Lesson 2 of 10
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1. Algebraic Expressions
Algebra uses letters (variables) to represent unknown numbers. Before solving equations, you must be able to simplify, expand, and factorise algebraic expressions.
Simplifying — Collecting Like Terms
Like terms have exactly the same variable(s) and power(s). Only like terms can be added or subtracted.
e.g. 3x² and −5x² are like terms. 3x² and 3x are NOT like terms.
e.g. 3x² and −5x² are like terms. 3x² and 3x are NOT like terms.
📐 Worked Example 1 — Simplify
Simplify: 5x² − 3x + 2x² + 7x − 4
1
Group like terms:
x² terms: 5x² + 2x² = 7x²
x terms: −3x + 7x = 4x
constants: −4
x² terms: 5x² + 2x² = 7x²
x terms: −3x + 7x = 4x
constants: −4
2
Answer: 7x² + 4x − 4
Expanding Brackets
Single bracket: Multiply every term inside the bracket by the term outside.
a(b + c) = ab + ac
Double brackets (FOIL): Multiply each term in the first bracket by each term in the second.
(a + b)(c + d) = ac + ad + bc + bd
a(b + c) = ab + ac
Double brackets (FOIL): Multiply each term in the first bracket by each term in the second.
(a + b)(c + d) = ac + ad + bc + bd
📐 Worked Example 2 — Expanding Double Brackets
Expand and simplify: (2x + 3)(x − 5)
1
Multiply each term in the first bracket by each term in the second:
2x × x = 2x²
2x × (−5) = −10x
3 × x = 3x
3 × (−5) = −15
2x × x = 2x²
2x × (−5) = −10x
3 × x = 3x
3 × (−5) = −15
2
Collect like terms:
2x² − 10x + 3x − 15 = 2x² − 7x − 15
2x² − 10x + 3x − 15 = 2x² − 7x − 15
Special Expansions to Memorise
Perfect Square (+)
(a + b)² = a² + 2ab + b²
Perfect Square (−)
(a − b)² = a² − 2ab + b²
Difference of Squares
(a + b)(a − b) = a² − b²
Factorisation
Factorisation is the reverse of expansion. There are four key methods:
| Method | When to Use | Example |
|---|---|---|
| Common Factor | All terms share a common factor. | 6x² + 9x = 3x(2x + 3) |
| Grouping | Four terms that can be grouped in pairs. | ax + ay + bx + by = a(x+y) + b(x+y) = (a+b)(x+y) |
| Quadratic Trinomial | Expression of the form ax² + bx + c. | x² + 5x + 6 = (x + 2)(x + 3) |
| Difference of Squares | Expression of the form a² − b². | 9x² − 16 = (3x + 4)(3x − 4) |
📐 Worked Example 3 — Factorising a Quadratic
Factorise: 2x² + 7x + 3
1
Multiply a × c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7.
1 × 6 = 6 and 1 + 6 = 7 ✓
1 × 6 = 6 and 1 + 6 = 7 ✓
2
Split the middle term using these numbers:
2x² + x + 6x + 3
2x² + x + 6x + 3
3
Group and factorise:
x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1)
x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1)
4
Check: (x + 3)(2x + 1) = 2x² + x + 6x + 3 = 2x² + 7x + 3 ✓
2. Indices (Laws of Indices)
Index (power/exponent): In aⁿ, a is the base and n is the index. The index tells you how many times the base is multiplied by itself.
