A4 - Algebraic fractions (Higher) | AQA GCSE Mathematics

Algebraic fractions follow the fraction rules you already use with numbers. You will simplify them by factorising, multiply and divide them, and add and subtract them using common denominators. The useful habit is to see whole factors before doing any cancelling.

Cancelling common factors (Higher)

Can $\dfrac{x^2-9}{x-3}$ really be replaced by $x+3$ ? At $x=5$ , both give $8$ . At $x=3$ , one of them has a problem. By the end, you will know exactly when the replacement works.

You already know

Fractions simplify by dividing the numerator and denominator by the same non-zero factor. Factorising rewrites a sum as a product, for example $x^2-9=(x-3)(x+3)$ .

Start with $\frac{12}{18}=\frac{6\times2}{6\times3}=\frac23$ . Cancelling the factor $6$ means dividing the whole numerator and denominator by $6$ . It does not mean deleting matching digits. Letters follow the same rule.

Algebraic fraction

A fraction whose numerator or denominator contains an algebraic expression. For example, $\frac{3x}{x+2}$ is an algebraic fraction. Any value making its denominator zero is excluded.

The restriction belongs to the original expression. Cancelling may hide a zero denominator, but it cannot make that original value allowed.

Cancelling a common factor

\frac{ac}{bc}=\frac ab \qquad b\ne0,\quad c\ne0

The cancelled expression must be a factor of the whole numerator and the whole denominator.

Not on the formulae sheet: learn it.

Factorising first makes each whole factor visible. Both examples are non-calculator.

Worked example 1: a common linear factor

Question. Simplify $\dfrac{x^2+5x}{x+5}$ . State the excluded value.

Plan. Factorise the numerator, then divide out the common factor.

\begin{aligned} \frac{x^2+5x}{x+5}&=\frac{x(x+5)}{x+5}&&\text{take out }x\\ &=x&&\text{divide by }x+5 \end{aligned}

Why this step? The complete numerator contains the factor $x+5$ . This factor can be cancelled only when it is non-zero.

Answer. $x$ , with $x\ne-5$ .

Check. At $x=2$ , the original fraction is $14/7=2$ , agreeing with $x$ .

The next example changes the numerator to a quadratic with two linear factors. The decision is still to factorise before cancelling.

Worked example 2: factorising a quadratic

Question. Simplify $\dfrac{x^2+7x+10}{x+5}$ . State the excluded value.

Plan. Find two numbers with product $10$ and sum $7$ , then cancel the common factor.

\begin{aligned} x^2+7x+10&=(x+2)(x+5)&&2\times5=10,\quad2+5=7\\ \frac{x^2+7x+10}{x+5}&=\frac{(x+2)(x+5)}{x+5}&&\text{factorise first}\\ &=x+2&&\text{cancel the whole factor} \end{aligned}

Why this step? Cancelling acts on products. Expanding would hide the factor that makes this fraction simple.

Answer. $x+2$ , with $x\ne-5$ .

Check. At $x=1$ , $18/6=3=1+2$ .

A letter appearing above and below the line does not by itself give a common factor.

Watch out

Wrong: $\frac{x+6}{x}=6$ . The numerator is a sum, not $6x$ . Right: $\frac{x+6}{x}=1+\frac6x$ , for $x\ne0$ . At $x=2$ , the original gives $4$ , not $6$ .

Change the constant terms again, and keep the original restriction even after the denominator disappears.

Multiplying and dividing fractions (Higher)

Multiplication joins the numerators and joins the denominators. Division becomes multiplication by the reciprocal of the second fraction.

For numbers, $\frac23\div\frac45=\frac23\times\frac54=\frac56$ . The same reciprocal step works when the numerators and denominators contain letters.

Products and quotients

\frac ab\times\frac cd=\frac{ac}{bd} \qquad \frac ab\div\frac cd=\frac ab\times\frac dc

For multiplication, $b\ne0$ and $d\ne0$ . For division, also $c\ne0$ , because the divisor must not be zero.

Not on the formulae sheet: learn it.

