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LET Secondary Mathematics — Elementary AlgebraDetailed Explanation

Summary Concepts Notes Revision Deep Dive Slides Flashcards Mistakes Cheat Mnemonics Templates

A detailed, step-by-step explanation of Elementary Algebra for LET Secondary aspirants. This page goes deeper than the summary and study notes, walking through the reasoning behind each concept so you understand why Professional Regulation Commission (PRC) tests it the way it does in the LET Secondary Mathematics subtest.

Exam context

The Licensure Examination for Professional Teachers — Secondary is conducted by Professional Regulation Commission (PRC) and is scheduled for Bi-annual. The Mathematics subtest is marked as "Core" in the official pattern, and Elementary Algebra appears in position 3rd of 7 in the LET Secondary Mathematics review rotation. Passing mark: Weighted average of 75% with no grade below 50%. Recent LET Secondary 2026 papers have drawn roughly a meaningful share of questions from this subject.

Elementary Algebra - Detailed Explanation

Elementary Algebra is one of the highest-yield topics in the Mathematics component of the Licensure Examination for Teachers (Elementary Level). The PRC-LET tests your ability to reason with symbols, manipulate algebraic expressions, solve equations and inequalities, and — most critically — translate word problems into algebraic sentences. As a future elementary teacher, mastering algebra also means you will be equipped to teach the foundational algebraic thinking embedded in the K-12 Basic Education Curriculum (BEC) for Grades 4-6, where pupils begin working with patterns, expressions, and simple equations under the Mathematics framework of DepEd. This guide covers all six major sub-topics: algebraic expressions and laws of exponents, linear equations and inequalities, systems of equations, special products and factoring, quadratic equations, and word-problem translation. Each section includes worked examples, common LET question patterns, and pitfalls to avoid. The single most important study habit for this chapter is slowing down on sign operations — the majority of errors on the LET algebra section are sign mistakes, not conceptual gaps. Work every example by hand, check every answer, and treat each word problem with the same careful reading you would apply to a comprehension passage.

Concepts

Algebraic Expressions and Laws of Exponents

An algebraic expression is a mathematical phrase that combines numbers, variables (letters representing unknown or changing values), and operations. A single piece of an expression separated by addition or subtraction is called a TERM. For example, in 5x² - 3xy + 7, there are three terms: 5x², -3xy, and 7. The number in front of the variable is called the COEFFICIENT (5 and -3 above), and the constant term (7) stands alone. LIKE TERMS share exactly the same variable part — same variables raised to the same powers — and can be combined by adding or subtracting their coefficients: 4x² + 3x² = 7x², but 4x² + 3x cannot be combined because x² and x are different. The DEGREE of a term is the sum of all the exponents on its variables: the degree of 5x²y³ is 2+3 = 5. The degree of the entire polynomial is the degree of its highest-degree term. The LAWS OF EXPONENTS are rules for simplifying expressions with powers. They are essential for the LET and must be memorized perfectly: 1. PRODUCT RULE: When multiplying powers of the same base, ADD the exponents. a^m × a^n = a^(m+n). Example: x³ × x⁴ = x^(3+4) = x⁷. 2. QUOTIENT RULE: When dividing powers of the same base, SUBTRACT the exponents. a^m ÷ a^n = a^(m-n). Example: x⁵ ÷ x² = x^(5-2) = x³. 3. POWER OF A POWER RULE: When raising a power to another power, MULTIPLY the exponents. (a^m)^n = a^(mn). Example: (x²)³ = x^(2×3) = x⁶. 4. POWER OF A PRODUCT RULE: Distribute the exponent to each factor. (ab)^n = a^n × b^n. Example: (2x)³ = 2³ × x³ = 8x³. 5. ZERO EXPONENT RULE: Any nonzero base raised to the zero power equals 1. a⁰ = 1 (where a ≠ 0). Example: 7⁰ = 1, (5xy)⁰ = 1. 6. NEGATIVE EXPONENT RULE: A negative exponent means the reciprocal. a^(-n) = 1/a^n. Example: 2^(-3) = 1/2³ = 1/8. A classic LET combination problem: Simplify (2x³)² × 3x. Apply the power of a product rule first: (2x³)² = 2² × (x³)² = 4x⁶. Then multiply: 4x⁶ × 3x = 12x^(6+1) = 12x⁷.

Examples

The key move is recognizing that the exponent 3 outside the parentheses must be distributed to ALL factors inside: the coefficient 3, the variable x², and the variable y. Then apply the power-of-a-power rule to x².

Scenario

Simplify: (3x²y)³

Solution

Step 1: Apply the power of a product rule — distribute the exponent 3 to each factor. (3x²y)³ = 3³ × (x²)³ × y³ Step 2: Compute each part. 3³ = 27 (x²)³ = x^(2×3) = x⁶ (power of a power rule) y³ = y³ Step 3: Combine. Answer: 27x⁶y³

Apply the quotient rule variable by variable. When the exponent in the denominator is larger, the result is a negative exponent, which becomes a fraction.

Scenario

Simplify: (4a³b²) ÷ (2a b⁴)

Solution

Step 1: Separate the coefficients and each variable group. = (4/2) × (a³/a¹) × (b²/b⁴) Step 2: Divide the coefficients: 4 ÷ 2 = 2. Step 3: Apply the quotient rule to each variable. a³ ÷ a¹ = a^(3-1) = a² b² ÷ b⁴ = b^(2-4) = b^(-2) = 1/b² Step 4: Combine. Answer: 2a²/b² (or equivalently 2a²b^(-2))

Placing every substituted value in parentheses is the technique that prevents sign errors. Notice that -2(2)(-3) = -12 is negative, so subtracting it gives +12.

Scenario

Evaluate: 5x² - 2xy + 3 when x = 2 and y = -3

Solution

Step 1: Substitute, placing each substituted value in parentheses to protect signs. = 5(2)² - 2(2)(-3) + 3 Step 2: Evaluate powers first (order of operations). (2)² = 4 = 5(4) - 2(2)(-3) + 3 Step 3: Multiply. = 20 - (-12) + 3 Step 4: Apply subtraction of a negative = addition. = 20 + 12 + 3 Answer: 35

Applications

Computing areas and perimeters expressed algebraically (common in LET word problems involving geometry).
Simplifying formulas used in science and engineering, such as compound interest or area formulas.
Expanding expressions in preparation for factoring in higher-level algebra.
Grade 5-6 K-12 BEC patterns and algebra strand: teachers use exponent rules to explain scientific notation and large-number contexts.
Evaluating expressions is the basis of the function concept in Grade 7 and beyond.

Misconceptions

WRONG: x² × x³ = x⁶ (multiplying exponents instead of adding). CORRECT: x² × x³ = x⁵.
WRONG: (2x)³ = 2x³ (forgetting to apply the exponent to the coefficient). CORRECT: (2x)³ = 8x³.
WRONG: 3x² + 4x = 7x³ (combining unlike terms). These CANNOT be combined.
WRONG: -2² = 4 (thinking the negative is inside the exponent). CORRECT: -2² = -(2²) = -4. But (-2)² = 4.
WRONG: a⁰ = 0. CORRECT: a⁰ = 1 for any nonzero a.
WRONG: 2^(-3) = -8. CORRECT: 2^(-3) = 1/8 (positive fraction, not negative).

Related Concepts

Order of Operations (GEMDAS/PEMDAS)
Properties of Real Numbers (Distributive, Commutative, Associative)
Polynomial Operations
Scientific Notation
Function Evaluation

Common Exam Questions

Example

Simplify (2x²)³ ÷ (4x⁴). Answer: 8x⁶ ÷ 4x⁴ = 2x².

Approach

Apply exponent laws one at a time in order: first handle parentheses (power of a product/power), then multiplication (product rule), then division (quotient rule). Write every intermediate step.

Question Type

Simplification of expressions with multiple exponent laws

Example

Simplify 3(2x - 1) + 4(x + 5). = 6x - 3 + 4x + 20 = 10x + 17.

Approach

Distribute first, then group and collect like terms. List each variable category separately before adding.

Question Type

Combining like terms after expanding

Example

If f(x) = x² - 3x, find f(-2). = (-2)² - 3(-2) = 4 + 6 = 10.

Approach

Wrap every substituted value in parentheses. Follow PEMDAS/GEMDAS strictly: exponents first, then multiplication, then addition/subtraction.

