14.2 Solving Linear Equations and Inequalities in One Variable
Key Takeaways
- Solving linear equations involves systematic inverse operations: eliminating grouping symbols via distribution, clearing fractional denominators using the Least Common Denominator (LCD), collecting variables on one side, and isolating the unknown.
- Equations produce three distinct structural outcomes: conditional equations (exactly one unique solution), identities (infinitely many solutions, simplifying to universally true identities like 0 = 0), and contradictions (no solution, simplifying to false statements like 0 = 7).
- Solving linear inequalities follows standard equation properties with one critical exception: whenever multiplying or dividing both sides by a negative number, the inequality sign must be immediately reversed (< becomes >, ≤ becomes ≥).
- Number line representations use open circles (○) for strict inequalities (<, >) and closed circles (●) for inclusive inequalities (≤, ≥); compound 'AND' inequalities describe bounded intersections, while compound 'OR' inequalities describe disjoint unions.
- Absolute value equations |ax + b| = c split into two distinct linear cases (ax + b = c or ax + b = -c for c ≥ 0); absolute value inequalities split into bounded sandwiches for 'less than' (|u| < c <=> -c < u < c) and disjoint disjunctions for 'greater than' (|u| > c <=> u > c or u < -c).
Solving Linear Equations and Inequalities in One Variable
Quick Answer: On the WEST-B (Objective 0017), linear equations and inequalities test your algebraic precision. For multi-step equations: Clear fractions by multiplying every term by the LCD; Distribute to eliminate grouping; Combine like terms; and Isolate the variable. Watch out for special classifications: if variables cancel leaving a true statement (e.g., 5 = 5), it is an Identity (Infinite Solutions); if variables cancel leaving a false statement (e.g., 0 = 8), it is a Contradiction (No Solution). For inequalities, always reverse the inequality direction when multiplying or dividing by a negative number. For absolute values: |ax + b| = c splits into ax + b = c or ax + b = -c; |u| < c becomes -c < u < c; and |u| > c becomes u > c or u < -c.
1. Multi-Step Linear Equation Solving Protocol
A linear equation in one variable can be written in the standard form ax + b = 0 (where a != 0). Solving it requires undoing operations in reverse order using the Addition, Subtraction, Multiplication, and Division Properties of Equality.
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| THE 6-STEP LINEAR EQUATION ALGORITHM |
| |
| Step 1: Clear Fractions / Decimals |
| Multiply every term on both sides by the LCD (or by 10^k for decimals). |
| Step 2: Eliminate Grouping Symbols |
| Apply the Distributive Property to expand all parentheses, brackets, and braces. |
| Step 3: Combine Like Terms on Each Side |
| Simplify the left side and right side independently (combine variable and constant terms)|
| Step 4: Collect Variable Terms on One Side |
| Use addition or subtraction to shift all variable terms to one side of the equals sign. |
| Step 5: Collect Constant Terms on the Other Side |
| Use addition or subtraction to isolate the variable term. |
| Step 6: Isolate the Variable (Divide or Multiply) |
| Divide both sides by the variable's coefficient to obtain x = value. Check solution! |
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Clearing Fractions with the LCD
Attempting to solve equations with mixed fractional coefficients without clearing denominators frequently causes arithmetic errors. Multiply every term by the Least Common Denominator (LCD):
Solve: (2x - 3)/4 - (x - 1)/6 = 5/3
- Determine the LCD: The denominators are 4, 6, and 3. The LCD = 12.
- Multiply every single term by 12: 12[(2x - 3)/4] - 12[(x - 1)/6] = 12(5/3)
- Simplify before expanding: 3(2x - 3) - 2(x - 1) = 4(5)
- Distribute: 6x - 9 - 2x + 2 = 20
- Combine like terms: 4x - 7 = 20
- Isolate x: 4x = 27 => x = 27/4 = 6.75
Clearing Decimals
Multiply all terms by 10, 100, or 1000 based on the maximum decimal places: 0.15(x - 20) + 0.05x = 0.20(30) Multiply entire equation by 100: 15(x - 20) + 5x = 20(30) => 15x - 300 + 5x = 600 => 20x = 900 => x = 45
2. Solution Classifications: Conditional, Identity, Contradiction
Not all linear equations yield a single numerical answer. On the WEST-B exam, you will encounter equations that produce all real numbers or no solution.
