Algebraic Fractions

1. What Are Algebraic Fractions?

An algebraic fraction is a fraction where the numerator, the denominator, or both contain algebraic expressions (expressions with variables).

Examples of algebraic fractions:

$$\frac{x}{3}, \quad \frac{2x+1}{x-4}, \quad \frac{x^2-9}{x^2+x-6}, \quad \frac{3}{x(x+2)}$$

They follow the same rules as ordinary numerical fractions. The key difference is that the variable can take many values — but it cannot take any value that makes the denominator equal to zero (more on this in Section 10).

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2. Simplification of Algebraic Fractions

Method: Factorise the numerator and denominator completely, then cancel any common factors.

Important: You can only cancel factors (things that are multiplied), never terms (things that are added or subtracted).

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Example 2.1 — Simple cancellation

Simplify $\dfrac{6x^2}{9x}$.

$$\frac{6x^2}{9x} = \frac{6 \times x \times x}{9 \times x} = \frac{6x}{9} = \frac{2x}{3}$$

Cancel the common factor of $x$, and simplify $\dfrac{6}{9} = \dfrac{2}{3}$.

$$\boxed{\dfrac{2x}{3}}$$

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Example 2.2 — Factorising a linear expression

Simplify $\dfrac{2x+6}{x^2+3x}$.

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

Factorise numerator: $2x + 6 = 2(x+3)$

Factorise denominator: $x^2 + 3x = x(x+3)$

Cancel the common factor $(x+3)$.

$$\boxed{\dfrac{2}{x}}$$

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Example 2.3 — Difference of two squares

Simplify $\dfrac{x^2-16}{x^2+5x+4}$.

Factorise numerator (difference of two squares): $x^2 - 16 = (x+4)(x-4)$

Factorise denominator: $x^2 + 5x + 4 = (x+1)(x+4)$

$$\frac{x^2-16}{x^2+5x+4} = \frac{(x+4)(x-4)}{(x+1)(x+4)} = \frac{x-4}{x+1}$$

$$\boxed{\dfrac{x-4}{x+1}}$$

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Example 2.4 — Factorising both numerator and denominator (quadratics)

Simplify $\dfrac{x^2-x-6}{x^2-4x+3}$.

Factorise numerator: $x^2 - x - 6 = (x-3)(x+2)$

Factorise denominator: $x^2 - 4x + 3 = (x-1)(x-3)$

$$\frac{x^2-x-6}{x^2-4x+3} = \frac{(x-3)(x+2)}{(x-1)(x-3)} = \frac{x+2}{x-1}$$

$$\boxed{\dfrac{x+2}{x-1}}$$

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3. Multiplication of Algebraic Fractions

Method:

1. Factorise all numerators and denominators.
2. Multiply numerators together and denominators together.
3. Cancel common factors and simplify.

$$\frac{A}{B} \times \frac{C}{D} = \frac{A \times C}{B \times D}$$

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Example 3.1

Simplify $\dfrac{3x}{4} \times \dfrac{8}{x^2}$.

$$\frac{3x}{4} \times \frac{8}{x^2} = \frac{3x \times 8}{4 \times x^2} = \frac{24x}{4x^2} = \frac{6}{x}$$

Cancel: $24 \div 4 = 6$, and $x \div x^2 = \dfrac{1}{x}$.

$$\boxed{\dfrac{6}{x}}$$

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Example 3.2

Simplify $\dfrac{x^2-9}{x+2} \times \dfrac{x^2+5x+6}{x-3}$.

Factorise each part:

• $x^2 - 9 = (x+3)(x-3)$
• $x^2 + 5x + 6 = (x+2)(x+3)$

$$\frac{(x+3)(x-3)}{x+2} \times \frac{(x+2)(x+3)}{x-3}$$

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

Cancel $(x-3)$ and $(x+2)$:

$$= (x+3)(x+3) = (x+3)^2$$

$$\boxed{(x+3)^2}$$

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Example 3.3

Simplify $\dfrac{2x^2+4x}{x^2-1} \times \dfrac{x+1}{6x}$.

