3.1 Linear Equations, Linear Inequalities & Absolute Value Equations

3.1 Linear Equations, Linear Inequalities & Absolute Value Equations

Algebraic equations and inequalities form the bedrock of college mathematics and quantitative reasoning. Mastery of linear relations, compound inequalities, and absolute value equations enables candidates to formulate and solve mathematical models with speed and precision.

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1. Solving Single-Variable Linear Equations

A linear equation in one variable is an equation that can be written in the standard form:

$ax + b = 0$

where $a$ and $b$ are real numbers and $a \neq 0$. The exponent on the variable is always 1.

Systematic 5-Step Solution Procedure

1. Clear Fractions and Decimals: Multiply every term on both sides of the equation by the Least Common Denominator (LCD) of all fractions or by an appropriate power of 10 for decimals.
2. Apply the Distributive Property: Eliminate all grouping symbols (parentheses, brackets) by multiplying: $a(b + c) = ab + ac$.
3. Combine Like Terms: Group and combine variable terms with variable terms and constant terms with constant terms on each side independently.
4. Isolate Variable Terms: Use the Addition and Subtraction Properties of Equality ($A = B \iff A \pm C = B \pm C$) to gather all terms containing the variable on one side and all constant terms on the opposite side.
5. Isolate the Variable: Use the Multiplication and Division Properties of Equality ($A = B \iff A \cdot C = B \cdot C$ for $C \neq 0$) to obtain a variable coefficient of 1.

Worked Example: Solving Linear Equations with Fractions

Solve the equation for $x$:

$\frac{2}{3}(x - 4) - \frac{1}{2}(3x + 1) = \frac{5}{6} - x$

Step 1: The denominators are 3, 2, and 6. The LCD is 6. Multiply every term by 6:

$6 \cdot \left[\frac{2}{3}(x - 4)\right] - 6 \cdot \left[\frac{1}{2}(3x + 1)\right] = 6 \cdot \left(\frac{5}{6}\right) - 6 \cdot x$

$4(x - 4) - 3(3x + 1) = 5 - 6x$

Step 2: Distribute across parentheses:

$4x - 16 - 9x - 3 = 5 - 6x$

Step 3: Combine like terms on the left side:

$-5x - 19 = 5 - 6x$

Step 4: Add $6x$ to both sides, then add 19 to both sides:

$-5x + 6x - 19 = 5$

$x - 19 = 5$

$x = 24$

Step 5: Check by substitution into the original equation:

• Left Side: $\frac{2}{3}(24 - 4) - \frac{1}{2}(3(24) + 1) = \frac{2}{3}(20) - \frac{1}{2}(73) = \frac{40}{3} - \frac{73}{2} = \frac{80 - 219}{6} = -\frac{139}{6}$
• Right Side: $\frac{5}{6} - 24 = \frac{5 - 144}{6} = -\frac{139}{6}$

Both sides evaluate to $-\frac{139}{6}$, confirming $x = 24$.

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2. Classification of Linear Equations

When simplifying a linear equation, three structural outcomes can occur:

[!NOTE] On the CLEP exam, if solving an equation causes the variable to cancel out completely, check the resulting constant statement:

• If the statement is true (e.g., $7 = 7$ or $0 = 0$), the solution set is all real numbers, ${x \mid x \in \mathbb{R}}$.
• If the statement is false (e.g., $0 = 9$ or $-4 = 5$), the equation has no solution, denoted as the empty set $\emptyset$.

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3. Linear Inequalities & Interval Notation

A linear inequality contains relational operators: $<$, $\le$, $>$, or $\ge$. Solving linear inequalities mirrors solving linear equations, with one vital distinction:

[!IMPORTANT] The Negative Multiplication / Division Rule: Whenever you multiply or divide both sides of an inequality by a negative number, you MUST reverse (flip) the direction of the inequality sign.

