Algebraic Expressions
Expanding brackets · Factorising · Algebraic fractions · Index laws
Expanding Brackets
\(a(b + c) = ab + ac\)
\((a+b)(c+d) = ac + ad + bc + bd\)
\((a+b)^2 = a^2 + 2ab + b^2\)
\((a-b)^2 = a^2 - 2ab + b^2\)
\((a+b)(a-b) = a^2 - b^2\) ← difference of squares
Factorising Methods
HCF
\(6x^2 + 9x = 3x(2x + 3)\) — take out highest common factor
Grouping
\(ax + ay + bx + by = (a+b)(x+y)\)
Quadratic
\(x^2 + 5x + 6 = (x+2)(x+3)\) — find two numbers that multiply to c and add to b
Diff of sq²
\(x^2 - 9 = (x+3)(x-3)\)
Factorising check: Always expand your answer back to verify it matches the original expression.
Index Laws
\(x^m \cdot x^n = x^{m+n}\)
\(\dfrac{x^m}{x^n} = x^{m-n}\)
\((x^m)^n = x^{mn}\)
\(x^0 = 1\)
\(x^{-n} = \dfrac{1}{x^n}\)
\(x^{1/n} = \sqrt[n]{x}\)
Algebraic Fractions
Simplify by cancelling common factors:
\(\dfrac{6x^2}{9x} = \dfrac{2x}{3}\)
Add/Subtract using LCM of denominators:
\(\dfrac{1}{x} + \dfrac{2}{x+1} = \dfrac{x+1+2x}{x(x+1)} = \dfrac{3x+1}{x(x+1)}\)
Multiply: multiply tops & bottoms, cancel after
Divide: flip second fraction then multiply
\(\dfrac{6x^2}{9x} = \dfrac{2x}{3}\)
Add/Subtract using LCM of denominators:
\(\dfrac{1}{x} + \dfrac{2}{x+1} = \dfrac{x+1+2x}{x(x+1)} = \dfrac{3x+1}{x(x+1)}\)
Multiply: multiply tops & bottoms, cancel after
Divide: flip second fraction then multiply
Common trap: You cannot cancel across a + or − sign. \(\dfrac{x+3}{3} \neq x+1\)
Changing the Subject of a Formula
Goal
Rearrange to get the target variable alone on one side.
Steps
Expand brackets → Collect target variable terms on one side → Factorise if needed → Divide
Key rule
Do the same operation to both sides. Inverse of +/− is ∓; inverse of ×/÷ is ÷/×; inverse of square is square root.
Worked Example
Make \(r\) the subject of \(V = \dfrac{4}{3}\pi r^3\)
÷ const
\(\dfrac{3V}{4\pi} = r^3\)
Cube root
\(r = \sqrt[3]{\dfrac{3V}{4\pi}}\)
Answer: \(r = \sqrt[3]{\dfrac{3V}{4\pi}}\)
Equations & Inequalities
Linear · Simultaneous (3 methods) · Quadratic · Inequalities
Simultaneous Equations — Which Method?
METHOD 1
Substitution
Use when one equation is linear and the other is non-linear (quadratic, etc.).
METHOD 2
Elimination
Use when both equations are linear. Multiply to match one variable, then add/subtract.
METHOD 3
Graphical
Plot both lines. Solutions = intersection points. Only gives approximate answers.
Method 2 — Elimination (Full Worked Example)
Solve: \(3x + 2y = 16\) and \(5x - 2y = 8\)
1
Line up equations clearly:
\(3x + 2y = 16 \quad\) (Eq1)
\(5x - 2y = 8 \quad\) (Eq2)
\(3x + 2y = 16 \quad\) (Eq1)
\(5x - 2y = 8 \quad\) (Eq2)
2
The \(y\) coefficients are already equal in magnitude. Add Eq1 + Eq2 to eliminate \(y\):
\(8x = 24 \Rightarrow x = 3\)
\(8x = 24 \Rightarrow x = 3\)
3
Substitute \(x=3\) into Eq1:
\(9 + 2y = 16 \Rightarrow y = 3.5\)
\(9 + 2y = 16 \Rightarrow y = 3.5\)
Answer: \(x = 3,\; y = 3.5\)
Add when signs are opposite · Subtract when signs are the same.
