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3.3 Basic Algebra, Sequences, and Patterns

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Philippines Computer-Based National Career Assessment Examination CB-NCAE

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1 Chapter 1: Overview, Architecture, and Testing Policies

1.1 CB-NCAE Overview, Governance, and Purpose 1.2 Computer-Based Platform, Delivery, and Accommodations 1.3 Senior High School Track Integration and Career Guidance

2 Chapter 2: GSA — Verbal Reasoning (English & Filipino)

2.1 Reading Comprehension and Passage Analysis 2.2 Vocabulary in Context, Context Clues, and Idiomatic Usage 2.3 Verbal Analogies and Word Relationships 2.4 Critical Evaluation, Tone, and Argument Synthesis

3 Chapter 3: GSA — Numerical Reasoning

3.1 Arithmetic Operations, Fractions, Decimals, and Percentages 3.2 Ratios, Proportions, and Rates 3.3 Basic Algebra, Sequences, and Patterns 3.4 Geometry, Measurement, and Data Interpretation

4 Chapter 4: GSA — Logical Reasoning

4.1 Categorical Syllogisms and Deductive Reasoning 4.2 Conditional Logic, Truth Tables, and Contrapositives 4.3 Logical Fallacies and Argument Flaws 4.4 Analytical Puzzles, Linear Ordering, and Grouping

5 Chapter 5: GSA — Abstract Reasoning

5.1 Linear Figure Series and Sequential Progression 5.2 Matrix Completion and Multi-Variable Grids 5.3 Figure Classification and Odd-One-Out 5.4 Rule Induction, Pattern Analogy, and Complex Transformations

6 Chapter 6: GSA — Spatial Reasoning

6.1 2D to 3D Mental Folding and Surface Nets 6.2 Orthographic Projection and Multiview Visualization 6.3 3D Spatial Rotation and Block Visualization 6.4 Mirror Images, Reflections, and Spatial Symmetry

7 Chapter 7: GSA — Mechanical Reasoning

7.1 Simple Machines: Levers, Moments, and Mechanical Advantage 7.2 Pulleys, Wheels and Axles, and Inclined Planes 7.3 Gears, Belt Drives, and Rotational Mechanics 7.4 Fluid Mechanics, Pressure, and Basic Electrical Circuits

8 Chapter 8: GSA — Perceptual Accuracy and Response Efficiency (PARE)

8.1 Alphanumeric Code and String Matching 8.2 Clerical Filing, Indexing, and Alphabetical Sorting 8.3 Symbol Counting, Visual Inspection, and Transposition Detection

9 Chapter 9: CORE — Career and Occupational Roles Explorer & Track Guidance

9.1 CORE-A: Holland's RIASEC Typology 9.2 CORE-B: Thurstone Model of Occupational Interests 9.3 Profile Synthesis and Senior High School Track Selection

3.3 Basic Algebra, Sequences, and Patterns

Key Takeaways Explain This Topic AI

Word problem translation requires identifying operational triggers and grouping phrases; order of terms is vital for phrases like 'subtracted from' or 'less than'.

Systems of two linear equations are efficiently solved by elimination when matching coefficients can be created, or by substitution when a variable is easily isolated.

Arithmetic sequences feature a constant common difference ( a n = a 1 + ( n − 1 ) d a_n = a_1 + (n - 1)d a n ​ = a 1 ​ + ( n − 1 ) d ), while geometric sequences feature a constant common ratio ( a n = a 1 r n − 1 a_n = a_1 r^{n-1} a n ​ = a 1 ​ r n − 1 ).

Number pattern detection follows a systematic diagnostic protocol: evaluate first differences, second differences, ratios, alternating sequences, and Fibonacci-type additive rules.

Last updated: October 2026

3.3 Basic Algebra, Sequences, and Patterns

Algebraic reasoning provides the symbolic toolkit required to model complex numerical relationships compactly. In the CB-NCAE Numerical Reasoning subtest, algebra questions test your ability to convert verbal scenarios into mathematical equations, solve for unknown quantities, analyze progressions, and deduce underlying sequences.

Translating Verbal Statements into Algebraic Equations

Translating English or Filipino verbal descriptions into mathematical expressions requires vigilance regarding word order and operation triggers.

The Operational Translation Dictionary

Addition ( + + + ): sum, plus, increased by, more than, total of, exceeds by.

Subtraction ( − - − ): difference, decreased by, less, diminished by, subtracted from, reduced by.

Multiplication ( × \times × ): product, times, of, twice ( 2 x 2x 2 x ), thrice ( 3 x 3x 3 x ).

