6.3 Linear Equations, Inequalities & Absolute Value

6.3 Linear Equations, Inequalities & Absolute Value

Linear equations and inequalities represent fundamental mathematical relationships where variables are raised exclusively to the first power. The AFQT Mathematics Knowledge subtest requires rapid, accurate execution of algebraic techniques to isolate variables, handle inequality flips, and solve absolute value conditions.

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Solving Single-Variable Linear Equations

A linear equation in one variable can be written in the standard form $ax + b = c$, where $a \neq 0$. Isolating $x$ requires applying properties of equality (doing the exact same mathematical operation to both sides of the equation).

Systematic 5-Step Equation Solver Framework

1. Clear Fractions and Decimals: Multiply every term by the Least Common Denominator (LCD) of all fractions or by an appropriate power of 10 for decimals.
2. Simplify Both Sides Independently: Expand parentheses using the Distributive Property ($a(b+c) = ab + ac$) and combine like terms on each side.
3. Isolate Variable Terms: Use addition or subtraction properties of equality to collect all terms containing the variable on one side.
4. Isolate Constant Terms: Use addition or subtraction properties of equality to move all numerical constants to the opposite side.
5. Isolate the Variable: Multiply or divide by the variable's coefficient to achieve $x = \text{value}$, then check by back-substituting into the original equation.

Comprehensive Multi-Step Example

Solve for $x$: $\frac{3(x - 2)}{4} - \frac{x + 1}{3} = \frac{1}{2}$

1. Clear Fractions: The denominators are 4, 3, and 2. The LCD is 12. Multiply every term on both sides by 12: $12 \cdot \left[\frac{3(x - 2)}{4}\right] - 12 \cdot \left[\frac{x + 1}{3}\right] = 12 \cdot \left[\frac{1}{2}\right]$ $3 \cdot [3(x - 2)] - 4 \cdot [x + 1] = 6$ $9(x - 2) - 4(x + 1) = 6$
2. Distribute & Combine Like Terms: $9x - 18 - 4x - 4 = 6$ $5x - 22 = 6$
3. Isolate Constant & Variable: $5x = 6 + 22 \implies 5x = 28$
4. Solve for $x$: $x = \frac{28}{5} \quad \text{or} \quad 5.6$

Special Solution Outcomes

• Single Unique Solution: Occurs when $x$ equals a specific numerical value (e.g., $x = 4$).
• No Solution (Inconsistent Equation): Occurs when variable terms cancel completely, leaving a false statement (e.g., $0 = 7$).
• Infinitely Many Solutions (Identity Equation): Occurs when variable terms cancel completely, leaving a true statement (e.g., $5 = 5$ or $0 = 0$).

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Linear Inequalities in One Variable

Linear inequalities use order symbols ($<, \le, >, \ge$) to describe a range of solutions rather than a single fixed point.

⚡ THE GOLDEN RULE OF INEQUALITIES: Whenever you multiply or divide both sides of an inequality by a negative number, you MUST reverse (flip) the direction of the inequality symbol.

Why Does the Inequality Sign Flip?

Consider the true numerical statement: $3 < 7$.

• Multiply both sides by $+2 \implies 6 < 14$ (True).
• Multiply both sides by $-2 \implies -6 \text{ vs } -14$. On a number line, $-6$ lies to the right of $-14$, so $-6 > -14$. The sign must flip to remain true.

Multi-Step Inequality Example

Solve for $x$: $-4(2x - 3) \ge 36$

1. Distribute $-4$: $-8x + 12 \ge 36$
2. Subtract 12 from both sides: $-8x \ge 24$
3. Divide both sides by $-8$ (MUST FLIP INEQUALITY SIGN FROM $\ge$ TO $\le$): $x \le \frac{24}{-8} \implies x \le -3$

Solution Set = All real numbers less than or equal to $-3$ (Interval: $(-\infty, -3]$).

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Absolute Value Equations & Inequalities

The absolute value of a real number, written $|x|$, represents its non-negative distance from zero on the real number line. Because distance is never negative, $|x| \ge 0$ for all real numbers.

Absolute Value Equations: $|ax + b| = c$

• If $c < 0$: No Solution (distance cannot be negative).
• If $c = 0$: Single solution $ax + b = 0$.
• If $c > 0$: Split into two separate linear equations: $ax + b = c \quad \text{OR} \quad ax + b = -c$

Example: Solve $|3x - 5| = 16$

3x - 5 = 16 & \text{OR} & 3x - 5 = -16 \\ 3x = 21 & & 3x = -11 \\ x = 7 & & x = -\frac{11}{3} \end{array}$$ Both solutions must be checked in the original absolute value equation. ### Absolute Value Inequalities To solve absolute value inequalities where $c > 0$, memorize the split rules: 1. **Less Than / Less Than or Equal To ("Less thAND")**: $|ax + b| \le c$ Translates into a **three-part continuous compound conjunction**: $$-c \le ax + b \le c$$ - *Example*: Solve $|2x - 7| < 9$ $$-9 < 2x - 7 < 9 \implies -2 < 2x < 16 \implies -1 < x < 8$$ 2. **Greater Than / Greater Than or Equal To ("GreatOR")**: $|ax + b| \ge c$ Translates into a **two-part disjunction joined by OR**: $$ax + b \le -c \quad \text{OR} \quad ax + b \ge c$$ - *Example*: Solve $|4x + 3| \ge 15$ $$4x + 3 \le -15 \quad \text{OR} \quad 4x + 3 \ge 15$$ $$4x \le -18 \implies x \le -4.5 \quad \text{OR} \quad 4x \ge 12 \implies x \ge 3$$