Lesson 3: Algebra — Expressions and Equations

Algebra is mathematics with letters — and it is the gateway to everything higher: GCSE, A-level, physics, economics, and computer science all run on it. This lesson builds the core manipulative skills: collecting like terms, expanding brackets, factorising, and solving linear equations. The approach is always the same: understand what the letters mean, learn the moves, then practise until they are automatic. UK exam boards (Edexcel, AQA, OCR) test these skills in almost every paper.

Expressions, terms, and collecting like terms

An expression like 3x + 2y − x + 7 combines terms: 3x, 2y, −x, 7. Like terms have identical letter parts: 3x and −x are like terms; 3x and 3y are not; 2ab and 5ab are like, but 2a and 2ab are not. Collecting like terms: 3x + 2y − x + 7 = 2x + 2y + 7. The 2y and 7 have no partners, so they stay.

Worked example: simplify 5a − 3b + 2a² − a + 4b − 6. Group the terms: a terms give 5a − a = 4a; b terms give −3b + 4b = b; the a² term is 2a² (nothing combines with it); the constant is −6. Result: 2a² + 4a + b − 6. Exam tip: underline like terms in different colours (mentally or on paper) before combining — it prevents the classic error of combining 2a² with 4a.

In context: a taxi in Glasgow charges £3 fixed plus £2 per mile. For m miles, the fare is 3 + 2m pounds. For a 7-mile trip: 3 + 2(7) = £17. Substituting numbers into expressions — evaluation — is the first step toward formulas, which dominate science and finance.

Expanding brackets

Expanding means removing brackets by multiplication: 3(x + 4) = 3x + 12 — the 3 multiplies everything inside. With double brackets, use FOIL (First, Outer, Inner, Last) or the grid method: (x + 5)(x + 3) = x² + 3x + 5x + 15 = x² + 8x + 15. Every term in the first bracket multiplies every term in the second.

Worked example: expand (2x − 3)(x + 4). First: 2x × x = 2x². Outer: 2x × 4 = 8x. Inner: −3 × x = −3x. Last: −3 × 4 = −12. Collect: 2x² + 8x − 3x − 12 = 2x² + 5x − 12. Watch the signs — the minus before the 3 applies to both of its products. The most common GCSE error lives here; slow down on negatives. (Foundation tier stops at single brackets and simple quadratics; double brackets with coefficients are Higher-tier territory.)

Special cases to memorise: (a + b)² = a² + 2ab + b² (not a² + b²!); (a − b)² = a² − 2ab + b²; (a + b)(a − b) = a² − b² (difference of two squares). So (x + 7)² = x² + 14x + 49, and (2x + 5)(2x − 5) = 4x² − 25. These identities appear constantly in factorising and quadratics.

Worked example: expand and simplify (x + 4)(x − 6). FOIL: x² − 6x + 4x − 24 = x² − 2x − 24. Check with x = 10: (14)(4) = 56, and 100 − 20 − 24 = 56. ✓ A quick numeric check like this catches most expansion slips.

Factorising — expanding in reverse

Factorising is the reverse of expanding: 6x + 9 = 3(2x + 3) — take out the highest common factor. For 4x² + 10x the HCF is 2x, so 4x² + 10x = 2x(2x + 5). Always check by re-expanding mentally. Factorising quadratics reverses double-bracket expansion: x² + 8x + 15 = (x + 5)(x + 3) — find two numbers with product 15 and sum 8.

Worked example: factorise x² − 7x + 12. You need two numbers with product 12 and sum −7. Both must be negative: −3 and −4 (product 12, sum −7). So x² − 7x + 12 = (x − 3)(x − 4). For x² + 2x − 15: product −15, sum +2 → +5 and −3: (x + 5)(x − 3).

Difference of two squares: x² − 49 = (x + 7)(x − 7). Spot the pattern: a square minus a square. So 4x² − 25 = (2x + 5)(2x − 5), and 9x² − 16y² = (3x + 4y)(3x − 4y). GCSE Higher also tests factorising quadratics whose x² coefficient is not 1, e.g. 2x² + 7x + 3 = (2x + 1)(x + 3) — use trial of factor pairs.

Worked example: factorise fully 6x³ − 9x². The HCF of the terms is 3x², giving 3x²(2x − 3). 'Fully' means take out everything common — stopping at 3(2x³ − 3x²) loses a mark. Always look for the highest common factor of both the coefficients and the letter parts.

Solving linear equations

An equation states that two expressions are equal; solving finds the unknown. The golden rule: do the same to both sides. For 3x + 5 = 20: subtract 5 from both sides → 3x = 15; divide by 3 → x = 5. Check: 3(5) + 5 = 20. ✓ Always check — it takes seconds and catches sign errors.

