Algebraic Fractions

Algebraic fractions stop being scary the moment you treat them like ordinary fractions: factorise, cancel whole brackets, and use a common denominator. This page covers the full Edexcel IGCSE Maths skill set: simplifying by factorising, multiplying and dividing, and adding and subtracting with the lowest common denominator, each with worked examples. When you are ready, jump down to the auto-marked practice rooms and drill every skill with fresh randomly generated questions.

Prior Knowledge This page builds directly on Simplifying Algebraic Fractions, Factorising Quadratic Expressions and Simple Factorising. Factorising is the engine of this whole topic, so warm up there first if you need to.

⇩ Jump to Practice Questions ⇩

Factorise first

Before anything can cancel, write the top and bottom as products: common factors, quadratics, difference of two squares.

Cancel whole brackets

\(\dfrac{(x+2)(x-7)}{(x+2)(x+1)} = \dfrac{x-7}{x+1}\)

A bracket cancels only when the entire bracket appears top and bottom.

LCD to add or subtract

Rewrite both fractions over the lowest common denominator, then combine the numerators.

Flip to divide

\(\dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{A}{B} \times \dfrac{D}{C}\)

Keep the first fraction, flip the second, then multiply.

How to Simplify Algebraic Fractions

Algebraic fractions follow exactly the same rules as number fractions. A number fraction cancels when the top and bottom share a factor; an algebraic fraction is no different, except that the shared factor is usually a bracket. That gives a two-step method that never changes:

Factorise the numerator and the denominator completely. Look for a common factor first, then factorise any quadratic, and keep an eye out for the difference of two squares.
Cancel whole brackets that appear on both the top and the bottom, then write down what is left.

The single most important rule: you can only cancel a factor of the whole top with a factor of the whole bottom. A term hiding inside a bracket is not a factor, so it can never be crossed out on its own.

✅ Right: cancel the whole bracket

The bracket \((x+3)\) is a factor of the whole top and the whole bottom, so it cancels.

❌ Wrong: crossing out part of a bracket

\(x+3\) and \(x+5\) share no common factor. Crossing out the two \(x\) terms changes the value of the fraction.

Quick test of the rule

Try \(x = 1\) in \(\dfrac{x+3}{x+5}\): the fraction equals \(\dfrac{4}{6} = \dfrac{2}{3}\), but the wrong cancel above claims it equals \(\dfrac{3}{5}\). One test value is enough to prove a cancel was illegal.

Multiplying and Dividing Algebraic Fractions

To multiply, you could multiply the tops and bottoms straight away, but the working stays far cleaner if you factorise everything first and cancel before multiplying. Whatever survives the cancelling is your answer.

To divide, keep the first fraction, flip the second fraction and multiply, exactly as with number fractions. After the flip it is the same factorise-and-cancel routine.

🔄 Dividing: keep the first, flip the second

Only the second fraction changes: \(C\) and \(D\) trade places, and the divide becomes a multiply. Then factorise and cancel as usual.

The routine

\(\text{factorise} \; \Rightarrow \; \text{cancel whole brackets} \; \Rightarrow \; \text{multiply what is left}\)

Cancelling is allowed between any top and any bottom in a multiplication, not just within one fraction. That is why factorising first is so powerful here.

Adding and Subtracting: Use the Lowest Common Denominator

Adding and subtracting algebraic fractions works like number fractions: rewrite each fraction over the lowest common denominator (LCD), combine the numerators, then simplify.

With number denominators, the LCD is the lowest common multiple of the numbers.
With \(x\) in the denominators, the LCD of \(x+2\) and \(x-5\) is the product \((x+2)(x-5)\).
If a denominator is a quadratic, factorise it first: one of its brackets may already be the other denominator, and the real LCD is smaller than the full product.

🔗 Where the lowest common denominator comes from

Each fraction is multiplied by the other denominator over itself. That is multiplying by 1, so the value never changes, and both fractions end up over the same bottom line.

Keep the numerator in brackets until the very last step, especially when subtracting: the minus sign must reach every term of the second numerator. It is usually best to leave the denominator in its factorised form; the answer often cancels at the end.

Worked Examples

💡 Example 1: Simplify a quadratic over a quadratic

Simplify fully \(\dfrac{x^2+7x+12}{x^2+x-12}\)

\[ \begin{array}{rcl} \dfrac{x^2+7x+12}{x^2+x-12} &=& \dfrac{(x+3)(x+4)}{(x+4)(x-3)} \\[6pt] &=& \dfrac{x+3}{x-3} \end{array} \]

What's happening?

Factorise both quadratics, then cancel the whole bracket \((x+4)\). Nothing else is a shared factor, so the answer is fully simplified.

💡 Example 2: Multiply with factorising

Simplify \(\dfrac{x^2+5x}{x-4} \times \dfrac{2x-8}{x+5}\)

\[ \begin{array}{rcl} \dfrac{x^2+5x}{x-4} \times \dfrac{2x-8}{x+5} &=& \dfrac{x(x+5)}{x-4} \times \dfrac{2(x-4)}{x+5} \\[6pt] &=& 2x \end{array} \]

What's happening?

Factorise every top and bottom first. The brackets \((x+5)\) and \((x-4)\) each appear once on a top and once on a bottom, so both cancel, leaving \(x \times 2 = 2x\).

