AL2 - Simplifying algebraic fractions

Algebraic fractions obey the same rules as numerical fractions, but their factors can be hidden inside brackets and quadratics. You will simplify, combine, multiply and divide them while keeping track of values that would make the original expression undefined. The aim is an equal expression that is easier to read and use.

Cancelling factors and keeping restrictions

Could you add $\frac1{1\times2}+\frac1{2\times3}+\cdots+\frac1{99\times100}$ exactly, without entering $99$ fractions into a calculator? A little algebra will make almost every term disappear. First we need to distinguish two very different kinds of disappearing act: cancelling factors is valid; crossing out terms is not.

You already know

For a non-zero number $k$ , $\frac{ka}{kb}=\frac ab$ , provided $b\ne0$ . GCSE factorisation reverses expansion, and a difference of squares gives $x^2-a^2=(x-a)(x+a)$ .

A fraction is undefined whenever its denominator is zero. For example, $\frac{x^2-9}{x-3}$ has no value at $x=3$ , even though it simplifies to $x+3$ everywhere else. Simplifying an expression does not add an input that the original expression excluded.

Algebraic fraction

An algebraic fraction is a quotient of algebraic expressions, such as $\frac{2x+1}{x-4}$ . Its denominator must be non-zero. A common factor multiplies the whole numerator and the whole denominator; only such non-zero factors can be cancelled.

For instance, $\frac{x(x+2)}{x}=x+2$ for $x\ne0$ , because the numerator is a product containing $x$ . By contrast, $\frac{x+2}{x}$ has no common factor $x$ : its numerator is a sum, so crossing out its $x$ would change the value. At $x=2$ , the fraction is $2$ , not $3$ .

Cancelling a common factor

\frac{A(x)C(x)}{B(x)C(x)}=\frac{A(x)}{B(x)}

This equality applies where $B(x)\ne0$ and $C(x)\ne0$ . Keep all exclusions from the original denominator, including those hidden by cancellation.
Not on the formulae sheet: learn it.

Factorise before deciding what can cancel. A bracket can be one complete factor, just like a number.

Worked example 1: a hidden common bracket

Question. Simplify $\frac{x^2+5x+6}{x^2-4}$ , stating the values excluded by the original denominator.

Plan. Factorise the top and bottom, record the restrictions, then cancel their common bracket.

\begin{aligned} x^2+5x+6&=(x+2)(x+3) && \text{factorise the numerator}\\ x^2-4&=(x-2)(x+2) && \text{difference of squares}\\ \frac{x^2+5x+6}{x^2-4}&=\frac{(x+2)(x+3)}{(x-2)(x+2)} && x\ne-2, 2\\ &=\frac{x+3}{x-2} && \text{cancel the non-zero factor }x+2 \end{aligned}

Why keep $x\ne-2$ ? Although the simplified fraction can be evaluated there, the original fraction cannot. They agree on the original permitted inputs.

Answer. $\frac{x+3}{x-2}$ , with $x\ne-2, 2$ .

Check. At $x=4$ , the original gives $\frac{42}{12}=\frac72$ , and the simplified fraction gives $\frac72$ .

The next example changes the order of the difference of squares in the denominator. Reversing a subtraction introduces a minus sign, which must survive the cancellation.

Worked example 2: a factor with the opposite sign

Question. Simplify $\frac{x^2+5x+6}{4-x^2}$ , stating the excluded values.

Plan. Write the denominator as the negative of $x^2-4$ , then cancel the matching factors.

\begin{aligned} \frac{x^2+5x+6}{4-x^2} &=\frac{(x+2)(x+3)}{-(x-2)(x+2)} && \text{factorise with the sign visible}\\ &=-\frac{x+3}{x-2} && \text{cancel }x+2\\ &=\frac{x+3}{2-x} && \text{move the minus sign} \end{aligned}

Why write the minus sign first? $4-x^2$ is the opposite of $x^2-4$ , so using the latter without its minus sign would reverse the answer's sign.

Answer. $-\frac{x+3}{x-2}$ , or $\frac{x+3}{2-x}$ , with $x\ne-2, 2$ .

Check. At $x=0$ , both forms give $\frac64=\frac32$ .

