Mathematics · AQA 8300 §N2
Fractions
- Simplify fractions to their lowest terms using the highest common factor
- Convert between mixed numbers and improper fractions fluently
- Add, subtract, multiply and divide fractions including mixed numbers
- Find a fraction of a quantity and solve reverse fraction problems
- Solve multi-step and algebraic problems involving fractions at Grade 9 level
🔑 Core Concepts
What is a Fraction?
A fraction represents a part of a whole. The numerator (top number) tells you how many parts you have; the denominator (bottom number) tells you how many equal parts the whole is divided into.
DEFINITION — Fraction
A fraction $\dfrac{a}{b}$ represents $a$ out of $b$ equal parts, where $b \neq 0$.
The numerator is $a$ and the denominator is $b$.
COMMON MISTAKE — Zero Denominator
A fraction with denominator zero is undefined. Never write $\frac{5}{0}$ — it has no meaning in standard mathematics.
Equivalent Fractions
Two fractions are equivalent if they represent the same proportion of a whole. You can create equivalent fractions by multiplying or dividing both the numerator and denominator by the same non-zero number.
DEFINITION — Equivalent Fractions
$\dfrac{a}{b} = \dfrac{a \times k}{b \times k}$ for any non-zero integer $k$.
Equally, $\dfrac{a}{b} = \dfrac{a \div k}{b \div k}$ provided $k$ divides both $a$ and $b$.
Equivalent Fraction Rule
$$\frac{a}{b} = \frac{a \times k}{b \times k} = \frac{a \div k}{b \div k}$$
$a$ = numerator
$b$ = denominator
$k$ = any non-zero integer
EXAM TIP — Proving Equivalence
To prove $\frac{3}{4} = \frac{15}{20}$, show that $15 = 3 \times 5$ and $20 = 4 \times 5$ — the same multiplier $k = 5$ applies to both. Always state the multiplier explicitly for full marks.
Simplifying Fractions Using HCF
A fraction is in its simplest form (lowest terms) when the numerator and denominator share no common factor other than 1. To simplify, divide both by their Highest Common Factor (HCF).
Find HCF of numerator and denominator
→
Divide both numerator and denominator by HCF
→
Check: numerator and denominator share no common factor
WORKED EXAMPLE — Simplifying
Simplify $\dfrac{36}{48}$.HCF(36, 48) = 12 (since $36 = 12 \times 3$ and $48 = 12 \times 4$).
$\dfrac{36}{48} = \dfrac{36 \div 12}{48 \div 12} = \dfrac{3}{4}$
EXAM TIP — One-Step Simplification
Always divide by the HCF, not just any common factor. Dividing by a smaller factor means you need multiple steps and risk making an error. Find the HCF first.
Mixed Numbers and Improper Fractions
An improper fraction has a numerator greater than or equal to its denominator (e.g. $\frac{11}{4}$). A mixed number combines a whole number and a proper fraction (e.g. $2\frac{3}{4}$). Both represent the same value.
DEFINITION — Mixed Number & Improper Fraction
Mixed number $n\dfrac{a}{b}$ means $n + \dfrac{a}{b}$.To convert to improper: $n\dfrac{a}{b} = \dfrac{n \times b + a}{b}$
To convert to mixed: divide numerator by denominator; quotient is whole part, remainder is new numerator.
Mixed Number → Improper Fraction
$$n\frac{a}{b} = \frac{n \times b + a}{b}$$
$n$ = whole number part
$a$ = fractional numerator
$b$ = denominator
MEMORY TRICK — Mixed to Improper
"Multiply the whole, add the top, keep the bottom." For $3\frac{2}{5}$: $(3 \times 5) + 2 = 17$, so $\frac{17}{5}$.
Adding and Subtracting Fractions
You can only add or subtract fractions directly when they share the same denominator. If they do not, you must first find a common denominator — typically the Lowest Common Multiple (LCM) of the denominators — and create equivalent fractions.
Find LCM of both denominators
→
Convert each fraction to an equivalent fraction with the LCM as denominator
→
Add or subtract the numerators; keep the denominator
→
Simplify the result if possible
Adding Fractions
$$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$$
Works when $b$ and $d$ share no common factor
Otherwise use LCM($b$,$d$) as denominator for a simpler result
EXAM TIP — Use LCM Not Product
Using $b \times d$ as your common denominator always works but often gives large numbers you then need to simplify. Using LCM($b$, $d$) keeps numbers manageable and saves time.