Division ( ÷ \div ÷ ): quotient, ratio, divided by, per.

Equality ( = = = ): is, equals, results in, is equivalent to, yields, represents.

Caution

The phrase "subtracted from" or "less than" reverses the written order of operands:

"6 less than x x x " translates to x − 6 x - 6 x − 6 , not 6 − x 6 - x 6 − x .

" y y y subtracted from 15" translates to 15 − y 15 - y 15 − y , not y − 15 y - 15 y − 15 .

"Twice the sum of x x x and 5" requires parentheses: 2 ( x + 5 ) 2(x + 5) 2 ( x + 5 ) , whereas "the sum of twice x x x and 5" is 2 x + 5 2x + 5 2 x + 5 .

  • Modeling Consecutive Integers
  • Consecutive integers: n , n + 1 , n + 2 , … n, n+1, n+2, \dots n , n + 1 , n + 2 , …

Consecutive even integers: n , n + 2 , n + 4 , … n, n+2, n+4, \dots n , n + 2 , n + 4 , … (where n n n is even)

  • Consecutive odd integers: n , n + 2 , n + 4 , … n, n+2, n+4, \dots n , n + 2 , n + 4 , … (where n n n is odd)
  • Age Problem Formulation

Age problems involve relationships across different time frames (past, present, future). Always organize the information into a structured reference grid:

  • Person Past ( − k -k − k years) Present Age Future ( + m +m + m years)
  • Parent P − k P - k P − k P P P P + m P + m P + m
  • Child C − k C - k C − k C C C C + m C + m C + m

If the parent is currently 3 times as old as the child ( P = 3 C P = 3C P = 3 C ), in 10 years the parent's age will be P + 10 = 3 C + 10 P + 10 = 3C + 10 P + 10 = 3 C + 10 , and the child's age will be C + 10 C + 10 C + 10 . Equating their future relationship yields an easily solvable single-variable linear equation.

Solving Linear Equations and Systems of Equations

A linear equation in one variable can always be placed in the standard form a x + b = c ax + b = c a x + b = c . Isolate the variable by applying inverse operations systematically.

Inequalities and the Sign Reversal Rule

When solving linear inequalities, all standard balancing operations apply, with one critical rule: Multiplying or dividing both sides by a negative number reverses the inequality direction. \text{Multiplying or dividing both sides by a negative number reverses the inequality direction.} Multiplying or dividing both sides by a negative number reverses the inequality direction. Example: Solve − 3 x + 7 ≤ 22 -3x + 7 \le 22 − 3 x + 7 ≤ 22 : − 3 x ≤ 22 − 7 ⟹ − 3 x ≤ 15 ⟹ x ≥ 15 − 3 ⟹ x ≥ − 5 -3x \le 22 - 7 \implies -3x \le 15 \implies x \ge \frac{15}{-3} \implies x \ge -5 − 3 x ≤ 22 − 7 ⟹ − 3 x ≤ 15 ⟹ x ≥ − 3 15 ​ ⟹ x ≥ − 5

Systems of Two Linear Equations

Many word problems feature two unknowns constrained by two distinct conditions. Standard forms are solved via substitution or elimination:

The Elimination Method: Multiply one or both equations by constants so that the coefficients of one variable become additive inverses, then add the equations together: { 2 x + 3 y = 31 5 x − y = 18 \begin{cases} 2x + 3y = 31 \\ 5x - y = 18 \end{cases} { 2 x + 3 y = 31 5 x − y = 18 ​ Multiply the second equation by 3: 15 x − 3 y = 54 15x - 3y = 54 15 x − 3 y = 54 . Add to the first equation: ( 2 x + 3 y ) + ( 15 x − 3 y ) = 31 + 54 ⟹ 17 x = 85 ⟹ x = 5 (2x + 3y) + (15x - 3y) = 31 + 54 \implies 17x = 85 \implies x = 5 ( 2 x + 3 y ) + ( 15 x − 3 y ) = 31 + 54 ⟹ 17 x = 85 ⟹ x = 5 Substitute back: 2 ( 5 ) + 3 y = 31 ⟹ 3 y = 21 ⟹ y = 7 2(5) + 3y = 31 \implies 3y = 21 \implies y = 7 2 ( 5 ) + 3 y = 31 ⟹ 3 y = 21 ⟹ y = 7 . Check in the second equation: 5 ( 5 ) − 7 = 18 5(5) - 7 = 18 5 ( 5 ) − 7 = 18 .

The Substitution Method: Isolate one variable with coefficient 1 and substitute into the second equation. Best suited when an equation is already configured as y = m x + b y = mx + b y = m x + b .