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| EQUATION CLASSIFICATION MATRIX |
| |
| +-------------------+----------------------------+-----------------------+------------------+ |
| | CLASSIFICATION | ALGEBRAIC RESULT | SOLUTION SET | GEOMETRIC MEANING| |
| +-------------------+----------------------------+-----------------------+------------------+ |
| | Conditional | x = c (single unique value)| Exactly one solution | Single point of | |
| | | e.g., 2x + 3 = 11 -> x = 4 | { c } | intersection | |
| +-------------------+----------------------------+-----------------------+------------------+ |
| | Identity | Variables cancel, leaving | Infinitely many | Coincident lines | |
| | | a TRUE identity (0 = 0) | solutions (all reals) | (same exact line)| |
| | | e.g., 2(x+3) = 2x + 6 | (-∞, ∞) or ℝ | | |
| +-------------------+----------------------------+-----------------------+------------------+ |
| | Contradiction / | Variables cancel, leaving | No solution (null set)| Parallel lines | |
| | Inconsistent | a FALSE statement (0 = 8) | ∅ or { } | (never intersect)| |
| | | e.g., 3x + 5 = 3x - 2 | | | |
| +-------------------+----------------------------+-----------------------+------------------+ |
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Detailed Algebraic Examples
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Identity Example: 4(2x - 3) + 5 = 8x - 7 => 8x - 12 + 5 = 8x - 7 => 8x - 7 = 8x - 7 => 0 = 0 Because 0 = 0 is universally true, the equation is an Identity, and the solution is all real numbers (ℝ).
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Contradiction Example: 3(x + 4) - x = 2(x - 1) + 9 => 3x + 12 - x = 2x - 2 + 9 => 2x + 12 = 2x + 7 => 12 = 7 Because 12 = 7 is false, the equation is a Contradiction, and there is no solution (∅).
3. Solving Linear Inequalities & The Negative Sign-Flip Rule
A linear inequality involves relational inequality operators (<, >, ≤, ≥). The algebraic mechanics mirror linear equations, with one crucial exception.
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| THE NEGATIVE SIGN-FLIP RULE FOR INEQUALITIES |
| |
| When you MULTIPLY or DIVIDE both sides of an inequality by a NEGATIVE number, |
| you MUST REVERSE (FLIP) the direction of the inequality symbol: |
| |
| < flips to > > flips to < |
| ≤ flips to ≥ ≥ flips to ≤ |
| |
| Why? Consider true statement: -2 < 5 |
| Multiply both sides by -1: (-2)(-1) ? (5)(-1) ==> +2 > -5 (Sign must flip!) |
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Multi-Step Inequality Example
Solve: 5 - 3(2x - 4) ≥ 41
- Distribute -3: 5 - 6x + 12 ≥ 41
- Combine like terms: -6x + 17 ≥ 41
- Subtract 17: -6x ≥ 24
- Divide by -6 (FLIP THE SIGN!): x ≤ 24 / (-6) => x ≤ -4
4. Number Line Representation & Compound Inequalities
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| NUMBER LINE GRAPHING CONVENTIONS |
| |
| +-------------------+--------------------+------------------------+-------------------------+ |
| | SYMBOL | CIRCLE TYPE | MEANING / INCLUSION | INTERVAL NOTATION | |
| +-------------------+--------------------+------------------------+-------------------------+ |
| | < (less than) | Open Circle (○) | Value is NOT included | (-∞, a) | |
| +-------------------+--------------------+------------------------+-------------------------+ |
| | > (greater than) | Open Circle (○) | Value is NOT included | (a, ∞) | |
| +-------------------+--------------------+------------------------+-------------------------+ |
| | ≤ (less or equal) | Closed Circle (●) | Value IS included | (-∞, a] | |
| +-------------------+--------------------+------------------------+-------------------------+ |
| | ≥ (great or equal)| Closed Circle (●) | Value IS included | [a, ∞) | |
| +-------------------+--------------------+------------------------+-------------------------+ |
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Compound Inequalities: "AND" vs. "OR"
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Compound "AND" (Conjunction / Intersection ∩):
- Condition: Both inequalities must be satisfied simultaneously.
- Graph: A single bounded segment between two endpoints.
- Example: -3 < x ≤ 5 (Open circle at -3, closed circle at 5, shaded between).
- Solving Double Inequality: Keep all three parts balanced: -7 ≤ 2x + 3 < 15 => -10 ≤ 2x < 12 => -5 ≤ x < 6
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Compound "OR" (Disjunction / Union ∪):
- Condition: At least one inequality must be satisfied.
- Graph: Two separate rays pointing outward in opposite directions.
- Example: x < -2 or x ≥ 4 (Ray left from -2 with open circle; ray right from 4 with closed circle).
Compound Inequality Visual Graphs
Conjunction (AND): -3 < x ≤ 5 ----○===================●--------->
-3 5
Disjunction (OR): x < -2 or x ≥ 4 <===○-------------●=============>
-2 4
5. Absolute Value Equations and Inequalities
The absolute value |x| measures the distance between x and 0 on a number line. Distance is always non-negative (|x| ≥ 0).