COMMON MISTAKE — Adding Denominators
Never add the denominators: $\dfrac{1}{3} + \dfrac{1}{4} \neq \dfrac{2}{7}$. The correct answer is $\dfrac{4}{12} + \dfrac{3}{12} = \dfrac{7}{12}$.
Multiplying Fractions
Multiplying fractions is the most straightforward operation: multiply the numerators together and multiply the denominators together. No common denominator is needed.
Multiplying Fractions
$$\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d} = \frac{ac}{bd}$$
$a, c$ = numerators
$b, d$ = denominators ($b,d \neq 0$)
EXAM TIP — Cross-Cancel Before Multiplying
Before multiplying, look for common factors between any numerator and any denominator (diagonally). Cancel them first to keep numbers small.
E.g. $\dfrac{4}{9} \times \dfrac{3}{8}$: cancel 4 and 8 by 4, cancel 3 and 9 by 3 → $\dfrac{1}{3} \times \dfrac{1}{2} = \dfrac{1}{6}$.
Dividing Fractions — KFC: Keep, Flip, Change
Dividing by a fraction is equivalent to multiplying by its reciprocal. The reciprocal of $\frac{c}{d}$ is $\frac{d}{c}$ (flip the fraction). The memory aid KFC reminds you: Keep the first fraction, Flip the second, Change the division to multiplication.
Dividing Fractions (KFC)
$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}$$
Keep: $\frac{a}{b}$ unchanged
Flip: $\frac{c}{d}$ becomes $\frac{d}{c}$ (reciprocal)
Change: $\div$ becomes $\times$
MEMORY TRICK — KFC
Keep the first fraction · Flip the second · Change ÷ to ×Just like ordering at KFC — you keep your order, flip the menu, change your mind about the sauce.
COMMON MISTAKE — Flipping the Wrong Fraction
Only flip the fraction you are dividing by (the second one). Students often flip both, or flip the first fraction instead. Keep $\frac{a}{b}$ exactly as it is.
Fraction of a Quantity
To find a fraction of a quantity: divide by the denominator (to find one part), then multiply by the numerator (to find the required number of parts).
Fraction of a Quantity
$$\frac{a}{b} \text{ of } Q = Q \div b \times a = \frac{Q \times a}{b}$$
$Q$ = the whole quantity
$b$ = denominator (number of equal parts)
$a$ = numerator (parts required)
EXAM TIP — Reverse Fraction Problems
If "$\frac{3}{5}$ of a number is 24, find the whole," divide by the numerator first, then multiply by the denominator: $24 \div 3 = 8$, then $8 \times 5 = 40$.
Ordering Fractions
To order fractions, convert all fractions to a common denominator, then compare numerators. Alternatively, convert to decimals, but the common-denominator method is exact and earns more marks.
EXAM TIP — Ordering Strategy
For ordering 3 or more fractions: find the LCM of all denominators, convert each fraction, then rank numerators. Always write the fractions in their original form in your final answer — not the equivalent forms you used for comparison.
Algebraic Fractions (Grade 9)
At Grade 9, fractions may have algebraic numerators or denominators. The same rules apply: simplify by factorising and cancelling common factors; find common denominators using algebraic expressions.
DEFINITION — Algebraic Fraction
An algebraic fraction has one or more algebraic expressions in the numerator and/or denominator. Example: $\dfrac{x^2 - 4}{x - 2} = \dfrac{(x+2)(x-2)}{x-2} = x + 2$ (provided $x \neq 2$).
Adding Algebraic Fractions
$$\frac{1}{x} + \frac{1}{x+1} = \frac{(x+1) + x}{x(x+1)} = \frac{2x+1}{x(x+1)}$$
Common denominator = $x(x+1)$
Expand numerator carefully; do not expand denominator unless required
IMPORTANT — Restrictions on Algebraic Fractions
When simplifying algebraic fractions by cancelling factors, state any restrictions. For $\dfrac{(x+2)(x-2)}{x-2} = x+2$, you must note $x \neq 2$, otherwise the original expression is undefined.
