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E2.3 Algebraic Fractions: Manipulating Rational Expressions

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An algebraic fraction behaves exactly like a numerical fraction — the extra step is that before you can simplify, add, or cancel anything, you almost always need to factorise first so that matching brackets become visible.

Why must I always factorise before I simplify?

A fraction can only be simplified by cancelling a factor that multiplies the entire numerator and the entire denominator. Written in expanded form, $\dfrac{x^2-2x}{x^2-5x+6}$ looks like it has no common factor at all. Only once both parts are factorised into brackets — $\dfrac{x(x-2)}{(x-2)(x-3)}$ — does the shared bracket $(x-2)$ become visible, and only then is it safe to cancel it.

How do I simplify an algebraic fraction?

Factorise the numerator and denominator completely, then cancel any bracket that appears in both.

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

Exam Tip

You can only cancel a factor that multiplies the whole of the top and bottom. In $\dfrac{x^2-2x}{x^2-5x+6}$ , the $x^2$ terms cannot be cancelled directly, because they aren't standalone factors — factorising first is what makes the common bracket $(x-2)$ visible.

What if the numerator or denominator is a difference of two squares?

Recognise the pattern $a^2-b^2=(a+b)(a-b)$ before you try any other method — it often creates the common bracket you need to cancel.

\frac{x^2-9}{x^2+5x+6} = \frac{(x+3)(x-3)}{(x+2)(x+3)} = \frac{x-3}{x+2}

How do I add or subtract two algebraic fractions?

Find the lowest common denominator (usually the product of the two denominators, unless they share a factor), rewrite each fraction over it, then combine the numerators into one.

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

How do I multiply or divide algebraic fractions?

For multiplication, factorise everything first, cancel any common bracket diagonally, then multiply what's left. For division, keep the first fraction as it is, flip the second fraction upside down, change $\div$ to $\times$ , and multiply as before.

\frac{3a}{4}\times\frac{9a}{10} = \frac{27a^2}{40} \qquad\qquad \frac{3a}{4}\div\frac{9a}{10} = \frac{3a}{4}\times\frac{10}{9a} = \frac{30a}{36a} = \frac{5}{6}

Worked Example 1

Express as a single fraction in its simplest form: $\dfrac{x+1}{x-2} - \dfrac{x^2-2x}{x^2-5x+6}$

Simplify the second fraction first (from the earlier example): $\dfrac{x^2-2x}{x^2-5x+6} = \dfrac{x}{x-3}$
Find the common denominator of $\dfrac{x+1}{x-2}$ and $\dfrac{x}{x-3}$ : it is $(x-2)(x-3)$ .
Rewrite both fractions over this denominator:

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

Expand both numerators: $(x+1)(x-3) = x^2-2x-3$ and $x(x-2)=x^2-2x$ .
Subtract the numerators, being careful with the minus sign in front of the whole second bracket:

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

Exam Tip

When one fraction is subtracted from another, the minus sign applies to every term in the second numerator, not just the first term. Writing the subtracted numerator in its own bracket, as in step 5, is the safest way to avoid a sign error.

Worked Example 2

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

Factorise the difference of two squares: $x^2-4 = (x+2)(x-2)$ , so the expression becomes $\dfrac{(x+2)(x-2)}{3x} \div \dfrac{x+2}{9x^2}$ .
Turn the division into multiplication by the reciprocal: $\dfrac{(x+2)(x-2)}{3x} \times \dfrac{9x^2}{x+2}$ .
Cancel the common bracket $(x+2)$ : $\dfrac{(x-2)}{3x} \times \dfrac{9x^2}{1}$ .
Cancel the numerical and $x$ factors ( $9x^2 \div 3x = 3x$ ): $(x-2)\times 3x = 3x(x-2)$ .

Exam Tip

Restrictions like $x\neq -2$ , $x\neq 2$ , and $x\neq 0$ exist here because those values would make an original denominator zero — IGCSE questions rarely ask you to state them explicitly, but recognising why they arise builds the same care needed to avoid dividing by zero elsewhere.