Algebra — Study Notes for HSSC CET
Overview
Algebra forms the foundation of quantitative reasoning in competitive exams. For HSSC CET, you need to master algebraic expressions, standard identities, and solving linear equations. These concepts appear directly as standalone questions and indirectly when solving problems on percentages, profit-loss, time-work, and other arithmetic topics.
The typical CET algebra question tests your ability to simplify expressions quickly, apply the right identity without expanding everything, or solve equations in one or two variables. Speed matters—most algebra questions can be solved in under 90 seconds if you know the shortcuts. Memorizing identities and practising their reverse application (factorization) is non-negotiable for scoring well.
This topic connects heavily to Data Interpretation and Mensuration, where algebraic manipulation is required to extract answers from given data or formulas.
Key Concepts
- Algebraic Expression: A combination of constants, variables, and operations (e.g., 3x² + 5x − 7). Terms are separated by + or − signs.
- Polynomial Degree: The highest power of the variable. In 4x³ + 2x − 1, the degree is 3.
- Like Terms: Terms with identical variable parts (e.g., 3xy and −5xy). Only like terms can be added or subtracted directly.
- Identity vs Equation: An identity holds true for all values of variables (e.g., (a + b)² = a² + 2ab + b²). An equation is true only for specific values.
- Linear Equation in One Variable: Form ax + b = 0. Solution: x = −b/a.
- Linear Equations in Two Variables: Form a₁x + b₁y = c₁ and a₂x + b₂y = c₂. Solved using substitution, elimination, or cross-multiplication.
- Factorization: Breaking an expression into product of simpler expressions. Reverse of expansion.
- Condition for Unique Solution: Two linear equations have a unique solution when a₁/a₂ ≠ b₁/b₂.
Formulas / Key Facts
Standard Algebraic Identities (must memorize)
| Identity | Expansion |
|---|---|
| (a + b)² | a² + 2ab + b² |
| (a − b)² | a² − 2ab + b² |
| (a + b)(a − b) | a² − b² |
| (a + b)³ | a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a + b) |
| (a − b)³ | a³ − 3a²b + 3ab² − b³ = a³ − b³ − 3ab(a − b) |
| a³ + b³ | (a + b)(a² − ab + b²) |
| a³ − b³ | (a − b)(a² + ab + b²) |
| (a + b + c)² | a² + b² + c² + 2ab + 2bc + 2ca |
Derived Results (frequently tested)
- 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)
- If a + b + c = 0, then a³ + b³ + c³ = 3abc
Linear Equations — Solution Conditions
For equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂:
- Unique solution: a₁/a₂ ≠ b₁/b₂
- Infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂
- No solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Worked Examples
Example 1: Using Identity for Quick Calculation
If x + 1/x = 5, find x² + 1/x².
Solution: Use the identity (a + b)² = a² + 2ab + b²
(x + 1/x)² = x² + 2(x)(1/x) + 1/x² 25 = x² + 2 + 1/x² x² + 1/x² = 25 − 2 = 23
Example 2: Factorization Using a³ − b³
Simplify: (125x³ − 27) ÷ (5x − 3)
Solution: Recognize 125x³ = (5x)³ and 27 = 3³
Using a³ − b³ = (a − b)(a² + ab + b²): 125x³ − 27 = (5x − 3)(25x² + 15x + 9)
So, (125x³ − 27) ÷ (5x − 3) = 25x² + 15x + 9
Example 3: Solving Simultaneous Equations
Solve: 3x + 4y = 18 and 5x − 2y = 4
Solution (Elimination Method): Multiply second equation by 2: 10x − 4y = 8
Add to first equation: 3x + 4y + 10x − 4y = 18 + 8 13x = 26 x = 2
Substitute in first equation: 3(2) + 4y = 18 4y = 12 y = 3
Answer: x = 2, y = 3
Example 4: Applying (a + b + c)² Identity
If a + b + c = 10 and a² + b² + c² = 38, find ab + bc + ca.
Solution: (a + b + c)² = a² + b² + c² + 2(ab + bc + ca) 100 = 38 + 2(ab + bc + ca) 2(ab + bc + ca) = 62 ab + bc + ca = 31
Common Mistakes
- Forgetting the middle term in squares: Students write (a + b)² = a² + b², missing 2ab. → Always remember: square of sum = sum of squares + twice the product.
- Sign errors in (a − b)²: Writing (a − b)² = a² + 2ab + b² instead of a² − 2ab + b². → The middle term takes the sign from the bracket.
- Confusing a³ + b³ with (a + b)³: These are different expressions. a³ + b³ = (a + b)(a² − ab + b²), not (a + b)³. → Check by substituting a = 1, b = 1 to verify your formula.
- Dividing by zero in equations: When simplifying, ensure you're not dividing by a variable that could be zero. → Always check if the denominator can equal zero.
- Wrong condition recall for no solution: Mixing up the conditions for infinite vs no solution. → No solution requires the constant ratio to be different from the coefficient ratios.
Quick Reference
- (a + b)² = a² + 2ab + b² — never forget the middle term
- a² − b² = (a + b)(a − b) — difference of squares always factors
- If x + 1/x = k, then x² + 1/x² = k² − 2
- If a + b + c = 0, then a³ + b³ + c³ = 3abc — appears often in exams
- Unique solution: a₁/a₂ ≠ b₁/b₂ — ratios must be unequal
- Cross-multiplication for two linear equations: x = (b₁c₂ − b₂c₁)/(a₁b₂ − a₂b₁)