Question Type

Evaluation of expressions with negative substitution values

Key Points To Remember

Like terms must have identical variable parts — same variables AND same exponents — before you can add or subtract them.
Product rule: MULTIPLY bases → ADD exponents (a^m × a^n = a^(m+n)).
Quotient rule: DIVIDE bases → SUBTRACT exponents (a^m ÷ a^n = a^(m-n)).
Power of a power: RAISE power to power → MULTIPLY exponents ((a^m)^n = a^(mn)).
Any nonzero number to the zero power is 1; a⁰ = 1.
Negative exponent means reciprocal, NOT a negative number: 2^(-3) = 1/8, which is positive.
When applying a power to a product like (2x)³, the exponent applies to BOTH the coefficient and the variable.
The degree of a polynomial is determined by its highest-degree term.
Coefficients are multiplied; exponents are added (not multiplied) in the product rule.
Always simplify step by step — do not try to combine unlike terms.

Linear Equations and Inequalities

A LINEAR EQUATION is an equation in which the variable appears only to the first power (no x², no x³, etc.). It can always be written in the form ax + b = c, where a ≠ 0. The SOLUTION (or root) is the value of x that makes the equation true. The core strategy for solving any linear equation is to ISOLATE the variable using inverse (opposite) operations, while keeping the equation BALANCED — whatever operation you perform on one side, you must perform on the other side. STEPS FOR SOLVING A LINEAR EQUATION: 1. Clear parentheses by distributing. 2. Combine like terms on each side separately. 3. Move all variable terms to one side (add or subtract). 4. Move all constant terms to the other side. 5. Divide both sides by the coefficient of the variable. 6. Check by substituting back into the original equation. Example: Solve 3(2x - 4) = 2x + 4. Step 1: Distribute → 6x - 12 = 2x + 4. Step 2: Subtract 2x from both sides → 4x - 12 = 4. Step 3: Add 12 to both sides → 4x = 16. Step 4: Divide by 4 → x = 4. Check: 3(2·4 - 4) = 3(4) = 12; and 2·4 + 4 = 12. ✓ LINEAR INEQUALITIES use the symbols <, >, ≤, ≥ instead of =. They are solved the same way as equations with ONE CRITICAL EXCEPTION: when you MULTIPLY or DIVIDE both sides by a NEGATIVE NUMBER, you must REVERSE (flip) the inequality symbol. This is the most tested rule for inequalities on the LET. Example: Solve -2x < 6. Divide both sides by -2 (negative!) → REVERSE the symbol. x > -3. The solution is x > -3, meaning any number greater than -3 satisfies the inequality. Solutions to inequalities are RANGES of values (not a single answer) and can be represented on a number line: open circle (○) for strict inequalities (< or >) and closed circle (●) for ≤ or ≥, with an arrow extending in the direction of the solution.

Examples

The most critical step is dividing by -2 and remembering to flip the inequality. This is the most common error on LET items for inequalities. Always verify with a test value from your solution set.

Scenario

Solve: 5 - 2x ≥ 11

Solution

Step 1: Subtract 5 from both sides. 5 - 2x - 5 ≥ 11 - 5 -2x ≥ 6 Step 2: Divide both sides by -2. Since we are dividing by a NEGATIVE, REVERSE the inequality symbol (≥ becomes ≤). -2x ÷ (-2) ≤ 6 ÷ (-2) x ≤ -3 Answer: x ≤ -3 (all numbers less than or equal to -3). Check with x = -4: 5 - 2(-4) = 5 + 8 = 13 ≥ 11 ✓ Check that x = 0 does NOT work: 5 - 2(0) = 5, which is NOT ≥ 11 ✓ (correctly excluded).

When fractions appear, multiply through by the LCD first to eliminate all fractions. This transforms the problem into a simpler integer equation.

Scenario

Solve: (x + 3)/2 - 1 = (2x - 1)/3

Solution

Step 1: Multiply every term by the LCD = 6 to clear fractions. 6·(x+3)/2 - 6·1 = 6·(2x-1)/3 3(x + 3) - 6 = 2(2x - 1) Step 2: Distribute. 3x + 9 - 6 = 4x - 2 3x + 3 = 4x - 2 Step 3: Subtract 3x from both sides. 3 = x - 2 Step 4: Add 2 to both sides. x = 5 Check: (5+3)/2 - 1 = 8/2 - 1 = 4 - 1 = 3; and (2·5-1)/3 = 9/3 = 3. ✓

Solutions can be fractions — do not assume the answer must be a whole number. Always check in the ORIGINAL equation, not a simplified version.

Scenario

Solve: -3(x - 2) + 5 = 2(x + 1) - 10

Solution

Step 1: Distribute on both sides. -3x + 6 + 5 = 2x + 2 - 10 Step 2: Combine like terms on each side. -3x + 11 = 2x - 8 Step 3: Add 3x to both sides. 11 = 5x - 8 Step 4: Add 8 to both sides. 19 = 5x Step 5: Divide by 5. x = 19/5 (or 3.8) Check: -3(19/5 - 2) + 5 = -3(9/5) + 5 = -27/5 + 25/5 = -2/5 2(19/5 + 1) - 10 = 2(24/5) - 10 = 48/5 - 50/5 = -2/5 ✓

Applications

Setting up and solving real-world problems: finding unknown quantities (e.g., number of students, price of items).
Grade 6 K-12 Mathematics: simple equations with one unknown, a direct prerequisite to Grade 7 algebra.
Budget planning problems on the LET: 'A teacher spends ₱x on materials and receives ₱500 change from ₱1,000...'
Determining eligibility thresholds: 'A student needs a score greater than 75 to pass...' (inequality application).
Ratio and proportion problems are essentially linear equations in disguise.

Misconceptions

WRONG: When solving -3x = 9, writing x = -27 (multiplying instead of dividing). CORRECT: x = 9 ÷ (-3) = -3.
WRONG: Forgetting to reverse the inequality symbol when dividing by a negative: -2x > 6 → x > -3. CORRECT: x < -3.
WRONG: Distributing incorrectly — 3(x - 4) = 3x - 4. CORRECT: 3(x - 4) = 3x - 12.
WRONG: Cancelling a variable that appears on both sides before moving all terms: treating 2x = x + 3 as if 2 = 1 + 3. CORRECT: subtract x from both sides to get x = 3.
WRONG: Forgetting to check the solution, missing extraneous answers or arithmetic errors.

Related Concepts

Properties of Equality (Addition, Subtraction, Multiplication, Division)
Distributive Property
Number Line Representation
Absolute Value Equations
Linear Functions and Graphs

Common Exam Questions

Example

Solve 4(x - 3) = 2x + 2. → 4x - 12 = 2x + 2 → 2x = 14 → x = 7.

Approach

Distribute all parentheses first, combine like terms on each side, then collect variables on one side and constants on the other. Divide by the coefficient last. Check your answer.

Question Type

Solve a linear equation with parentheses and variables on both sides

Example

Solve 3 - 4x > 11. → -4x > 8 → x < -2 (symbol reversed because we divided by -4).

Approach

Solve exactly like an equation, BUT reverse the inequality sign when multiplying or dividing by a negative. State the solution as a range (e.g., x > 3) and optionally show on a number line.

Question Type

Solve a linear inequality and identify the solution set

Example

If 2x + 3 = 11, find 3x - 1. Solution: x = 4; 3(4) - 1 = 11.

Approach

Solve for x first, then substitute the found value into the given expression. This is a two-part LET item format.

Question Type

Find the value of x that satisfies the equation, then evaluate an expression

Key Points To Remember

A linear equation has the variable to the FIRST POWER only.
Use inverse operations to isolate the variable: addition/subtraction undone by subtraction/addition; multiplication/division undone by division/multiplication.
Keep the equation balanced — any operation done to one side must be done to the other.
CRITICAL INEQUALITY RULE: Multiplying or dividing both sides by a NEGATIVE number REVERSES the inequality symbol.
Linear equation solutions are SINGLE VALUES; inequality solutions are RANGES.
Always check your solution by substituting it back into the original equation.
For inequalities, checking with a test value in the solution range confirms correctness.
Equations with the variable on both sides: collect variable terms on one side first.
If you get a false statement (e.g., 3 = 5), there is NO solution.
If you get an always-true statement (e.g., 5 = 5), there are INFINITELY MANY solutions.

