GCSE Maths Formulae, functions & proof revision

Substitution means replacing each letter by its value. Bracket the values — especially negatives: if a = −2 then a² = (−2)² = 4, not −4. For v = u + at with u = 3, a = 1.5 and t = 4: v = 3 + 1.5 × 4 = 9, multiplication before addition. Scientific formulas bring powers and fractions: kinetic energy E = ½mv² with m = 4 and v = 3 gives E = ½ × 4 × 9 = 18.

Fractions in formulas are divisions: the whole of the denominator is underneath, so E = (mv²)⁄2 — compute mv² first, then halve. Rooted formulas reverse the reading: if d = √(2s⁄a), square both sides mentally to see the structure d² = 2s⁄a, which is what later rearranging undoes.

The subject sits alone on one side. To change it, undo the operations acting on it in reverse order: from v = u + at, subtract u then divide by t, giving a = (v − u)⁄t. Whole-bracket moves help: P = 2(l + w) divides by 2 to give l + w = P⁄2, then subtracts l, so w = P⁄2 − l.

Roots and powers undo each other: from A = πr², divide by π then square-root, giving r = √(A⁄π); from V = ⁴⁄₃πr³, multiply by ¾, divide by π, then cube-root: r = ∛(3V⁄4π). If the letter appears twice, collect it first: 3x + 5 = y − 2x becomes 5x = y − 5, then x = (y − 5)⁄5.

Higher. A function f maps each input x to one output: f(x) = 3x − 2 sends 4 to 10. The inverse f⁻¹ undoes it: swap x and y in y = 3x − 2 to get x = 3y − 2, then rearrange to y = (x + 2)⁄3, so f⁻¹(x) = (x + 2)⁄3. Check: f⁻¹(10) = 4 ✓. Graphically the inverse is the reflection in the line y = x.

A composite runs two functions in sequence: fg(x) means g first, then f. With f(x) = x² and g(x) = 2x + 1, fg(3) = f(7) = 49, while gf(3) = g(9) = 19 — order matters. For algebra: fg(x) = f(2x + 1) = (2x + 1)². Solve-with-functions questions use the same reading: “fg(x) = 49” gives (2x + 1)² = 49.

Higher. To show two expressions are equivalent, expand or factorise one into the other: (x + 1)(x + 3) = x² + 4x + 3 — the expansion is the proof, no numbers needed. Number proofs translate facts into algebra: even numbers are 2n, odd numbers 2n + 1, consecutive integers n, n + 1, n + 2. Their sum is 3n + 3 = 3(n + 1), which is a multiple of 3 — that final sentence (“a multiple of 3, as required”) is the proof.

Useful starting facts: the sum of two even numbers 2n + 2m = 2(n + m) is even; two odd numbers (2n + 1) + (2m + 1) = 2(n + m + 1) is even; an odd times an odd, (2n + 1)(2m + 1) = 4nm + 2n + 2m + 1 = 2(2nm + n + m) + 1, is odd. Never prove with examples — testing three cases convinces nobody; the algebra covers every case.

Higher. To solve x³ + 4x = 30 numerically, rearrange to x = ∛(30 − 4x). Starting from x₀ = 2: x₁ = ∛(30 − 8) = ∛22 ≈ 2.802, x₂ = ∛(30 − 4 × 2.802) ≈ 2.659, then 2.685, 2.680, 2.681… The values converge, so the root is 2.681 to 3 decimal places. Check by substitution: 2.681³ + 4 × 2.681 = 29.994…, close to 30 ✓.

Iteration questions give you the rearranged formula and a starting value — the skill is accurate feeding: keep several decimal places in the calculator, write each iterate to the asked accuracy, and stop when the question’s accuracy stops changing (two identical rounded values usually signal convergence). Substituting the final rounded value back into the original equation is the check examiners reward.