The method also handles cubic numerators: use the factor theorem and division from the previous lesson to expose their factors, then apply exactly the same cancellation rule. You do not need a new fraction rule for a higher power.

Watch out

Wrong: $\frac{x+6}{x+2}=\frac62=3$ , from cancelling the two $x$ terms. They are terms in sums, not factors of the whole top and bottom. At $x=2$ , the actual value is $\frac84=2$ , disproving that cancellation. Factorise first; if no factor is shared, nothing cancels.

Keep both the reversed subtraction and the original restrictions in the next simplification.

Adding and subtracting algebraic fractions

To combine two fractions, first express them in units of the same size: that means a common denominator.

Just as $\frac13+\frac14=\frac4{12}+\frac3{12}$ , an algebraic numerator must change when its denominator changes. For denominators $x-1$ and $x+2$ , their product is a convenient common denominator. Multiply each numerator by the bracket missing from its denominator. If denominators already share factors, use each needed factor only as often as necessary; for example, denominators $x$ and $x(x+1)$ need only $x(x+1)$ .

A common denominator

\frac ab+\frac cd=\frac{ad+bc}{bd} \qquad \frac ab-\frac cd=\frac{ad-bc}{bd}

Here $b\ne0$ and $d\ne0$ . The letters may represent algebraic expressions. Collect the numerator, then cancel any common factors if they exist.
Not on the formulae sheet: learn it.

Keep the numerator products in brackets until the operation between them has been applied.

Worked example 3: adding two fractions

Question. Simplify $\frac3{x-1}+\frac2{x+2}$ .

Plan. Use $(x-1)(x+2)$ as the common denominator and collect the numerator.

\begin{aligned} \frac3{x-1}+\frac2{x+2} &=\frac{3(x+2)+2(x-1)}{(x-1)(x+2)} && \text{scale both numerators}\\ &=\frac{3x+6+2x-2}{(x-1)(x+2)} && \text{expand the top}\\ &=\frac{5x+4}{(x-1)(x+2)} && \text{collect like terms} \end{aligned}

Why does $3$ multiply $x+2$ ? The first denominator $x-1$ needs that extra factor to become the common denominator; multiplying top and bottom by it preserves the fraction.

Answer. $\frac{5x+4}{(x-1)(x+2)}$ , for $x\ne1, -2$ .

Check. At $x=2$ , the original is $3+\frac12=\frac72$ ; the answer is $\frac{14}{4}=\frac72$ .

Now change only the operation to subtraction. The entire second numerator must be subtracted, including its negative constant term.

Worked example 4: subtracting a complete numerator

Question. Simplify $\frac3{x-1}-\frac2{x+2}$ .

Plan. Form the same common denominator and keep brackets round the subtracted numerator.

\begin{aligned} \frac3{x-1}-\frac2{x+2} &=\frac{3(x+2)-2(x-1)}{(x-1)(x+2)} && \text{common denominator}\\ &=\frac{3x+6-2x+2}{(x-1)(x+2)} && \text{subtract both terms}\\ &=\frac{x+8}{(x-1)(x+2)} && \text{collect the top} \end{aligned}

Why is the last constant $+2$ ? The term $-2(x-1)$ expands to $-2x+2$ , since the outer minus affects both terms.

Answer. $\frac{x+8}{(x-1)(x+2)}$ , for $x\ne1, -2$ .

Check. At $x=2$ , $3-\frac12=\frac52$ , matching $\frac{10}{4}$ .

Combining fractions also reveals the opening sum's hidden pattern:

\begin{aligned} \frac1n-\frac1{n+1} &=\frac{(n+1)-n}{n(n+1)} && \text{common denominator}\\ &=\frac1{n(n+1)} && \text{the numerator is }1 \end{aligned}

For $n=1, 2, \ldots, 99$ , adding these differences gives

\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1{99}-\frac1{100}\right)=1-\frac1{100}=\frac{99}{100}

Every middle fraction is added once and subtracted once. Ninety-nine fractions collapse to two end terms. This identity, valid for $n\ne0, -1$ , is not on the formulae sheet: derive it by combining fractions.