Factorise before multiplying out. You may then cancel factors across the product. Both examples are non-calculator.

Worked example 3: multiplying factorised fractions

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

Plan. Factorise the difference of two squares, then cancel common factors in the product.

\begin{aligned} \frac{3x}{x+2}\times\frac{x^2-4}{6x} &=\frac{3x(x-2)(x+2)}{6x(x+2)}&&\text{factorise }x^2-4\\ &=\frac{x-2}{2}&&\text{cancel }3x(x+2) \end{aligned}

Why this step? The numerator and denominator are products, so cancellation can happen before any expansion.

Answer. $\dfrac{x-2}{2}$ , for $x\ne0,-2$ .

Check. At $x=4$ , the original is $2\times\frac12=1$ ; the answer gives $(4-2)/2=1$ .

Now change only the operation to division. The second fraction must be turned upside down before cancelling.

Worked example 4: dividing the same fractions

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

Plan. Multiply by the reciprocal, then factorise the denominator.

\begin{aligned} \frac{3x}{x+2}\div\frac{x^2-4}{6x} &=\frac{3x}{x+2}\times\frac{6x}{x^2-4}&&\text{use the reciprocal}\\ &=\frac{18x^2}{(x+2)(x-2)(x+2)}&&\text{factorise }x^2-4\\ &=\frac{18x^2}{(x-2)(x+2)^2}&&\text{collect equal factors} \end{aligned}

Why this step? Dividing asks how many copies of the second fraction fit into the first. Multiplying by its reciprocal performs that division.

Answer. $\dfrac{18x^2}{(x-2)(x+2)^2}$ , for $x\ne0,-2,2$ . The extra exclusion $2$ makes the divisor zero.

Check. At $x=4$ , $2\div\frac12=4$ , and $288/(2\times36)=4$ .

The restrictions need checking before and after the reciprocal step.

Watch out

Wrong: $\frac ab\div\frac cd=\frac ba\times\frac dc$ . Only the divisor is inverted. Right: keep $\frac ab$ and multiply it by $\frac dc$ .

The next division changes the numerical coefficient. Show the reciprocal step before simplifying.

Adding with common denominators (Higher)

You can add thirds to thirds. You cannot add thirds to fifths until both are written in the same sized pieces.

For example, $\frac13+\frac15=\frac5{15}+\frac3{15}=\frac8{15}$ . Algebra changes the numerators, but it does not change this reasoning.

Adding and subtracting fractions

\frac ab\pm\frac cd=\frac{ad\pm bc}{bd} \qquad b\ne0,\quad d\ne0

A smaller common denominator may be available. Change both numerator and denominator of each fraction by the same factor.

Not on the formulae sheet: learn it.

Start with denominators that are numbers. The letters remain in the numerators. Both examples are non-calculator.

Worked example 5: adding linear numerators

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

Plan. Use the common denominator $12$ , then expand and collect the numerator.

\begin{aligned} \frac{x+1}{3}+\frac{x-2}{4} &=\frac{4(x+1)+3(x-2)}{12}&&\text{make twelfths}\\ &=\frac{4x+4+3x-6}{12}&&\text{expand both brackets}\\ &=\frac{7x-2}{12}&&\text{collect like terms} \end{aligned}

Why this step? Each whole numerator must be multiplied. The brackets keep $x+1$ and $x-2$ together.

Answer. $\dfrac{7x-2}{12}$ .

Check. At $x=2$ , the original is $1+0=1$ ; the answer is $12/12=1$ .

Change only addition to subtraction. This makes the brackets around the second numerator essential.

Worked example 6: subtracting a whole numerator

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

Plan. Use twelfths, then distribute the subtraction across the second bracket.

\begin{aligned} \frac{x+1}{3}-\frac{x-2}{4} &=\frac{4(x+1)-3(x-2)}{12}&&\text{make twelfths}\\ &=\frac{4x+4-3x+6}{12}&&\text{subtract both terms}\\ &=\frac{x+10}{12}&&\text{collect like terms} \end{aligned}

Why this step? Subtracting $3x-6$ means adding its opposite, $-3x+6$ .