Systems of Linear Equations

A SYSTEM OF LINEAR EQUATIONS is a set of two or more equations involving the same variables. For two equations in two unknowns (x and y), the SOLUTION is an ordered pair (x, y) that satisfies BOTH equations simultaneously. Geometrically, each equation represents a straight line, and the solution is the POINT OF INTERSECTION of those lines. There are three possible outcomes: 1. ONE SOLUTION (Consistent and Independent): The lines intersect at exactly one point. This is the most common case on the LET. 2. NO SOLUTION (Inconsistent): The lines are PARALLEL — they never meet. The equations have the same slope but different y-intercepts. 3. INFINITELY MANY SOLUTIONS (Consistent and Dependent): The lines are IDENTICAL — one is a multiple of the other. THREE METHODS OF SOLVING: METHOD 1 — SUBSTITUTION: Best when one variable is already isolated or easy to isolate. Step 1: Solve one equation for one variable. Step 2: Substitute that expression into the other equation. Step 3: Solve the resulting single-variable equation. Step 4: Substitute back to find the other variable. Step 5: Check in BOTH original equations. Example: Solve x + 2y = 11 and 3x - y = 5. From the first equation: x = 11 - 2y. Substitute into the second: 3(11 - 2y) - y = 5 → 33 - 6y - y = 5 → 33 - 7y = 5 → 7y = 28 → y = 4. Back-substitute: x = 11 - 2(4) = 3. Solution: (3, 4). Check: 3 + 8 = 11 ✓ and 9 - 4 = 5 ✓. METHOD 2 — ELIMINATION (Addition/Subtraction): Best when the coefficients of one variable are equal or easily made equal. Step 1: Multiply one or both equations by constants so that the coefficients of one variable are opposites. Step 2: Add the equations to eliminate that variable. Step 3: Solve for the remaining variable. Step 4: Substitute back to find the other. Step 5: Check. Example: Solve 2x + 3y = 12 and 4x - 3y = 6. The coefficients of y are already opposites (+3 and -3). Add the equations: (2x + 3y) + (4x - 3y) = 12 + 6 6x = 18, so x = 3. Substitute: 2(3) + 3y = 12 → 6 + 3y = 12 → 3y = 6 → y = 2. Solution: (3, 2). METHOD 3 — GRAPHING: Graph both lines and identify the intersection point. Used more for conceptual understanding on the LET than for computation.

Examples

When one equation already has y (or x) isolated, substitution is the fastest method. The key is to substitute into the OTHER equation, not back into the same one.

Scenario

Solve by Substitution: y = 2x - 1 and 3x + y = 9

Solution

Step 1: y is already isolated in the first equation: y = 2x - 1. Step 2: Substitute into the second equation. 3x + (2x - 1) = 9 Step 3: Combine like terms. 5x - 1 = 9 Step 4: Add 1 to both sides. 5x = 10 Step 5: Divide by 5. x = 2 Step 6: Back-substitute. y = 2(2) - 1 = 3 Solution: (2, 3). Check in Equation 1: 3 = 2(2) - 1 = 3 ✓ Check in Equation 2: 3(2) + 3 = 6 + 3 = 9 ✓

When y-coefficients are equal and opposite, elimination by addition is immediate. This is often the fastest method when the coefficients are set up this way.

Scenario

Solve by Elimination: 3x + 2y = 16 and 5x - 2y = 0

Solution

Step 1: Observe that the y-coefficients are +2 and -2 (already opposites). Step 2: Add the two equations. (3x + 2y) + (5x - 2y) = 16 + 0 8x = 16 Step 3: Divide by 8. x = 2 Step 4: Substitute into the first equation. 3(2) + 2y = 16 6 + 2y = 16 2y = 10 y = 5 Solution: (2, 5). Check in Equation 2: 5(2) - 2(5) = 10 - 10 = 0 ✓

Coin problems always generate TWO equations: one for the COUNT (number of items) and one for the TOTAL VALUE. This is the standard template for all LET coin/collection problems.

Scenario

Coin Problem (System Application): A teacher collects ₱5 and ₱10 coins from a class activity — 12 coins worth ₱95 total. How many of each coin?

Solution

Step 1: Define variables. Let x = number of ₱5 coins Let y = number of ₱10 coins Step 2: Write the system. Count equation: x + y = 12 Value equation: 5x + 10y = 95 Step 3: From the count equation: x = 12 - y. Step 4: Substitute into the value equation. 5(12 - y) + 10y = 95 60 - 5y + 10y = 95 60 + 5y = 95 5y = 35 y = 7 Step 5: x = 12 - 7 = 5. Answer: 5 five-peso coins and 7 ten-peso coins. Check: 5(5) + 10(7) = 25 + 70 = 95 ✓ and 5 + 7 = 12 ✓

Applications

Mixture problems: combining two solutions of different concentrations.
Age problems: relating present ages to ages in the future or past.
Finance problems: splitting a total into two categories with different unit values.
Motion problems: two vehicles traveling toward or away from each other.
School supply/classroom resource allocation scenarios common in teacher education contexts.

Misconceptions

WRONG: Checking the solution in only one equation. CORRECT: Check in BOTH equations.
WRONG: In elimination, multiplying only one equation when you need to scale both to create opposite coefficients. CORRECT: Scale equations carefully so that the variable being eliminated has equal and opposite coefficients.
WRONG: Concluding that 0 = 0 means no solution. CORRECT: 0 = 0 is always true and means INFINITE solutions (dependent system).
WRONG: Substituting back into the equation you already manipulated, not the original. CORRECT: Always substitute into an original equation.
WRONG: In a word problem, writing two equations that say the same thing — only one condition is captured. CORRECT: Each equation must represent a DIFFERENT condition of the problem.

Related Concepts

Linear Equations in One Variable
Graphing Linear Equations
Slope and y-intercept
Word Problem Translation
Matrix Methods (advanced, not typically on elementary LET)

Common Exam Questions

Example

Solve: 2x + y = 7 and x - y = 2. Add the equations: 3x = 9, x = 3; y = 7 - 2(3) = 1. Solution: (3, 1).

Approach

Choose substitution if one variable has coefficient 1; choose elimination if coefficients line up for easy cancellation. Always check in both equations.

Question Type

Solve a system of two equations in two unknowns

Example

Two numbers sum to 20 and differ by 4. Find the numbers. x + y = 20, x - y = 4. Adding: 2x = 24, x = 12, y = 8.

Approach

Identify the two unknowns, write one equation for each condition stated in the problem. Solve by substitution or elimination. Verify the answer satisfies the original word problem conditions.

Question Type

Word problem requiring a system of equations

Example

Are 2x + y = 5 and 4x + 2y = 8 consistent? Rewrite: y = -2x + 5 and y = -2x + 4. Same slope, different intercepts → Inconsistent (no solution).

Approach

Rewrite both equations in slope-intercept form (y = mx + b). Compare slopes and intercepts: same slope, different intercept = inconsistent; same slope, same intercept = dependent; different slopes = one solution.

Question Type

Identify the nature of a system (consistent/inconsistent/dependent)

Key Points To Remember

A system solution is an ordered pair (x, y) that satisfies ALL equations in the system.
Three outcomes: one solution (consistent independent), no solution (inconsistent/parallel lines), infinite solutions (dependent/same line).
Substitution method: best when one variable has a coefficient of 1 or -1.
Elimination method: best when a variable can be eliminated by simply adding or subtracting the equations.
After finding the solution, always CHECK in BOTH original equations, not just one.
If elimination gives 0 = 0 (always true), the system is DEPENDENT (infinite solutions).
If elimination gives 0 = non-zero (e.g., 0 = 5, always false), the system is INCONSISTENT (no solution).
The LET often presents coin, age, or mixture problems that require setting up a system of two equations.
Parallel lines have the same slope — if you compute slopes and they are equal, there is no solution.
Label your variables clearly before writing the system to avoid confusion in word problems.

