Algebra
Concise revision notes aligned to the O Level syllabus.
Expressions vs Equations vs Formulae
An expression is a combination of terms with no equals sign. It can be simplified or evaluated, but not "solved".
- Examples:
3x + 2,x² − 4,2a + 3b
An equation states that two expressions are equal. It contains an equals sign and can be solved to find the value(s) of the unknown.
- Example:
3x + 2 = 11→ solving gives x = 3
A formula is an equation that shows a relationship between two or more variables, where one variable is expressed in terms of the others.
- Example:
A = πr²— the area A depends on the radius r
An identity is true for all values of the variable. The symbol ≡ means "identical to".
- Example:
(x + 1)² ≡ x² + 2x + 1— this holds for every value of x
The key difference: an expression has no equals sign and cannot be solved. An equation has an equals sign and is only true for specific values. An identity is true for all values.
Types of Equations
Understanding what type of equation you have tells you immediately which method to use.
| Type | Form | Highest power | Method |
|---|---|---|---|
| Linear | ax + b = c | x¹ | Rearrange and isolate x |
| Quadratic | ax² + bx + c = 0 | x² | Factorise, complete the square, or formula |
| Simultaneous | two equations, two unknowns | x¹ (usually) | Elimination or substitution |
| Inequality | ax + b > c | x¹ | Same as linear — but flip sign when multiplying/dividing by a negative |
Linear equations have one solution (one value of x that makes it true).
Quadratic equations have up to two solutions (roots). The discriminant b² − 4ac tells you how many:
- Positive → two distinct real roots
- Zero → one repeated root (the parabola just touches the x-axis)
- Negative → no real roots (the parabola does not cross the x-axis)
Higher-degree equations (cubic x³, quartic x⁴, etc.) are not required at O Level, but the pattern continues — degree = maximum number of roots.
Systems of Equations
A system of equations (or simultaneous equations) is a set of two or more equations that must all be satisfied at the same time. The solution is the set of values that makes every equation true simultaneously.
Graphical interpretation: each linear equation in two variables describes a straight line. The solution to a system is the point where the lines intersect.
| Number of solutions | What it means geometrically |
|---|---|
| One solution | Lines cross at exactly one point |
| No solution | Lines are parallel (same gradient, different intercepts) |
| Infinitely many | Lines are identical (same equation in disguise) |
Two methods for solving simultaneous linear equations:
Elimination — multiply one or both equations so a variable has the same coefficient, then add or subtract to eliminate it.
Substitution — rearrange one equation to express one variable in terms of the other, then substitute into the second equation.
For a linear–quadratic system (one linear, one quadratic), substitution is almost always the method: substitute the linear expression into the quadratic to get a quadratic in one variable, then solve.
A system of n equations can (in general) solve for n unknowns. Two equations → two unknowns. One equation with two unknowns has infinitely many solutions — you need the second equation to pin down a unique answer.
Substitution
To evaluate an expression, replace each letter with its given value and then calculate.
Example: If x = 3, find 2x² − x + 1.
- 2(3)² − 3 + 1 = 2(9) − 3 + 1 = 18 − 3 + 1 = 16
Watch out: x² means (x)², so (−2)² = 4, not −4.
Always substitute before you simplify. Write out the substitution step clearly — most marks in exam questions are awarded for showing working.
Simplifying Expressions
- Like terms share the same variable and power: 3x and 5x are like terms; 3x and 3x² are not.
- Collect like terms: 4x + 2y − x + 3y = 3x + 5y
- Multiply variables: add exponents — x² × x³ = x⁵
- Divide variables: subtract exponents — x⁶ ÷ x² = x⁴
Expanding Brackets
Single bracket: multiply every term inside by the term outside.
- 3(2x + 5) = 6x + 15
- −2(x − 4) = −2x + 8 (note: minus × minus = plus)
Double brackets (FOIL): multiply each term in the first bracket by each term in the second.
- (x + 3)(x + 2) = x² + 2x + 3x + 6 = x² + 5x + 6
Special patterns to recognise:
- Perfect square: (a + b)² = a² + 2ab + b²
- Difference of squares: (a + b)(a − b) = a² − b²
Example: (x + 5)² = x² + 10x + 25; (x + 3)(x − 3) = x² − 9.
Expand fully before collecting like terms. A common mistake is writing (x + 3)² = x² + 9 — the middle term 2(x)(3) = 6x is missing.
Factorising
HCF factorising: take the highest common factor outside.
- 6x + 9 = 3(2x + 3)
- 4x² − 8x = 4x(x − 2)
Difference of two squares: a² − b² = (a + b)(a − b)
- x² − 25 = (x + 5)(x − 5)
- 3x² − 75 = 3(x² − 25) = 3(x + 5)(x − 5)
Quadratic trinomials x² + bx + c: find two numbers that add to b and multiply to c.