🗺️ Visual Notes
Fractions
Simplifying
- Find HCF of numerator & denominator
- Divide both by HCF
- Result is in lowest terms
- Factorisation needed for algebraic fractions
+ and −
- Find LCM of denominators
- Convert to equivalent fractions
- Add/subtract numerators only
- Simplify the result
× and ÷
- Multiply: numerator × numerator, denominator × denominator
- Divide: KFC — Keep, Flip, Change
- Cross-cancel before multiplying
- Convert mixed numbers first
Mixed Numbers
- Mixed → improper: $(n \times b + a) / b$
- Improper → mixed: divide and find remainder
- Always convert before operating
- Convert back at end if required
Fraction of Quantity
- ÷ by denominator, × by numerator
- Reverse: ÷ by numerator, × by denominator
- "Of" means multiply
- Used in ratio and percentage problems
Grade 9 — Algebraic
- Factorise numerator and denominator
- Cancel common factors
- State restrictions on variable
- Combine over common algebraic denominator
Operations at a Glance
| Operation | Rule | Example | Result |
|---|---|---|---|
| Add | Common denominator; add numerators | $\dfrac{1}{3} + \dfrac{1}{4}$ | $\dfrac{7}{12}$ |
| Subtract | Common denominator; subtract numerators | $\dfrac{3}{4} - \dfrac{1}{6}$ | $\dfrac{7}{12}$ |
| Multiply | Multiply numerators; multiply denominators | $\dfrac{2}{3} \times \dfrac{3}{4}$ | $\dfrac{1}{2}$ |
| Divide | KFC — multiply by reciprocal | $\dfrac{2}{3} \div \dfrac{4}{5}$ | $\dfrac{5}{6}$ |
| Simplify | Divide numerator and denominator by HCF | $\dfrac{18}{24}$ | $\dfrac{3}{4}$ |
| Mixed → Improper | $(n \times b + a) / b$ | $2\dfrac{3}{5}$ | $\dfrac{13}{5}$ |
Which Method? — Decision Process
Are the fractions mixed numbers?
→
YES: Convert to improper fractions first
→
Is the operation + or − ?
→
YES: Find LCM, convert, operate on numerators
→
Is the operation × or ÷ ?
→
× : multiply straight across | ÷ : KFC then multiply
→
Simplify result; convert back to mixed number if required
✏️ Worked Examples
Grade 4–5
Calculate $\dfrac{3}{4} + \dfrac{2}{5}$. Give your answer as a mixed number in its simplest form.
1
Find the LCM of the denominators
LCM(4, 5) = 20. This will be our common denominator.
2
Convert to equivalent fractions
$\dfrac{3}{4} = \dfrac{3 \times 5}{4 \times 5} = \dfrac{15}{20}$ and $\dfrac{2}{5} = \dfrac{2 \times 4}{5 \times 4} = \dfrac{8}{20}$
3
Add the numerators
$\dfrac{15}{20} + \dfrac{8}{20} = \dfrac{23}{20}$
4
Convert to mixed number and simplify
$\dfrac{23}{20} = 1\dfrac{3}{20}$. Check: HCF(3, 20) = 1, so already in simplest form.
Answer: $1\dfrac{3}{20}$
Grade 6–7
$= \dfrac{7}{2} \times \dfrac{4}{7}$
Cancel 4 and 2: $\dfrac{1}{\cancel{2}} \times \dfrac{\cancel{4}}{1} = \dfrac{1}{1} \times \dfrac{2}{1} = 2$
Calculate $3\dfrac{1}{2} \div 1\dfrac{3}{4}$. Give your answer as a fraction in its simplest form.
1
Convert both mixed numbers to improper fractions
$3\dfrac{1}{2} = \dfrac{3 \times 2 + 1}{2} = \dfrac{7}{2}$ and $1\dfrac{3}{4} = \dfrac{1 \times 4 + 3}{4} = \dfrac{7}{4}$
2
Apply KFC — Keep, Flip, Change
$\dfrac{7}{2} \div \dfrac{7}{4}$: Keep $\dfrac{7}{2}$, Flip $\dfrac{7}{4}$ to get $\dfrac{4}{7}$, Change ÷ to ×.$= \dfrac{7}{2} \times \dfrac{4}{7}$
3
Cross-cancel and multiply
Cancel the 7s: $\dfrac{\cancel{7}}{2} \times \dfrac{4}{\cancel{7}} = \dfrac{1}{2} \times \dfrac{4}{1}$Cancel 4 and 2: $\dfrac{1}{\cancel{2}} \times \dfrac{\cancel{4}}{1} = \dfrac{1}{1} \times \dfrac{2}{1} = 2$
4
State the final answer
The result is a whole number. $3\frac{1}{2} \div 1\frac{3}{4} = 2$.