Factorise each part:

• $2x^2 + 4x = 2x(x+2)$
• $x^2 - 1 = (x+1)(x-1)$

$$\frac{2x(x+2)}{(x+1)(x-1)} \times \frac{x+1}{6x}$$

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

Cancel $2x$, $(x+1)$ (note: $2x \div 6x = \dfrac{1}{3}$):

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

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

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4. Division of Algebraic Fractions

Method: Multiply the first fraction by the reciprocal of the second fraction, then simplify.

$$\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C} = \frac{A \times D}{B \times C}$$

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Example 4.1

Simplify $\dfrac{5x}{6} \div \dfrac{10x^2}{3}$.

Flip the second fraction and multiply:

$$\frac{5x}{6} \times \frac{3}{10x^2} = \frac{5x \times 3}{6 \times 10x^2} = \frac{15x}{60x^2} = \frac{1}{4x}$$

$$\boxed{\dfrac{1}{4x}}$$

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Example 4.2

Simplify $\dfrac{x^2-4}{x+3} \div \dfrac{x-2}{x^2+6x+9}$.

Flip and multiply:

$$\frac{x^2-4}{x+3} \times \frac{x^2+6x+9}{x-2}$$

Factorise:

• $x^2 - 4 = (x+2)(x-2)$
• $x^2 + 6x + 9 = (x+3)^2$

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

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

Cancel $(x-2)$ and one $(x+3)$:

$$= (x+2)(x+3)$$

$$\boxed{(x+2)(x+3)}$$

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Example 4.3

Simplify $\dfrac{2x^2+6x}{x^2-4} \div \dfrac{x+3}{x+2}$.

Flip and multiply:

$$\frac{2x^2+6x}{x^2-4} \times \frac{x+2}{x+3}$$

Factorise:

• $2x^2 + 6x = 2x(x+3)$
• $x^2 - 4 = (x+2)(x-2)$

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

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

Cancel $(x+3)$ and $(x+2)$:

$$= \frac{2x}{x-2}$$

$$\boxed{\dfrac{2x}{x-2}}$$

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5. Addition and Subtraction with the SAME Denominator

Method: When denominators are the same, simply add or subtract the numerators and keep the denominator.

$$\frac{A}{C} + \frac{B}{C} = \frac{A+B}{C}, \qquad \frac{A}{C} - \frac{B}{C} = \frac{A-B}{C}$$

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Example 5.1

Simplify $\dfrac{3x}{x+1} + \dfrac{5}{x+1}$.

Same denominator $(x+1)$, so add numerators:

$$\frac{3x+5}{x+1}$$

This cannot be simplified further.

$$\boxed{\dfrac{3x+5}{x+1}}$$

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Example 5.2

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

Same denominator $(x-2)$, so subtract numerators:

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

Watch out: Distribute the negative sign to BOTH terms in $(x+1)$.

$$\boxed{\dfrac{x+2}{x-2}}$$

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6. Addition and Subtraction with DIFFERENT Denominators

Method:

1. Factorise each denominator.
2. Find the Lowest Common Multiple (LCM) of the denominators.
3. Convert each fraction to an equivalent fraction with the LCM as denominator.
4. Add or subtract the numerators.
5. Simplify if possible.

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Example 6.1 — Simple different denominators

Simplify $\dfrac{3}{x} + \dfrac{2}{x+1}$.

LCM of $x$ and $(x+1)$ is $x(x+1)$.

$$\frac{3}{x} + \frac{2}{x+1} = \frac{3(x+1)}{x(x+1)} + \frac{2x}{x(x+1)}$$

$$= \frac{3(x+1) + 2x}{x(x+1)} = \frac{3x+3+2x}{x(x+1)} = \frac{5x+3}{x(x+1)}$$

$$\boxed{\dfrac{5x+3}{x(x+1)}}$$

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Example 6.2 — Subtraction with different denominators

Simplify $\dfrac{4}{x+2} - \dfrac{1}{x-3}$.