Laws of Indices — Must Know All Seven:
1. aᵐ × aⁿ = aᵐ⁺ⁿ (multiply → add powers)
2. aᵐ ÷ aⁿ = aᵐ⁻ⁿ (divide → subtract powers)
3. (aᵐ)ⁿ = aᵐⁿ (power of a power → multiply)
4. a⁰ = 1 (anything to the power 0 = 1, a ≠ 0)
5. a⁻ⁿ = 1/aⁿ (negative power = reciprocal)
6. a^(1/n) = ⁿ√a (fractional power = root)
7. a^(m/n) = ⁿ√(aᵐ) = (ⁿ√a)ᵐ (fractional power = root and power)
1. aᵐ × aⁿ = aᵐ⁺ⁿ (multiply → add powers)
2. aᵐ ÷ aⁿ = aᵐ⁻ⁿ (divide → subtract powers)
3. (aᵐ)ⁿ = aᵐⁿ (power of a power → multiply)
4. a⁰ = 1 (anything to the power 0 = 1, a ≠ 0)
5. a⁻ⁿ = 1/aⁿ (negative power = reciprocal)
6. a^(1/n) = ⁿ√a (fractional power = root)
7. a^(m/n) = ⁿ√(aᵐ) = (ⁿ√a)ᵐ (fractional power = root and power)
📐 Worked Example 4 — Indices
Evaluate: (a) 27^(2/3) (b) 16^(−3/4)
1
Part (a): 27^(2/3) — cube root first, then square.
³√27 = 3, then 3² = 9
Answer: 27^(2/3) = 9
³√27 = 3, then 3² = 9
Answer: 27^(2/3) = 9
2
Part (b): 16^(−3/4) — negative means take the reciprocal.
16^(3/4) = (⁴√16)³ = 2³ = 8
So: 16^(−3/4) = 1/8
16^(3/4) = (⁴√16)³ = 2³ = 8
So: 16^(−3/4) = 1/8
Strategy for Fractional Indices: Always take the ROOT first (denominator), then apply the POWER (numerator). Root first keeps numbers smaller and easier to work with.
Simplifying Expressions with Indices
📐 Worked Example 5 — Simplify Algebraic Indices
Simplify: (a) (3x²y)³ (b) (4a³b²) ÷ (2a⁻¹b⁴)
1
Part (a): Raise each factor to the power 3.
3³ × (x²)³ × y³ = 27x⁶y³
3³ × (x²)³ × y³ = 27x⁶y³
2
Part (b): Divide coefficients, subtract indices.
(4 ÷ 2) × a³⁻⁽⁻¹⁾ × b²⁻⁴
= 2 × a⁴ × b⁻²
= 2a⁴/b²
(4 ÷ 2) × a³⁻⁽⁻¹⁾ × b²⁻⁴
= 2 × a⁴ × b⁻²
= 2a⁴/b²
3. Linear Equations
A linear equation contains variables to the first power only (no x², x³, etc.). The goal is to isolate the unknown on one side.
Golden Rule of Equations: Whatever you do to one side, you must do to the other side. The equation remains balanced.
📐 Worked Example 6 — Solving Linear Equations
Solve: (a) 3(2x − 1) = 4x + 7 (b) (x+3)/4 − (2x−1)/3 = 1
1
Part (a): Expand brackets first.
6x − 3 = 4x + 7
6x − 3 = 4x + 7
2
Collect x terms on left, numbers on right:
6x − 4x = 7 + 3
2x = 10 → x = 5
6x − 4x = 7 + 3
2x = 10 → x = 5
3
Part (b): Multiply throughout by the LCM of 4 and 3 = 12.
3(x+3) − 4(2x−1) = 12
3x + 9 − 8x + 4 = 12
−5x + 13 = 12
−5x = −1 → x = 1/5
3(x+3) − 4(2x−1) = 12
3x + 9 − 8x + 4 = 12
−5x + 13 = 12
−5x = −1 → x = 1/5
Inequalities
Inequality: A mathematical statement showing that two expressions are not equal. Solved like an equation with one critical rule: when multiplying or dividing by a negative number, reverse the inequality sign.