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| ABSOLUTE VALUE RELATION RULES (c > 0) |
| |
| +-----------------------+-----------------------------+-------------------------------------+ |
| | RELATION TYPE | SPLIT RULE / EXPANSION | GRAPHICAL INTERPRETATION | |
| +-----------------------+-----------------------------+-------------------------------------+ |
| | |u| = c | u = c OR u = -c | Exactly two points at distance c | |
| +-----------------------+-----------------------------+-------------------------------------+ |
| | |u| < c (or ≤) | -c < u < c (bounded 'AND')| Distance from center is LESS than c | |
| | "Less th-AND" | | (Single continuous segment) | |
| +-----------------------+-----------------------------+-------------------------------------+ |
| | |u| > c (or ≥) | u > c OR u < -c | Distance from center is MORE than c | |
| | "Great-OR" | (disjoint 'OR') | (Two outward pointing rays) | |
| +-----------------------+-----------------------------+-------------------------------------+ |
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Absolute Value Equations: |ax + b| = c
Step 1: Isolate the absolute value expression on one side of the equation. Step 2: Check the sign of c:
- If c < 0: No solution (∅), since absolute value cannot equal a negative number.
- If c = 0: Solve the single linear equation ax + b = 0.
- If c > 0: Split into two separate linear equations: ax + b = c or ax + b = -c.
Example: Solve 3|2x - 5| + 7 = 28
- Isolate absolute value: 3|2x - 5| = 21 => |2x - 5| = 7
- Split into cases: 2x - 5 = 7 => 2x = 12 => x = 6 2x - 5 = -7 => 2x = -2 => x = -1
- Solution set: { -1, 6 }
Absolute Value Inequalities: Special Edge Cases
- Case 1: |u| < -5 => No Solution (∅) (Distance cannot be strictly less than a negative number).
- Case 2: |u| > -5 => All Real Numbers (ℝ) (Distance is always ≥ 0, which is always > -5).
6. Step-by-Step Worked Problems & Exact Derivations
Problem 1: Multi-Step Equation with Fractions and Distributive Grouping
Problem: Solve the linear equation for x: (3x - 1)/4 - (x + 3)/3 = 1/2
Step-by-Step Solution:
- Find the Least Common Denominator (LCD): LCD(4, 3, 2) = 12.
- Multiply every term by 12 to eliminate fractions: 12[(3x - 1)/4] - 12[(x + 3)/3] = 12(1/2)
- Simplify each term: 3(3x - 1) - 4(x + 3) = 6
- Distribute constants (be careful with the negative sign on -4): 9x - 3 - 4x - 12 = 6
- Combine like terms on the left: 5x - 15 = 6
- Add 15 to both sides: 5x = 21
- Divide by 5: x = 21/5 = 4.2
Problem 2: Compound Inequality with Fractions and Negative Inversion
Problem: Solve the compound inequality and express the solution in interval notation: -3 ≤ (5 - 2x)/3 < 7
Step-by-Step Solution:
- Multiply all three parts of the inequality by 3 (positive multiplier, no flip): 3(-3) ≤ 3[(5 - 2x)/3] < 3(7) -9 ≤ 5 - 2x < 21
- Subtract 5 from all three parts: -9 - 5 ≤ -2x < 21 - 5 -14 ≤ -2x < 16
- Divide all three parts by -2 (REVERSE BOTH INEQUALITY SIGNS!): -14 / (-2) ≥ -2x / (-2) > 16 / (-2) 7 ≥ x > -8
- Rewrite in standard ascending order (from smallest to largest): -8 < x ≤ 7
- Express in interval notation: (-8, 7]
Problem 3: Absolute Value Inequality with Isolation Step
Problem: Solve the absolute value inequality: |3x - 7| + 4 ≥ 15.
Step-by-Step Solution:
- Isolate the absolute value by subtracting 4: |3x - 7| ≥ 11
- Identify rule: This is a "Great-OR" inequality (|u| ≥ c). Split into two disjoint compound inequalities: 3x - 7 ≥ 11 OR 3x - 7 ≤ -11
- Solve Case 1: 3x - 7 ≥ 11 => 3x ≥ 18 => x ≥ 6
- Solve Case 2: 3x - 7 ≤ -11 => 3x ≤ -4 => x ≤ -4/3
- Combine the solution set: x ≤ -4/3 or x ≥ 6 In interval notation: (-∞, -4/3] ∪ [6, ∞)
What is the solution to the linear equation (3x - 1)/4 - (x + 3)/3 = 1/2?
Which of the following linear equations is an identity that possesses infinitely many solutions?
What is the complete solution set for the absolute value inequality |2x - 5| ≤ 9?
What is the solution to the linear inequality -4(2x - 3) + 7 > 35?