LCM is $(x+2)(x-3)$.

$$\frac{4(x-3)}{(x+2)(x-3)} - \frac{1(x+2)}{(x+2)(x-3)}$$

$$= \frac{4(x-3) - (x+2)}{(x+2)(x-3)}$$

$$= \frac{4x - 12 - x - 2}{(x+2)(x-3)}$$

$$= \frac{3x-14}{(x+2)(x-3)}$$

$$\boxed{\dfrac{3x-14}{(x+2)(x-3)}}$$

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Example 6.3 — One denominator is a factor of the other

Simplify $\dfrac{5}{x(x+3)} + \dfrac{2}{x}$.

The LCM of $x(x+3)$ and $x$ is $x(x+3)$, since $x$ is already a factor of $x(x+3)$.

$$\frac{5}{x(x+3)} + \frac{2(x+3)}{x(x+3)}$$

$$= \frac{5 + 2(x+3)}{x(x+3)}$$

$$= \frac{5 + 2x + 6}{x(x+3)}$$

$$= \frac{2x+11}{x(x+3)}$$

$$\boxed{\dfrac{2x+11}{x(x+3)}}$$

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Example 6.4 — Quadratic denominators

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

First factorise each denominator:

• $x^2 + 3x + 2 = (x+1)(x+2)$
• $x^2 - 1 = (x+1)(x-1)$

LCM is $(x+1)(x+2)(x-1)$.

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

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

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

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

$$= \frac{2x-5}{(x+1)(x+2)(x-1)}$$

$$\boxed{\dfrac{2x-5}{(x+1)(x+2)(x-1)}}$$

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7. Complex Fractions (Fraction Within a Fraction)

Not in Sec 2 2026: Complex fractions (fractions within fractions) are not explicitly listed in the 2026 Sec 2 syllabus. The notes below are kept for completeness and challenge practice.

A complex fraction has a fraction in the numerator, denominator, or both. Simplify by treating it as a division problem.

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Example 7.1

Simplify $\dfrac{\dfrac{x}{3}}{\dfrac{x^2}{6}}$.

Rewrite as division:

$$\frac{x}{3} \div \frac{x^2}{6} = \frac{x}{3} \times \frac{6}{x^2} = \frac{6x}{3x^2} = \frac{2}{x}$$

$$\boxed{\dfrac{2}{x}}$$

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Example 7.2

Simplify $\dfrac{\dfrac{1}{x} + \dfrac{1}{y}}{\dfrac{1}{xy}}$.

First simplify the numerator $\dfrac{1}{x} + \dfrac{1}{y}$:

$$\frac{1}{x} + \frac{1}{y} = \frac{y}{xy} + \frac{x}{xy} = \frac{x+y}{xy}$$

Now divide by $\dfrac{1}{xy}$:

$$\frac{x+y}{xy} \div \frac{1}{xy} = \frac{x+y}{xy} \times \frac{xy}{1} = x+y$$

$$\boxed{x + y}$$

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8. Solving Equations with Algebraic Fractions

Not in Sec 2 2026: Solving equations with algebraic fractions is not explicitly listed in the 2026 Sec 2 syllabus as a standalone topic. Where the cleared equation is linear, this is an extension of linear equation skills. Where the cleared equation is quadratic (as in Example 8.3 below), this goes beyond Sec 2 scope. These notes are kept for completeness; focus on linear outcomes for Sec 2 practice.

Method:

1. Find the LCM of all denominators.
2. Multiply every term on both sides of the equation by the LCM to clear all fractions.
3. Solve the resulting equation (usually linear or quadratic).
4. Check your answer does not make any denominator zero.

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Example 8.1 — Linear equation

Solve $\dfrac{x}{3} + \dfrac{x}{4} = 7$.