Solve: \(2x + 3y = 12\) and \(x - y = 1\) — coefficients differ
1
Multiply Eq2 by 3 to match the \(y\) coefficient:
\(3x - 3y = 3\quad\) (Eq2')
\(3x - 3y = 3\quad\) (Eq2')
2
Add Eq1 + Eq2':
\((2x+3y)+(3x-3y) = 12+3\)
\(5x = 15 \Rightarrow x = 3\)
\((2x+3y)+(3x-3y) = 12+3\)
\(5x = 15 \Rightarrow x = 3\)
3
Sub into \(x - y = 1\): \(3 - y = 1 \Rightarrow y = 2\)
Answer: \(x = 3,\; y = 2\)
Solving Quadratic Equations
METHOD A
Factorisation
Rearrange to 0. Find two numbers to factorise. Set each bracket = 0.
\(x^2 - 5x + 6 = 0\)
\((x-2)(x-3)=0\)
\(x=2\) or \(x=3\)
\(x^2 - 5x + 6 = 0\)
\((x-2)(x-3)=0\)
\(x=2\) or \(x=3\)
METHOD B
Quadratic Formula
\[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]Use when it doesn't factorise neatly.
METHOD C
Completing the Square
Rewrite as \((x+p)^2 = q\). Take square root of both sides. Good for finding vertex.
Inequalities
Rule 1
Solve like an equation — but keep the inequality sign.
Rule 2
When you multiply or divide by a negative number, FLIP the inequality.
Example
\(-2x > 6 \Rightarrow x < -3\) ← sign flipped!
Number line
Open circle \(\circ\) for strict \(>\) or \(<\); filled circle \(\bullet\) for \(\geq\) or \(\leq\).
Worked Example
Solve \(3x - 5 \leq x + 7\) and show on a number line.
Collect
\(3x - x \leq 7 + 5\)
Simplify
\(2x \leq 12\)
Divide
\(x \leq 6\) — number line: arrow pointing left from filled circle at 6.
Sequences
nth term · Arithmetic · Geometric · Quadratic sequences
Arithmetic Sequence (AP)
Has a constant common difference \(d\) between consecutive terms.
\(T_n = a + (n-1)d\)
\(a\) = first term \(d\) = common difference \(n\) = term number
Sum of AP: \(S_n = \dfrac{n}{2}[2a+(n-1)d]\)
Geometric Sequence (GP)
Has a constant common ratio \(r\) between consecutive terms.
\(T_n = ar^{n-1}\)
\(a\) = first term \(r\) = common ratio
Identify type: If the difference between terms is constant → AP. If the ratio is constant → GP.
Quadratic Sequences — Finding the nth Term
1
Find the first differences (gap between consecutive terms).
2
Find the second differences (gap between first differences). If constant → quadratic.
3
Coefficient of \(n^2\) = (second difference) ÷ 2.
4
Subtract \(an^2\) from original sequence. Find the linear nth term of the remainder.
5
Combine: \(T_n = an^2 + bn + c\).
Worked Example
Find nth term of: 3, 9, 19, 33, 51, …
1st diff
6, 10, 14, 18 (not constant)
2nd diff
4, 4, 4 → constant! Quadratic.
Coeff n²
\(4 \div 2 = 2\) → \(2n^2\)
Remainder
\(3-2=1,\; 9-8=1,\; 19-18=1\) → constant 1
Answer: \(T_n = 2n^2 + 1\)
Special Sequences to Recognise
SQUARE NUMBERS
1, 4, 9, 16, 25, … → \(T_n = n^2\)
CUBE NUMBERS
1, 8, 27, 64, 125, … → \(T_n = n^3\)
TRIANGULAR NUMBERS
1, 3, 6, 10, 15, … → \(T_n = \dfrac{n(n+1)}{2}\)
Practice Questions
3 Easy · 3 Medium · 3 Hard — attempt first, then reveal full solution
Easy 1
[2]
Factorise fully: \(12x^2y - 8xy^2\).
STEP-BY-STEP SOLUTION
HCF
Common factors of 12 and 8 → HCF is 4.
Both terms contain \(xy\) → take \(xy\) out.
Both terms contain \(xy\) → take \(xy\) out.
Factorise
\(12x^2y - 8xy^2 = 4xy(3x - 2y)\)
Answer
\(\boldsymbol{4xy(3x-2y)}\)
Check: expand \(4xy(3x-2y) = 12x^2y - 8xy^2\) ✓
Easy 2
[3]
Solve simultaneously by elimination: \(3x + 2y = 11\) and \(x - y = 2\).
