation: $(-\infty, -18] \cup [6, \infty)$.

Summary of Special Cases and Key Rules

• Let $A$ be an algebraic expression and $k > 0$ be a positive real number:

  • Equation $|A| = -k$: No Solution (Absolute value cannot yield a negative value).
  • Inequality $|A| < -k$: No Solution (Absolute value is always $\ge 0$, so it cannot be less than a negative number).
  • Inequality $|A| > -k$: All Real Numbers, $(-\infty, \infty)$ (Non-negative numbers are always strictly greater than negative numbers).

• Summary Table of Solution Structures:

  • $|A| < k \implies -k < A < k$ (Yields a single bounded interval solution).
  • $|A| > k \implies A < -k \text{ or } A > k$ (Yields two disjoint unbounded intervals joined by a union $\cup$ symbol).

• Questions & Discussion:

  • Question: Why solve for negative values when the isolated equation equals a positive number?
  • Response: Values inside the absolute value expression can be negative prior to applying absolute value; both positive and negative values map to the same distance from zero.
  • Question: Difference between undefined and empty set?
  • Response: Undefined represents an invalid operational state (e.g., division by zero). An empty set is a well-defined mathematical set containing zero elements.