Special Products and Factoring

SPECIAL PRODUCTS are shortcut formulas for multiplying certain binomials. They are called 'special' because they produce predictable, patterned results. You must be able to use them in BOTH directions: expanding (multiplying out) and factoring (working backward). The LET tests both directions heavily. SPECIAL PRODUCT FORMULAS TO MEMORIZE: 1. SQUARE OF A SUM: (a + b)² = a² + 2ab + b² Example: (x + 5)² = x² + 2(x)(5) + 5² = x² + 10x + 25 2. SQUARE OF A DIFFERENCE: (a - b)² = a² - 2ab + b² Example: (2x - 3)² = (2x)² - 2(2x)(3) + 3² = 4x² - 12x + 9 3. PRODUCT OF SUM AND DIFFERENCE (Difference of Two Squares): (a + b)(a - b) = a² - b² Example: (x + 7)(x - 7) = x² - 49 4. PRODUCT OF TWO BINOMIALS (FOIL method): (x + p)(x + q) = x² + (p + q)x + pq Example: (x + 3)(x + 5) = x² + 8x + 15 [since 3 + 5 = 8 and 3 × 5 = 15] FACTORING is the reverse of multiplication — breaking a polynomial into its factors. The GOLDEN RULE: always look for and remove the GREATEST COMMON MONOMIAL FACTOR (CMF) FIRST before applying any other factoring technique. FACTORING TECHNIQUES IN ORDER: 1. COMMON MONOMIAL FACTOR: Find the GCF of all terms and factor it out. Example: 6x² + 9x = 3x(2x + 3) [GCF is 3x] 2. DIFFERENCE OF TWO SQUARES: a² - b² = (a + b)(a - b) Example: x² - 25 = (x + 5)(x - 5); 4x² - 9 = (2x + 3)(2x - 3) 3. PERFECT SQUARE TRINOMIAL: a² + 2ab + b² = (a + b)² or a² - 2ab + b² = (a - b)² Example: x² + 6x + 9 = (x + 3)²; x² - 10x + 25 = (x - 5)² 4. GENERAL TRINOMIAL x² + bx + c: Find two integers whose PRODUCT is c and SUM is b. Example: x² - 5x + 6 → need two numbers: product = 6, sum = -5 → use -2 and -3. Answer: (x - 2)(x - 3) 5. GENERAL TRINOMIAL ax² + bx + c (a ≠ 1): Use trial-and-error with FOIL or the AC method. Example: 2x² + 7x + 3 → (2x + 1)(x + 3) SIMPLIFYING RATIONAL EXPRESSIONS: Factor the numerator and denominator, then cancel common factors. Example: (x² - 9)/(x - 3) = (x + 3)(x - 3)/(x - 3) = x + 3 (for x ≠ 3)

Examples

Map each part of the formula to the specific terms in the problem. The middle term is NEGATIVE because of the minus sign in (a - b)². A common mistake is writing +12x instead of -12x.

Scenario

Expand: (3x - 2)²

Solution

Identify: a = 3x, b = 2. Apply (a - b)² = a² - 2ab + b². Step 1: a² = (3x)² = 9x² Step 2: 2ab = 2(3x)(2) = 12x Step 3: b² = 2² = 4 Answer: (3x - 2)² = 9x² - 12x + 4

Never skip the GCF step. After factoring out 2x, we got a difference of squares that could be further factored. ALWAYS check if factors can be factored further.

Scenario

Factor completely: 2x³ - 8x

Solution

Step 1: Look for the Greatest Common Monomial Factor. GCF of 2x³ and 8x is 2x. Factor out 2x: 2x(x² - 4) Step 2: Can x² - 4 be factored further? Yes! It is a difference of two squares: x² - 4 = x² - 2² = (x + 2)(x - 2). Final Answer: 2x(x + 2)(x - 2) Check by expanding: 2x(x + 2)(x - 2) = 2x(x² - 4) = 2x³ - 8x ✓

Since c is positive and b is negative, BOTH factors must be negative. This narrows the search significantly. The product = c and sum = b rule is the key technique for factoring general trinomials.

Scenario

Factor: x² - 7x + 12

Solution

This is a general trinomial of the form x² + bx + c where b = -7, c = 12. Step 1: Find two integers whose PRODUCT is 12 and SUM is -7. Consider pairs that multiply to 12: 1 × 12 = 12 (sum = 13) — No 2 × 6 = 12 (sum = 8) — No 3 × 4 = 12 (sum = 7) — No (-3) × (-4) = 12 (sum = -7) — YES! Step 2: Write the factored form. x² - 7x + 12 = (x - 3)(x - 4) Check by FOIL: x² - 4x - 3x + 12 = x² - 7x + 12 ✓

Applications

Solving quadratic equations by factoring (sets each factor equal to zero).
Simplifying algebraic fractions (rational expressions) by canceling common factors.
Geometry: finding dimensions of rectangles when area is given as a polynomial.
Number sense: quickly computing products like 99 × 101 = (100-1)(100+1) = 10000 - 1 = 9999 using difference of squares.
Algebraic proofs and identities in higher mathematics.

Misconceptions

WRONG: (a + b)² = a² + b² (forgetting the middle term 2ab). CORRECT: (a + b)² = a² + 2ab + b².
WRONG: Factoring a sum of two squares: x² + 9 = (x + 3)(x - 3). INCORRECT — x² + 9 is PRIME over the reals.
WRONG: Canceling TERMS in a fraction: (x² + 4)/(x²) simplifying to 1 + 4 = 5. CORRECT: You can only cancel FACTORS, not terms.
WRONG: Stopping factoring too early, e.g., leaving 2(x² - 4) without noticing x² - 4 is a difference of squares.
WRONG: In (a - b)², computing the middle term as +2ab instead of -2ab.

Related Concepts

Greatest Common Factor (GCF)
Polynomial Multiplication
Quadratic Equations (factoring method)
Rational Expressions
Zero Product Property

Common Exam Questions

Example

Expand (2x + 3)². = (2x)² + 2(2x)(3) + 3² = 4x² + 12x + 9.

Approach

Identify which special product formula applies (square of sum/difference, difference of squares, or FOIL). Apply the formula, being careful with coefficients and signs.

Question Type

Expand a special product expression

Example

Factor 3x² - 12. GCF first: 3(x² - 4). Then difference of squares: 3(x+2)(x-2).

Approach

Step 1: Remove GCF. Step 2: Count the number of terms — two terms may be difference of squares; three terms may be perfect square trinomial or general trinomial. Step 3: Check each factor for further factoring.

Question Type

Factor a polynomial completely

Example

Simplify (x² + 5x + 6)/(x + 2). Factor numerator: (x+2)(x+3). Cancel (x+2): x + 3, for x ≠ -2.

Approach

Factor numerator and denominator completely. Cancel common factors (not common terms). State the restriction. This is a common two-step LET item.

Question Type

Simplify a rational algebraic expression

Key Points To Remember

ALWAYS remove the Greatest Common Monomial Factor first before other factoring.
The middle term of (a + b)² is POSITIVE 2ab; the middle term of (a - b)² is NEGATIVE 2ab.
Difference of two squares (a² - b²) always has two factors: (a + b)(a - b). There is NO 'sum of two squares' factoring over real numbers.
For a perfect square trinomial, check: is the first term a perfect square? Is the last term a perfect square? Is the middle term twice the product of the square roots?
For trinomial x² + bx + c: the two numbers in the factors must MULTIPLY to c and ADD to b.
FOIL stands for First, Outer, Inner, Last — the four products when multiplying two binomials.
Factoring completely means applying all possible factoring techniques until no factor can be further factored.
A polynomial is PRIME (irreducible) if it cannot be factored over the integers.
When simplifying rational expressions, factor BEFORE canceling — never cancel terms that are added or subtracted.
Always state the restriction (x ≠ value that makes the denominator zero) when simplifying rational expressions.