$-3x \le 15 \implies x \ge \frac{15}{-3} \implies x \ge -5$

Interval Notation Reference Table

Worked Example: Multi-Step Inequality with Sign Reversal

Solve the inequality and state the solution set in interval notation:

$-4(2x - 5) \ge 3(x + 3) + 22$

Step 1: Distribute on both sides:

$-8x + 20 \ge 3x + 9 + 22$

$-8x + 20 \ge 3x + 31$

Step 2: Collect variable terms on the left by subtracting $3x$:

$-11x + 20 \ge 31$

Step 3: Subtract 20 from both sides:

$-11x \ge 11$

Step 4: Divide by $-11$ and reverse the inequality symbol from $\ge$ to $\le$:

$x \le \frac{11}{-11} \implies x \le -1$

In interval notation, the solution set is $(-\infty, -1]$.

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4. Compound Inequalities: Conjunctions vs. Disjunctions

A compound inequality consists of two separate inequalities joined by either the connective word AND or OR.

+-----------------------------------------------------------------------------+
|                     COMPOUND INEQUALITY ARCHITECTURE                        |
|                                                                             |
|   CONJUNCTION (AND / Bounded)             DISJUNCTION (OR / Union)          |
|   ---------------------------             ------------------------          |
|   Condition: BOTH must be true            Condition: AT LEAST ONE is true   |
|   Operation: INTERSECTION (∩)             Operation: UNION (∪)              |
|   Graph: Overlapping region               Graph: Separate outer branches    |
|   Form: a < x < b                         Form: x < a OR x > b              |
|                                                                             |
|   Example: -3 <= 2x + 1 < 9               Example: 3x < -6 OR 2x + 5 >= 13  |
|   ===> -2 <= x < 4                        ===> x < -2 OR x >= 4             |
|   Interval: [-2, 4)                       Interval: (-inf, -2) U [4, inf)   |
+-----------------------------------------------------------------------------+

Conjunctions (AND / Bounded 3-Part Inequalities)

A conjunction requires both conditions to hold simultaneously. A common format is the 3-part inequality $a \le f(x) \le b$. To solve, perform identical operations across all three parts.

Example: Solve $-5 < 3x + 4 \le 19$

1. Subtract 4 from all three parts: $-9 < 3x \le 15$
2. Divide all three parts by 3: $-3 < x \le 5$
3. Interval notation: $(-3, 5]$

Disjunctions (OR / Unions)

A disjunction requires either the first condition, the second condition, or both to hold. Solve each inequality separately, then take the union ($\cup$) of their solution sets.

Example: Solve $4x - 1 < -13$ or $2(x - 3) \ge 4$

1. Solve left branch: $4x < -12 \implies x < -3 \implies (-\infty, -3)$
2. Solve right branch: $2x - 6 \ge 4 \implies 2x \ge 10 \implies x \ge 5 \implies [5, \infty)$
3. Combine via union: $(-\infty, -3) \cup [5, \infty)$

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5. Absolute Value Equations

The absolute value of a real number, denoted $|x|$, represents the geometric distance between $x$ and 0 on the real number line. Because distance is never negative, $|x| \ge 0$ for all $x \in \mathbb{R}$.

Solving $|ax + b| = c$

To solve an absolute value equation:

1. Isolate the absolute value expression on one side: $|ax + b| = c$.
2. Evaluate the sign of $c$:
   • If $c < 0$: There is no solution ($\emptyset$). An absolute value cannot equal a negative number.
   • If $c = 0$: There is exactly one linear equation: $ax + b = 0$.
   • If $c > 0$: Split into two distinct linear equations: $ax + b = c \quad \text{OR} \quad ax + b = -c$

Worked Example: Absolute Value Equation

Solve $3|2x - 5| - 7 = 14$

Step 1: Isolate the absolute value expression:

$3|2x - 5| = 21 \implies |2x - 5| = 7$

Step 2: Because $7 > 0$, split into two linear cases:

