Algebraic Fractions · Study Notes · Edexcel IGCSE 4MA1 Higher

Manipulate fractions where numerators and denominators are algebraic expressions. Always factorise first.

At a glance

Always factorise FIRST — both numerator and denominator.
Simplify by cancelling common FACTORS (not terms).
Add/subtract: common denominator (LCM).
Multiply: factorise, cancel, then multiply.
Divide: keep, change, flip.
Final answer: simplest form.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

2.5 — Simplify algebraic fractions by factoring.
2.5 — Add, subtract, multiply, divide algebraic fractions.
2.5 — Express two or more fractions as a single fraction.

Simplify by cancelling common factors

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Factor numerator and denominator. Cancel common factors only — never individual terms.

Cancellation rule: can only cancel COMMON FACTORS in numerator and denominator.

You CANNOT cancel terms that aren't factors:

$\dfrac{x + 3}{x + 5} \ne \dfrac{3}{5}$ . The $x$ 's aren't factors of the whole.

Always factorise first.

Only whole factors cancel — an x buried inside a sum is not a factor.

Example. $\dfrac{x^{2} - 9}{x^{2} + 5x + 6}$ .

Factor numerator: $x^{2} - 9 = (x - 3)(x + 3)$ .
Factor denominator: $x^{2} + 5x + 6 = (x + 2)(x + 3)$ .
Cancel common factor $(x + 3)$ : $\dfrac{(x - 3)}{(x + 2)}$ .

More examples:

Original	Factor	Simplified
$\dfrac{x^{2} - 4}{x - 2}$	$\dfrac{(x-2)(x+2)}{x-2}$	$x + 2$
$\dfrac{2x^{2} + 6x}{x}$	$\dfrac{2x(x+3)}{x}$	$2(x + 3)$
$\dfrac{x^{2} - 1}{x^{2} - 2x - 3}$	$\dfrac{(x-1)(x+1)}{(x-3)(x+1)}$	$\dfrac{x-1}{x-3}$

Edexcel tip. Cancel only AFTER factoring. Show every step explicitly.

Cancel common FACTORS only.
Factor numerator + denominator first.
Final answer in simplest form.
$\dfrac{(a)(b)}{(c)(b)} = \dfrac{a}{c}$ — $b$ 's cancel.

Add and subtract algebraic fractions

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Common denominator (LCM). Adjust numerators. Combine.

Method:

Factor each denominator if helpful.
Find the LCM of denominators.
Express each fraction with the LCM as denominator.
Add/subtract numerators (CAREFULLY with signs).
Simplify if possible.

Example. $\dfrac{2}{x - 1} + \dfrac{3}{x + 2}$ .

LCM: $(x - 1)(x + 2)$ .
Adjust: $\dfrac{2(x + 2)}{(x-1)(x+2)} + \dfrac{3(x - 1)}{(x-1)(x+2)}$ .
Combine numerators: $\dfrac{2(x + 2) + 3(x - 1)}{(x-1)(x+2)} = \dfrac{2x + 4 + 3x - 3}{(x-1)(x+2)} = \dfrac{5x + 1}{(x-1)(x+2)}$ .

With factoring needed.

$\dfrac{1}{x^{2} - 4} - \dfrac{1}{x + 2}$ .

Factor: $x^{2} - 4 = (x - 2)(x + 2)$ .
LCM: $(x - 2)(x + 2)$ .
Adjust: $\dfrac{1}{(x-2)(x+2)} - \dfrac{1 \cdot (x - 2)}{(x-2)(x+2)} = \dfrac{1 - (x - 2)}{(x-2)(x+2)} = \dfrac{3 - x}{(x-2)(x+2)}$ .

Worked qualitative. Why subtract carefully?

$1 - (x - 2) = 1 - x + 2 = 3 - x$ . The minus distributes to BOTH terms in the bracket.
A common error is writing $1 - x - 2 = -1 - x$ , which is wrong.

Edexcel tip. Show the adjusted fractions before combining. Mark schemes credit each step.

LCM of denominators.
Adjust numerators consistently.
Distribute minus carefully.
Simplify final result.

Multiply and divide algebraic fractions

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Multiply: factorise, cancel, multiply. Divide: keep, change, flip.

Multiplication.

Factorise everything.
Cancel common factors across numerators and denominators.
Multiply remaining numerators; multiply remaining denominators.