LCM of 3 and 4 is 12. Multiply every term by 12:

$$12 \times \frac{x}{3} + 12 \times \frac{x}{4} = 12 \times 7$$

$$4x + 3x = 84$$

$$7x = 84$$

$$x = 12$$

$$\boxed{x = 12}$$

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Example 8.2 — Algebraic denominators

Solve $\dfrac{3}{x} + \dfrac{1}{2} = \dfrac{5}{x}$.

LCM of $x$, $2$, and $x$ is $2x$. Multiply every term by $2x$:

$$2x \times \frac{3}{x} + 2x \times \frac{1}{2} = 2x \times \frac{5}{x}$$

$$6 + x = 10$$

$$x = 4$$

Check: $x = 4 \neq 0$, so the denominator is not zero. Answer is valid.

$$\boxed{x = 4}$$

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Example 8.3 — Results in a quadratic

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

LCM is $(x-2)(x+1)$. Multiply every term by $(x-2)(x+1)$:

$$x(x+1) - 3(x-2) = (x-2)(x+1)$$

Expand the left side:

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

$$x^2 - 2x + 6 = x^2 - x - 2$$

Subtract $x^2$ from both sides:

$$-2x + 6 = -x - 2$$

$$6 + 2 = -x + 2x$$

$$8 = x$$

$$x = 8$$

Check: $x = 8$ gives denominators $8-2 = 6 \neq 0$ and $8+1 = 9 \neq 0$. Answer is valid.

$$\boxed{x = 8}$$

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9. Common Errors to Avoid

Error 1: Cancelling Terms Instead of Factors

Wrong: [\frac{x+3}{x+5} \neq \frac{3}{5} \quad \text{(You CANNOT cancel the } x \text{ here!)}]

The $x$ in the numerator and the $x$ in the denominator are terms (connected by addition), not factors. You can only cancel when the entire numerator or denominator shares a common factor.

Correct: $\dfrac{x+3}{x+5}$ cannot be simplified further.

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Error 2: Forgetting to Factorise Before Simplifying

Wrong: [\frac{x^2-4}{x-2} \neq x^2-2 \quad \text{(Incorrect — you cannot just cancel like this)}]

Correct: Factorise first: [\frac{x^2-4}{x-2} = \frac{(x+2)(x-2)}{x-2} = x+2]

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Error 3: Sign Errors When Subtracting Fractions

Wrong: [\frac{2x+3}{x-1} - \frac{x+5}{x-1} = \frac{2x+3-x+5}{x-1} = \frac{x+8}{x-1} \quad \text{(Incorrect!)}]

The negative sign must be distributed to all terms in the second numerator.

Correct: [\frac{2x+3}{x-1} - \frac{x+5}{x-1} = \frac{(2x+3)-(x+5)}{x-1} = \frac{2x+3-x-5}{x-1} = \frac{x-2}{x-1}]

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Error 4: Multiplying Only One Side When Solving Equations

When you multiply by the LCM to clear fractions, you must multiply every term on both sides of the equation.

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10. Restrictions on Variables

The denominator of a fraction can never be zero (division by zero is undefined). You must always state the values of the variable that are not allowed.

How to find restrictions: Set each denominator equal to zero and solve.

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Example 10.1

For $\dfrac{3x}{x-5}$, the restriction is:

$$x - 5 \neq 0 \implies x \neq 5$$

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Example 10.2

For $\dfrac{x+1}{x^2-4}$, factorise the denominator first:

$$x^2 - 4 = (x+2)(x-2)$$

Restrictions: $(x+2)(x-2) \neq 0$, so $x \neq -2$ and $x \neq 2$.

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Example 10.3

For $\dfrac{2}{x^2+x-6}$, factorise the denominator:

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

Restrictions: $x \neq -3$ and $x \neq 2$.

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In examinations: Unless the question specifically asks for restrictions, you do not always need to state them — but always check that your solutions to equations do not violate the restrictions.

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11. Practice Questions

Try these on your own. Show all working clearly.

