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How do you simplify, add, subtract and solve algebraic fractions, and prove results algebraically?

Manipulate algebraic expressions and algebraic fractions: simplify, add, subtract, multiply and divide them, solve equations involving them, and construct simple algebraic proofs.

A CCEA GCSE Further Mathematics answer on algebraic manipulation and algebraic fractions, covering simplifying by factorising, the four operations on fractions, solving fractional equations, and writing algebraic proofs in the compulsory Pure unit.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. Simplifying algebraic fractions
  3. The four operations on fractions
  4. Solving equations with algebraic fractions
  5. Algebraic proof
  6. Why this matters

What this dot point is asking

Algebraic manipulation is the foundation of the whole Pure unit, and CCEA GCSE Further Mathematics pushes it well beyond ordinary GCSE. You must be fluent with algebraic fractions: simplifying them by factorising and cancelling, combining them with the four operations, solving equations that contain them, and using algebra to prove general results. Every later Pure topic, from quadratics to calculus, leans on these skills, so they have to be automatic and accurate.

Simplifying algebraic fractions

An algebraic fraction simplifies only through common factors, so the first move is always to factorise the numerator and denominator completely. Look for a common factor, a difference of two squares a2βˆ’b2=(aβˆ’b)(a+b)a^2 - b^2 = (a - b)(a + b), or a quadratic that factorises into two brackets. Once both parts are in factor form, cancel any factor that appears top and bottom.

For example, 6x2βˆ’63x+3=6(x2βˆ’1)3(x+1)=6(xβˆ’1)(x+1)3(x+1)=2(xβˆ’1)\dfrac{6x^2 - 6}{3x + 3} = \dfrac{6(x^2 - 1)}{3(x + 1)} = \dfrac{6(x - 1)(x + 1)}{3(x + 1)} = 2(x - 1). The danger is cancelling parts of a sum: in x+2x+4\dfrac{x + 2}{x + 4} nothing cancels, because xx is a term, not a factor.

The four operations on fractions

The rules mirror ordinary number fractions, but you must keep brackets intact.

  • Multiply: multiply numerators together and denominators together, then cancel. Cancelling before multiplying keeps the numbers small.
  • Divide: turn the second fraction upside down and multiply (multiply by the reciprocal).
  • Add or subtract: rewrite both fractions over a common denominator, usually the product of the denominators or their lowest common multiple, then combine the numerators, expanding carefully.

Solving equations with algebraic fractions

When an equation contains fractions, clear them in one step by multiplying every term by the lowest common denominator. This turns the equation into a linear or quadratic one you already know how to solve.

Algebraic proof

Proof questions ask you to show a statement is always true, not just to check examples. The technique is to write the general object algebraically, manipulate it, and reach a form that makes the conclusion obvious. An even number is 2n2n, an odd number is 2n+12n + 1, and consecutive integers are n,n+1,n+2n, n + 1, n + 2, where nn is any integer.

For example, to prove the sum of three consecutive integers is a multiple of 33, write them as n+(n+1)+(n+2)=3n+3=3(n+1)n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1). Because the result is 33 times an integer, it is always a multiple of 33. The key is that a single algebraic argument covers every case at once, which is exactly the reasoning CCEA rewards.

Why this matters

These manipulation skills are the language of the entire Pure unit. Algebraic fractions reappear when you differentiate, integrate, work with functions, and rearrange formulae; difference-of-two-squares and quadratic factorising drive the polynomial and calculus topics; and the discipline of arguing generally underlies every proof you will meet. Marks here are method marks elsewhere, so building speed and accuracy now pays off across the whole paper.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA Unit 1 (style)3 marksSimplify fully x2βˆ’9x2+7x+12\dfrac{x^2 - 9}{x^2 + 7x + 12}.
Show worked answer β†’

Factorise the numerator as a difference of two squares: x2βˆ’9=(xβˆ’3)(x+3)x^2 - 9 = (x - 3)(x + 3).

Factorise the denominator: two numbers multiplying to 1212 and adding to 77 are 33 and 44, so x2+7x+12=(x+3)(x+4)x^2 + 7x + 12 = (x + 3)(x + 4).

The fraction is (xβˆ’3)(x+3)(x+3)(x+4)\dfrac{(x - 3)(x + 3)}{(x + 3)(x + 4)}. Cancel the common factor (x+3)(x + 3) to get xβˆ’3x+4\dfrac{x - 3}{x + 4}.

Marks are for factorising each part and for cancelling. The classic error is to cancel terms that are not factors, for example cancelling the x2x^2 or the 99 separately.

CCEA Unit 1 (style)4 marksExpress 2xβˆ’1+3x+2\dfrac{2}{x - 1} + \dfrac{3}{x + 2} as a single fraction in its simplest form.
Show worked answer β†’

The common denominator is the product (xβˆ’1)(x+2)(x - 1)(x + 2).

Write each fraction over it: 2(x+2)(xβˆ’1)(x+2)+3(xβˆ’1)(xβˆ’1)(x+2)\dfrac{2(x + 2)}{(x - 1)(x + 2)} + \dfrac{3(x - 1)}{(x - 1)(x + 2)}.

Add the numerators: 2(x+2)+3(xβˆ’1)=2x+4+3xβˆ’3=5x+12(x + 2) + 3(x - 1) = 2x + 4 + 3x - 3 = 5x + 1.

So the answer is 5x+1(xβˆ’1)(x+2)\dfrac{5x + 1}{(x - 1)(x + 2)}. Marks are for the common denominator, for both numerator products, and for the simplified numerator. A sign slip in 3(xβˆ’1)3(x - 1) is the common loss.

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