With unknowns on both sides: 5x − 3 = 2x + 12. Collect unknowns on one side, numbers on the other: subtract 2x → 3x − 3 = 12; add 3 → 3x = 15; x = 5. With brackets: 3(x − 4) = 2(x + 5) → 3x − 12 = 2x + 10 → x = 22. With fractions: x/3 + x/4 = 7 — multiply everything by 12 (the LCM): 4x + 3x = 84 → 7x = 84 → x = 12.

Worked example: solve 4(x + 3) − 2(x − 5) = 28. Expand carefully (note −2 × −5 = +10): 4x + 12 − 2x + 10 = 28 → 2x + 22 = 28 → 2x = 6 → x = 3. Check: 4(6) − 2(−2) = 24 + 4 = 28. ✓

Worked example (UK context): a mobile plan in Birmingham costs £8 per month plus £0.15 per GB of data. A bill is £14. How many GB were used? 8 + 0.15d = 14 → 0.15d = 6 → d = 6 ÷ 0.15 = 40 GB. Forming and solving equations from word problems — define the variable, write the equation, solve, check in context — is the most valuable GCSE skill of all.

Simultaneous equations — a first look

Two unknowns need two equations. Worked example: 3 apples and 2 bananas cost £2.90; 5 apples and 4 bananas cost £5.30. Find the cost of each. In pence: 3a + 2b = 290 and 5a + 4b = 530. Double the first equation: 6a + 4b = 580. Subtract the second: a = 50. Substitute back: 3(50) + 2b = 290 → 2b = 140 → b = 70. An apple costs 50p, a banana 70p. Check in the second equation: 5(50) + 4(70) = 250 + 280 = 530. ✓ Elimination — making one coefficient match, then adding or subtracting — is the standard method.

Inequalities and rearranging formulas

Inequalities use <, >, ≤, ≥ instead of =. Solve them like equations, with one twist: multiplying or dividing by a negative reverses the sign. So −2x > 10 → x < −5. (With positive multipliers nothing changes: 3x ≤ 21 → x ≤ 7.)

Rearranging formulas — changing the subject — is algebra's most practical skill: it is how every science formula is used. Make t the subject of v = u + at: subtract u → v − u = at; divide by a → t = (v − u)/a. Make x the subject of y = 3x − 7: add 7 → y + 7 = 3x; divide by 3 → x = (y + 7)/3.

Worked example: the formula for converting Celsius to Fahrenheit is F = 9C/5 + 32. Make C the subject (useful for interpreting American weather reports!). Subtract 32: F − 32 = 9C/5. Multiply by 5: 5(F − 32) = 9C. Divide by 9: C = 5(F − 32)/9. Treat the target letter as the 'x' in an equation and isolate it step by step.

A stretch example: making r the subject of P = 2πr + πr² is awkward because r appears twice — rearranged, it becomes the quadratic πr² + 2πr − P = 0, solvable with the quadratic formula (Lesson 4). GCSE rarely goes this far, but recognising 'the letter appears twice' as the warning sign is a powerful habit.

Common mistakes

• (a + b)² = a² + b². Wrong — the middle term 2ab is missing: (x + 7)² = x² + 14x + 49, not x² + 49.
• Combining unlike terms. 2a² + 4a cannot be simplified to 6a³ or 6a². Only identical letter parts combine.
• Sign slips in expansion. −3(x − 4) = −3x + 12, not −3x − 12. The multiplier hits every term, sign included.
• Factorising that isn't full. 6x³ − 9x² = 3x²(2x − 3); leaving an x inside a partly-factored answer loses the mark.
• Forgetting to flip an inequality. Dividing −2x > 10 by −2 reverses the sign: x < −5.
• One-sided operations. Whatever you do to one side of an equation, do to the other — every time, written down.

Exam Experience

Question 1. Expand and simplify (x − 4)(x + 7). [2 marks]

Question 2. Solve 5(x − 2) = 3x + 14. [3 marks]

Question 3. n is an integer. Show that (n + 3)² − (n − 3)² is always a multiple of 12. [3 marks]

Mark scheme.

• Q1: M1 for an expansion with at least three of the four terms correct (x² + 7x − 4x − 28). A1 for x² + 3x − 28.
• Q2: M1 for expanding the bracket (5x − 10 = 3x + 14). M1 for collecting terms correctly (2x = 24). A1 for x = 12.
• Q3: M1 for expanding one square correctly, e.g. (n + 3)² = n² + 6n + 9. M1 for expanding both and subtracting: (n² + 6n + 9) − (n² − 6n + 9) = 12n. A1 for the conclusion: 12n is a multiple of 12 for every integer n.