Quadratic Equations

A QUADRATIC EQUATION is any equation that can be written in the STANDARD FORM ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0 (if a = 0, the equation would be linear, not quadratic). The solutions to a quadratic equation are called ROOTS or ZEROS. A quadratic equation may have two real roots, one repeated real root, or no real roots (two complex roots), depending on the value of the DISCRIMINANT. THREE METHODS FOR SOLVING QUADRATIC EQUATIONS: METHOD 1 — FACTORING (most efficient when applicable): Step 1: Write the equation in standard form (ax² + bx + c = 0). Step 2: Factor the left side completely. Step 3: Apply the ZERO PRODUCT PROPERTY: if AB = 0, then A = 0 or B = 0. Step 4: Set each factor equal to zero and solve. Example: Solve x² - 5x + 6 = 0. Factor: (x - 2)(x - 3) = 0. Set each to zero: x - 2 = 0 → x = 2; x - 3 = 0 → x = 3. Roots: x = 2 or x = 3. METHOD 2 — SQUARE ROOT METHOD (only when there is no bx term, i.e., b = 0): Step 1: Isolate the x² term. Step 2: Take the square root of both sides. Remember the ± sign! Example: Solve x² - 49 = 0. x² = 49 x = ±√49 = ±7 Roots: x = 7 or x = -7. METHOD 3 — QUADRATIC FORMULA (always works for any quadratic): For ax² + bx + c = 0: x = [-b ± √(b² - 4ac)] / (2a) This formula ALWAYS gives the correct roots regardless of whether the equation factors nicely. Example: Solve 2x² + 3x - 5 = 0. a = 2, b = 3, c = -5 Discriminant: b² - 4ac = 9 - 4(2)(-5) = 9 + 40 = 49 x = [-3 ± √49] / (2·2) = (-3 ± 7) / 4 x = (-3 + 7)/4 = 4/4 = 1 or x = (-3 - 7)/4 = -10/4 = -5/2 Roots: x = 1 or x = -5/2. THE DISCRIMINANT (b² - 4ac) tells the NATURE OF THE ROOTS: - If b² - 4ac > 0: TWO distinct real roots - If b² - 4ac = 0: ONE repeated real root (two equal roots) - If b² - 4ac < 0: NO real roots (two complex roots) This is often tested directly on the LET: 'Without solving, determine the nature of the roots.'

Examples

The key is finding the pair of integers that MULTIPLY to c (-12) and ADD to b (+1). Since the product is negative, one integer is positive and one is negative.

Scenario

Solve by factoring: x² + x - 12 = 0

Solution

Step 1: The equation is already in standard form. Step 2: Factor. Find two numbers with product = -12 and sum = +1. Pairs: (-3)(4) = -12 and -3 + 4 = 1 ✓ Factored form: (x - 3)(x + 4) = 0 Step 3: Apply Zero Product Property. x - 3 = 0 → x = 3 x + 4 = 0 → x = -4 Roots: x = 3 or x = -4. Check x = 3: 9 + 3 - 12 = 0 ✓ Check x = -4: 16 + (-4) - 12 = 16 - 16 = 0 ✓

When b is negative, -b becomes positive in the formula. D = 100 is a perfect square, confirming rational roots. This also means the equation could have been factored: 3x² - 2x - 8 = (3x + 4)(x - 2).

Scenario

Use the Quadratic Formula: 3x² - 2x - 8 = 0

Solution

Step 1: Identify a = 3, b = -2, c = -8. Step 2: Compute the discriminant. D = b² - 4ac = (-2)² - 4(3)(-8) = 4 + 96 = 100 Step 3: Apply the formula. x = [-(-2) ± √100] / (2·3) x = [2 ± 10] / 6 Root 1: x = (2 + 10)/6 = 12/6 = 2 Root 2: x = (2 - 10)/6 = -8/6 = -4/3 Roots: x = 2 or x = -4/3. Check x = 2: 3(4) - 2(2) - 8 = 12 - 4 - 8 = 0 ✓

The LET frequently asks only about the nature of roots, not the actual roots. Computing only the discriminant saves time. D = 0 means the equation is a perfect square trinomial.

Scenario

Determine the nature of roots WITHOUT solving: 4x² - 4x + 1 = 0

Solution

Step 1: Identify a = 4, b = -4, c = 1. Step 2: Compute the discriminant only. D = b² - 4ac = (-4)² - 4(4)(1) = 16 - 16 = 0 Step 3: Interpret. Since D = 0, the equation has EXACTLY ONE repeated real root. (In fact, 4x² - 4x + 1 = (2x - 1)² = 0, giving x = 1/2 as a repeated root.)

Applications

Finding dimensions of geometric figures when the area is given (length × width = area leads to a quadratic).
Projectile motion problems: height = -5t² + v₀t + h₀, solving for t when height = 0.
Revenue/profit problems in economics: Revenue = price × quantity, where one depends on the other.
Age problems that result in quadratic equations.
Number problems: 'The product of two consecutive integers is 72' leads to n(n+1) = 72.

Misconceptions

WRONG: Solving x² = 25 as x = 5 only. CORRECT: x = ±5 (both +5 and -5 are solutions).
WRONG: In the formula, computing -b as -(b) without considering the sign of b. If b = -3, then -b = 3 (not -3 again).
WRONG: Dividing the formula incorrectly — only dividing part of the numerator by 2a. CORRECT: BOTH -b and ±√D must be in the numerator, divided by 2a.
WRONG: Forgetting standard form — applying the formula to x² + 3 = 2x without first rewriting as x² - 2x + 3 = 0.
WRONG: Interpreting D < 0 as having 'negative roots' rather than 'no real roots'.

Related Concepts

Factoring General Trinomials
Zero Product Property
Square Roots and Radicals
Parabolas and Graphs of Quadratic Functions
Discriminant and Nature of Roots

Common Exam Questions

Example

Solve x² - 9 = 0. (x+3)(x-3) = 0. x = 3 or x = -3.

Approach

Write in standard form, factor, apply zero product property, list both roots. Check both roots.

Question Type

Solve a quadratic equation by factoring

Example

For x² + 2x + 5 = 0: D = 4 - 20 = -16 < 0 → no real roots.

Approach

Identify a, b, c from standard form. Compute D = b² - 4ac. Interpret: D > 0 (two real), D = 0 (one real), D < 0 (no real roots).

Question Type

Find the nature of roots using the discriminant

Example

Solve x² + 6x + 7 = 0. D = 36 - 28 = 8. x = (-6 ± 2√2)/2 = -3 ± √2.

Approach

Write in standard form, identify a, b, c (careful with signs), compute discriminant, substitute into formula, simplify. Present both roots.

Question Type

Apply the quadratic formula

Key Points To Remember

Standard form of a quadratic equation: ax² + bx + c = 0, where a ≠ 0.
Always write the equation in standard form BEFORE solving or applying the quadratic formula.
Zero Product Property: if AB = 0, then A = 0 OR B = 0. This is the foundation of the factoring method.
Square root method: when b = 0, take ±√ of both sides. Do NOT forget the ± sign.
The Quadratic Formula always works — use it when factoring is not obvious.
The discriminant D = b² - 4ac: D > 0 → two real roots; D = 0 → one real root; D < 0 → no real roots.
Always identify a, b, and c correctly before applying the quadratic formula. Watch signs especially for negative b and c.
Verify roots by substituting them back into the original equation.
A quadratic equation has AT MOST two roots.
If the discriminant is a perfect square, the roots are rational and the equation can be factored.

Translating Words into Equations (Word Problems)

The ability to translate verbal descriptions into algebraic equations is the most frequently tested skill in the LET algebra section. Every word problem follows the same four-step process based on Polya's problem-solving model: STEP 1 — UNDERSTAND: Read carefully. Identify what is asked and what information is given. STEP 2 — PLAN: Define your variable(s). Let x = the unknown quantity (be specific: 'Let x = the age of the son today'). STEP 3 — EXECUTE: Translate the words into an equation and solve. STEP 4 — CHECK: Verify your numerical answer against the ORIGINAL WORD PROBLEM (not just the equation). KEY TRANSLATION VOCABULARY (must be memorized): - 'the sum of a number and 7' → n + 7 - 'a number increased by 5' → n + 5 - '5 less than a number' → n - 5 (NOT 5 - n — order matters!) - 'a number decreased by 3' → n - 3 - 'twice a number' → 2n - 'the product of 4 and a number' → 4n - 'the quotient of a number and 2' → n/2 - 'a number squared' → n² - 'three times a number decreased by 5' → 3n - 5 - 'five decreased by three times a number' → 5 - 3n - 'is', 'equals', 'is the same as' → = CONSECUTIVE INTEGER TEMPLATES: - Consecutive integers: n, n+1, n+2, ... - Consecutive EVEN integers: n, n+2, n+4, ... (n must be even) - Consecutive ODD integers: n, n+2, n+4, ... (n must be odd — same form, different condition) AGE PROBLEM TEMPLATE: - Present ages: let person A = x, person B = y (or express one in terms of the other) - In 'k' years: add k to each age - 'k' years ago: subtract k from each age MOTION PROBLEM TEMPLATE: Use D = R × T (Distance = Rate × Time) - Same direction, different rates: set up expressions for each object - Opposite directions: their distances ADD up to the total distance - Same distance (round trip): set distances equal

Examples

Always simplify the left side by combining like terms before solving. The check must verify the sum equals 72 using the ACTUAL numbers, not just the variable value.