💡 Example 3: Divide (flip the second fraction)

Simplify \(\dfrac{x^2-16}{3x+6} \div \dfrac{x+4}{x+2}\)

\[ \begin{array}{rcl} \dfrac{x^2-16}{3x+6} \div \dfrac{x+4}{x+2} &=& \dfrac{(x-4)(x+4)}{3(x+2)} \times \dfrac{x+2}{x+4} \\[6pt] &=& \dfrac{x-4}{3} \end{array} \]

What's happening?

Flip the second fraction and multiply. \(x^2-16\) is a difference of two squares; after factorising, \((x+4)\) and \((x+2)\) both cancel.

💡 Example 4: Add with number denominators

Write as a single fraction \(\dfrac{2x+1}{4} + \dfrac{x-3}{6}\)

\[ \begin{array}{rcl} \dfrac{2x+1}{4} + \dfrac{x-3}{6} &=& \dfrac{3(2x+1) + 2(x-3)}{12} \\[6pt] &=& \dfrac{6x+3+2x-6}{12} \\[6pt] &=& \dfrac{8x-3}{12} \end{array} \]

What's happening?

The LCD of 4 and 6 is 12. Keep the numerators in brackets until the second line so no sign or term gets lost when expanding.

💡 Example 5: Subtract with x in the denominators

Write as a single fraction \(\dfrac{5}{x+3} - \dfrac{2}{x-1}\)

\[ \begin{array}{rcl} \dfrac{5}{x+3} - \dfrac{2}{x-1} &=& \dfrac{5(x-1) - 2(x+3)}{(x+3)(x-1)} \\[6pt] &=& \dfrac{5x-5-2x-6}{(x+3)(x-1)} \\[6pt] &=& \dfrac{3x-11}{(x+3)(x-1)} \end{array} \]

What's happening?

The LCD is the product \((x+3)(x-1)\). Watch the subtraction: \(-2(x+3)\) becomes \(-2x-6\), with the minus reaching both terms. Leave the denominator factorised.

💡 Example 6: Factorise the denominator to find the LCD

Write as a single fraction \(\dfrac{4}{x+1} - \dfrac{8}{x^2+4x+3}\)

\[ \begin{array}{rcl} \dfrac{4}{x+1} - \dfrac{8}{x^2+4x+3} &=& \dfrac{4}{x+1} - \dfrac{8}{(x+1)(x+3)} \\[6pt] &=& \dfrac{4(x+3) - 8}{(x+1)(x+3)} \\[6pt] &=& \dfrac{4x+4}{(x+1)(x+3)} \\[6pt] &=& \dfrac{4(x+1)}{(x+1)(x+3)} \\[6pt] &=& \dfrac{4}{x+3} \end{array} \]

What's happening?

Factorising \(x^2+4x+3\) shows it already contains \((x+1)\), so the LCD is just \((x+1)(x+3)\), not the full product of both denominators. Factorise the finished numerator too: the answer cancels right down.

🔑 Key Points

Factorise the top and bottom completely before you try to cancel anything.
Only whole brackets (whole factors) cancel, never individual terms inside them.
To multiply: factorise, cancel any top with any bottom, then multiply what is left.
To divide: keep the first fraction, flip the second, and multiply.
To add or subtract: rewrite over the lowest common denominator, then combine numerators.
Factorise quadratic denominators first; the LCD is often smaller than the full product.
Finish by checking your answer still simplifies; factorise the final numerator.

⚠️ Common Pitfalls

Cancelling terms instead of factors, such as crossing out the \(x\) in \(\dfrac{x+3}{x+5}\). Test a number to see why this fails.
Trying to cancel before factorising; nothing is a factor until the expression is a product.
Sign errors when subtracting numerators. Keep brackets until the end: the minus applies to every term.
Multiplying out the denominators too early. Leave them factorised so the final cancel is visible.
Cancelling across a \(+\) or \(-\) sign, for example crossing the \(x^2\) terms in \(\dfrac{x^2+1}{x^2+5}\).

⇩ Practice Questions ⇩

Comfortable simplifying, multiplying, dividing, adding and subtracting? The next step is to use these exact skills to solve equations that contain algebraic fractions.

Solving Equations with Algebraic Fractions →

Algebraic Fractions: Practice Rooms

These algebraic fractions practice rooms give you unlimited auto-marked questions for Edexcel IGCSE Maths (Higher). Room 1 simplifies by factorising first and cancelling whole brackets; Room 2 multiplies and divides (flip the second fraction to divide); Room 3 adds and subtracts with the lowest common denominator; Room 4 mixes every type, and Room 0 is an optional warm-up on ordinary fraction skills. Difficulty rises from Starter to Master left to right across each grid. Type answers with / for a fraction and ^ for a power; brackets are fine, so (x+1)/(x-2), 2x/(x+3) and x^2+3x all work, and your answer renders as real maths in the box. Give every answer fully simplified: a correct but uncancelled fraction is marked wrong. Factorised and expanded forms are both accepted when they are genuinely the same simplified answer.

✓ Correct 0

✗ Re-attempts 0

🔥 Streak 0

🏆 Best 0

Keep denominators factorised if you like: (3x+1)/((x+2)(x-5)) and the expanded form are both accepted. In Room 0, give number answers as fractions in simplest form (exact decimals are also accepted where they exist).