| Symbol | Meaning | Example |
|---|---|---|
| > | Greater than | x > 3 means x can be 3.1, 4, 100, ... |
| < | Less than | x < −2 means x can be −3, −100, ... |
| ≥ | Greater than or equal to | x ≥ 5 includes 5 itself |
| ≤ | Less than or equal to | x ≤ 0 includes 0 itself |
📐 Worked Example 7 — Solving Inequalities
Solve and list the integer solutions: −3 < 2x − 1 ≤ 7
1
Split into two inequalities or work on all parts simultaneously. Add 1 throughout:
−3 + 1 < 2x ≤ 7 + 1
−2 < 2x ≤ 8
−3 + 1 < 2x ≤ 7 + 1
−2 < 2x ≤ 8
2
Divide throughout by 2:
−1 < x ≤ 4
−1 < x ≤ 4
3
Integer solutions (x > −1 AND x ≤ 4):
x = 0, 1, 2, 3, 4
x = 0, 1, 2, 3, 4
4. Simultaneous Equations (Linear)
Two equations with two unknowns. We find the values of both unknowns that satisfy both equations at the same time. Two methods: elimination and substitution.
📐 Worked Example 8 — Elimination Method
Solve: 3x + 2y = 16 and 5x − 2y = 8
1
The y coefficients are already equal (2y and −2y). Add the equations to eliminate y:
(3x + 2y) + (5x − 2y) = 16 + 8
8x = 24 → x = 3
(3x + 2y) + (5x − 2y) = 16 + 8
8x = 24 → x = 3
2
Substitute x = 3 into the first equation:
3(3) + 2y = 16 → 9 + 2y = 16 → 2y = 7 → y = 3.5
3(3) + 2y = 16 → 9 + 2y = 16 → 2y = 7 → y = 3.5
3
Check in equation 2: 5(3) − 2(3.5) = 15 − 7 = 8 ✓
Answer: x = 3, y = 3.5
Answer: x = 3, y = 3.5
📐 Worked Example 9 — Substitution Method
Solve: y = 2x − 1 and 3x + 4y = 18
1
Since y is already expressed in terms of x, substitute into equation 2:
3x + 4(2x − 1) = 18
3x + 8x − 4 = 18
11x = 22 → x = 2
3x + 4(2x − 1) = 18
3x + 8x − 4 = 18
11x = 22 → x = 2
2
Substitute back: y = 2(2) − 1 = y = 3
3
Check in eq 2: 3(2) + 4(3) = 6 + 12 = 18 ✓
Which Method to Choose?
Use elimination when coefficients of one variable can be made equal by multiplying.
Use substitution when one equation already has a variable isolated (e.g. y = ... or x = ...).
Always check your answer in BOTH original equations.
Use elimination when coefficients of one variable can be made equal by multiplying.
Use substitution when one equation already has a variable isolated (e.g. y = ... or x = ...).
Always check your answer in BOTH original equations.
Word Problems — Setting Up Simultaneous Equations
📐 Worked Example 10 — Real World Problem
5 pens and 3 notebooks cost $14.10. 2 pens and 4 notebooks cost $11.40. Find the cost of one pen and one notebook.
1
Let pen = p cents and notebook = n cents. Set up equations:
5p + 3n = 1410 ... (1)
2p + 4n = 1140 ... (2)
5p + 3n = 1410 ... (1)
2p + 4n = 1140 ... (2)
2
Eliminate p: multiply (1) by 2 and (2) by 5:
10p + 6n = 2820
10p + 20n = 5700
Subtract: 14n = 2880 → n = 180 (180 cents = $1.80)
10p + 6n = 2820
10p + 20n = 5700
Subtract: 14n = 2880 → n = 180 (180 cents = $1.80)
3
Substitute into (1): 5p + 3(180) = 1410 → 5p = 870 → p = 174 cents = $1.74
Pen = $1.74, Notebook = $1.80
Pen = $1.74, Notebook = $1.80
5. Formulae — Substitution and Rearrangement
Substituting into Formulae
📐 Worked Example 11 — Substitution
Given that v² = u² + 2as, find v when u = 3, a = −4, s = 5.
1
Substitute the values directly:
v² = 3² + 2(−4)(5) = 9 − 40 = −31
v² = 3² + 2(−4)(5) = 9 − 40 = −31
2
v² = −31. Since v² cannot be negative for real values, there is no real solution with these values. (This is a valid exam answer — always state it.)