Answer. $\dfrac{x+10}{12}$ .

Check. At $x=5$ , the original is $2-\frac34=\frac54$ ; the answer is $15/12=\frac54$ .

A common denominator describes the size of the pieces. It stays the same when their numerators are added.

Watch out

Wrong: $\frac x3+\frac x4=\frac{2x}{7}$ . Adding denominators changes the pieces. Right: $\frac x3+\frac x4=\frac{4x+3x}{12}=\frac{7x}{12}$ .

Keep the same subtraction method with new constants in the numerators.

Working with algebraic denominators (Higher)

A denominator containing a letter still represents one whole expression. The factors you need for a common denominator are now algebraic ones.

For $\frac1x+\frac1{x+1}$ , a common denominator is $x(x+1)$ . Neither fraction already has both factors. Notice the neat consequence: at $x=1$ the sum is $3/2$ , at $x=2$ it is $5/6$ , and at $x=3$ it is $7/12$ . The odd numerators are no coincidence: the combined numerator is $2x+1$ .

Both examples below are non-calculator. Each new denominator needs only the factors it is missing.

Worked example 7: two different linear factors

Question. Write $\dfrac2x+\dfrac3{x+1}$ as a single fraction.

Plan. Use $x(x+1)$ as the common denominator.

\begin{aligned} \frac2x+\frac3{x+1} &=\frac{2(x+1)+3x}{x(x+1)}&&\text{supply missing factors}\\ &=\frac{2x+2+3x}{x(x+1)}&&\text{expand the numerator}\\ &=\frac{5x+2}{x(x+1)}&&\text{collect like terms} \end{aligned}

Why this step? Multiplying the first fraction by $(x+1)/(x+1)$ and the second by $x/x$ changes neither value.

Answer. $\dfrac{5x+2}{x(x+1)}$ , for $x\ne0,-1$ .

Check. At $x=2$ , $1+1=2$ , while $12/(2\times3)=2$ .

The next example changes the second denominator to a quadratic that factorises. This time one denominator already contains the other.

Worked example 8: recognising a shared denominator factor

Question. Simplify $\dfrac2{x-3}+\dfrac3{x^2-9}$ .

Plan. Factorise $x^2-9$ , then give the first fraction its missing factor.

\begin{aligned} \frac2{x-3}+\frac3{x^2-9} &=\frac{2(x+3)+3}{(x-3)(x+3)}&&\text{use }x^2-9=(x-3)(x+3)\\ &=\frac{2x+9}{(x-3)(x+3)}&&\text{collect the numerator} \end{aligned}

Why this step? Using $(x-3)(x+3)$ avoids an unnecessary second copy of $x-3$ . The second fraction already has the full common denominator.

Answer. $\dfrac{2x+9}{x^2-9}$ , for $x\ne3,-3$ .

Check. At $x=4$ , $2+3/7=17/7$ , matching $(8+9)/(16-9)$ .

The opening puzzle uses this same factorisation: $(x^2-9)/(x-3)=x+3$ for every allowed $x$ , but the original is undefined at $3$ . A simplified expression can look more welcoming than the original; keep its restriction with it.

Exam tip

Show the factorised denominator and the common-denominator line. They reveal a valid method even if a later sign error changes your answer. In a “Show that” question, the algebraic working is essential.

Now change the subtraction and the constants. Factorise before selecting a denominator.

Quick Recap

Factorise before cancelling; only whole common factors can cancel.
Multiply numerators and denominators. For division, invert only the second fraction.
Add and subtract using a common denominator, with brackets around whole numerators.
Preserve every restriction from the original denominators and any zero divisor.

The opening fraction equals $x+3$ whenever $x\ne3$ . At $x=3$ , its denominator is zero, so the fraction has no value.

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A4 - Algebraic fractions (Higher)

Cancelling common factors (Higher)

Multiplying and dividing fractions (Higher)

Adding with common denominators (Higher)

Working with algebraic denominators (Higher)

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