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5.2 Linear Equations, Quadratic Equations, and Inequalities
CAT-ASVAB Computer Adaptive ASVAB
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1 Chapter 1: CAT-ASVAB Overview, Adaptive Mechanics & Test Strategy
1.1 CAT-ASVAB Architecture, MEPS Administration & Subtest Structure 1.2 Computer Adaptive Testing (CAT), IRT Engine & Pacing Rules 1.3 AFQT Scoring Formula, VE Weighting & Service Branch Minimums 1.4 Military Composite Line Scores & MOS/AFSC/Rating Classification
2 Chapter 2: Word Knowledge (WK)
2.1 Greek and Latin Roots, Prefixes, and Suffixes 2.2 Context Clues, Secondary Meanings, and Nuanced Synonyms 2.3 Military, Technical, and Formal Vocabulary in Context 2.4 High-Frequency Word Bank and the 36-Second Word-Attack Protocol
3 Chapter 3: Paragraph Comprehension (PC)
3.1 Identifying Main Ideas, Central Themes, and Author Purpose 3.2 Supporting Details, Factual Extraction, and Structural Analysis 3.3 Inferences, Contextual Meaning, and Tone Evaluation
4 Chapter 4: Arithmetic Reasoning (AR)
4.1 Translating Complex Word Problems into Algebraic Expressions 4.2 Ratios, Proportions, Rates, and Combined Work Problems 4.3 Percentages, Interest, Profit, and Financial Math 4.4 Applied Geometric Word Problems (Area, Perimeter, Volume)
5 Chapter 5: Mathematics Knowledge (MK)
5.1 Number Properties, Exponents, Radicals, and Polynomials 5.2 Linear Equations, Quadratic Equations, and Inequalities 5.3 Geometric Principles, Coordinate Geometry, and Trigonometry Fundamentals 5.4 Basic Probability, Permutations, Combinations, and Statistics 5.5 Fractions, Rational Expressions, and Non-Calculator Computation
6 Chapter 6: General Science (GS)
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7 Chapter 7: Electronics Information (EI)
7.1 Voltage, Current, Resistance, Ohm's Law, and Power 7.2 Circuit Analysis: Series, Parallel, and Combination Circuits with Kirchhoff's Laws 7.3 Electronic Components, Semiconductor Devices, and Schematic Symbols 7.4 Magnetism, Electromagnetism, Motors, and Generators 7.5 Circuit Protection, Wiring, Measurement Instruments, and Electrical Safety
8 Chapter 8: Auto Information (AI)
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9 Chapter 9: Shop Information (SI)
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10 Chapter 10: Mechanical Comprehension (MC)
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11 Chapter 11: Assembling Objects (AO)
11.1 Spatial Relationship Fundamentals & 2D Silhouette Synthesis 11.2 Connection Point Problems, Interior Seams, and Labeled Points 11.3 High-Speed Pacing, Elimination Tactics, and Visual Pattern Recognition
5.2 Linear Equations, Quadratic Equations, and Inequalities
Key Takeaways Explain This Topic AI
Solving multi-step linear equations requires systematic variable isolation using inverse operations, with clearing fractions via the Least Common Denominator (LCD) serving as the fastest initial simplification step.
The mandatory sign reversal rule dictates that multiplying or dividing both sides of an inequality by a negative value immediately flips the inequality symbol (< becomes >, <= becomes >=).
Absolute value equations (|ax+b|=c) split into two distinct linear equations (ax+b=c or ax+b=-c), while |ax+b| <= c forms a bounded conjunction (-c <= ax+b <= c) and |ax+b| >= c forms a disjoint union (ax+b >= c or ax+b <= -c).
Quadratic equations (ax²+bx+c=0) can be solved via trinomial factoring with the Zero Product Property, taking square roots, or applying the Quadratic Formula, where the discriminant (b²-4ac) reveals the number and nature of roots.