Answer: $2$
Grade 9
Denominator: $x^2 + 5x + 6 = (x+2)(x+3)$ — look for two numbers that multiply to 6 and add to 5.
Both fractions share denominator $(x+2)$, so combine:
$\dfrac{x - 3 + 1}{x+2} = 2 \Rightarrow \dfrac{x-2}{x+2} = 2$
$x - 2 = 2x + 4 \Rightarrow -6 = x$
Check restriction: $x \neq -2$ and $x \neq -3$; $x = -6$ is valid.
Simplify fully: $\dfrac{x^2 - 9}{x^2 + 5x + 6}$. Hence, or otherwise, solve $\dfrac{x^2 - 9}{x^2 + 5x + 6} + \dfrac{1}{x + 2} = 2$.
1
Factorise numerator and denominator
Numerator: $x^2 - 9 = (x+3)(x-3)$ — difference of two squares.Denominator: $x^2 + 5x + 6 = (x+2)(x+3)$ — look for two numbers that multiply to 6 and add to 5.
2
Cancel the common factor
$\dfrac{(x+3)(x-3)}{(x+2)(x+3)} = \dfrac{x-3}{x+2}$ provided $x \neq -3$ and $x \neq -2$.
3
Substitute into the equation and find common denominator
$\dfrac{x-3}{x+2} + \dfrac{1}{x+2} = 2$Both fractions share denominator $(x+2)$, so combine:
$\dfrac{x - 3 + 1}{x+2} = 2 \Rightarrow \dfrac{x-2}{x+2} = 2$
4
Multiply both sides by $(x+2)$ and solve
$x - 2 = 2(x + 2) = 2x + 4$$x - 2 = 2x + 4 \Rightarrow -6 = x$
Check restriction: $x \neq -2$ and $x \neq -3$; $x = -6$ is valid.
Simplified form: $\dfrac{x-3}{x+2}$ ($x \neq -2, x \neq -3$) | Solution: $x = -6$
❓ Exam Questions
Q1
1 mark
Write $\dfrac{24}{36}$ in its simplest form.
Mark Scheme:
HCF(24, 36) = 12.
$\dfrac{24 \div 12}{36 \div 12} = \dfrac{2}{3}$ ✓ (1 mark)
Accept any method that correctly identifies the HCF and applies it.
HCF(24, 36) = 12.
$\dfrac{24 \div 12}{36 \div 12} = \dfrac{2}{3}$ ✓ (1 mark)
Accept any method that correctly identifies the HCF and applies it.
Q2
2 marks
Calculate $\dfrac{5}{6} - \dfrac{3}{8}$. Give your answer as a fraction in its simplest form.
Mark Scheme:
LCM(6, 8) = 24. ✓ (1 mark for correct common denominator or equivalent fractions)
$\dfrac{5}{6} = \dfrac{20}{24}$, $\quad \dfrac{3}{8} = \dfrac{9}{24}$
$\dfrac{20}{24} - \dfrac{9}{24} = \dfrac{11}{24}$ ✓ (1 mark for correct answer)
HCF(11, 24) = 1, so $\dfrac{11}{24}$ is already in simplest form.
LCM(6, 8) = 24. ✓ (1 mark for correct common denominator or equivalent fractions)
$\dfrac{5}{6} = \dfrac{20}{24}$, $\quad \dfrac{3}{8} = \dfrac{9}{24}$
$\dfrac{20}{24} - \dfrac{9}{24} = \dfrac{11}{24}$ ✓ (1 mark for correct answer)
HCF(11, 24) = 1, so $\dfrac{11}{24}$ is already in simplest form.
Q3
3 marks
A recipe requires $2\dfrac{1}{3}$ cups of flour and $1\dfrac{3}{4}$ cups of sugar. Sarah wants to make $1\dfrac{1}{2}$ times the recipe. How many cups of flour and sugar does she need in total? Give your answer as a mixed number.