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Q1. Simplify $\dfrac{4x^3}{10x}$.

Q2. Simplify $\dfrac{x^2-25}{x^2+3x-10}$.

Q3. Simplify $\dfrac{x^2-2x-8}{x^2+x-2}$.

Q4. Find $\dfrac{3x^2}{x+4} \times \dfrac{x^2+5x+4}{9x}$.

Q5. Find $\dfrac{x^2-9}{x^2+2x} \div \dfrac{x-3}{x+2}$.

Q6. Simplify $\dfrac{5}{2x} + \dfrac{3}{4x}$.

Q7. Simplify $\dfrac{2}{x+3} - \dfrac{1}{x-1}$.

Q8. Simplify $\dfrac{3}{x^2-1} + \dfrac{2}{x^2+x}$.

Q9. Solve $\dfrac{2}{x} + \dfrac{1}{3} = 1$.

Q10. Solve $\dfrac{x+1}{x-1} - \dfrac{2}{x+2} = 2$.

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Answers to Practice Questions

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A1.

$$\frac{4x^3}{10x} = \frac{4x^2}{10} = \frac{2x^2}{5}$$

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A2.

Factorise: $x^2 - 25 = (x+5)(x-5)$, and $x^2 + 3x - 10 = (x+5)(x-2)$

$$\frac{(x+5)(x-5)}{(x+5)(x-2)} = \frac{x-5}{x-2}$$

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A3.

Factorise: $x^2 - 2x - 8 = (x-4)(x+2)$, and $x^2 + x - 2 = (x+2)(x-1)$

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

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A4.

Factorise: $x^2 + 5x + 4 = (x+4)(x+1)$

$$\frac{3x^2}{x+4} \times \frac{(x+4)(x+1)}{9x} = \frac{3x^2(x+4)(x+1)}{9x(x+4)} = \frac{3x(x+1)}{9} = \frac{x(x+1)}{3}$$

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A5.

Factorise: $x^2 - 9 = (x+3)(x-3)$, $x^2 + 2x = x(x+2)$

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

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A6.

LCM of $2x$ and $4x$ is $4x$.

$$\frac{10}{4x} + \frac{3}{4x} = \frac{13}{4x}$$

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A7.

LCM is $(x+3)(x-1)$.

$$\frac{2(x-1) - 1(x+3)}{(x+3)(x-1)} = \frac{2x-2-x-3}{(x+3)(x-1)} = \frac{x-5}{(x+3)(x-1)}$$

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A8.

Factorise: $x^2 - 1 = (x+1)(x-1)$, $x^2 + x = x(x+1)$

LCM is $x(x+1)(x-1)$.

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

$$= \frac{3x + 2x - 2}{x(x+1)(x-1)} = \frac{5x-2}{x(x+1)(x-1)}$$

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A9.

LCM is $3x$. Multiply through:

$$6 + x = 3x$$

$$6 = 2x$$

$$x = 3$$

Check: $x = 3 \neq 0$. Valid. $\boxed{x = 3}$

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A10.

LCM is $(x-1)(x+2)$. Multiply through:

$$(x+1)(x+2) - 2(x-1) = 2(x-1)(x+2)$$

Expand left side: $x^2 + 3x + 2 - 2x + 2 = x^2 + x + 4$

Expand right side: $2(x^2 + x - 2) = 2x^2 + 2x - 4$

$$x^2 + x + 4 = 2x^2 + 2x - 4$$

$$0 = x^2 + x - 8$$

Using the quadratic formula:

$$x = \frac{-1 \pm \sqrt{1 + 32}}{2} = \frac{-1 \pm \sqrt{33}}{2}$$

$$\boxed{x = \frac{-1 + \sqrt{33}}{2} \approx 2.37 \quad \text{or} \quad x = \frac{-1 - \sqrt{33}}{2} \approx -3.37}$$

Check restrictions: neither value equals $1$ or $-2$, so both solutions are valid.

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End of Algebraic Fractions Study Notes