Arithmetic Sequences and Series

An arithmetic sequence (or arithmetic progression) is a succession of numbers wherein the difference between any consecutive terms is a constant value known as the common difference ( d d d ): d = a n − a n − 1 d = a_n - a_{n-1} d = a n ​ − a n − 1 ​

Core Formulas for Arithmetic Sequences

General ( n n n -th) Term: a n = a 1 + ( n − 1 ) d a_n = a_1 + (n - 1)d a n ​ = a 1 ​ + ( n − 1 ) d where a 1 a_1 a 1 ​ is the first term, n n n is the term index, and d d d is the common difference.

Position Formula (finding how many terms are in a sequence): n = a n − a 1 d + 1 n = \frac{a_n - a_1}{d} + 1 n = d a n ​ − a 1 ​ ​ + 1

Arithmetic Mean: The arithmetic mean of two numbers x x x and y y y is simply their midpoint: M = x + y 2 M = \frac{x + y}{2} M = 2 x + y ​ .

Sum of the First n n n Terms ( S n S_n S n ​ ): S n = n 2 ( a 1 + a n ) = n 2 [ 2 a 1 + ( n − 1 ) d ] S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}[2a_1 + (n - 1)d] S n ​ = 2 n ​ ( a 1 ​ + a n ​ ) = 2 n ​ [ 2 a 1 ​ + ( n − 1 ) d ]

Quick Mental Application: What is the sum of all integers from 1 to 50? S 50 = 50 2 ( 1 + 50 ) = 25 × 51 = 1 , 275 S_{50} = \frac{50}{2}(1 + 50) = 25 \times 51 = 1,275 S 50 ​ = 2 50 ​ ( 1 + 50 ) = 25 × 51 = 1 , 275 .

Geometric Sequences and Series

A geometric sequence is a sequence in which each term after the first is obtained by multiplying the preceding term by a non-zero constant called the common ratio ( r r r ): r = a n a n − 1 r = \frac{a_n}{a_{n-1}} r = a n − 1 ​ a n ​ ​

Core Formulas for Geometric Sequences

General ( n n n -th) Term: a n = a 1 r n − 1 a_n = a_1 r^{n-1} a n ​ = a 1 ​ r n − 1

Geometric Mean: The geometric mean between two positive numbers x x x and y y y is G = x y G = \sqrt{xy} G = x y ​ .

Sum of the First n n n Terms ( S n S_n S n ​ , where r ≠ 1 r \neq 1 r  = 1 ): S n = a 1 ( 1 − r n ) 1 − r S_n = \frac{a_1(1 - r^n)}{1 - r} S n ​ = 1 − r a 1 ​ ( 1 − r n ) ​

Comparison: An arithmetic sequence grows linearly (constant rate of change, graph is a straight line), while a geometric sequence grows exponentially (multiplicative rate of change, curve steepens rapidly).

Number Pattern Recognition and Series Diagnostics

In the GSA, number series questions evaluate inductive reasoning. When confronting an unfamiliar number sequence, follow this systematic 5-step diagnostic protocol: Step 1: Check First Differences (Δ1 = a_{k+1} - a_k)

  • Is Δ1 constant? -> YES -> Arithmetic Sequence (+d)
  • | NO
  • Step 2: Check Quotients (r = a_{k+1} / a_k)
  • Is r constant? -> YES -> Geometric Sequence (*r)
  • | NO
  • Step 3: Check Second Differences (Δ2 = Δ1_{k+1} - Δ1_k)
  • Is Δ2 constant? -> YES -> Quadratic Sequence (an^2 + bn + c)
  • | NO
  • Step 4: Check for Interleaved (Alternating) Series

Do odd terms (1, 3, 5) and even terms (2, 4, 6) follow separate rules?

  • | NO
  • Step 5: Check Additive / Fibonacci Patterns or Powers

Is a_n = a_{n-1} + a_{n-2}, or based on n^2, n^3, n^2 ± 1, 2^n?

Pattern Examples

Alternating Series: 4 , 20 , 7 , 17 , 10 , 14 , 13 , 11 , … 4, 20, 7, 17, 10, 14, 13, 11, \dots 4 , 20 , 7 , 17 , 10 , 14 , 13 , 11 , …

Odd positions: 4 , 7 , 10 , 13 , … 4, 7, 10, 13, \dots 4 , 7 , 10 , 13 , … ( + 3 +3 + 3 arithmetic step).

Even positions: 20 , 17 , 14 , 11 , … 20, 17, 14, 11, \dots 20 , 17 , 14 , 11 , … ( − 3 -3 − 3 arithmetic step).