Simultaneous 2x2 linear systems are solved using elimination or substitution, corresponding geometrically to intersecting lines (one unique solution), parallel lines (no solution), or coincident lines (infinite solutions).
Last updated: August 2026
5.2 Linear Equations, Quadratic Equations, and Inequalities
Core Principle: Algebraic equations and inequalities constitute approximately 40% to 50% of all Mathematics Knowledge items on the CAT-ASVAB. Mastery of algebraic structures—from solving fractional linear equations and compound inequalities to computing simultaneous 2 × 2 2 \times 2 2 × 2 systems and solving quadratics via factoring or the quadratic formula—is the single most decisive factor in achieving a top-tier MK score.
1. Multi-Step Linear Equations & Literal Equation Rearrangement
To solve a single-variable linear equation, systematically isolate the target variable using inverse operations in reverse operational order: +-----------------------------------------------------------------------------------------+
- | 5-STEP LINEAR EQUATION SOLVER ALGORITHM |
- +-----------------------------------------------------------------------------------------+
- | Step 1: Clear Fractions/Decimals | Multiply every term on both sides by the LCD. |
- | Step 2: Distribute | Expand all parentheses: a(b + c) = ab + ac. |
- | Step 3: Combine Like Terms | Simplify algebraic expressions on each side. |
- | Step 4: Isolate Variable Terms | Add/subtract variable terms to gather on one side. |
- | Step 5: Isolate Constant & Solve | Divide by the numerical coefficient. |
- +-----------------------------------------------------------------------------------------+
Step-by-Step Worked Example (Fraction Clearing): Solve for x : 3 x − 1 4 − 2 x + 3 5 = 1 2 \text{Solve for } x: \quad \frac{3x - 1}{4} - \frac{2x + 3}{5} = \frac{1}{2} Solve for x : 4 3 x − 1 − 5 2 x + 3 = 2 1
Multiply every term by the LCD = 20 \text{LCD} = 20 LCD = 20 : 20 ⋅ ( 3 x − 1 4 ) − 20 ⋅ ( 2 x + 3 5 ) = 20 ⋅ ( 1 2 ) 20 \cdot \left(\frac{3x - 1}{4}\right) - 20 \cdot \left(\frac{2x + 3}{5}\right) = 20 \cdot \left(\frac{1}{2}\right) 20 ⋅ ( 4 3 x − 1 ) − 20 ⋅ ( 5 2 x + 3 ) = 20 ⋅ ( 2 1 )
Simplify multipliers: 5 ( 3 x − 1 ) − 4 ( 2 x + 3 ) = 10 5(3x - 1) - 4(2x + 3) = 10 5 ( 3 x − 1 ) − 4 ( 2 x + 3 ) = 10
Distribute carefully (watching negative signs): 15 x − 5 − 8 x − 12 = 10 15x - 5 - 8x - 12 = 10 15 x − 5 − 8 x − 12 = 10
Combine like terms: 7 x − 17 = 10 7x - 17 = 10 7 x − 17 = 10
Add 17 to both sides and divide by 7: 7 x = 27 ⟹ x = 27 7 7x = 27 \implies x = \frac{27}{7} 7 x = 27 ⟹ x = 7 27
Literal Equations (Formula Rearrangement)
ASVAB items regularly ask you to solve a multi-variable physical formula for a specified variable.