\text{Case 1: } 2x - 5 &= 7 & \text{Case 2: } 2x - 5 &= -7 \\ 2x &= 12 & 2x &= -2 \\ x &= 6 & x &= -1 \end{aligned}$$ The solution set is **$\{-1, 6\}$**. --- ## 6. Absolute Value Inequalities: "Less Than" vs. "Greater Than" Solving absolute value inequalities depends entirely on the direction of the inequality symbol relative to the isolated absolute value expression: | Inequality Form ($c > 0$) | Equivalent Compound Form | Algebraic Relationship | Interval Notation | | :--- | :--- | :--- | :--- | | $|u| < c$ | $-c < u < c$ | **Conjunction (AND)** — Distance is *less than* $c$ | $(-c, c)$ | | $|u| \le c$ | $-c \le u \le c$ | **Conjunction (AND)** — Distance is *at most* $c$ | $[-c, c]$ | | $|u| > c$ | $u > c \text{ or } u < -c$ | **Disjunction (OR)** — Distance is *greater than* $c$ | $(-\infty, -c) \cup (c, \infty)$ | | $|u| \ge c$ | $u \ge c \text{ or } u \le -c$ | **Disjunction (OR)** — Distance is *at least* $c$ | $(-\infty, -c] \cup [c, \infty)$ | > [!TIP] > **Memory Mnemonic:** > - $|u| < c$ $\rightarrow$ **Less thAND** $\rightarrow$ Conjunction (Sandwich: $-c < u < c$) > - $|u| > c$ $\rightarrow$ **GreatOR** $\rightarrow$ Disjunction (Two outer wings: $u < -c$ or $u > c$) ### Worked Example: Less Than (Bounded Conjunction) Solve $|3x - 4| \le 11$ **Step 1:** Express as a 3-part inequality: $$-11 \le 3x - 4 \le 11$$ **Step 2:** Add 4 across all three parts: $$-7 \le 3x \le 15$$ **Step 3:** Divide by 3: $$-\frac{7}{3} \le x \le 5$$ In interval notation, the solution set is **$\left[-\frac{7}{3}, 5\right]$**. ### Worked Example: Greater Than (Disjunction) Solve $|2x + 7| > 15$ **Step 1:** Split into two separate inequalities with the word OR: $$2x + 7 > 15 \quad \text{or} \quad 2x + 7 < -15$$ **Step 2:** Solve each inequality independently: - $2x > 8 \implies x > 4$ - $2x < -22 \implies x < -11$ In interval notation, the solution set is **$(-\infty, -11) \cup (4, \infty)$**. ### Special Boundary Cases with Negative Constants - **$|u| < -4$:** No solution ($\emptyset$). An absolute value is always non-negative ($\ge 0$), so it can never be less than a negative number. - **$|u| > -4$:** All real numbers $(-\infty, \infty)$. Since $|u| \ge 0$, it is strictly greater than any negative number for all real $u$. --- ## 7. Common CLEP Traps & Strategic Checkpoints - **Trap 1: Forgetting to Reverse the Inequality Sign:** When dividing by a negative coefficient (e.g., $-5x \le 20$), failing to flip $\le$ to $\ge$ produces the opposite incorrect half-line. - **Trap 2: Splitting Absolute Values Before Isolating:** Never split an equation like $2|x - 3| + 5 = 13$ into cases before subtracting 5 and dividing by 2. Always isolate the bars first: $|x - 3| = 4$. - **Trap 3: Blind Splitting with Negative Constants:** If you encounter $|2x + 1| = -5$, do not set up $2x + 1 = -5$ and $2x + 1 = 5$. The correct answer is immediately **no solution ($\emptyset$)**. - **Trap 4: Mixing up Parentheses and Brackets in Interval Notation:** Strict inequalities ($<, >$) require round parentheses $(, )$, while non-strict inequalities ($\le, \ge$) require square brackets $[, ]$. Inifinities ($-\infty, \infty$) always take round parentheses.