Next term (position 9): 13 + 3 = 16 13 + 3 = 16 13 + 3 = 16 .

Fibonacci-Type Series: 3 , 4 , 7 , 11 , 18 , 29 , … 3, 4, 7, 11, 18, 29, \dots 3 , 4 , 7 , 11 , 18 , 29 , …

Each term is the sum of the two preceding terms ( 7 = 3 + 4 7 = 3 + 4 7 = 3 + 4 ; 11 = 4 + 7 11 = 4 + 7 11 = 4 + 7 ; 18 = 7 + 11 18 = 7 + 11 18 = 7 + 11 ).

Next term: 18 + 29 = 47 18 + 29 = 47 18 + 29 = 47 .

  • Step-by-Step Worked Problems
  • Worked Example 1: Age Relationship Word Problem

Problem: A father is currently three times as old as his daughter. Ten years from now, the father's age will be 10 years more than twice his daughter's age at that time. What are the present ages of the father and the daughter?

Step-by-Step Solution:

Define variables for the present: Let d d d represent the daughter's current age. The father's current age is f = 3 d f = 3d f = 3 d .

Express their ages in 10 years:

  • Daughter's future age: d + 10 d + 10 d + 10
  • Father's future age: 3 d + 10 3d + 10 3 d + 10

Set up the linear equation based on the future relationship: 3 d + 10 = 2 ( d + 10 ) + 10 3d + 10 = 2(d + 10) + 10 3 d + 10 = 2 ( d + 10 ) + 10

Expand and simplify: 3 d + 10 = 2 d + 20 + 10 3d + 10 = 2d + 20 + 10 3 d + 10 = 2 d + 20 + 10 3 d + 10 = 2 d + 30 3d + 10 = 2d + 30 3 d + 10 = 2 d + 30

Isolate variable d d d : 3 d − 2 d = 30 − 10 ⟹ d = 20 3d - 2d = 30 - 10 \implies d = 20 3 d − 2 d = 30 − 10 ⟹ d = 20

Calculate father's current age: f = 3 × 20 = 60 f = 3 \times 20 = 60 f = 3 × 20 = 60 Verification: Present ages: Father 60, Daughter 20 ( 60 = 3 × 20 60 = 3 \times 20 60 = 3 × 20 ). In 10 years: Father is 70, Daughter is 30. Twice daughter's age plus 10: 2 ( 30 ) + 10 = 60 + 10 = 70 2(30) + 10 = 60 + 10 = 70 2 ( 30 ) + 10 = 60 + 10 = 70 . The condition is confirmed.

Worked Example 2: Sequence with Second-Order Differences

Problem: Determine the next number in the sequence: 4 , 7 , 12 , 19 , 28 , … 4, 7, 12, 19, 28, \dots 4 , 7 , 12 , 19 , 28 , …

Step-by-Step Solution:

Calculate first differences between consecutive terms ( Δ 1 \Delta_1 Δ 1 ​ ):

  • 7 − 4 = 3 7 - 4 = 3 7 − 4 = 3
  • 12 − 7 = 5 12 - 7 = 5 12 − 7 = 5
  • 19 − 12 = 7 19 - 12 = 7 19 − 12 = 7

28 − 19 = 9 28 - 19 = 9 28 − 19 = 9 The first differences are 3 , 5 , 7 , 9 3, 5, 7, 9 3 , 5 , 7 , 9 .

Calculate second differences ( Δ 2 \Delta_2 Δ 2 ​ ):

  • 5 − 3 = 2 5 - 3 = 2 5 − 3 = 2
  • 7 − 5 = 2 7 - 5 = 2 7 − 5 = 2

9 − 7 = 2 9 - 7 = 2 9 − 7 = 2 The second difference is a constant value of + 2 +2 + 2 .

Project the next first difference: 9 + 2 = 11 9 + 2 = 11 9 + 2 = 11 .

Calculate the next term in the sequence: 28 + 11 = 39 28 + 11 = 39 28 + 11 = 39 . (Alternative pattern recognition: Each term is n 2 + 3 n^2 + 3 n 2 + 3 : 1 2 + 3 = 4 1^2+3=4 1 2 + 3 = 4 , 2 2 + 3 = 7 2^2+3=7 2 2 + 3 = 7 , 3 2 + 3 = 12 3^2+3=12 3 2 + 3 = 12 , 4 2 + 3 = 19 4^2+3=19 4 2 + 3 = 19 , 5 2 + 3 = 28 5^2+3=28 5 2 + 3 = 28 , 6 2 + 3 = 39 6^2+3=39 6 2 + 3 = 39 .)