STEP-BY-STEP — ELIMINATION
Multiply
Multiply Eq2 by 2: \(\quad 2x - 2y = 4\)
Add
Add to Eq1: \((3x+2y)+(2x-2y)=11+4\)
\(5x=15 \Rightarrow x=3\)
\(5x=15 \Rightarrow x=3\)
Find y
Sub \(x=3\) into \(x-y=2\): \(3-y=2 \Rightarrow y=1\)
Answer
\(\boldsymbol{x=3,\; y=1}\)
Check: \(3(3)+2(1)=11\) ✓ and \(3-1=2\) ✓
Easy 3
[2]
Find the 8th term of the arithmetic sequence: 5, 11, 17, 23, …
STEP-BY-STEP SOLUTION
Identify
\(a = 5,\quad d = 6\) (common difference)
Formula
\(T_n = a + (n-1)d = 5 + (n-1)\times 6\)
Substitute
\(T_8 = 5 + 7 \times 6 = 5 + 42 = \boldsymbol{47}\)
Medium 1
[4]
Solve simultaneously: \(2x + y = 7\) and \(x^2 + y^2 = 25\).
STEP-BY-STEP — SUBSTITUTION (linear into non-linear)
Rearrange
From linear equation: \(y = 7 - 2x\)
Substitute
\(x^2 + (7-2x)^2 = 25\)
\(x^2 + 49 - 28x + 4x^2 = 25\)
\(5x^2 - 28x + 24 = 0\)
\(x^2 + 49 - 28x + 4x^2 = 25\)
\(5x^2 - 28x + 24 = 0\)
Factorise
\((5x - 4)(x - 6) = 0\)
\(x = \dfrac{4}{5}\) or \(x = 6\)
\(x = \dfrac{4}{5}\) or \(x = 6\)
Find y
When \(x=\frac{4}{5}\): \(y = 7 - \frac{8}{5} = \frac{27}{5}\)
When \(x=6\): \(y = 7 - 12 = -5\)
When \(x=6\): \(y = 7 - 12 = -5\)
Answer
\(\boldsymbol{\left(\tfrac{4}{5},\,\tfrac{27}{5}\right)}\) and \(\boldsymbol{(6,\,-5)}\)
⚠ Always sub back into the linear equation to find y — avoids square root sign errors.
Medium 2
[3]
The nth term of a sequence is \(3n^2 - n + 2\). Find the 10th term and the first term greater than 200.
STEP-BY-STEP SOLUTION
10th term
\(T_{10} = 3(100) - 10 + 2 = 292\)
Inequality
\(3n^2 - n + 2 > 200\)
\(3n^2 - n - 198 > 0\)
\(3n^2 - n - 198 > 0\)
Estimate n
Try \(n = 8\): \(3(64)-8+2 = 186\) < 200
Try \(n = 9\): \(3(81)-9+2 = 236\) > 200
Try \(n = 9\): \(3(81)-9+2 = 236\) > 200
Answer
First term > 200 is at \(n = 9\): \(\boldsymbol{T_9 = 236}\)
Medium 3
[3]
Make \(x\) the subject of \(\;t = \sqrt{\dfrac{x-3}{2x+1}}\).
STEP-BY-STEP — CHANGING THE SUBJECT
Square
\(t^2 = \dfrac{x-3}{2x+1}\)
Cross-multiply
\(t^2(2x+1) = x - 3\)
\(2t^2 x + t^2 = x - 3\)
\(2t^2 x + t^2 = x - 3\)
Collect x
\(2t^2 x - x = -3 - t^2\)
\(x(2t^2 - 1) = -3 - t^2\)
\(x(2t^2 - 1) = -3 - t^2\)
Answer
\(\boldsymbol{x = \dfrac{-3-t^2}{2t^2-1}}\)
⚠ Collect ALL x terms before factorising. Never cancel x from both sides of an equation.
Hard 1
[5]
Two numbers \(x\) and \(y\) satisfy \(x + y = 8\) and \(x^2 + y^2 = 40\). Find the values of \(x\) and \(y\), and hence find \(xy\).