Example 1: Solve Fahrenheit to Celsius formula for F F F : C = 5 9 ( F − 32 ) ⟹ 9 5 C = F − 32 ⟹ F = 9 5 C + 32 C = \frac{5}{9}(F - 32) \implies \frac{9}{5}C = F - 32 \implies F = \frac{9}{5}C + 32 C = 9 5 ( F − 32 ) ⟹ 5 9 C = F − 32 ⟹ F = 5 9 C + 32
Example 2: Solve Cylinder Volume for height h h h : V = π r 2 h ⟹ h = V π r 2 V = \pi r^2 h \implies h = \frac{V}{\pi r^2} V = π r 2 h ⟹ h = π r 2 V
2. Linear Inequalities & Compound Intervals
Solving single-variable linear inequalities mirrors equation solving with one mandatory rule :
The Sign Inversion 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 ( < < < becomes > > > , ≤ \le ≤ becomes ≥ \ge ≥ , and vice versa). Worked Example: -4x + 9 ≤ 29
- 4x ≤ 20
- Divide by -4 (FLIP SIGN!): x ≥ -5
Number Line Graphing Conventions:
< < < or > > > (Strict Inequalities): Indicated by an Open Circle ( ∘ ) (\circ) ( ∘ ) at the boundary point.
≤ \le ≤ or ≥ \ge ≥ (Inclusive Inequalities): Indicated by a Closed/Solid Circle ( ∙ ) (\bullet) ( ∙ ) at the boundary point.
Compound Inequalities: "AND" vs. "OR"
Conjunction ("AND" / Bounded Double Inequality): Solution is the overlapping intersection of both conditions. − 5 ≤ 3 x + 1 < 13 ⟹ − 6 ≤ 3 x < 12 ⟹ − 2 ≤ x < 4 -5 \le 3x + 1 < 13 \implies -6 \le 3x < 12 \implies -2 \le x < 4 − 5 ≤ 3 x + 1 < 13 ⟹ − 6 ≤ 3 x < 12 ⟹ − 2 ≤ x < 4 (Graph: A continuous line segment between closed circle at − 2 -2 − 2 and open circle at 4 4 4 .)
Disjunction ("OR" / Disjoint Rays): Solution is the union of two separate conditions pointing in opposite directions. 2 x − 7 < − 11 or 4 x + 3 ≥ 19 2x - 7 < -11 \quad \text{or} \quad 4x + 3 \ge 19 2 x − 7 < − 11 or 4 x + 3 ≥ 19 2 x < − 4 ⟹ x < − 2 or 4 x ≥ 16 ⟹ x ≥ 4 2x < -4 \implies x < -2 \quad \text{or} \quad 4x \ge 16 \implies x \ge 4 2 x < − 4 ⟹ x < − 2 or 4 x ≥ 16 ⟹ x ≥ 4
3. Absolute Value Equations & Inequalities
Geometrically, ∣ u ∣ |u| ∣ u ∣ measures distance from zero on the number line. Because distance is non-negative, equations and inequalities split into specific cases:
1. Absolute Value Equations: ∣ a x + b ∣ = c |ax + b| = c ∣ a x + b ∣ = c
If c < 0 c < 0 c < 0 , there is no solution ( ∣ u ∣ = − 5 |u| = -5 ∣ u ∣ = − 5 is impossible).
If c ≥ 0 c \ge 0 c ≥ 0 , split into two linear equations: a x + b = c or a x + b = − c ax + b = c \quad \text{or} \quad ax + b = -c a x + b = c or a x + b = − c .
Example: ∣ 2 x − 3 ∣ = 9 ⟹ 2 x − 3 = 9 ⟹ 2 x = 12 ⟹ x = 6 |2x - 3| = 9 \implies 2x - 3 = 9 \implies 2x = 12 \implies x = 6 ∣2 x − 3∣ = 9 ⟹ 2 x − 3 = 9 ⟹ 2 x = 12 ⟹ x = 6 , or 2 x − 3 = − 9 ⟹ 2 x = − 6 ⟹ x = − 3 2x - 3 = -9 \implies 2x = -6 \implies x = -3 2 x − 3 = − 9 ⟹ 2 x = − 6 ⟹ x = − 3 .
