Algebra

Manipulation and indices

Factorising is expanding run backwards — every technique here pulls a product apart or puts it back together.

Expanding and factorising

Expanding multiplies every term in one bracket by every term in the other. Factorising reverses it — highest common factor out first, always. Two key patterns: a² − b² = (a + b)(a − b), the difference of two squares; and (a ± b)² = a² ± 2ab + b², the perfect square.

Order of attack: common factor first, then two squares, then two brackets (ax² + bx + c), then grouping in pairs for four terms.

The laws of indices

The laws apply only when the bases match: aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ, a⁰ = 1, a⁻ⁿ = 1/aⁿ, and a^(m/n) = (ⁿ√a)ᵐ. A negative index is a reciprocal; a fractional index is a root. Both are examined at Extended.

Worked example: factorise fully 12x²y − 27y. Both terms share 3y: 12x²y − 27y = 3y(4x² − 9). Then 4x² − 9 is two squares split by a minus, so it becomes (2x + 3)(2x − 3). "Fully" is the word that costs marks — 3y(4x² − 9) is correct but not complete. Answer: 3y(2x + 3)(2x − 3).

Algebraic fractions and changing the subject

An algebraic fraction behaves exactly like a numerical one — but you must factorise before you can see what cancels.

Simplifying algebraic fractions

Factorise numerator and denominator completely, then cancel any bracket appearing in both. Nothing cancels until both parts are products. To multiply, factorise and cancel across; to divide, invert the second fraction and multiply; to add or subtract, use a common denominator.

Worked example: simplify (x² − 9) / (x² + x − 12). The numerator is (x + 3)(x − 3); the denominator factorises with two numbers multiplying to −12 and adding to 1, namely 4 and −3, giving (x + 4)(x − 3). The (x − 3) cancels, leaving (x + 3) / (x + 4).

Changing the subject

Rearranging uses the same inverse operations as solving — you are solving for a letter instead of a number. Undo operations in reverse order; treat every other letter as a known constant. A subject trapped under a square introduces ± when rooted; a subject appearing twice must be collected on one side and factorised out.

Worked example: make r the subject of V = πr²h / 3. Multiply by 3: 3V = πr²h. Divide by πh: r² = 3V / πh. Square-root both sides — the ± is required at this line: r = ±√(3V / πh). Context discards the negative root, but the ± must still be written to earn the mark.

Equations and inequalities

An inequality is solved exactly like an equation, with one exception — and that exception carries most of the marks.

Solving linear inequalities

Add, subtract, multiply and divide as you would with an equals sign. The one rule that differs: multiplying or dividing by a negative reverses the inequality. Worked example: solve 3(2x − 1) ≤ 4x + 7. Expand: 6x − 3 ≤ 4x + 7. Collect: 2x ≤ 10, so x ≤ 5. No step divided by a negative, so the sign never reversed — and ≤ includes 5, the largest integer satisfying it.

Inequality regions in two dimensions

A linear inequality in x and y splits the plane in two. Draw the boundary, decide whether it is included (solid line) or not (broken line), then shade the half you do not want. A region defined by several inequalities is the part left unshaded by all of them.

Simultaneous equations

Two unknowns need two equations. The whole method is a single idea repeated: get rid of one letter, solve for the other, then go back for the one you eliminated.

Two linear equations

Elimination scales one or both equations until the coefficients of one variable match, then adds or subtracts to remove it — usually faster when both equations are in the form ax + by = c. Substitution rearranges one equation to make a variable the subject and puts that expression into the other — the better choice the moment one equation already has a variable on its own, and the only route once a quadratic is involved.

One linear, one quadratic

Substitution is compulsory here. Make a variable the subject of the linear equation, substitute it into the quadratic, and a single quadratic in one unknown falls out. Solve it, then substitute each root back into the linear equation. Expect two solution pairs; report them as pairs.

Worked example: solve y = x + 2 and y = x². Substitute: x + 2 = x², so x² − x − 2 = 0. Factorise: (x − 2)(x + 1) = 0, so x = 2 or x = −1. Substitute into y = x + 2: x = 2 gives y = 4; x = −1 gives y = 1. The solutions are the pairs (2, 4) and (−1, 1) — two pairs, not four values.

Quadratic equations

Three methods, one equation. Factorising is fastest but only works when the roots are rational; completing the square exposes the turning point; the formula never fails. Choosing well is a marked skill in itself.

Which method, and when

Try factorising first — if a question is worth three marks and the numbers are small, it almost certainly factorises. Use completing the square when the question asks for a turning point or a minimum value, or says so explicitly. Use the formula whenever factorising fails, and always when the answer is wanted to three significant figures — that instruction signals the roots are not whole numbers.

The quadratic formula is x = (−b ± √(b² − 4ac)) / 2a, requiring ax² + bx + c = 0 with a ≠ 0. Completing the square writes ax² + bx + c as a(x + b/2a)² + (c − b²/4a); for a(x + p)² + q the turning point sits at (−p, q) and the axis of symmetry is x = −p.

Worked example: solve 2x² − 7x + 3 = 0. Read off a = 2, b = −7, c = 3. Then b² − 4ac = 49 − 24 = 25. Substitute: x = (7 ± √25) / 4 = (7 ± 5) / 4, so x = 3 or x = 0.5. A whole number under the root shows a faster route existed — here (2x − 1)(x − 3) = 0.

Sequences

Differencing tells you what you are holding — constant first differences mean linear, second quadratic, third cubic.

Linear sequences

If the first differences are constant, the rule is linear: Tₙ = dn + (a − d), where a is the first term and d the common difference (the coefficient of n).

Quadratic and cubic sequences

Constant second differences mean a quadratic rule Tₙ = an² + bn + c. The second difference equals 2a, fixing the leading coefficient at once; b and c follow by substituting two known terms.

Worked example: find the nth term of 3, 8, 15, 24, 35. First differences 5, 7, 9, 11 — not constant. Second differences 2, 2, 2 — constant, so quadratic. Second difference = 2a, so a = 1. Substituting n = 1 and n = 2 into n² + bn + c = 3 and 4 + 2b + c = 8 gives b = 2, c = 0. Test n = 4: 16 + 8 = 24, which matches. So Tₙ = n² + 2n.

Proportion

Proportion questions are a three-step ritual, and the steps never vary: write the relationship with a k in it, use the given pair of values to find k, then rewrite the equation with k replaced by a number.

The forms you must recognise

The symbol ∝ reads "is proportional to". It becomes an equals sign only when the constant of proportionality k is introduced. Cambridge examines proportionality to a square, a cube and a square root, not only to x itself: y = kx, y = kx², y = k√x (direct), and y = k/x, y = k/x² (inverse).

The three-step method

Translate the sentence into one of the forms. Substitute the given pair of values and solve for k. Then write the completed equation and use it to answer whatever was asked — a value of y, or a value of x found by working the equation backwards.

Worked example: y is directly proportional to x². When x = 3, y = 45, so 45 = k × 9 and k = 5, giving y = 5x². For x = 4: y = 5 × 16 = 80. For y = 125: 125 = 5x², so x² = 25 and x = ±5; the positive value is x = 5, but the ± is still written down.