STEP-BY-STEP — SUBSTITUTION + IDENTITY TRICK
Method
From \(x+y=8\): \(y = 8 - x\). Substitute:
Substitute
\(x^2 + (8-x)^2 = 40\)
\(x^2 + 64 - 16x + x^2 = 40\)
\(2x^2 - 16x + 24 = 0 \Rightarrow x^2 - 8x + 12 = 0\)
\(x^2 + 64 - 16x + x^2 = 40\)
\(2x^2 - 16x + 24 = 0 \Rightarrow x^2 - 8x + 12 = 0\)
Factorise
\((x-2)(x-6)=0 \Rightarrow x=2\) or \(x=6\)
Pairs
\(x=2, y=6\) or \(x=6, y=2\)
Find xy
\(xy = 2 \times 6 = \boldsymbol{12}\)
Identity check: \((x+y)^2 = x^2+2xy+y^2 \Rightarrow 64 = 40 + 2xy \Rightarrow xy=12\) ✓
Hard 2
[4]
Solve \(\dfrac{2x+1}{x-3} - \dfrac{x+2}{x+1} = 1\).
STEP-BY-STEP — ALGEBRAIC FRACTIONS
Multiply through
Multiply every term by \((x-3)(x+1)\):
\((2x+1)(x+1) - (x+2)(x-3) = (x-3)(x+1)\)
\((2x+1)(x+1) - (x+2)(x-3) = (x-3)(x+1)\)
Expand LHS
\((2x^2+3x+1) - (x^2-x-6) = x^2+4x+7\)
Expand RHS
\(x^2-2x-3\)
Solve
\(x^2+4x+7 = x^2-2x-3\)
\(6x = -10 \Rightarrow x = -\dfrac{5}{3}\)
\(6x = -10 \Rightarrow x = -\dfrac{5}{3}\)
Answer
\(\boldsymbol{x = -\dfrac{5}{3}}\)
⚠ State any excluded values: \(x \neq 3\) and \(x \neq -1\) (denominators can't be zero).
Hard 3
[5]
Find the nth term of the quadratic sequence: 2, 11, 26, 47, 74, …
Hence find the value of n for which \(T_n = 362\).
Hence find the value of n for which \(T_n = 362\).
STEP-BY-STEP — QUADRATIC nth TERM
Differences
1st diff: 9, 15, 21, 27
2nd diff: 6, 6, 6 → constant → quadratic
2nd diff: 6, 6, 6 → constant → quadratic
Coeff of n²
\(6 \div 2 = 3\) → start with \(3n^2\)
Remainder
\(T_n - 3n^2\): \(2-3=-1,\; 11-12=-1,\; 26-27=-1\) → constant −1
nth term
\(T_n = 3n^2 - 1\)
Solve for n
\(3n^2 - 1 = 362 \Rightarrow 3n^2 = 363 \Rightarrow n^2 = 121 \Rightarrow \boldsymbol{n = 11}\)
Past Year Paper Style Questions
Cambridge IGCSE 0580 style — multi-part, show all working
0580 Style
Simultaneous Equations
[9]
(a) Solve by elimination: \(4x + 3y = 18\) and \(2x - y = 4\). [3]
(b) A café sells coffee for $\(c\) and tea for $\(t\). 3 coffees and 2 teas cost $14.50. 2 coffees and 5 teas cost $16. Form two simultaneous equations and solve to find \(c\) and \(t\). [4]
(c) Explain why the simultaneous equations \(2x + y = 5\) and \(4x + 2y = 7\) have no solution. [2]
(b) A café sells coffee for $\(c\) and tea for $\(t\). 3 coffees and 2 teas cost $14.50. 2 coffees and 5 teas cost $16. Form two simultaneous equations and solve to find \(c\) and \(t\). [4]
(c) Explain why the simultaneous equations \(2x + y = 5\) and \(4x + 2y = 7\) have no solution. [2]
FULL WORKED SOLUTION
(a) Multiply
Multiply Eq2 by 3: \(6x - 3y = 12\). Add to Eq1:
\(10x = 30 \Rightarrow x = 3\)
\(10x = 30 \Rightarrow x = 3\)
(a) Find y
Sub \(x=3\) into \(2(3)-y=4\): \(y=2\)
\(\boldsymbol{x=3,\; y=2}\)
\(\boldsymbol{x=3,\; y=2}\)
(b) Set up
\(3c + 2t = 14.50\) (Eq1)
\(2c + 5t = 16.00\) (Eq2)
\(2c + 5t = 16.00\) (Eq2)
(b) Eliminate c
Eq1×2: \(6c+4t=29\). Eq2×3: \(6c+15t=48\). Subtract:
\(11t = 19 \Rightarrow t = \frac{19}{11} \approx 1.73\)
\(11t = 19 \Rightarrow t = \frac{19}{11} \approx 1.73\)
(b) Find c
\(3c = 14.5 - 2(\frac{19}{11}) = 14.5 - \frac{38}{11} = \frac{121.5}{11}\)
\(\boldsymbol{c \approx \$3.68,\; t \approx \$1.73}\)
\(\boldsymbol{c \approx \$3.68,\; t \approx \$1.73}\)
(c) Explain
Divide Eq2 by 2: \(2x+y=3.5\). This contradicts Eq1 (\(2x+y=5\)).