Example. $\dfrac{x^{2} - 1}{2x} \times \dfrac{4x^{2}}{x + 1}$ .

Factor: $x^{2} - 1 = (x-1)(x+1)$ .
Substitute: $\dfrac{(x-1)(x+1)}{2x} \times \dfrac{4x^{2}}{x+1}$ .
Cancel: $(x+1)$ across numerator and denominator. $4x^{2}/2x = 2x$ .
Result: $(x - 1) \times 2x = 2x(x - 1) = 2x^{2} - 2x$ .

Division — keep, change, flip.

$\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$ .

Then multiply as above.

Example. $\dfrac{x^{2} + 3x}{x^{2} - 4} \div \dfrac{x + 3}{x - 2}$ .

Flip: $\dfrac{x^{2} + 3x}{x^{2} - 4} \times \dfrac{x - 2}{x + 3}$ .
Factor: $x^{2} + 3x = x(x+3)$ . $x^{2} - 4 = (x-2)(x+2)$ .
Substitute: $\dfrac{x(x+3)}{(x-2)(x+2)} \times \dfrac{x-2}{x+3}$ .
Cancel $(x+3)$ and $(x-2)$ : $\dfrac{x}{x+2}$ .

Edexcel tip. Don't expand the brackets unless asked. Cancel factors first to simplify.

Multiply: factorise, cancel, multiply.
Divide: keep, change, flip.
Cancel BEFORE multiplying.
Don't expand prematurely.

Quick recap

Always factorise first.
Cancel common FACTORS only.
Add/subtract: LCM denominator.
Multiply: factorise, cancel, multiply.
Divide: keep, change, flip.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Cancel common factors only — not terms.
Add/sub: LCM, adjust, combine.
Multiply: factorise, cancel, multiply.
Divide: keep, change, flip.

How it’s examined

Algebraic fractions appear Paper 1H (3-5 marks: simplify or single fraction). Examiner reports flag (1) cancelling terms instead of factors, (2) sign errors when subtracting, (3) cross-multiplying for division.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on algebraic fractions, written exactly the way a tutor would explain them at the board.

1Simplify by factoring
Higher• Adapted from 4MA1/1H May/Jun 2024 Q15• simplify
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Question
Simplify $\dfrac{x^{2} - 9}{x^{2} + 5x + 6}$ .
Step-by-step solution
1. Step 1
  Factor numerator: $x^{2} - 9 = (x - 3)(x + 3)$ (difference of squares).
2. Step 2
  Factor denominator: $x^{2} + 5x + 6 = (x + 2)(x + 3)$ .
3. Step 3
  Cancel common factor $(x + 3)$ :
  $\frac{(x - 3)(x + 3)}{(x + 2)(x + 3)} = \frac{x - 3}{x + 2}$
Answer
$\dfrac{x - 3}{x + 2}$
Examiner tip
ALWAYS factorise before simplifying. Mark schemes give M1 for factorising, M1 for cancelling, A1 for the answer.
2Add algebraic fractions
Higher• addition
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Question
Express as a single fraction: $\dfrac{2}{x - 1} + \dfrac{3}{x + 2}$ .
Step-by-step solution
1. Step 1
  Common denominator: $(x - 1)(x + 2)$ .
2. Step 2
  Adjust each fraction:
  $\frac{2(x + 2)}{(x - 1)(x + 2)} + \frac{3(x - 1)}{(x - 1)(x + 2)}$
3. Step 3
  Combine: $\dfrac{2(x + 2) + 3(x - 1)}{(x - 1)(x + 2)}$ .
4. Step 4
  Expand numerator: $2x + 4 + 3x - 3 = 5x + 1$ .
5. Step 5
  Final:
  $\frac{5x + 1}{(x - 1)(x + 2)}$
Answer
$\dfrac{5x + 1}{(x - 1)(x + 2)}$
Examiner tip
Always find LCM of denominators. Edexcel mark schemes credit the common-denominator step explicitly.
3Multiply algebraic fractions
Higher• multiplication
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Question
Simplify $\dfrac{x^{2} - 1}{2x} \times \dfrac{4x^{2}}{x + 1}$ .
Step-by-step solution
1. Step 1
  Factor: $x^{2} - 1 = (x - 1)(x + 1)$ .
2. Step 2
  Multiply:
  $\frac{(x - 1)(x + 1)}{2x} \times \frac{4x^{2}}{x + 1}$
3. Step 3
  Cancel common factors: $(x + 1)$ cancels, $2x$ cancels with $4x^{2}$ leaving $2x$ :
  $\frac{(x - 1)}{1} \times \frac{2x}{1} = 2x(x - 1)$
Answer
$2x(x - 1)$ or $2x^{2} - 2x$
Examiner tip
Factorise FIRST, then multiply, then cancel. Don't expand prematurely.
4Divide algebraic fractions
Higher• division
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Question
Simplify $\dfrac{x^{2} + 3x}{x^{2} - 4} \div \dfrac{x + 3}{x - 2}$ .
Step-by-step solution
1. Step 1
  Keep, change, flip:
  $\frac{x^{2} + 3x}{x^{2} - 4} \times \frac{x - 2}{x + 3}$
2. Step 2
  Factor: $x^{2} + 3x = x(x + 3)$ . $x^{2} - 4 = (x - 2)(x + 2)$ .
3. Step 3
  Substitute and cancel:
  $\frac{x(x + 3)}{(x - 2)(x + 2)} \times \frac{x - 2}{x + 3}$
4. Step 4
  Cancel $(x + 3)$ and $(x - 2)$ :
  $\frac{x}{x + 2}$
Answer
$\dfrac{x}{x + 2}$
5Cancelling a whole bracket [1 mark]
Getting started• Paper 1H/2H style — 1 mark• simplifying, factorising
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Question
Simplify $\dfrac{3x - 6}{x - 2}$ [1]
Step-by-step solution
1. Step 1
  Take out the common factor $3$ from the numerator.
  $3x - 6 = 3(x - 2)$
2. Step 2
  B1 The bracket $(x - 2)$ is a factor of both parts, so it cancels.
  $\frac{3(x - 2)}{x - 2} = 3$
Answer
$3$
Examiner tip
Examiner tip: An algebraic fraction can simplify to a plain number. If the answer looks too simple, check it with a value: $x = 5$ gives $\dfrac{9}{3} = 3$ .