- x² + 7x + 12 → numbers: 3 and 4 → (x + 3)(x + 4)
- x² − x − 6 → numbers: −3 and 2 → (x − 3)(x + 2)
For 2x² + bx + c, use trial and error or the AC method.
- 2x² + 5x + 3 → (2x + 3)(x + 1)
Always check your factorisation by re-expanding. If you get back to the original expression, you are correct.
Solving Linear Equations
The goal is to isolate the unknown by doing the same operation to both sides.
Steps for 3(x − 2) = 12:
- Expand: 3x − 6 = 12
- Add 6: 3x = 18
- Divide by 3: x = 6
For equations with the unknown on both sides, collect all x-terms on one side:
- 4x + 1 = 2x + 9 → 2x = 8 → x = 4
Simultaneous Equations
Elimination method: add or subtract the equations to cancel one variable.
Example: 3x + 2y = 16 and x + y = 6
- Multiply the second equation by 2: 2x + 2y = 12
- Subtract: (3x + 2y) − (2x + 2y) = 16 − 12 → x = 4
- Substitute back: 4 + y = 6 → y = 2
Substitution method: express one variable in terms of the other and substitute.
Always substitute your answer back into one of the original equations to verify it is correct. Write the check explicitly in exam answers.
Quadratic Equations
By factorising: rearrange to = 0, factorise, then set each bracket to zero.
- x² + 5x + 6 = 0 → (x + 2)(x + 3) = 0 → x = −2 or x = −3
By the quadratic formula: for ax² + bx + c = 0,
x = (−b ± √(b² − 4ac)) / 2a
The discriminant b² − 4ac tells you about the roots:
- b² − 4ac > 0 → two distinct real roots
- b² − 4ac = 0 → one repeated root
- b² − 4ac < 0 → no real roots
The quadratic formula always works, even when factorising is not obvious. You are expected to give answers to 3 significant figures unless the roots are exact integers.
Laws of Indices
| Law | Rule |
|---|---|
| Multiply | xᵐ × xⁿ = xᵐ⁺ⁿ |
| Divide | xᵐ ÷ xⁿ = xᵐ⁻ⁿ |
| Power of a power | (xᵐ)ⁿ = xᵐⁿ |
| Zero index | x⁰ = 1 |
| Negative index | x⁻ⁿ = 1/xⁿ |
| Fractional index | x^(1/n) = ⁿ√x |
| Combined fractional | x^(m/n) = (ⁿ√x)ᵐ |
Example: 8^(2/3) = (∛8)² = 2² = 4
Algebraic Fractions
Simplify by factorising the numerator and denominator, then cancel common factors.
- (x² − 4) / (x + 2) = (x − 2)(x + 2) / (x + 2) = x − 2
Add/Subtract: find a common denominator.
- 1/2 + 1/x: common denominator is 2x → x/(2x) + 2/(2x) = (x + 2)/2x
Multiply: multiply numerators together and denominators together, then simplify.
Divide: multiply by the reciprocal of the divisor.
Inequalities
Solve exactly like equations, except: when you multiply or divide by a negative number, flip the inequality sign.
- 2x + 1 > 7 → 2x > 6 → x > 3
- −2x > 6 → x < −3 (sign flipped when dividing by −2)
On a number line: open circle ○ for strict inequality (<, >); closed circle ● for ≤ or ≥.
Sequences and nth Term
Arithmetic sequence (constant difference d): nth term = a + (n − 1)d
- Sequence 5, 8, 11, 14, … → d = 3, a = 5 → nth term = 3n + 2
Check: 1st term = 3(1) + 2 = 5 ✓ ; 4th term = 3(4) + 2 = 14 ✓
Geometric sequence (constant ratio r): nth term = arⁿ⁻¹
- Sequence 2, 4, 8, 16, … → r = 2 → nth term = 2ⁿ
Quadratic sequence (second difference is constant): nth term contains n².
Changing the Subject of a Formula
Rearrange to make the required variable the subject by isolating it.
Example: Make r the subject of A = πr²
- Divide by π: A/π = r²
- Square root: r = √(A/π)
Example with the variable in two places: Make x the subject of y = (x + 1)/(x − 2)
- y(x − 2) = x + 1
- yx − 2y = x + 1
- yx − x = 1 + 2y
- x(y − 1) = 2y + 1
- x = (2y + 1)/(y − 1)
When the unknown appears in two places after substitution, collect all its terms on one side, factorise out the unknown, then divide.
Functions
Notation: f(x) means "function of x". f(3) means evaluate f at x = 3.
Inverse function f⁻¹(x): the function that undoes f. To find it, write y = f(x), rearrange for x, then replace y with x.
- f(x) = 3x + 1 → y = 3x + 1 → x = (y − 1)/3 → f⁻¹(x) = (x − 1)/3
Composite function fg(x): apply g first, then f.
- f(x) = 2x, g(x) = x + 3 → fg(x) = f(g(x)) = f(x + 3) = 2(x + 3) = 2x + 6