Scenario

Consecutive Integer Problem: The sum of three consecutive integers is 72. Find the integers.

Solution

Step 1: Define variables. Let n = first integer n + 1 = second integer n + 2 = third integer Step 2: Write the equation from the condition 'sum is 72'. n + (n + 1) + (n + 2) = 72 Step 3: Simplify and solve. 3n + 3 = 72 3n = 69 n = 23 Step 4: State the answer. The three consecutive integers are 23, 24, and 25. Check: 23 + 24 + 25 = 72 ✓

The two time frames ('now' and 'in 12 years') generate the two expressions. Express Maria's age in terms of the son's age to work with only one variable, making this a one-equation, one-variable problem.

Scenario

Age Problem: Maria is 3 times as old as her son. In 12 years, Maria will be twice as old as her son. Find their present ages.

Solution

Step 1: Define variables. Let x = son's present age 3x = Maria's present age Step 2: In 12 years: Son's age = x + 12 Maria's age = 3x + 12 Step 3: Write the condition 'Maria will be twice as old as her son'. 3x + 12 = 2(x + 12) Step 4: Solve. 3x + 12 = 2x + 24 3x - 2x = 24 - 12 x = 12 Step 5: State the answer. Son is 12 years old; Maria is 3(12) = 36 years old. Check: In 12 years, son = 24, Maria = 48. Is 48 = 2 × 24? ✓

Having two conditions naturally leads to a two-equation system, but Condition 1 already expresses y in terms of x, so direct substitution into Condition 2 is the most efficient path.

Scenario

Number Problem: One number is 5 more than twice another. Their sum is 35. Find both numbers.

Solution

Step 1: Define variables. Let x = the smaller number Let y = the larger number Step 2: Translate the conditions. Condition 1: 'One number is 5 more than twice another' → y = 2x + 5 Condition 2: 'Their sum is 35' → x + y = 35 Step 3: Substitute (y is already isolated in Condition 1). x + (2x + 5) = 35 3x + 5 = 35 3x = 30 x = 10 Step 4: Find y. y = 2(10) + 5 = 25 The numbers are 10 and 25. Check: 25 = 2(10) + 5 = 25 ✓; 10 + 25 = 35 ✓

Geometry word problems on the LET typically involve setting up an algebraic equation from a geometric formula (perimeter, area). Express one dimension in terms of the other to use only one variable.

Scenario

Rectangular Perimeter Problem: A classroom bulletin board's length is 3 dm more than its width. Its perimeter is 26 dm. Find its dimensions.

Solution

Step 1: Define variables. Let w = width w + 3 = length Step 2: Use the perimeter formula: P = 2(l + w). 2[(w + 3) + w] = 26 Step 3: Simplify and solve. 2(2w + 3) = 26 4w + 6 = 26 4w = 20 w = 5 Step 4: State dimensions. Width = 5 dm, Length = 5 + 3 = 8 dm. Check: P = 2(5 + 8) = 2(13) = 26 ✓

Applications

Real DepEd classroom scenarios: computing number of pupils, distributing materials, budgeting for classroom supplies.
Age problems involving teacher and student age comparisons (context-rich LET items).
Motion problems: scheduling field trips, calculating travel times.
Mixture/solution problems in Science class contexts.
Financial literacy applications: computing total costs, change, and budgets.

Misconceptions

WRONG: Translating '5 less than x' as 5 - x. CORRECT: x - 5. The number x is being reduced, so it comes first.
WRONG: For age problems, writing 'In 5 years, Maria's age is 3x + 5' when Maria's age was already defined as 3x — CORRECT, but forgetting to ALSO update the son's age: this leads to an incorrect equation.
WRONG: For consecutive integers, using n, n+1, n+3 instead of n, n+1, n+2.
WRONG: Solving for x and stopping without answering the actual question — e.g., the problem asks for the LARGER of two numbers but you report x (the smaller).
WRONG: Checking only the equation and not the original word problem — a sign or translation error can produce an equation whose solution doesn't match the story.

Related Concepts

Linear Equations (one and two variables)
Systems of Linear Equations
Quadratic Equations (for more complex word problems)
Polya's Problem-Solving Heuristics
Ratio and Proportion

Common Exam Questions

Example

Sum of three consecutive odd integers is 57. n + (n+2) + (n+4) = 57 → 3n + 6 = 57 → n = 17. Integers: 17, 19, 21.

Approach

Use n, n+1, n+2 for consecutive integers. Write the equation from the given condition (sum, product, etc.). Solve for n and state all integers. Verify by computing the sum/product with actual numbers.

Question Type

Consecutive integer sum or product problem

Example

Father is 4 times as old as son. In 8 years he will be twice as old. Son: x, Father: 4x. 4x + 8 = 2(x + 8) → x = 4. Son is 4, Father is 16.

Approach

Define variables for present ages. Add or subtract the time difference for future/past ages. Write the equation from the given ratio or relationship. Solve and verify by checking ages at BOTH time frames.

Question Type

Age problem

Example

The sum of two numbers is 28, and their difference is 4. x + y = 28, x - y = 4. Add: 2x = 32, x = 16, y = 12.

Approach

Carefully translate each phrase. Watch 'less than' and 'more than' for correct subtraction order. Write one equation per condition. Solve using substitution or elimination.

Question Type

Number relationship problem

Key Points To Remember

'5 less than n' = n - 5 (NOT 5 - n). Order of subtraction follows the direction of language.
'5 decreased by n' = 5 - n. The subject of 'decreased' is what comes first.
Consecutive even AND odd integers both use n, n+2, n+4 as the template.
For age problems, set up clearly which person is which variable, and be consistent with 'now', 'in X years', or 'X years ago'.
For coin problems: always write TWO equations — one for COUNT, one for TOTAL VALUE.
For motion problems: D = RT. Organize in a table with columns for Rate, Time, and Distance.
Check your answer against the ORIGINAL PROBLEM STATEMENT, not just the equation you set up.
If the answer is a fraction or negative when the problem asks for a number of people or objects, recheck your equation.
Label your variable clearly: 'Let x = the SMALLER number' is better than just 'Let x = a number'.
For mixture/investment problems, the total amount equation and the value/interest equation form the system.

Functions, Slope, and the Cartesian Plane

A FUNCTION is a special relationship where every input (x-value) has EXACTLY ONE output (y-value). Functions are written as f(x), read as 'f of x.' Evaluating a function means substituting a specific value for x. For f(x) = 2x² - 3x + 1, finding f(-2) means replacing every x with -2 (in parentheses!) and computing: f(-2) = 2(-2)² - 3(-2) + 1 = 2(4) + 6 + 1 = 15. The CARTESIAN PLANE (coordinate plane) is formed by two perpendicular number lines: the horizontal x-axis and the vertical y-axis, meeting at the ORIGIN (0, 0). Any point is located by an ORDERED PAIR (x, y) where x is the horizontal coordinate (positive = right, negative = left) and y is the vertical coordinate (positive = up, negative = down). The plane is divided into four QUADRANTS: - Quadrant I (upper right): x > 0, y > 0 - Quadrant II (upper left): x < 0, y > 0 - Quadrant III (lower left): x < 0, y < 0 - Quadrant IV (lower right): x > 0, y < 0 A LINEAR FUNCTION graphs as a straight line and is written in SLOPE-INTERCEPT FORM: y = mx + b, where m is the SLOPE and b is the Y-INTERCEPT (the y-value where the line crosses the y-axis). SLOPE (m) measures the STEEPNESS and DIRECTION of a line: m = (y₂ - y₁) / (x₂ - x₁) = rise / run Types of slope: - Positive slope: line rises from left to right - Negative slope: line falls from left to right - Zero slope: horizontal line (y = constant) - Undefined slope: vertical line (x = constant) Example: Find slope through (1, 2) and (4, 11). m = (11 - 2) / (4 - 1) = 9 / 3 = 3. The line rises 3 units for each 1 unit to the right.

Examples

The parentheses around -1 ensure that the exponent is applied to the negative value correctly: (-1)² = 1, not -1. Without parentheses, -1² would be incorrectly computed as -1.