Common mistake: Cancelling the $x$ terms only, as $\dfrac{3 - 6}{-2} = 1.5$ . The $x$ in $3x$ is part of a term, not a factor of the whole numerator.
6Subtracting fractions with x in the denominator [2 marks]
Getting started• Paper 1H/2H style — 2 marks• subtracting, common denominator
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Question
Express $\dfrac{5}{2x} - \dfrac{1}{3x}$ as a single fraction in its simplest form. [2]
Step-by-step solution
1. Step 1
  M1 The lowest common denominator is $6x$ . Multiply the first fraction by $\dfrac{3}{3}$ and the second by $\dfrac{2}{2}$ .
  $\frac{15}{6x} - \frac{2}{6x}$
2. Step 2
  A1 Subtract the numerators and keep the denominator.
  $\frac{15 - 2}{6x} = \frac{13}{6x}$
Answer
$\dfrac{13}{6x}$
Examiner tip
Examiner tip: The lowest common denominator of $2x$ and $3x$ is $6x$ , not $6x^{2}$ . Using $6x^{2}$ works but you then have to cancel an $x$ .

Common mistake: Writing $\dfrac{13}{6}$ , having dropped the $x$ from the denominator. The June 2024 examiner report noted students who reached the correct numerator but were careless in writing the final answer, omitting the $x$ from the denominator.
7Solving with the unknown in the denominator [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• solving equations, clearing fractions
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Question
Solve $\dfrac{2x + 3}{x - 1} = 5$

Show clear algebraic working. [3]
Step-by-step solution
1. Step 1
  M1 Multiply both sides by $(x - 1)$ , keeping it in a bracket.
  $2x + 3 = 5(x - 1)$
2. Step 2
  M1 Expand and collect the $x$ terms on one side.
  $2x + 3 = 5x - 5 \;\Rightarrow\; 8 = 3x$
3. Step 3
  A1 Divide by $3$ .
  $x = \frac{8}{3}$
Answer
$x = \dfrac{8}{3}$ (or $2\dfrac{2}{3}$ , or $2.67$ to 3 s.f.)
Examiner tip
Examiner tip: Check: $\dfrac{2 \times \frac{8}{3} + 3}{\frac{8}{3} - 1} = \dfrac{\frac{25}{3}}{\frac{5}{3}} = 5$ .