Rearranging Formulae (Changing the Subject)
To make a different variable the subject, apply the same rules as solving equations — inverse operations, maintaining balance.
📐 Worked Example 12 — Rearranging Formulae
Make r the subject of: A = πr²h
1
Divide both sides by πh:
A/(πh) = r²
A/(πh) = r²
2
Take the square root of both sides:
r = √(A/πh)
r = √(A/πh)
📐 Worked Example 13 — Rearranging (variable appears twice)
Make x the subject of: 3(x + a) = b(x − 2)
1
Expand both sides:
3x + 3a = bx − 2b
3x + 3a = bx − 2b
2
Collect all x terms on one side:
3x − bx = −2b − 3a
3x − bx = −2b − 3a
3
Factorise the left side (x is common):
x(3 − b) = −2b − 3a
x(3 − b) = −2b − 3a
4
Divide by (3 − b):
x = (−2b − 3a) / (3 − b)
x = (−2b − 3a) / (3 − b)
When the variable appears in TWO places: Expand → collect all x terms on one side → factorise → divide. This is a very common exam question at Grade A/A* level.
6. Algebraic Fractions
Algebraic fractions follow the same rules as numerical fractions — but require factorisation to simplify and find common denominators.
Simplifying Algebraic Fractions
📐 Worked Example 14 — Simplify Algebraic Fractions
Simplify: (x² − 9) / (x² + x − 6)
1
Factorise the numerator (difference of squares):
x² − 9 = (x + 3)(x − 3)
x² − 9 = (x + 3)(x − 3)
2
Factorise the denominator (quadratic trinomial):
x² + x − 6 = (x + 3)(x − 2)
x² + x − 6 = (x + 3)(x − 2)
3
Cancel the common factor (x + 3):
(x + 3)(x − 3) / (x + 3)(x − 2) = (x − 3)/(x − 2)
(x + 3)(x − 3) / (x + 3)(x − 2) = (x − 3)/(x − 2)
Adding and Subtracting Algebraic Fractions
📐 Worked Example 15 — Adding Algebraic Fractions
Simplify: 3/(x+1) + 2/(x−2)
1
Common denominator = (x+1)(x−2).
3(x−2)/[(x+1)(x−2)] + 2(x+1)/[(x+1)(x−2)]
3(x−2)/[(x+1)(x−2)] + 2(x+1)/[(x+1)(x−2)]
2
Expand numerators:
= [3x − 6 + 2x + 2] / [(x+1)(x−2)]
= (5x − 4) / [(x+1)(x−2)]
= [3x − 6 + 2x + 2] / [(x+1)(x−2)]
= (5x − 4) / [(x+1)(x−2)]
📝 Exam Practice Questions
Q1 [2 marks] — Expand and simplify: (3x − 4)²
Working:
(3x − 4)² = (3x)² − 2(3x)(4) + 4²
= 9x² − 24x + 16
(3x − 4)² = (3x)² − 2(3x)(4) + 4²
= 9x² − 24x + 16
Exam Tip: Use the formula (a−b)² = a²−2ab+b². Students who expand term by term often get the sign wrong on the middle term.
Q2 [2 marks] — Factorise completely: 12x²y − 8xy²
Working:
HCF of 12x²y and 8xy² = 4xy
4xy(3x − 2y)
HCF of 12x²y and 8xy² = 4xy
4xy(3x − 2y)
Q3 [2 marks] — Evaluate: 125^(−2/3)
Working:
125^(2/3) = (³√125)² = 5² = 25
125^(−2/3) = 1/25 = 0.04
125^(2/3) = (³√125)² = 5² = 25
125^(−2/3) = 1/25 = 0.04
Exam Tip: Root first (denominator of the fraction), then power (numerator). Negative index = take the reciprocal at the end.
Q4 [3 marks] — Solve the inequality 3 − 2x < 7. List all integer solutions where x is also greater than −3.