2. Absolute Value Inequalities: "Less ThAND" vs. "GreatOR" +-----------------------------------------------------------------------------------------+
- | ABSOLUTE VALUE INEQUALITY RULES |
- +------------------------------------+----------------------------------------------------+
- | LESS THAN: |u| ≤ c ("Less thAND") | Bounded Conjunction: -c ≤ u ≤ c |
- | Example: |3x - 5| ≤ 16 | -16 ≤ 3x - 5 ≤ 16 ⇒ -11 ≤ 3x ≤ 21 ⇒ -11/3 ≤ x ≤ 7 |
- +------------------------------------+----------------------------------------------------+
- | GREATER THAN: |u| ≥ c ("GreatOR") | Disjoint Union: u ≥ c OR u ≤ -c |
- | Example: |2x + 7| > 13 | 2x + 7 > 13 ⇒ 2x > 6 ⇒ x > 3 |
- | | OR: 2x + 7 < -13 ⇒ 2x < -20 ⇒ x < -10 |
- +------------------------------------+----------------------------------------------------+
- 4. Systems of Two Linear Equations ( 2 × 2 2 \times 2 2 × 2 )
A simultaneous system of two linear equations represents two straight lines in the Cartesian coordinate plane. The geometric relationship determines the solution:
Geometric Relationship Slopes & Intercepts Number of Solutions System Classification
Intersecting Lines Different Slopes ( m 1 ≠ m 2 m_1 \neq m_2 m 1 = m 2 ) Exactly One ( x , y ) (x, y) ( x , y ) Consistent & Independent
Parallel Lines Same Slope ( m 1 = m 2 , b 1 ≠ b 2 m_1 = m_2, b_1 \neq b_2 m 1 = m 2 , b 1 = b 2 ) No Solution Inconsistent
Coincident Lines Identical ( m 1 = m 2 , b 1 = b 2 m_1 = m_2, b_1 = b_2 m 1 = m 2 , b 1 = b 2 ) Infinitely Many Consistent & Dependent
Method 1: Substitution (Best when a variable coefficient is ± 1 \pm 1 ± 1 ) { x − 3 y = 4 2 x + 5 y = 19 \begin{cases} x - 3y = 4 \\ 2x + 5y = 19 \end{cases} { x − 3 y = 4 2 x + 5 y = 19
Isolate x x x : x = 3 y + 4 x = 3y + 4 x = 3 y + 4
Substitute into equation 2: 2 ( 3 y + 4 ) + 5 y = 19 ⟹ 6 y + 8 + 5 y = 19 ⟹ 11 y = 11 ⟹ y = 1 2(3y + 4) + 5y = 19 \implies 6y + 8 + 5y = 19 \implies 11y = 11 \implies y = 1 2 ( 3 y + 4 ) + 5 y = 19 ⟹ 6 y + 8 + 5 y = 19 ⟹ 11 y = 11 ⟹ y = 1
Back-substitute: x = 3 ( 1 ) + 4 = 7 x = 3(1) + 4 = 7 x = 3 ( 1 ) + 4 = 7 . Solution: ( 7 , 1 ) (7, 1) ( 7 , 1 ) .
Method 2: Elimination / Addition (Best when coefficients can be aligned) { 3 x + 4 y = 7 5 x − 2 y = 29 \begin{cases} 3x + 4y = 7 \\ 5x - 2y = 29 \end{cases} { 3 x + 4 y = 7 5 x − 2 y = 29
Multiply equation 2 by 2 to align y y y -coefficients: 2 ( 5 x − 2 y = 29 ) ⟹ 10 x − 4 y = 58 2(5x - 2y = 29) \implies 10x - 4y = 58 2 ( 5 x − 2 y = 29 ) ⟹ 10 x − 4 y = 58
Add to equation 1 to eliminate y y y : ( 3 x + 10 x ) + ( 4 y − 4 y ) = 7 + 58 ⟹ 13 x = 65 ⟹ x = 5 (3x + 10x) + (4y - 4y) = 7 + 58 \implies 13x = 65 \implies x = 5 ( 3 x + 10 x ) + ( 4 y − 4 y ) = 7 + 58 ⟹ 13 x = 65 ⟹ x = 5
Substitute x = 5 x = 5 x = 5 into equation 1: 3 ( 5 ) + 4 y = 7 ⟹ 15 + 4 y = 7 ⟹ 4 y = − 8 ⟹ y = − 2 3(5) + 4y = 7 \implies 15 + 4y = 7 \implies 4y = -8 \implies y = -2 3 ( 5 ) + 4 y = 7 ⟹ 15 + 4 y = 7 ⟹ 4 y = − 8 ⟹ y = − 2
Solution: ( 5 , − 2 ) (5, -2) ( 5 , − 2 ) .