Common mistake: Writing $2x + 3 = 5x - 1$ . Without the bracket only the $x$ is multiplied by $5$ , which is the error examiners describe as a failure to use brackets when multiplying both sides.
8Show that a difference of fractions cancels down [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• show that, subtracting, difference of two squares
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Question
Show that $\dfrac{1}{x - 3} - \dfrac{6}{x^{2} - 9} = \dfrac{1}{x + 3}$ [3]
Step-by-step solution
1. Step 1
  M1 Factorise $x^{2} - 9 = (x - 3)(x + 3)$ and write both fractions over this common denominator.
  $\frac{x + 3}{(x - 3)(x + 3)} - \frac{6}{(x - 3)(x + 3)}$
2. Step 2
  M1 Subtract the numerators.
  $\frac{x + 3 - 6}{(x - 3)(x + 3)} = \frac{x - 3}{(x - 3)(x + 3)}$
3. Step 3
  A1 Cancel the common factor $(x - 3)$ to reach the printed result.
  $\frac{x - 3}{(x - 3)(x + 3)} = \frac{1}{x + 3}$
Answer
$\dfrac{x - 3}{(x - 3)(x + 3)} = \dfrac{1}{x + 3}$ , as required
Examiner tip
Examiner tip: In a "Show that" the last line is printed, so the A mark is for the line before it. Here that is $\dfrac{x - 3}{(x - 3)(x + 3)}$ , which must appear.

Common mistake: Using $(x - 3)(x^{2} - 9)$ as the denominator because the $x^{2} - 9$ was not factorised. The algebra then gets much longer and rarely reaches the printed answer.
9An equation that becomes a factorisable quadratic [4 marks]
Stretch• Paper 1H/2H style — 4 marks• solving equations, quadratics
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Question
Solve $\dfrac{3}{x} - \dfrac{2}{x + 1} = \dfrac{1}{2}$

Show clear algebraic working. [4]
Step-by-step solution
1. Step 1
  M1 Multiply every term by $2x(x + 1)$ .
  $6(x + 1) - 4x = x(x + 1)$
2. Step 2
  A1 Expand and rearrange to a quadratic equal to zero.
  $6x + 6 - 4x = x^{2} + x \;\Rightarrow\; x^{2} - x - 6 = 0$
3. Step 3
  M1 Factorise: two numbers with product $-6$ and sum $-1$ are $-3$ and $2$ .
  $(x - 3)(x + 2) = 0$
4. Step 4
  A1 Neither root is $0$ or $-1$ , so both are valid.
  $x = 3 \quad \text{or} \quad x = -2$
Answer
$x = 3$ or $x = -2$
Examiner tip
Examiner tip: Check both roots: $x = 3$ gives $1 - \dfrac{1}{2} = \dfrac{1}{2}$ , and $x = -2$ gives $-\dfrac{3}{2} + 2 = \dfrac{1}{2}$ .

Common mistake: Multiplying by $x(x + 1)$ only and leaving the right-hand side as $\dfrac{1}{2}$ , or giving only the positive root. Both solutions are needed for the final mark.
10Write as a single fraction, hence solve [5 marks]
Stretch• Paper 1H/2H style — 5 marks• adding, solving equations, hence
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Question
(a) Express $\dfrac{2}{x + 1} + \dfrac{3}{x + 4}$ as a single fraction in its simplest form. [2]

(b) Hence solve $\dfrac{2}{x + 1} + \dfrac{3}{x + 4} = 1$

Give your solutions in surd form. [3]
Step-by-step solution
1. Step 1
  M1 (a) Use the common denominator $(x + 1)(x + 4)$ .
  $\frac{2(x + 4) + 3(x + 1)}{(x + 1)(x + 4)}$
2. Step 2
  A1 Expand and collect the numerator.
  $\frac{2x + 8 + 3x + 3}{(x + 1)(x + 4)} = \frac{5x + 11}{(x + 1)(x + 4)}$
3. Step 3
  M1 (b) "Hence" means use part (a). Set the single fraction equal to $1$ and multiply by the denominator.
  $5x + 11 = (x + 1)(x + 4)$
4. Step 4
  M1 Expand and simplify. The $5x$ terms cancel.
  $5x + 11 = x^{2} + 5x + 4 \;\Rightarrow\; x^{2} = 7$
5. Step 5
  A1 Take both square roots.
  $x = \sqrt{7} \quad \text{or} \quad x = -\sqrt{7}$
Answer
(a) $\dfrac{5x + 11}{(x + 1)(x + 4)}$ (b) $x = \pm\sqrt{7}$
Examiner tip
Examiner tip: "Hence" tells you part (a) is the quickest route. Starting part (b) from scratch still works but wastes time and invites new errors.

