Algebra — Study Notes for Bihar TET Paper II

Overview

Algebra forms the backbone of upper-primary mathematics and carries significant weight in the Bihar TET Paper II Mathematics section. This topic tests your ability to manipulate symbols, simplify expressions, and solve equations—skills that directly connect arithmetic to higher mathematics.

For the exam, you must demonstrate two competencies: (1) fluency in algebraic operations and standard identities, and (2) the ability to set up and solve linear equations from word problems. Questions typically range from straightforward simplification to application-based problems involving age, money, or geometric relationships.

Mastering algebra is also essential for teaching Classes 6–8, where students transition from concrete numbers to abstract symbols. A teacher who understands common student errors in algebra can design better lessons and assessments.

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Key Concepts

**Variable and Constant**: A variable (like x, y) represents an unknown or changing quantity; a constant (like 5, −3) has a fixed value.

**Algebraic Expression**: A combination of variables, constants, and operations (e.g., 3x + 5y − 7). Unlike equations, expressions have no equality sign.

**Terms, Coefficients, and Like Terms**: In 4x² + 3x − 2, there are three terms; 4 is the coefficient of x²; like terms share the same variable and power (e.g., 5x and −2x).

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

**Algebraic Identity**: An equation true for all values of the variable—distinct from an equation which is true only for specific values.

**Linear Equation in One Variable**: Has the form ax + b = 0, where a ≠ 0. The solution is a single value of x.

**Linear Equation in Two Variables**: Has the form ax + by + c = 0. Graphically represents a straight line; needs two equations to find a unique solution.

**Transposition Rule**: When moving a term across the equality sign, change its sign (+ becomes −, × becomes ÷).

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Formulas / Key Facts

### Standard Algebraic Identities (Must Memorise)

| Identity | Expanded Form | |----------|---------------| | (a + b)² | a² + 2ab + b² | | (a − b)² | a² − 2ab + b² | | a² − b² | (a + b)(a − b) | | (a + b + c)² | a² + b² + c² + 2ab + 2bc + 2ca | | (x + a)(x + b) | x² + (a + b)x + ab |

### Useful Facts

**Degree of a polynomial**: Highest exponent of the variable (e.g., degree of 5x³ − 2x + 1 is 3).
**Zero of a polynomial**: Value of x for which the polynomial equals zero.
**Solution of linear equation**: The value satisfying the equation. Verify by substituting back.
**Consistent system**: Two linear equations with at least one common solution.
**Unique solution condition**: For a₁x + b₁y = c₁ and a₂x + b₂y = c₂, unique solution exists when a₁/a₂ ≠ b₁/b₂.

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Worked Examples

### Example 1: Simplify using identity **Problem**: Find the value of 103².

**Solution**: Write 103 as (100 + 3). Using (a + b)² = a² + 2ab + b²: 103² = (100 + 3)² = 100² + 2 × 100 × 3 + 3² = 10000 + 600 + 9 = **10609**

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### Example 2: Solve a linear equation **Problem**: Solve 3x − 7 = 2x + 5

**Solution**: Step 1: Bring variable terms to one side. 3x − 2x = 5 + 7

Step 2: Simplify. x = 12

**Verification**: LHS = 3(12) − 7 = 36 − 7 = 29 RHS = 2(12) + 5 = 24 + 5 = 29 ✓

Answer: **x = 12**

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### Example 3: Word problem (Age) **Problem**: A father is 30 years older than his son. In 5 years, the father's age will be three times the son's age. Find their present ages.

**Solution**: Let son's present age = x years. Father's present age = (x + 30) years.

After 5 years: Son's age = x + 5 Father's age = x + 30 + 5 = x + 35

Given: Father's age = 3 × Son's age x + 35 = 3(x + 5) x + 35 = 3x + 15 35 − 15 = 3x − x 20 = 2x x = 10

Son's present age = **10 years** Father's present age = 10 + 30 = **40 years**

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Common Mistakes

1. **Sign errors during transposition**

*Wrong*: Moving +5 from RHS to LHS and keeping it +5.
*Correct*: When a term crosses the = sign, its sign reverses. +5 becomes −5.

2. **Confusing expression with equation**

*Wrong*: Trying to "solve" 3x + 5 for x.
*Correct*: An expression cannot be solved; only equations (with =) can be solved.

3. **Misapplying identities**

*Wrong*: Writing (a + b)² = a² + b².
*Correct*: (a + b)² = a² + **2ab** + b². The middle term is crucial.

4. **Forgetting to verify the solution**

*Wrong*: Accepting x = 5 without checking.
*Correct*: Always substitute back into the original equation to confirm.

5. **Combining unlike terms**

*Wrong*: Adding 3x² + 2x to get 5x³.
*Correct*: Unlike terms (different powers) cannot be combined. 3x² + 2x remains as is.

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Quick Reference

**(a + b)² = a² + 2ab + b²** — the middle term has coefficient 2.

**To solve ax + b = c**: isolate x = (c − b)/a.

**Like terms** share the same variable AND the same exponent.

**Degree** of 7x⁴ − 3x² + x is **4** (highest power).

**Verify every solution** by substituting back into the original equation.

**Identity vs Equation**: Identity holds for ALL values; equation holds for SPECIFIC values only.

Algebra