Mark Scheme:
Total ingredients in original recipe: $2\frac{1}{3} + 1\frac{3}{4} = \frac{7}{3} + \frac{7}{4} = \frac{28}{12} + \frac{21}{12} = \frac{49}{12} = 4\frac{1}{12}$ ✓ (M1 A1)
Multiply by $1\frac{1}{2} = \frac{3}{2}$:
$\dfrac{49}{12} \times \dfrac{3}{2} = \dfrac{147}{24} = \dfrac{49}{8} = 6\dfrac{1}{8}$ cups ✓ (A1)
M1: correct method for adding the original quantities; A1: $\frac{49}{12}$ or equivalent; A1: $6\frac{1}{8}$
Total ingredients in original recipe: $2\frac{1}{3} + 1\frac{3}{4} = \frac{7}{3} + \frac{7}{4} = \frac{28}{12} + \frac{21}{12} = \frac{49}{12} = 4\frac{1}{12}$ ✓ (M1 A1)
Multiply by $1\frac{1}{2} = \frac{3}{2}$:
$\dfrac{49}{12} \times \dfrac{3}{2} = \dfrac{147}{24} = \dfrac{49}{8} = 6\dfrac{1}{8}$ cups ✓ (A1)
M1: correct method for adding the original quantities; A1: $\frac{49}{12}$ or equivalent; A1: $6\frac{1}{8}$
Q4
4 marks
Show that $\dfrac{3}{n+1} - \dfrac{2}{n-1} = \dfrac{n-5}{(n+1)(n-1)}$, where $n \neq 1$ and $n \neq -1$.
Mark Scheme:
Common denominator is $(n+1)(n-1)$. ✓ (M1)
$\dfrac{3}{n+1} = \dfrac{3(n-1)}{(n+1)(n-1)}$ ✓ (A1)
$\dfrac{2}{n-1} = \dfrac{2(n+1)}{(n+1)(n-1)}$ ✓ (A1)
Subtract: $\dfrac{3(n-1) - 2(n+1)}{(n+1)(n-1)} = \dfrac{3n - 3 - 2n - 2}{(n+1)(n-1)} = \dfrac{n-5}{(n+1)(n-1)}$ ✓ (A1)
Full marks require clear algebraic working at each step. A "show that" question requires you to begin from the LHS and reach the RHS — do not start from the RHS.
Common denominator is $(n+1)(n-1)$. ✓ (M1)
$\dfrac{3}{n+1} = \dfrac{3(n-1)}{(n+1)(n-1)}$ ✓ (A1)
$\dfrac{2}{n-1} = \dfrac{2(n+1)}{(n+1)(n-1)}$ ✓ (A1)
Subtract: $\dfrac{3(n-1) - 2(n+1)}{(n+1)(n-1)} = \dfrac{3n - 3 - 2n - 2}{(n+1)(n-1)} = \dfrac{n-5}{(n+1)(n-1)}$ ✓ (A1)
Full marks require clear algebraic working at each step. A "show that" question requires you to begin from the LHS and reach the RHS — do not start from the RHS.
Q5
3 marks
Place these fractions in ascending order (smallest first): $\dfrac{7}{12}$, $\dfrac{3}{5}$, $\dfrac{5}{8}$, $\dfrac{11}{20}$
Mark Scheme:
LCM(12, 5, 8, 20) = 120. ✓ (M1 for finding common denominator)
$\dfrac{7}{12} = \dfrac{70}{120}$, $\dfrac{3}{5} = \dfrac{72}{120}$, $\dfrac{5}{8} = \dfrac{75}{120}$, $\dfrac{11}{20} = \dfrac{66}{120}$ ✓ (A1)
Ascending order: $\dfrac{11}{20}, \dfrac{7}{12}, \dfrac{3}{5}, \dfrac{5}{8}$ ✓ (A1)
M1: evidence of common denominator method; A1: all four equivalent fractions correct; A1: correct final order in original form.
LCM(12, 5, 8, 20) = 120. ✓ (M1 for finding common denominator)
$\dfrac{7}{12} = \dfrac{70}{120}$, $\dfrac{3}{5} = \dfrac{72}{120}$, $\dfrac{5}{8} = \dfrac{75}{120}$, $\dfrac{11}{20} = \dfrac{66}{120}$ ✓ (A1)
Ascending order: $\dfrac{11}{20}, \dfrac{7}{12}, \dfrac{3}{5}, \dfrac{5}{8}$ ✓ (A1)
M1: evidence of common denominator method; A1: all four equivalent fractions correct; A1: correct final order in original form.