5. Solving Quadratic Equations & The Quadratic Formula
A quadratic equation has standard form a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 (where a ≠ 0 a \neq 0 a = 0 ).
Method 1: Factoring and the Zero Product Property
If A ⋅ B = 0 A \cdot B = 0 A ⋅ B = 0 , then A = 0 A = 0 A = 0 or B = 0 B = 0 B = 0 . 2 x 2 − 7 x + 3 = 0 ⟹ ( 2 x − 1 ) ( x − 3 ) = 0 2x^2 - 7x + 3 = 0 \implies (2x - 1)(x - 3) = 0 2 x 2 − 7 x + 3 = 0 ⟹ ( 2 x − 1 ) ( x − 3 ) = 0 2 x − 1 = 0 ⟹ x = 1 2 , x − 3 = 0 ⟹ x = 3 2x - 1 = 0 \implies x = \frac{1}{2}, \quad x - 3 = 0 \implies x = 3 2 x − 1 = 0 ⟹ x = 2 1 , x − 3 = 0 ⟹ x = 3
Method 2: The Quadratic Formula
When a quadratic expression cannot be factored by inspection, use the Quadratic Formula : x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x = 2 a − b ± b 2 − 4 a c
Discriminant Analysis ( D = b 2 − 4 a c D = b^2 - 4ac D = b 2 − 4 a c )
The value under the radical ( D D D ) dictates the number and algebraic type of solutions: DISCRIMINANT: D = b² - 4ac
- |
- +------------------------+------------------------+
- | |
- D > 0 D = 0 D < 0
- Two Distinct Real Roots One Repeated Real Root No Real Roots
- (Rational if D is perfect square, x = -b / (2a) (Two Complex Conjugate
- Irrational if not perfect square) Roots: x = u ± vi)
Step-by-Step Worked Quadratic Example: Solve: 3 x 2 − 5 x − 2 = 0 \text{Solve: } 3x^2 - 5x - 2 = 0 Solve: 3 x 2 − 5 x − 2 = 0
Identify coefficients: a = 3 , b = − 5 , c = − 2 a = 3, b = -5, c = -2 a = 3 , b = − 5 , c = − 2
Calculate discriminant: D = ( − 5 ) 2 − 4 ( 3 ) ( − 2 ) = 25 + 24 = 49 D = (-5)^2 - 4(3)(-2) = 25 + 24 = 49 D = ( − 5 ) 2 − 4 ( 3 ) ( − 2 ) = 25 + 24 = 49 (Positive perfect square → \rightarrow → two rational roots)
Apply formula: x = − ( − 5 ) ± 49 2 ( 3 ) = 5 ± 7 6 x = \frac{-(-5) \pm \sqrt{49}}{2(3)} = \frac{5 \pm 7}{6} x = 2 ( 3 ) − ( − 5 ) ± 49 = 6 5 ± 7 x 1 = 5 + 7 6 = 12 6 = 2 , x 2 = 5 − 7 6 = − 2 6 = − 1 3 x_1 = \frac{5 + 7}{6} = \frac{12}{6} = 2, \quad x_2 = \frac{5 - 7}{6} = \frac{-2}{6} = -\frac{1}{3} x 1 = 6 5 + 7 = 6 12 = 2 , x 2 = 6 5 − 7 = 6 − 2 = − 3 1