The lines are parallel (same gradient, different intercepts) → no intersection → no solution.
The lines are parallel (same gradient, different intercepts) → no intersection → no solution.
0580 Style
Sequences & nth Term
[8]
The first four terms of a sequence are: 5, 14, 29, 50, …
(a) Show that this is a quadratic sequence and find the nth term. [4]
(b) Find the 20th term. [1]
(c) Is 320 a term in the sequence? Show your working. [3]
(a) Show that this is a quadratic sequence and find the nth term. [4]
(b) Find the 20th term. [1]
(c) Is 320 a term in the sequence? Show your working. [3]
FULL WORKED SOLUTION
(a) Differences
1st diff: 9, 15, 21 2nd diff: 6, 6 → constant → quadratic ✓
(a) nth term
Coeff of \(n^2 = 3\). Subtract \(3n^2\): \(5-3=2,\; 14-12=2,\; 29-27=2\) → linear part = 2.
\(\boldsymbol{T_n = 3n^2 + 2}\)
\(\boldsymbol{T_n = 3n^2 + 2}\)
(b)
\(T_{20} = 3(400)+2 = \boldsymbol{1202}\)
(c)
\(3n^2+2=320 \Rightarrow n^2 = \frac{318}{3} = 106\)
\(\sqrt{106} \approx 10.3\) — not an integer.
\(\boldsymbol{320 \text{ is NOT a term in the sequence.}}\)
\(\sqrt{106} \approx 10.3\) — not an integer.
\(\boldsymbol{320 \text{ is NOT a term in the sequence.}}\)
0580 Style
Algebraic Manipulation
[9]
(a) Simplify \(\dfrac{x^2-9}{x^2-x-6}\). [3]
(b) Make \(v\) the subject of \(E = \dfrac{1}{2}mv^2\). [2]
(c) Solve \(\dfrac{3}{x+2} + \dfrac{1}{x-1} = 2\). Give your answers correct to 2 decimal places. [4]
(b) Make \(v\) the subject of \(E = \dfrac{1}{2}mv^2\). [2]
(c) Solve \(\dfrac{3}{x+2} + \dfrac{1}{x-1} = 2\). Give your answers correct to 2 decimal places. [4]
FULL WORKED SOLUTION
(a)
Numerator: \(x^2-9 = (x+3)(x-3)\)
Denominator: \(x^2-x-6 = (x+2)(x-3)\)
Cancel \((x-3)\): \(\boldsymbol{\dfrac{x+3}{x+2}}\)
Denominator: \(x^2-x-6 = (x+2)(x-3)\)
Cancel \((x-3)\): \(\boldsymbol{\dfrac{x+3}{x+2}}\)
(b)
\(2E = mv^2 \Rightarrow v^2 = \dfrac{2E}{m} \Rightarrow \boldsymbol{v = \sqrt{\dfrac{2E}{m}}}\)
(c) Multiply
Multiply by \((x+2)(x-1)\):
\(3(x-1)+1(x+2) = 2(x+2)(x-1)\)
\(3x-3+x+2 = 2(x^2+x-2)\)
\(4x-1 = 2x^2+2x-4\)
\(3(x-1)+1(x+2) = 2(x+2)(x-1)\)
\(3x-3+x+2 = 2(x^2+x-2)\)
\(4x-1 = 2x^2+2x-4\)
(c) Solve
\(2x^2-2x-3=0\)
\(x = \dfrac{2\pm\sqrt{4+24}}{4} = \dfrac{2\pm\sqrt{28}}{4}\)
\(\boldsymbol{x \approx 1.82}\) or \(\boldsymbol{x \approx -0.82}\)
\(x = \dfrac{2\pm\sqrt{4+24}}{4} = \dfrac{2\pm\sqrt{28}}{4}\)
\(\boldsymbol{x \approx 1.82}\) or \(\boldsymbol{x \approx -0.82}\)