Q6
6 marks
Solve: $\dfrac{2x+1}{3} - \dfrac{x-2}{4} = 2\dfrac{1}{6}$
Mark Scheme:
Convert RHS: $2\frac{1}{6} = \frac{13}{6}$ ✓ (B1)
Common denominator of LHS is 12: ✓ (M1)
$\dfrac{4(2x+1)}{12} - \dfrac{3(x-2)}{12} = \dfrac{13}{6}$
Expand numerator: $\dfrac{8x+4 - 3x + 6}{12} = \dfrac{13}{6}$ ✓ (A1 for expansion)
$\dfrac{5x + 10}{12} = \dfrac{13}{6}$
Multiply both sides by 12: $5x + 10 = 26$ ✓ (M1)
$5x = 16$
$x = \dfrac{16}{5} = 3\dfrac{1}{5}$ ✓ (A1)
Check: LHS $= \frac{2(\frac{16}{5})+1}{3} - \frac{\frac{16}{5}-2}{4} = \frac{\frac{37}{5}}{3} - \frac{\frac{6}{5}}{4} = \frac{37}{15} - \frac{6}{20} = \frac{37}{15} - \frac{3}{10} = \frac{74-9}{30} = \frac{65}{30} = \frac{13}{6}$ ✓ (B1 for check)
Convert RHS: $2\frac{1}{6} = \frac{13}{6}$ ✓ (B1)
Common denominator of LHS is 12: ✓ (M1)
$\dfrac{4(2x+1)}{12} - \dfrac{3(x-2)}{12} = \dfrac{13}{6}$
Expand numerator: $\dfrac{8x+4 - 3x + 6}{12} = \dfrac{13}{6}$ ✓ (A1 for expansion)
$\dfrac{5x + 10}{12} = \dfrac{13}{6}$
Multiply both sides by 12: $5x + 10 = 26$ ✓ (M1)
$5x = 16$
$x = \dfrac{16}{5} = 3\dfrac{1}{5}$ ✓ (A1)
Check: LHS $= \frac{2(\frac{16}{5})+1}{3} - \frac{\frac{16}{5}-2}{4} = \frac{\frac{37}{5}}{3} - \frac{\frac{6}{5}}{4} = \frac{37}{15} - \frac{6}{20} = \frac{37}{15} - \frac{3}{10} = \frac{74-9}{30} = \frac{65}{30} = \frac{13}{6}$ ✓ (B1 for check)
⭐ Grade 9 Model Answers
Full Annotated Solution — Q6 (6 marks)
This question tests: algebraic manipulation of fractions with different denominators, handling mixed numbers, and multi-step equation solving. Here is a Grade 9 model response:
Grade 9 — Full Marks Strategy
$$\frac{4(2x+1)}{12} - \frac{3(x-2)}{12} = \frac{13}{6}$$ This earns the method mark for correctly establishing a common denominator.
$\dfrac{5x + 10}{12} = \dfrac{13}{6}$
Key error zone: $-3(x-2) = -3x + 6$ (not $-3x - 6$). The double negative is a common source of errors.
This is equivalent to multiplying through by the LCM of 12 and 6, which is 12.
Leave the answer as a mixed number or improper fraction — both are acceptable. Do not convert to a decimal unless instructed.
LHS $= \dfrac{2 \cdot \frac{16}{5} + 1}{3} - \dfrac{\frac{16}{5} - 2}{4} = \dfrac{\frac{37}{5}}{3} - \dfrac{\frac{6}{5}}{4} = \dfrac{37}{15} - \dfrac{6}{20} = \dfrac{37}{15} - \dfrac{3}{10}$
$= \dfrac{74}{30} - \dfrac{9}{30} = \dfrac{65}{30} = \dfrac{13}{6}$ = RHS ✓
Solve: $\dfrac{2x+1}{3} - \dfrac{x-2}{4} = 2\dfrac{1}{6}$
1
Convert the mixed number immediately [B1]
$2\dfrac{1}{6} = \dfrac{13}{6}$. Converting straight away prevents errors when balancing the equation.
2
Identify the LCM and create a common denominator for the LHS [M1]
LCM(3, 4) = 12. Multiply $\frac{2x+1}{3}$ by $\frac{4}{4}$ and $\frac{x-2}{4}$ by $\frac{3}{3}$:$$\frac{4(2x+1)}{12} - \frac{3(x-2)}{12} = \frac{13}{6}$$ This earns the method mark for correctly establishing a common denominator.
