Algebra — Study Notes for Railway Group D

Overview

Algebra is a compact but high-yield topic in Railway Group D mathematics, contributing 2–4 questions per paper. Unlike lengthy arithmetic problems, algebra questions test your ability to translate word problems into equations and solve them systematically. Mastery of linear equations, simultaneous equations, and basic identities is essential because these tools underpin many other topics — from age problems to ratio questions.

You must be comfortable setting up equations from statements, manipulating them using basic operations, and substituting values efficiently. The exam favours straightforward, single-variable or two-variable problems rather than complex polynomial manipulation. Speed matters: a well-practiced student can solve most algebra questions in under 90 seconds. Focus on forming equations correctly and avoiding sign errors during simplification.

Key Concepts

• **Variable and constant**: A variable (x, y, z) represents an unknown quantity; a constant is a fixed number. An algebraic expression combines variables and constants using operations.

• **Linear equation in one variable**: An equation of the form ax + b = 0, where a ≠ 0. It has exactly one solution: x = -b/a. The degree (highest power of the variable) is 1.

• **Simultaneous linear equations**: Two equations in two variables (x and y) that must both be true at the same time. Standard form: a₁x + b₁y = c₁ and a₂x + b₂y = c₂. They usually have one unique solution unless the lines are parallel or coincident.

• **Substitution method**: Solve one equation for one variable, then substitute that expression into the other equation. Useful when one equation is already isolated or easy to isolate.

• **Elimination method**: Multiply equations by suitable numbers so that adding or subtracting them cancels one variable. Often faster than substitution when coefficients align nicely.

• **Algebraic identities**: Pre-proved formulas that allow quick expansion or factorisation. The exam tests the three core square identities and their applications to simplify expressions and solve equations.

• **Transposition**: Moving a term from one side of an equation to the other by reversing its operation (addition ↔ subtraction, multiplication ↔ division). Maintain equality by performing the same operation on both sides.

• **Word-problem translation**: Keywords map to operations — "sum" → +, "difference" → −, "product" → ×, "is/equals" → =. Age, money, and number problems frequently appear disguised as algebra.

Formulas / Key Facts

**(a + b)² = a² + 2ab + b²** Expand the square of a sum; used to simplify (x + 3)², (2y + 5)², etc.

**(a − b)² = a² − 2ab + b²** Expand the square of a difference; beware the middle term is negative.

**(a + b)(a − b) = a² − b²** Difference of two squares; used for factorisation and quick multiplication (e.g., 103×97).

**a² + b² = (a + b)² − 2ab = (a − b)² + 2ab** Derived identities to find a² + b² when you know a + b or a − b and ab.

**Linear equation solution: ax + b = c ⟹ x = (c − b)/a** Isolate the variable by inverse operations.

**Simultaneous solution by elimination:** Multiply equation 1 by m, equation 2 by n, then add/subtract to eliminate one variable; solve for the other, substitute back.

**Checking your solution:** Substitute your answer back into the original equations to verify both are satisfied.

**Zero-product property:** If xy = 0, then x = 0 or y = 0. Useful in factorised equations.

Worked Examples

**Example 1: Solve for x: 3x − 7 = 2x + 5** *Step 1:* Bring variable terms to one side: 3x − 2x = 5 + 7 *Step 2:* Simplify: x = 12 *Check:* 3(12) − 7 = 36 − 7 = 29 and 2(12) + 5 = 24 + 5 = 29 ✓

**Example 2: Solve simultaneously: 2x + 3y = 13 and 3x − y = 3** *Method: Elimination* *Step 1:* Multiply equation 2 by 3 to match y-coefficients: 9x − 3y = 9 *Step 2:* Add to equation 1: (2x + 3y) + (9x − 3y) = 13 + 9 ⟹ 11x = 22 ⟹ x = 2 *Step 3:* Substitute x = 2 into equation 2: 3(2) − y = 3 ⟹ 6 − y = 3 ⟹ y = 3 *Answer:* x = 2, y = 3 *Check:* 2(2) + 3(3) = 4 + 9 = 13 ✓ and 3(2) − 3 = 6 − 3 = 3 ✓

**Example 3: Simplify (x + 4)² − (x − 4)² without expanding fully** *Step 1:* Use identity (a + b)² − (a − b)² = 4ab with a = x, b = 4 *Step 2:* = 4(x)(4) = 16x *Answer:* 16x (Direct expansion would take much longer; identity saves time.)

**Example 4: Age problem — Five years ago, a father was three times as old as his son. Five years hence, he will be twice as old. Find their present ages.** *Step 1:* Let present ages be f (father) and s (son). *Step 2:* Five years ago: f − 5 = 3(s − 5) ⟹ f − 5 = 3s − 15 ⟹ f = 3s − 10 … (1) *Step 3:* Five years hence: f + 5 = 2(s + 5) ⟹ f + 5 = 2s + 10 ⟹ f = 2s + 5 … (2) *Step 4:* Equate (1) and (2): 3s − 10 = 2s + 5 ⟹ s = 15 *Step 5:* Substitute into (1): f = 3(15) − 10 = 35 *Answer:* Father is 35 years old, son is 15 years old.

Common Mistakes

**Mistake 1: Sign error when transposing** *Wrong thinking:* 3x + 5 = 11 ⟹ 3x = 11 + 5 *Correct fix:* When moving +5 to the right, it becomes −5 ⟹ 3x = 11 − 5 = 6. Always reverse the sign.

**Mistake 2: Forgetting the middle term in (a − b)²** *Wrong thinking:* (x − 3)² = x² − 9 *Correct fix:* (x − 3)² = x² − 2(3)(x) + 9 = x² − 6x + 9. The middle term −2ab is essential.

**Mistake 3: Dividing both sides by a variable without considering it could be zero** *Wrong thinking:* xy = 5y ⟹ x = 5 by cancelling y *Correct fix:* If y = 0, the equation is 0 = 0, which is satisfied for any x. You must either specify y ≠ 0 or factor: y(x − 5) = 0, so x = 5 or y = 0.

**Mistake 4: Using wrong identity for difference of squares** *Wrong thinking:* x² − 16 = (x − 4)² *Correct fix:* x² − 16 = (x + 4)(x − 4), not a perfect square. Check the pattern: a² − b² = (a + b)(a − b).

**Mistake 5: Misinterpreting "difference" in word problems** *Wrong thinking:* "The difference of two numbers is 5" ⟹ x − y = 5 or y − x = 5 used interchangeably without consistency. *Correct fix:* Establish which is larger. If x > y, then x − y = 5. Stick to one convention throughout the problem to avoid confusion.

Quick Reference

• **One-variable linear**: ax + b = c ⟹ x = (c − b)/a in one step.
• **Simultaneous**: Use elimination if coefficients are neat; substitution if one variable is isolated.
• **(a + b)² = a² + 2ab + b²**; **(a − b)² = a² − 2ab + b²**; **(a + b)(a − b) = a² − b²** — memorise these three.
• **Always verify** your solution by substituting back into the original equation(s).
• **Word problems**: Define variables clearly, translate every sentence into an equation, then solve systematically.
• **Speed trick**: For expressions like (x + a)² − (x − a)², use 4ax directly instead of expanding.