Common mistake: Giving only $x = \sqrt{7}$ . Taking only the positive root is one of the errors named in the examiner reports, and it loses the final A mark.

Question 1
Paper 1H/2H style — 1 mark1 mark
Simplify $\dfrac{x^{2} - 6x}{3x}$ [1]
Model answer
$\dfrac{x^{2} - 6x}{3x} = \dfrac{x(x - 6)}{3x} = \dfrac{x - 6}{3}$

Answer: $\dfrac{x - 6}{3}$
Why this scores
Mark scheme:
- B1 for $\dfrac{x - 6}{3}$ oe, e.g. $\dfrac{x}{3} - 2$
Why this scores full marks: Taking out the common factor $x$ shows that the $x$ in the denominator cancels with a factor, not with one term. The answer $\dfrac{x - 6x}{3}$ , from cancelling only the $x^{2}$ , is the usual wrong answer.
Question 2
Paper 1H/2H style — 2 marks2 marks
Simplify fully $\dfrac{x^{2} - 10x + 25}{x^{2} - 25}$ [2]
Model answer
$\dfrac{x^{2} - 10x + 25}{x^{2} - 25} = \dfrac{(x - 5)(x - 5)}{(x - 5)(x + 5)}$

Answer: $\dfrac{x - 5}{x + 5}$
Why this scores
Mark scheme:
- M1 for $(x - 5)(x - 5)$ or $(x - 5)^{2}$ , or $(x - 5)(x + 5)$
- A1 for $\dfrac{x - 5}{x + 5}$ cao
Why this scores full marks: Writing the perfect square as two brackets makes it clear that only one $(x - 5)$ cancels, leaving one on top. Scripts that cancel both brackets reach $\dfrac{1}{x + 5}$ and lose the A mark.
Question 3
Paper 1H/2H style — 2 marks2 marks
Simplify fully $\dfrac{x + 1}{6} \times \dfrac{9}{x^{2} + x}$ [2]
Model answer
$\dfrac{x + 1}{6} \times \dfrac{9}{x(x + 1)} = \dfrac{9}{6x}$

Answer: $\dfrac{3}{2x}$
Why this scores
Mark scheme:
- M1 for $x(x + 1)$ seen, or $\dfrac{9(x + 1)}{6x(x + 1)}$ oe
- A1 for $\dfrac{3}{2x}$ oe in its simplest form
Why this scores full marks: Factorising $x^{2} + x$ shows the $(x + 1)$ that cancels, and the number part $\dfrac{9}{6}$ is then cancelled to $\dfrac{3}{2}$ . Stopping at $\dfrac{9}{6x}$ is not fully simplified.
Question 4
Paper 1H/2H style — 3 marks3 marks
Simplify fully $\dfrac{x^{2} - 4x}{x + 2} \div (x^{2} - 16)$ [3]
Model answer
$\dfrac{x^{2} - 4x}{x + 2} \div \dfrac{x^{2} - 16}{1} = \dfrac{x^{2} - 4x}{x + 2} \times \dfrac{1}{x^{2} - 16}$

$= \dfrac{x(x - 4)}{x + 2} \times \dfrac{1}{(x - 4)(x + 4)}$

Answer: $\dfrac{x}{(x + 2)(x + 4)}$
Why this scores
Mark scheme:
- M1 for multiplying by the reciprocal, $\dfrac{1}{x^{2} - 16}$ oe
- M1 for $x(x - 4)$ and $(x - 4)(x + 4)$
- A1 for $\dfrac{x}{(x + 2)(x + 4)}$ oe, e.g. $\dfrac{x}{x^{2} + 6x + 8}$
Why this scores full marks: Writing $x^{2} - 16$ as $\dfrac{x^{2} - 16}{1}$ makes its reciprocal obvious, so it ends up in the denominator. Scripts that multiply the numerator by $x^{2} - 16$ instead have divided the wrong way. After factorising, only $(x - 4)$ cancels, leaving $x$ over two brackets.
Question 5
Paper 1H/2H style — 3 marks3 marks
Solve $\dfrac{x}{x + 3} = 1 - \dfrac{2}{x}$