3
Expand numerator carefully, keeping denominator [A1]
$\dfrac{8x + 4 - 3x + 6}{12} = \dfrac{13}{6}$$\dfrac{5x + 10}{12} = \dfrac{13}{6}$
Key error zone: $-3(x-2) = -3x + 6$ (not $-3x - 6$). The double negative is a common source of errors.
4
Clear the fractions by multiplying both sides by 12 [M1]
$5x + 10 = \dfrac{13}{6} \times 12 = 26$This is equivalent to multiplying through by the LCM of 12 and 6, which is 12.
5
Solve the linear equation [A1]
$5x = 16 \Rightarrow x = \dfrac{16}{5} = 3\dfrac{1}{5}$Leave the answer as a mixed number or improper fraction — both are acceptable. Do not convert to a decimal unless instructed.
6
Verify by substitution [B1 — check mark]
Substitute $x = \frac{16}{5}$:LHS $= \dfrac{2 \cdot \frac{16}{5} + 1}{3} - \dfrac{\frac{16}{5} - 2}{4} = \dfrac{\frac{37}{5}}{3} - \dfrac{\frac{6}{5}}{4} = \dfrac{37}{15} - \dfrac{6}{20} = \dfrac{37}{15} - \dfrac{3}{10}$
$= \dfrac{74}{30} - \dfrac{9}{30} = \dfrac{65}{30} = \dfrac{13}{6}$ = RHS ✓
$x = \dfrac{16}{5} = 3\dfrac{1}{5}$ [6/6 marks]
WHY THIS EARNS FULL MARKS
- B1: Converting the mixed number correctly at the start
- M1: Demonstrating the correct method for a common denominator
- A1: Correct expansion (handling the double negative)
- M1: Clearing fractions systematically by multiplying by LCM
- A1: Correct final answer in acceptable form
- B1: Verification by substitution (shows examiner you're confident)
📋 Revision Sheet
Key Definitions
| Term | Meaning |
|---|---|
| Numerator | Top number — parts you have |
| Denominator | Bottom number — total equal parts |
| Proper fraction | Numerator < denominator (e.g. $\frac{3}{5}$) |
| Improper fraction | Numerator ≥ denominator (e.g. $\frac{7}{4}$) |
| Mixed number | Whole number + proper fraction (e.g. $1\frac{3}{4}$) |
| Equivalent fractions | Same value, different numerator/denominator |
| Reciprocal | Flip the fraction: reciprocal of $\frac{a}{b}$ is $\frac{b}{a}$ |
| HCF | Highest Common Factor — used to simplify |
| LCM | Lowest Common Multiple — used as common denominator |
Essential Formulae
$$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$$
$$\frac{a}{b} - \frac{c}{d} = \frac{ad - bc}{bd}$$
$$\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$$
$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}$$
$$n\frac{a}{b} = \frac{nb + a}{b}$$
$$\frac{a}{b} \text{ of } Q = \frac{aQ}{b}$$
Memory Hooks
- KFC — Keep · Flip · Change (for division)
- "Of means multiply" — $\frac{3}{4}$ of 20 = $\frac{3}{4} \times 20 = 15$
- Mixed to improper: Multiply whole, Add top, Keep bottom
- Adding fractions: Same denominator = add tops, same bottom
- HCF to simplify, LCM to add
- Cross-cancel before × — saves simplifying at the end
- Check algebraic fractions: Factorise first, cancel, then state restrictions
Exam Tips
- Always convert mixed numbers to improper fractions before multiplying or dividing
- Use LCM (not product) as common denominator — smaller numbers, fewer errors
- In "show that" proofs, start from LHS and arrive at RHS — never work backwards
- For algebraic fractions, always factorise first, then cancel — never cancel individual terms
- State restrictions ($x \neq ...$) when cancelling factors in algebraic fractions
- In fraction equations, multiply all terms by LCM of denominators to clear fractions
- Leave answers as fractions, not decimals, unless the question specifies otherwise
- In ordering questions, show the equivalent fractions — they earn the method mark
🔄 Flashcards
Click a card to reveal the answer. Use these to test your recall.