Watch out

Wrong: $3(x+2)-2(x-1)=3x+6-2x-2$ . The final sign is wrong. Right: $3x+6-2x+2=x+8$ . Writing the complete numerator before expanding makes that minus sign easier to control.

OCR's short simplification questions reward a collected numerator and a correct denominator. Do not leave an extra unsimplified sum in the numerator.

Multiplying and dividing algebraic fractions

Products of fractions can be simplified before you expand anything; division first becomes multiplication by a reciprocal.

The GCSE rules $\frac ab\times\frac cd=\frac{ac}{bd}$ and $\frac ab\div\frac cd=\frac ab\times\frac dc$ still apply. The original denominators must be non-zero; for division, the divisor fraction itself must also be non-zero, so its numerator cannot be zero. These rules are not on the formulae sheet: learn them.

Factor each polynomial and look across the whole product for common factors. This often avoids creating large expressions that would only need factorising again.

Worked example 5: cancelling within a product

Question. Simplify $\frac{x^2-9}{x^2+5x+6}\times\frac{x-1}{x+2}$ , stating the excluded values.

Plan. Factorise both quadratics and cancel common factors before multiplying.

\begin{aligned} \frac{x^2-9}{x^2+5x+6}\times\frac{x-1}{x+2} &=\frac{(x-3)(x+3)}{(x+2)(x+3)}\times\frac{x-1}{x+2} && \text{factorise}\\ &=\frac{(x-3)(x-1)}{(x+2)^2} && \text{cancel }x+3 \end{aligned}

Why leave the numerator in factors? It shows that neither factor matches $x+2$ , so there is no cancellation left. A factored answer is fully simplified.

Answer. $\frac{(x-3)(x-1)}{(x+2)^2}$ , with $x\ne-3, -2$ .

Check. At $x=4$ , the original is $\frac7{42}\times\frac36=\frac1{12}$ ; the simplified form gives $\frac{1\times3}{36}=\frac1{12}$ .

Now change only the multiplication sign to division. The second fraction must be inverted, and one additional input is forbidden because it would make the divisor zero.

Worked example 6: division and a forbidden zero

Question. Simplify $\frac{x^2-9}{x^2+5x+6}\div\frac{x-1}{x+2}$ , stating the excluded values.

Plan. Record the original denominator restrictions and the non-zero divisor condition, then multiply by the reciprocal.

\begin{aligned} \frac{x^2-9}{x^2+5x+6}\div\frac{x-1}{x+2} &=\frac{(x-3)(x+3)}{(x+2)(x+3)}\times\frac{x+2}{x-1} && \text{invert only the divisor}\\ &=\frac{x-3}{x-1} && \text{cancel the common factors} \end{aligned}

Why also exclude $x=1$ ? Both original denominators work there, but $\frac{x-1}{x+2}=0$ . Dividing by zero is undefined.

Answer. $\frac{x-3}{x-1}$ , with $x\ne-3, -2, 1$ .

Check. At $x=4$ , $\frac7{42}\div\frac36=\frac16\div\frac12=\frac13$ , matching $\frac{4-3}{4-1}$ .

The distinction between addition and multiplication now matters. Common denominators belong to sums and differences; common factors belong to cancellation in products. Choose the operation first, then simplify.

Watch out

Wrong: $\frac ab\div\frac cd=\frac ba\times\frac dc$ , inverting both fractions. Right: keep the first fraction and invert only the second. For example, $\frac23\div\frac45=\frac23\times\frac54=\frac56$ .

Use the division route once more, keeping every restriction from the original expression.

Quick Recap

Factorise before cancelling; cancel complete common factors, never terms inside sums.
Keep every excluded value from the original expression, even when its factor disappears.
Add or subtract using a common denominator, then expand and collect the numerator carefully.
Multiply fractions directly; divide by multiplying by the reciprocal of the second fraction. A divisor must be non-zero.

The opening sum is exactly $\frac{99}{100}$ , because $\frac1{n(n+1)}=\frac1n-\frac1{n+1}$ makes all its middle terms cancel.

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AL2 - Simplifying algebraic fractions

Cancelling factors and keeping restrictions

Adding and subtracting algebraic fractions

Multiplying and dividing algebraic fractions

Quick Recap

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