Show clear algebraic working. [3]
Model answer
Multiply every term by $x(x + 3)$ :

$x^{2} = x(x + 3) - 2(x + 3)$

$x^{2} = x^{2} + 3x - 2x - 6$

$0 = x - 6$

Answer: $x = 6$
Why this scores
Mark scheme:
- M1 for multiplying all three terms by $x(x + 3)$ oe
- M1 for $x^{2} = x^{2} + 3x - 2x - 6$ oe (correct expansion)
- A1 for $x = 6$ cao
Why this scores full marks: The $1$ on the right is multiplied by $x(x + 3)$ as well, and the $-2(x + 3)$ is expanded as $-2x - 6$ . The $x^{2}$ terms then cancel, so the equation is linear. Check: $\dfrac{6}{9} = 1 - \dfrac{2}{6} = \dfrac{2}{3}$ .
Question 6
Paper 1H/2H style — 3 marks3 marks
$\dfrac{3x^{2} + 11x - 4}{x^{2} - 16}$ can be written in the form $\dfrac{ax - 1}{x - b}$ , where $a$ and $b$ are integers.

Find the value of $a$ and the value of $b$ . [3]
Model answer
$3x^{2} + 11x - 4 = (3x - 1)(x + 4)$

$x^{2} - 16 = (x - 4)(x + 4)$

$\dfrac{(3x - 1)(x + 4)}{(x - 4)(x + 4)} = \dfrac{3x - 1}{x - 4}$

Answer: $a = 3$ , $b = 4$
Why this scores
Mark scheme:
- M1 for $(3x - 1)(x + 4)$
- M1 for $(x - 4)(x + 4)$
- A1 for $a = 3$ and $b = 4$
Why this scores full marks: The target form tells you the numerator must contain $(ax - 1)$ , and $3x^{2}$ with $-4$ points to $(3x - 1)(x + 4)$ . After cancelling $(x + 4)$ the values are read off. Giving $b = -4$ is a common slip: the form is $x - b$ , so $x - 4$ means $b = 4$ .
Question 7
Paper 1H/2H style — 4 marks4 marks
Express $\dfrac{2}{x - 1} - \dfrac{3}{x + 1} + \dfrac{6}{x^{2} - 1}$ as a single fraction in its simplest form. [4]
Model answer
$x^{2} - 1 = (x - 1)(x + 1)$

$\dfrac{2(x + 1) - 3(x - 1) + 6}{(x - 1)(x + 1)}$

$= \dfrac{2x + 2 - 3x + 3 + 6}{(x - 1)(x + 1)}$

Answer: $\dfrac{11 - x}{(x - 1)(x + 1)}$
Why this scores
Mark scheme:
- B1 for $(x - 1)(x + 1)$
- M1 for a common denominator of $(x - 1)(x + 1)$ with at least two numerators correct
- M1 for $2(x + 1) - 3(x - 1) + 6$ oe
- A1 for $\dfrac{11 - x}{(x - 1)(x + 1)}$ oe, e.g. $\dfrac{-x + 11}{x^{2} - 1}$
Why this scores full marks: Recognising $x^{2} - 1$ as the product of the other two denominators means one common denominator covers all three fractions, and the third fraction needs no change. The bracket in $-3(x - 1)$ gives $+3$ , not $-3$ .
Question 8
Paper 1H/2H style — 4 marks4 marks
Show that $\dfrac{x^{2} - 4}{x^{2} + x - 6} + \dfrac{1}{x + 3} = 1$ for all values of $x$ for which the expression is defined. [4]
Model answer
$x^{2} - 4 = (x - 2)(x + 2)$

