Elementary Algebra — RRB NTPC Study Notes

Overview

Elementary Algebra forms the backbone of quantitative reasoning in RRB NTPC Mathematics. Roughly 3–5 questions appear directly from this topic, and algebraic manipulation skills apply across Number System, Time-Work, Profit-Loss and other areas. Mastery means you can solve linear equations in seconds, handle simultaneous equations systematically, expand and factorise polynomials accurately, and apply standard identities without hesitation.

RRB NTPC tests algebra at a foundational level—no calculus, no complex polynomials. The focus is speed and accuracy with linear equations (one variable), pairs of simultaneous equations (two variables), recognising and applying the four basic polynomial identities, and factorising simple quadratic and cubic expressions. Questions often disguise algebra in word problems: "A number increased by 7 equals thrice the number decreased by 5—find the number." Your job is to translate English into equations and solve cleanly.

Expect a mix of direct computation ("Solve 3x − 5 = 2x + 7") and application problems. Time management is critical: you should spend 30–45 seconds per algebraic question. Practice until the identities and factorisation patterns become reflex.

Key Concepts

• **Linear equation in one variable**: An equation of the form ax + b = c, where a ≠ 0. Solution: isolate x by inverse operations—subtract, add, divide, multiply in reverse order of BODMAS.

• **Simultaneous linear equations**: Two equations in two unknowns (x and y). Three solution methods: substitution (solve one for x, plug into the other), elimination (add/subtract equations to cancel one variable), and cross-multiplication (determinant-based shortcut). All three yield the same answer; pick the fastest for the given problem.

• **Polynomial identity**: An equation true for all values of the variable. The four must-know identities let you expand or factorise instantly without multiplying term-by-term.

• **Factorisation**: Expressing a polynomial as a product of simpler polynomials. Techniques include taking out the common factor, grouping terms, and applying identities in reverse. Factorisation simplifies division, solves equations (set each factor = 0), and speeds up substitution.

• **Transposition**: Moving a term from one side of an equation to the other by changing its sign (addition ↔ subtraction, multiplication ↔ division). The golden rule: whatever you do to one side, do to the other to maintain equality.

• **Zero-product property**: If ab = 0, then a = 0 or b = 0. This is why factorising a quadratic lets you find roots immediately.

Formulas / Key Facts

1. **(a + b)² = a² + 2ab + b²** — Square of a sum. Memorise left-to-right and right-to-left.

2. **(a − b)² = a² − 2ab + b²** — Square of a difference. Watch the middle term sign.

3. **a² − b² = (a + b)(a − b)** — Difference of two squares. Works both ways: expanding and factorising.

4. **(a + b)³ = a³ + 3a²b + 3ab² + b³** or equivalently **a³ + b³ + 3ab(a + b)**. Use whichever form simplifies the problem.

5. **(a − b)³ = a³ − 3a²b + 3ab² − b³** or **a³ − b³ − 3ab(a − b)**.

6. **a³ + b³ = (a + b)(a² − ab + b²)** — Sum of cubes factorisation.

7. **a³ − b³ = (a − b)(a² + ab + b²)** — Difference of cubes factorisation.

8. **x² + (a + b)x + ab = (x + a)(x + b)** — Middle-term split for quadratics. If coefficient of x² is 1, find two numbers that multiply to constant term and add to coefficient of x.

Worked Examples

**Example 1: Solve 5x − 3 = 3x + 11**

Step 1: Bring like terms together. Subtract 3x from both sides: 5x − 3x − 3 = 11 → 2x − 3 = 11. Step 2: Add 3 to both sides: 2x = 14. Step 3: Divide by 2: x = 7. **Answer: x = 7.**

**Example 2: Solve the simultaneous equations 2x + 3y = 13 and 3x − y = 3 (elimination method)**

Step 1: Multiply second equation by 3 to match y coefficients: 9x − 3y = 9. Step 2: Add the two equations: (2x + 3y) + (9x − 3y) = 13 + 9 → 11x = 22 → x = 2. Step 3: Substitute x = 2 into first equation: 2(2) + 3y = 13 → 4 + 3y = 13 → 3y = 9 → y = 3. **Answer: x = 2, y = 3.**

**Example 3: Factorise x² − 5x + 6**

Step 1: Find two numbers that multiply to 6 and add to −5. They are −2 and −3. Step 2: Write as (x − 2)(x − 3). Check: (x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6 ✓ **Answer: (x − 2)(x − 3).**

**Example 4: Expand (2x + 3)² using identity**

Use (a + b)² = a² + 2ab + b² with a = 2x, b = 3. = (2x)² + 2(2x)(3) + 3² = 4x² + 12x + 9. **Answer: 4x² + 12x + 9.**

**Example 5: Simplify 97² using identity (useful for fast mental calculation)**

Write 97 = 100 − 3. Use (a − b)² = a² − 2ab + b². 97² = 100² − 2(100)(3) + 3² = 10000 − 600 + 9 = 9409. **Answer: 9409.** (This trick saves time when calculator isn't allowed.)

Common Mistakes

• **Sign errors in transposition**: Students write 3x − 5 = 10, then 3x = 10 − 5 (wrong: should be 3x = 10 + 5). → **Fix**: Remember transposition flips the sign. Term −5 becomes +5 when moved.

• **Forgetting to distribute the negative**: Expanding −(2x − 3) as −2x − 3 instead of −2x + 3. → **Fix**: Minus sign outside multiplies each term inside the bracket: −1 × 2x = −2x, −1 × (−3) = +3.

• **Misapplying (a + b)² as a² + b²**: Omitting the middle term 2ab. → **Fix**: Write the identity in full every time until it's muscle memory. (a + b)² always has three terms.

• **Wrong factor pairs in quadratic factorisation**: For x² + 5x + 6, writing (x + 2)(x + 4) because 2 + 4 = 6 (but 2 × 4 ≠ 6). → **Fix**: Check both conditions—sum AND product. Correct factors are (x + 2)(x + 3) since 2 + 3 = 5 and 2 × 3 = 6.

• **Incorrectly eliminating variables in simultaneous equations**: Adding equations when you should subtract, or vice versa, resulting in both variables surviving. → **Fix**: Align coefficients with opposite signs (one positive, one negative) before adding; or same sign before subtracting. Always verify one variable cancels.

Quick Reference

• **Linear solve**: Transpose terms → collect like terms → divide by coefficient of variable.
• **Simultaneous equations**: Substitution for simple cases; elimination when coefficients align easily.
• **(a ± b)² = a² ± 2ab + b²** (watch the middle-term sign).
• **a² − b² = (a + b)(a − b)** (instant factorisation/expansion).
• **Quadratic factorisation**: Find two numbers that multiply to constant and add to middle coefficient.
• **Always expand your answer mentally to verify factorisation** — takes 5 seconds, prevents silly mistakes.