Scenario

Evaluate the function f(x) = 3x² - 2x + 4 at x = -1

Solution

Step 1: Substitute x = -1 in parentheses. f(-1) = 3(-1)² - 2(-1) + 4 Step 2: Apply exponent first. (-1)² = 1 f(-1) = 3(1) - 2(-1) + 4 Step 3: Multiply. = 3 + 2 + 4 Step 4: Add. f(-1) = 9

The key is converting to slope-intercept form. Remember to divide ALL terms by -2, and the sign of b tells you whether the y-intercept is above or below the origin.

Scenario

Find the slope and y-intercept of 6x - 2y = 10

Solution

Step 1: Rewrite in slope-intercept form (y = mx + b) by isolating y. 6x - 2y = 10 -2y = -6x + 10 Divide every term by -2: y = 3x - 5 Step 2: Identify m and b. m = 3 (slope) b = -5 (y-intercept: the line crosses the y-axis at (0, -5)) Answer: Slope = 3, y-intercept = -5.

Applications

Graphing linear equations — a skill taught in Grade 7 and reviewed in teacher licensure exams.
Interpreting rates of change in real-world contexts (e.g., cost per pupil, distance per hour).
Setting up and solving systems of equations graphically.
Analyzing data trends in educational research and assessment contexts.
Understanding direct and inverse variation in Science integration.

Misconceptions

WRONG: f(x) means f times x. CORRECT: f(x) is the OUTPUT of the function f when the INPUT is x.
WRONG: slope = run/rise (inverting the formula). CORRECT: slope = rise/run = (y₂ - y₁)/(x₂ - x₁).
WRONG: A horizontal line has undefined slope. CORRECT: Horizontal lines have ZERO slope; VERTICAL lines have UNDEFINED slope.
WRONG: All four quadrants are equal in size and bounded by positive values. CORRECT: Quadrant II has negative x and positive y; Quadrant III has both negative.
WRONG: f(-2) = 2f(-1). Functions do not generally scale this way — only if f is linear and passes through the origin.

Related Concepts

Ordered Pairs and Graphing
Linear Equations in Two Variables
Slope-Intercept Form
Direct and Inverse Variation
Rate of Change

Common Exam Questions

Example

If g(x) = x² - 4x + 3, find g(2). = 4 - 8 + 3 = -1.

Approach

Substitute the given value for x (in parentheses), follow order of operations: exponents, multiplication, then addition/subtraction.

Question Type

Function evaluation at a specific value

Example

Slope through (-1, 3) and (2, -3): m = (-3-3)/(2-(-1)) = -6/3 = -2.

Approach

From two points: use m = (y₂ - y₁)/(x₂ - x₁). From equation: rewrite in y = mx + b form, then read off m directly.

Question Type

Finding slope from two points or from slope-intercept form

Example

In which quadrant is (-3, 5)? x is negative, y is positive → Quadrant II.

Approach

Check the signs of x and y: (+,+) → QI; (-,+) → QII; (-,-) → QIII; (+,-) → QIV.

Question Type

Identify the quadrant of a given point

Key Points To Remember

A function assigns EXACTLY ONE output to each input. Vertical Line Test: if a vertical line crosses the graph more than once, it is NOT a function.
f(x) notation means 'the function f evaluated at x' — it is NOT f times x.
Always substitute negative values in parentheses when evaluating functions to protect signs.
In y = mx + b: m is the slope (steepness), b is the y-intercept (where line crosses y-axis).
Slope formula: m = (y₂ - y₁) / (x₂ - x₁). Be consistent — use the same point as 'point 1' and 'point 2' in both numerator and denominator.
Horizontal lines (y = k) have slope ZERO; vertical lines (x = k) have UNDEFINED slope.
Parallel lines have EQUAL slopes; perpendicular lines have slopes that are NEGATIVE RECIPROCALS.
The quadrants are numbered counter-clockwise starting from the upper right (Quadrant I).
The x-axis and y-axis are NOT part of any quadrant — points on them are simply 'on the x-axis' or 'on the y-axis'.
Slope is the algebraic representation of RATE OF CHANGE — it connects algebra to real-world problems involving rates.

Practice Problems

Apply exponent laws in order: power of a product first, then the product rule (add exponents), then quotient rule (subtract exponents). Divide coefficients separately from variables.

Problem

Simplify: (3x²y)² × (2xy³) ÷ (6x³y⁴)

Solution

Step 1: Apply power of a product to (3x²y)². (3x²y)² = 9x⁴y² Step 2: Multiply by 2xy³. 9x⁴y² × 2xy³ = 18x⁵y⁵ Step 3: Divide by 6x³y⁴. 18x⁵y⁵ ÷ 6x³y⁴ = (18/6) × x^(5-3) × y^(5-4) = 3x²y Answer: 3x²y

Multiplying through by the LCD eliminates all fractions and transforms the problem into a simple linear equation. Always check in the ORIGINAL equation.

Problem

Solve for x: (3x - 1)/4 + 2 = (x + 3)/2

Solution

Step 1: Multiply all terms by LCD = 4 to clear fractions. 4 × (3x-1)/4 + 4 × 2 = 4 × (x+3)/2 (3x - 1) + 8 = 2(x + 3) Step 2: Simplify each side. 3x + 7 = 2x + 6 Step 3: Subtract 2x from both sides. x + 7 = 6 Step 4: Subtract 7. x = -1 Check: (3(-1)-1)/4 + 2 = (-4)/4 + 2 = -1 + 2 = 1; and ((-1)+3)/2 = 2/2 = 1. ✓

GCF first (3x), then factor the resulting quadratic trinomial. Since c = -5 is negative, the two integers have opposite signs. The sum being -4 means the larger absolute value is negative.

Problem

Factor completely: 3x³ - 12x² - 15x

Solution

Step 1: Find and remove the GCF. GCF of 3x³, 12x², and 15x is 3x. 3x³ - 12x² - 15x = 3x(x² - 4x - 5) Step 2: Factor the trinomial x² - 4x - 5. Find two numbers: product = -5, sum = -4. (-5)(1) = -5 and -5 + 1 = -4 ✓ x² - 4x - 5 = (x - 5)(x + 1) Step 3: Write the complete factored form. 3x(x - 5)(x + 1) Check by expanding: 3x(x² + x - 5x - 5) = 3x(x² - 4x - 5) = 3x³ - 12x² - 15x ✓

Not all systems have integer solutions — this one produces fractions. Multiplying the second equation by 4 creates +4y and -4y, which cancel on addition. Fractions are handled the same way as integers.

Problem

Solve the system: 3x + 4y = 10 and 2x - y = 1

Solution

Method: Elimination (eliminate y). Step 1: Multiply the second equation by 4 to make y-coefficients equal. 4(2x - y) = 4(1) → 8x - 4y = 4 Step 2: Add to the first equation. (3x + 4y) + (8x - 4y) = 10 + 4 11x = 14 x = 14/11 Step 3: Substitute into the simpler equation 2x - y = 1. 2(14/11) - y = 1 28/11 - y = 1 y = 28/11 - 11/11 = 17/11 Solution: (14/11, 17/11). Check in Equation 1: 3(14/11) + 4(17/11) = 42/11 + 68/11 = 110/11 = 10 ✓

D = 36 > 0 and is a perfect square, confirming two rational roots. Cross-verify using factoring — the roots match, confirming accuracy. The quadratic formula is reliable even when factoring is not obvious.

Problem

Solve using the quadratic formula: x² + 4x - 5 = 0

Solution

Step 1: Identify a = 1, b = 4, c = -5. Step 2: Compute the discriminant. D = b² - 4ac = 16 - 4(1)(-5) = 16 + 20 = 36 Step 3: Apply the quadratic formula. x = (-4 ± √36) / (2 × 1) = (-4 ± 6) / 2 Root 1: x = (-4 + 6)/2 = 2/2 = 1 Root 2: x = (-4 - 6)/2 = -10/2 = -5 Roots: x = 1 or x = -5. Verification by factoring: (x - 1)(x + 5) = x² + 5x - x - 5 = x² + 4x - 5 ✓

Consecutive even integers always differ by 2. Since 46 is even, 46 and 48 are both even and consecutive. If n were odd (e.g., n = 47), the answer would need re-examination — but 46 is even, so the answer is valid.

Problem

The sum of two consecutive even integers is 94. Find the integers.