Working:
3 − 2x < 7
−2x < 4
x > −2 (divide by −2, reverse sign)
Combined with x > −3: x > −2
Integer solutions: −1, 0, 1, 2, 3, ... (infinite, or state x > −2, x integer)
3 − 2x < 7
−2x < 4
x > −2 (divide by −2, reverse sign)
Combined with x > −3: x > −2
Integer solutions: −1, 0, 1, 2, 3, ... (infinite, or state x > −2, x integer)
Exam Tip: When dividing or multiplying an inequality by a NEGATIVE number, ALWAYS reverse the inequality sign. This is the most common error in inequality questions.
Q5 [4 marks] — Solve the simultaneous equations: 4x + 3y = 1 and 3x − y = 10
Working:
Multiply equation 2 by 3: 9x − 3y = 30
Add to equation 1: 4x + 3y + 9x − 3y = 1 + 30
13x = 31 → x = 31/13? Let me use substitution instead.
From eq 2: y = 3x − 10. Substitute into eq 1:
4x + 3(3x − 10) = 1 → 4x + 9x − 30 = 1 → 13x = 31
x = 31/13... Let me re-check: 4x + 3y = 1, 3x − y = 10.
From eq 2: y = 3x − 10. Sub: 4x + 3(3x−10)=1 → 13x−30=1 → 13x=31 → x = 31/13...
✦ Cleaner version using elimination:
Multiply eq 2 by 3: 9x − 3y = 30. Add to eq 1:
13x = 31 → x = 31/13. Sub back: y = 3(31/13) − 10 = 93/13 − 130/13 = −37/13
x = 31/13, y = −37/13
Multiply equation 2 by 3: 9x − 3y = 30
Add to equation 1: 4x + 3y + 9x − 3y = 1 + 30
13x = 31 → x = 31/13? Let me use substitution instead.
From eq 2: y = 3x − 10. Substitute into eq 1:
4x + 3(3x − 10) = 1 → 4x + 9x − 30 = 1 → 13x = 31
x = 31/13... Let me re-check: 4x + 3y = 1, 3x − y = 10.
From eq 2: y = 3x − 10. Sub: 4x + 3(3x−10)=1 → 13x−30=1 → 13x=31 → x = 31/13...
✦ Cleaner version using elimination:
Multiply eq 2 by 3: 9x − 3y = 30. Add to eq 1:
13x = 31 → x = 31/13. Sub back: y = 3(31/13) − 10 = 93/13 − 130/13 = −37/13
x = 31/13, y = −37/13
Exam Tip: Always check answers in BOTH equations. Non-integer answers are perfectly valid — do not assume something is wrong if x or y is a fraction.
Q6 [3 marks] — Make h the subject of: V = ⅓πr²h
Working:
Multiply both sides by 3: 3V = πr²h
Divide both sides by πr²:
h = 3V / (πr²)
Multiply both sides by 3: 3V = πr²h
Divide both sides by πr²:
h = 3V / (πr²)
Q7 [3 marks] — Make x the subject of: p = (x + q) / (x − r)
Working:
Multiply both sides by (x − r): p(x − r) = x + q
Expand: px − pr = x + q
Collect x terms: px − x = q + pr
Factorise: x(p − 1) = q + pr
x = (q + pr) / (p − 1)
Multiply both sides by (x − r): p(x − r) = x + q
Expand: px − pr = x + q
Collect x terms: px − x = q + pr
Factorise: x(p − 1) = q + pr
x = (q + pr) / (p − 1)
Exam Tip: When x appears in both numerator and denominator, always multiply out the denominator first, then collect and factorise.
Q8 [3 marks] — Simplify: (2x² + x − 3) / (x² − 1)
Working:
Factorise numerator: 2x² + x − 3 = (2x + 3)(x − 1)
Factorise denominator: x² − 1 = (x + 1)(x − 1)
Cancel (x − 1):
(2x + 3) / (x + 1)
Factorise numerator: 2x² + x − 3 = (2x + 3)(x − 1)
Factorise denominator: x² − 1 = (x + 1)(x − 1)
Cancel (x − 1):
(2x + 3) / (x + 1)