✗ Common Mistakes
MISTAKE 1 — Adding the Denominators
What students do wrong: $\dfrac{1}{3} + \dfrac{1}{5} = \dfrac{2}{8}$Why marks are lost: The denominator represents the size of each part. Adding denominators changes what size of part you are counting — it is meaningless.
How to avoid it: Always find a common denominator first. $\dfrac{1}{3} + \dfrac{1}{5} = \dfrac{5}{15} + \dfrac{3}{15} = \dfrac{8}{15}$.
MISTAKE 2 — Flipping the Wrong Fraction When Dividing
What students do wrong: $\dfrac{2}{3} \div \dfrac{4}{5}$ → flip the first: $\dfrac{3}{2} \times \dfrac{4}{5} = \dfrac{12}{10}$Why marks are lost: This gives $\frac{6}{5}$ instead of the correct $\frac{5}{6}$ — a completely different value, scoring 0 marks.
How to avoid it: KFC — you Keep the first fraction exactly as it is and only flip the second. $\dfrac{2}{3} \times \dfrac{5}{4} = \dfrac{10}{12} = \dfrac{5}{6}$.
MISTAKE 3 — Forgetting to Convert Mixed Numbers Before Operating
What students do wrong: $2\dfrac{1}{3} \times 1\dfrac{1}{2}$ treated as $2 \times 1 + \dfrac{1}{3} \times \dfrac{1}{2} = 2\dfrac{1}{6}$Why marks are lost: Multiplying the whole-number parts and fractional parts separately does not work because multiplication distributes over addition in a more complex way.
How to avoid it: Always convert to improper fractions first: $\dfrac{7}{3} \times \dfrac{3}{2} = \dfrac{7}{2} = 3\dfrac{1}{2}$.
MISTAKE 4 — Cancelling Terms Instead of Factors in Algebraic Fractions
What students do wrong: $\dfrac{x + 4}{x} = 4$ (cancelling the $x$ terms)Why marks are lost: You can only cancel factors, not terms. $x$ is a term added in the numerator, not a factor multiplying the whole numerator.
How to avoid it: Factorise the numerator first. If the numerator does not factorise to give a factor of $x$, no cancellation is possible. $\dfrac{x+4}{x}$ cannot be simplified further.
MISTAKE 5 — Sign Errors When Subtracting Fractions with Brackets
What students do wrong: $\dfrac{5}{12} - \dfrac{x-2}{4}$ → subtract $\dfrac{3(x-2)}{12}$ → get $5 - 3x - 6$ in the numeratorWhy marks are lost: Subtracting a bracket requires distributing the negative sign to all terms: $-(x-2) = -x+2$, not $-x-2$.
How to avoid it: Write the subtraction as $5 - 3(x-2) = 5 - 3x + 6 = 11 - 3x$. Expand brackets carefully before collecting terms.
MISTAKE 6 — Not Simplifying the Final Answer
What students do wrong: Leaving an answer as $\dfrac{8}{12}$ instead of $\dfrac{2}{3}$Why marks are lost: Most mark schemes require the simplest form. Leaving a fraction unsimplified when the question says "simplest form" or "lowest terms" will lose the final accuracy mark.
How to avoid it: Always check — does HCF(numerator, denominator) = 1? If not, divide both by the HCF. Make it a habit at the end of every fraction calculation.
✅ Final Checklist
Click each item as you master it. Your progress is saved automatically.
0 / 14- I can find the HCF of two numbers and use it to simplify a fraction in one step
- I can create equivalent fractions by multiplying or dividing numerator and denominator by the same value
- I can convert a mixed number to an improper fraction using the formula $\frac{nb+a}{b}$
- I can convert an improper fraction to a mixed number by dividing and finding the remainder
- I can add and subtract fractions by finding the LCM and converting to a common denominator
- I can add and subtract mixed numbers by converting to improper fractions first
- I can multiply two fractions by multiplying numerators and denominators separately
- I can apply KFC (Keep, Flip, Change) to divide fractions correctly
- I can find a fraction of a quantity and solve reverse fraction problems
- I can order three or more fractions using a common denominator
- I can simplify algebraic fractions by factorising and cancelling common factors
- I can add and subtract algebraic fractions using an algebraic common denominator
- I can solve equations involving fractions by multiplying through by the LCM of denominators
- I always state restrictions on variables when simplifying algebraic fractions