Solution

Step 1: Define variables. Let n = the first even integer n + 2 = the next consecutive even integer Step 2: Write the equation. n + (n + 2) = 94 Step 3: Solve. 2n + 2 = 94 2n = 92 n = 46 Step 4: The integers are 46 and 48. Check: 46 + 48 = 94 ✓; both are even ✓; they differ by 2 ✓

Always factor BOTH the numerator and denominator completely before canceling. State restrictions for BOTH values that make the original denominator zero (x = 2 and x = -2), not just the one canceled.

Problem

Simplify the rational expression: (x² + 3x - 10) / (x² - 4)

Solution

Step 1: Factor the numerator. x² + 3x - 10: find product = -10, sum = 3 → use 5 and -2. x² + 3x - 10 = (x + 5)(x - 2) Step 2: Factor the denominator. x² - 4 = x² - 2² = (x + 2)(x - 2) Step 3: Write the fraction with factored forms. [(x + 5)(x - 2)] / [(x + 2)(x - 2)] Step 4: Cancel the common factor (x - 2). = (x + 5) / (x + 2), for x ≠ 2 and x ≠ -2 Answer: (x + 5)/(x + 2) with restrictions x ≠ 2 and x ≠ -2.

This is a discriminant application problem — it tests conceptual understanding of the discriminant. Setting D = 0 directly gives a linear equation in k that is easy to solve.

Problem

Find the value of k such that kx² + 4x + 1 = 0 has exactly one real root.

Solution

Step 1: For exactly one real root, the discriminant D = 0. D = b² - 4ac = 0 Here a = k, b = 4, c = 1. Step 2: Set up the equation. (4)² - 4(k)(1) = 0 16 - 4k = 0 Step 3: Solve for k. 4k = 16 k = 4 Answer: k = 4. Verification: 4x² + 4x + 1 = (2x + 1)² = 0 → x = -1/2 (one repeated root) ✓

For objects moving in OPPOSITE directions, ADD their distances. For objects moving in the SAME direction, SUBTRACT their distances to find the gap. The key formula is D = RT.

Problem

Two buses leave a terminal at the same time, traveling in opposite directions. One travels at 60 kph and the other at 80 kph. After how many hours will they be 420 km apart?

Solution

Step 1: Let t = number of hours traveled. Step 2: Distance traveled by Bus 1 = 60t Distance traveled by Bus 2 = 80t Step 3: Since they travel in OPPOSITE DIRECTIONS, total separation = sum of distances. 60t + 80t = 420 Step 4: Combine like terms. 140t = 420 t = 3 hours. Answer: They will be 420 km apart after 3 hours. Check: 60(3) + 80(3) = 180 + 240 = 420 ✓

Area problems often lead to quadratic equations. Reject negative values since dimensions cannot be negative. This demonstrates the real-world importance of checking answers against the physical constraints of the problem.

Problem

The length of a rectangle is 2 cm more than three times its width. If the area is 56 cm², find the dimensions.

Solution

Step 1: Define variables. Let w = width 3w + 2 = length Step 2: Use the area formula: A = length × width. w(3w + 2) = 56 Step 3: Expand and write in standard form. 3w² + 2w = 56 3w² + 2w - 56 = 0 Step 4: Apply the quadratic formula. a = 3, b = 2, c = -56 D = 4 + 4(3)(56) = 4 + 672 = 676 = 26² w = (-2 ± 26) / 6 w = 24/6 = 4 or w = -28/6 (rejected, negative) Step 5: State dimensions. Width = 4 cm, Length = 3(4) + 2 = 14 cm. Check: Area = 4 × 14 = 56 cm² ✓

Exam Preparation Tips

SIGN DISCIPLINE IS EVERYTHING: The vast majority of algebra errors on the LET are sign mistakes. When distributing a negative (like -3(x - 4) = -3x + 12), slow down and check the sign of every term. Practice this deliberately.
MEMORIZE THE SIX EXPONENT LAWS WITH EXAMPLES: Write them on a card, practice with examples daily. Product law (add), quotient law (subtract), power law (multiply), zero exponent (=1), negative exponent (reciprocal), power of a product (distribute) — these appear directly in LET items.
MEMORIZE THE FOUR SPECIAL PRODUCT FORMULAS AND THEIR REVERSES: (a+b)², (a-b)², (a+b)(a-b), and FOIL must be automatic. Practice going from expanded form to factored form and back without hesitation.
FACTOR IN ORDER: (1) GCF always first. (2) Count the terms — two terms suggest difference of squares; three terms suggest perfect square trinomial or general trinomial. (3) Check if factors can be factored further.
FOR INEQUALITIES, FLAG THE REVERSAL RULE: Every time you solve an inequality problem, consciously ask yourself: 'Am I dividing or multiplying by a negative?' If yes, flip the symbol. Make this a habit.
PRACTICE WORD PROBLEM TRANSLATION DAILY: Take 5-10 minutes each day translating verbal expressions into algebraic ones. Focus especially on subtraction order ('5 less than n' vs. 'n less than 5'), age problems, and consecutive integer problems.
ALWAYS CHECK YOUR ANSWER IN THE ORIGINAL PROBLEM: Not the equation you set up — the original word problem or equation. This catches translation errors AND computation errors in one step.
USE POLYA'S FOUR STEPS FOR EVERY WORD PROBLEM: (1) Understand — what is given and asked; (2) Plan — define variables; (3) Execute — set up and solve; (4) Check — verify in original context. Make this your automatic procedure.
FOR THE QUADRATIC FORMULA, WRITE ALL THREE: a = ?, b = ?, c = ? before plugging in. Carefully note negative values of b and c. Compute the discriminant SEPARATELY before taking the square root.
PRACTICE SYSTEMS WITH BOTH METHODS: Some LET items give systems that are faster with substitution; others with elimination. Recognize when each is optimal: substitution when a coefficient is ±1; elimination when coefficients are already set up to cancel.
REVIEW FRACTION ARITHMETIC ALONGSIDE ALGEBRA: Many equation-solving steps involve fractions. Weak fraction skills cause cascading errors in algebra. Make sure you can add, subtract, multiply, and divide fractions fluently.
TIME MANAGEMENT ON THE LET: Algebra items can be solved in under 2 minutes each if you recognize patterns. Invest time learning to recognize — at a glance — whether a trinomial is a perfect square, whether a system is best solved by substitution or elimination, and which quadratic method is fastest.
CONNECT ALGEBRA TO K-12 CONTENT: Understanding these concepts pedagogically (how you would TEACH them to Grade 5-6 pupils) deepens your own understanding. Think about how the LET algebra content connects to the K-12 BEC Mathematics framework for the intermediate grades.
DO NOT SKIP STEPS IN PRACTICE: Many exam errors come from mental shortcuts that hide mistakes. When practicing, write every step — this builds the habit of organized work under exam conditions.
REVIEW ALL WORKED EXAMPLES IN THIS GUIDE AT LEAST TWICE: The ten chapter worked examples cover the most common LET item types. Master each one until you can reproduce the solution without looking.

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In summary

Elementary Algebra is the backbone of the Mathematics section of the LET, and mastery of it requires two parallel skills: technical accuracy in manipulation and conceptual understanding of problem translation. The key takeaways from this chapter are interconnected: exponent laws feed into simplification, which feeds into factoring, which feeds into solving equations, which feeds into word problem solutions. No subtopic stands alone. As a future elementary teacher under RA 7836 (the Philippine Teachers Professionalization Act), you are expected to demonstrate content mastery not only to pass the licensure examination but to serve as a competent, professional educator. The K-12 Basic Education Curriculum for Grades 4-6 introduces algebraic thinking through patterns, unknowns, and simple equations — the conceptual foundation you are solidifying in this review. Understanding these concepts deeply means you will be better prepared to scaffold your pupils' mathematical thinking in the classroom. Recap the non-negotiables: (1) Always look for the GCF before factoring. (2) Always reverse the inequality symbol when multiplying or dividing by a negative. (3) Always check your solution in the original equation or word problem. (4) Always write negative substitution values in parentheses. (5) 'Five less than n' is n - 5, not 5 - n. These five rules, consistently applied, prevent the majority of LET algebra errors. Approach your continued review with the same care and discipline you will bring to your classroom: systematic, thorough, and student-centered — in this case, you are your own most important student. Consistent daily practice of 10-15 algebra problems, with careful written work and self-checking, will build the automaticity and confidence you need on examination day. Maligayang pag-aaral at best of luck on your LET!

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