$x^{2} + x - 6 = (x + 3)(x - 2)$

$\dfrac{(x - 2)(x + 2)}{(x + 3)(x - 2)} = \dfrac{x + 2}{x + 3}$

$\dfrac{x + 2}{x + 3} + \dfrac{1}{x + 3} = \dfrac{x + 3}{x + 3} = 1$
Why this scores
Mark scheme:
- M1 for $(x - 2)(x + 2)$
- M1 for $(x + 3)(x - 2)$
- M1 for $\dfrac{x + 2}{x + 3}$ from correct cancelling
- A1 for a complete, correct argument showing $\dfrac{x + 3}{x + 3} = 1$
Why this scores full marks: Simplifying the first fraction before adding means both fractions already share the denominator $x + 3$ . The script shows every factorisation and the combined fraction $\dfrac{x + 3}{x + 3}$ , so the printed result follows from the working rather than being asserted.
Question 9
Paper 1H/2H style — 5 marks5 marks
Solve $\dfrac{2x + 1}{x + 2} = \dfrac{x}{x - 1}$

Show clear algebraic working. Give your solutions correct to 3 significant figures. [5]
Model answer
$(2x + 1)(x - 1) = x(x + 2)$

$2x^{2} - x - 1 = x^{2} + 2x$

$x^{2} - 3x - 1 = 0$

$x = \dfrac{3 \pm \sqrt{(-3)^{2} - 4 \times 1 \times (-1)}}{2 \times 1} = \dfrac{3 \pm \sqrt{13}}{2}$

Answer: $x = 3.30$ or $x = -0.303$
Why this scores
Mark scheme:
- M1 for $(2x + 1)(x - 1) = x(x + 2)$ oe
- M1 for correct expansion of both sides, $2x^{2} - x - 1 = x^{2} + 2x$
- A1 for $x^{2} - 3x - 1 = 0$ oe
- M1 for correct substitution into the quadratic formula (allow one sign error)
- A1 for $3.30$ and $-0.303$ (awrt)
Why this scores full marks: Writing $-4 \times 1 \times (-1)$ in full keeps the discriminant at $9 + 4 = 13$ ; misquoting it as $9 - 4$ is a common loss. Both roots are given to 3 significant figures, including the trailing zero in $3.30$ , and neither is $-2$ or $1$ , so both are valid.
Question 10
Paper 1H/2H style — 6 marks6 marks
Pipe A can fill a tank on its own in $x$ hours. Pipe B can fill the same tank on its own in $(x + 5)$ hours. Working together, the two pipes fill the tank in $6$ hours.

(a) Show that $x^{2} - 7x - 30 = 0$ [3]

(b) Work out how long pipe B takes to fill the tank on its own. Show clear algebraic working. [3]
Model answer
(a) In one hour, A fills $\dfrac{1}{x}$ of the tank and B fills $\dfrac{1}{x + 5}$ of it.

$\dfrac{1}{x} + \dfrac{1}{x + 5} = \dfrac{1}{6}$

$6(x + 5) + 6x = x(x + 5)$

$12x + 30 = x^{2} + 5x$

$x^{2} - 7x - 30 = 0$

(b) $(x - 10)(x + 3) = 0$

$x = 10$ ( $x = -3$ is impossible for a time)

Pipe B takes $10 + 5 = 15$ hours.

Answer: $15$ hours
Why this scores
Mark scheme:
- M1 (a) for $\dfrac{1}{x} + \dfrac{1}{x + 5} = \dfrac{1}{6}$ oe
- M1 for multiplying through by $6x(x + 5)$ oe
- A1 for a complete, correct argument leading to $x^{2} - 7x - 30 = 0$
- M1 (b) for $(x - 10)(x + 3)$ or correct substitution into the quadratic formula
- A1 for $x = 10$ (ignore $x = -3$ )
- A1 for $15$ hours cao
Why this scores full marks: The script works with the fraction of the tank filled in one hour, which is what makes the rates add. It rejects the negative root with a reason and then answers the question set: $x = 10$ is pipe A's time, so pipe B takes $15$ hours. Check: $\dfrac{1}{10} + \dfrac{1}{15} = \dfrac{5}{30} = \dfrac{1}{6}$ .

4MA1 Algebraic Fractions

At a glance

What you’ll learn

Simplify by cancelling common factors

Add and subtract algebraic fractions

Multiply and divide algebraic fractions

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Simplify by factoring

2Add algebraic fractions

3Multiply algebraic fractions

4Divide algebraic fractions

5Cancelling a whole bracket [1 mark]

6Subtracting fractions with x in the denominator [2 marks]

7Solving with the unknown in the denominator [3 marks]

8Show that a difference of fractions cancels down [3 marks]

9An equation that becomes a factorisable quadratic [4 marks]

10Write as a single fraction, hence solve [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10