Algebra — Study Notes for OTET Paper II

Overview

Algebra forms the backbone of upper-primary and secondary mathematics, bridging arithmetic with abstract mathematical reasoning. For OTET Paper II (classes VI–VIII), algebra questions test your understanding of algebraic expressions, standard identities, and the ability to solve linear equations—skills every mathematics teacher must demonstrate.

This topic carries direct weightage in the content section and also appears indirectly in pedagogy questions about teaching abstract concepts. Mastery here means you can simplify expressions confidently, apply identities without hesitation, and solve equations systematically. The exam typically presents straightforward problems, but careless sign errors and identity mix-ups cost marks.

Focus on three core areas: forming and simplifying algebraic expressions, memorising and applying the four standard identities, and solving linear equations in one variable. These skills transfer directly to classroom teaching at the elementary level.

Key Concepts

**Algebraic Expression**: A combination of constants and variables connected by operations (+, −, ×, ÷). Example: 3x² + 5x − 7. No equality sign—that makes it an equation.

**Terms, Coefficient, and Constant**: In 4x² − 3x + 2, there are three terms. The coefficient of x² is 4, the coefficient of x is −3, and 2 is the constant term.

**Like and Unlike Terms**: Like terms have identical variable parts (3x² and −5x² are like; 3x² and 3x are unlike). Only like terms can be added or subtracted directly.

**Polynomial Classification**: Monomial (1 term), Binomial (2 terms), Trinomial (3 terms). Degree = highest power of the variable.

**Identity vs Equation**: An identity is true for all values of variables; an equation is true only for specific values.

**Linear Equation in One Variable**: Standard form ax + b = 0 (a ≠ 0). The solution is a single value of x that satisfies the equation.

**Transposition Rule**: When a term moves across the equality sign, its sign changes. This is the most-used technique for solving linear equations.

**Verification**: Substitute the solution back into the original equation. If LHS = RHS, the answer is correct.

Formulas / Key Facts

**Standard Algebraic Identities (must memorise)**

1. (a + b)² = a² + 2ab + b² 2. (a − b)² = a² − 2ab + b² 3. (a + b)(a − b) = a² − b² 4. (x + a)(x + b) = x² + (a + b)x + ab

**Useful Derived Results**

a² + b² = (a + b)² − 2ab = (a − b)² + 2ab
(a + b)² − (a − b)² = 4ab
(a + b)² + (a − b)² = 2(a² + b²)

**Linear Equation Solution Steps**

1. Simplify both sides (remove brackets, combine like terms). 2. Transpose variable terms to one side, constants to the other. 3. Combine like terms. 4. Divide both sides by the coefficient of the variable. 5. Verify by substitution.

Worked Examples

### Example 1: Simplify (2x + 3)² − (2x − 3)²

**Step 1**: Use identity 3 in reverse form. (a + b)² − (a − b)² = 4ab, where a = 2x, b = 3

**Step 2**: Substitute. = 4 × (2x) × 3 = 24x

**Answer**: 24x

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### Example 2: If x + 1/x = 5, find the value of x² + 1/x²

**Step 1**: Square both sides of x + 1/x = 5. (x + 1/x)² = 25

**Step 2**: Expand using (a + b)² = a² + 2ab + b². x² + 2(x)(1/x) + 1/x² = 25 x² + 2 + 1/x² = 25

**Step 3**: Solve for x² + 1/x². x² + 1/x² = 25 − 2 = 23

**Answer**: 23

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### Example 3: Solve 3(x − 2) + 5 = 2(x + 1) − 4

**Step 1**: Expand brackets. 3x − 6 + 5 = 2x + 2 − 4 3x − 1 = 2x − 2

**Step 2**: Transpose 2x to LHS and −1 to RHS. 3x − 2x = −2 + 1 x = −1

**Step 3**: Verify. LHS = 3(−1 − 2) + 5 = 3(−3) + 5 = −9 + 5 = −4 RHS = 2(−1 + 1) − 4 = 2(0) − 4 = −4 LHS = RHS ✓

**Answer**: x = −1

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### Example 4: Factorise x² + 5x + 6

**Step 1**: Use identity 4: (x + a)(x + b) = x² + (a + b)x + ab. Find two numbers whose sum = 5 and product = 6. Numbers: 2 and 3 (since 2 + 3 = 5, 2 × 3 = 6)

**Step 2**: Write factors. x² + 5x + 6 = (x + 2)(x + 3)

**Answer**: (x + 2)(x + 3)

Common Mistakes

**Sign error during transposition** → Students write +5 on LHS becomes +5 on RHS. Fix: The sign always reverses when crossing the equality sign.

**Confusing (a − b)² with a² − b²** → Students expand (a − b)² as a² − b², forgetting the middle term 2ab. Fix: Say aloud "first squared, twice product, second squared" while writing.

**Adding unlike terms** → Writing 3x² + 2x = 5x³ or 5x². Fix: Check variable parts match exactly before combining.

**Forgetting to distribute the negative sign** → In 5 − (2x − 3), students write 5 − 2x − 3 instead of 5 − 2x + 3. Fix: Treat the minus before brackets as multiplying by −1.

**Not verifying solutions** → Students submit x = 3 without checking. In exams, a quick substitution catches careless errors and confirms your answer.

Quick Reference

(a + b)² = a² + 2ab + b² — square of sum
(a − b)² = a² − 2ab + b² — square of difference
(a + b)(a − b) = a² − b² — difference of squares
Transposing changes sign; multiplying/dividing does not change sign
Like terms: same variable, same power
Verify every linear equation solution by substituting back into the original

Algebra