2.9 - Adding and subtracting algebraic fractions

Adding algebraic fractions means making their pieces the same size before combining them. You will choose common denominators, handle subtraction brackets and use factorisation to combine fractions efficiently, including expressions with quadratic denominators.

Making denominators match

Why does $\dfrac1x-\dfrac1{x+1}$ become $\dfrac1{x(x+1)}$ ? Subtracting two fractions leaves a numerator of just $1$ , however large $x$ is. By the end you will see exactly where that $1$ comes from.

Start with $\frac12+\frac13=\frac36+\frac26=\frac56$ . Halves and thirds cannot be counted together until they have both become sixths. Algebraic denominators obey the same rule.

You already know

Multiplying a fraction's numerator and denominator by the same non-zero expression leaves its value unchanged. Factorise before cancelling, and keep values that make an original denominator zero excluded.

Unlike multiplication, addition and subtraction require matching denominators. A common denominator is one shared by equivalent forms of the fractions. Once they match, leave the denominator alone.

Adding and subtracting fractions

\frac AC\pm\frac BC=\frac{A\pm B}{C}\qquad C\ne0

For different denominators, first rewrite both fractions with a common denominator. Multiply each entire numerator by the same factor used on its denominator.
Not on the formulae sheet: learn it.

A numeric common denominator lets you concentrate on the numerator algebra first.

Worked example 1: adding with numeric denominators

Question. Write $\dfrac{2x+1}{6}+\dfrac{x-2}{4}$ as a single fraction in its simplest form.

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

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

Why this step? The denominator $6$ was multiplied by $2$ , so its whole numerator must also be multiplied by $2$ .

Answer. $\dfrac{7x-4}{12}$ .

Check. At $x=2$ , the original is $5/6+0=5/6$ , and the answer is $10/12=5/6$ .

The next example changes only addition to subtraction. The denominator stays the same, but the minus sign applies to a whole bracket.

Worked example 2: subtracting a whole numerator

Question. Write $\dfrac{2x+1}{6}-\dfrac{x-2}{4}$ as a single fraction in its simplest form.

Plan. Use denominator $12$ again, keeping the subtracted numerator in brackets.

\begin{aligned} \frac{2x+1}{6}-\frac{x-2}{4}&=\frac{2(2x+1)-3(x-2)}{12} &&\text{keep subtraction brackets}\\ &=\frac{4x+2-3x+6}{12} &&\text{distribute the minus sign}\\ &=\frac{x+8}{12} &&\text{collect like terms} \end{aligned}

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

Answer. $\dfrac{x+8}{12}$ .

Check. At $x=0$ , $1/6-(-2/4)=2/3$ , matching $8/12=2/3$ .

Common denominators also handle a whole-number term: for example, $1=12/12$ , so adding $1$ to Worked example 2 gives $(x+20)/12$ . With an algebraic denominator, use exactly the same idea: $1=x/x$ when $x\ne0$ .

Watch out

Wrong: $\frac12+\frac13=\frac25$ . Adding denominators changes the size of the pieces. Right: $\frac12+\frac13=\frac36+\frac26=\frac56$ .

Try the subtraction method with new numerators and a common denominator of $24$ .

Combining different linear denominators

When neither denominator is a factor of the other, their product is a common denominator. Treat each bracket as one factor, and write the missing factor above its fraction.

For example, denominators $x+1$ and $x-2$ both divide $(x+1)(x-2)$ . The first fraction needs an extra factor of $x-2$ , and the second needs an extra factor of $x+1$ .

Worked example 3: two different linear factors

Question. Write $\dfrac3{x+1}+\dfrac2{x-2}$ as a single fraction in its simplest form.

Plan. Use denominator $(x+1)(x-2)$ , then collect the numerator.

\begin{aligned} \frac3{x+1}+\frac2{x-2}&=\frac{3(x-2)+2(x+1)}{(x+1)(x-2)} &&\text{supply missing factors}\\ &=\frac{3x-6+2x+2}{(x+1)(x-2)} &&\text{expand the numerator}\\ &=\frac{5x-4}{(x+1)(x-2)} &&\text{collect like terms} \end{aligned}

Why this step? Multiplying the first denominator by $x-2$ requires multiplying its numerator by $x-2$ as well.

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

Check. At $x=0$ , the original is $3-1=2$ , and the answer is $(-4)/(-2)=2$ .

Change only the sign between the fractions. The common denominator is unaffected; the numerator must change.

Worked example 4: subtracting with linear factors

Question. Write $\dfrac3{x+1}-\dfrac2{x-2}$ as a single fraction in its simplest form.

Plan. Use the same denominator and subtract the entire second 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{expand with signs}\\ &=\frac{x-8}{(x+1)(x-2)} &&\text{collect like terms} \end{aligned}

Why this step? The minus sign belongs to the fraction as a whole, so it changes both terms of $2(x+1)$ .

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

Check. At $x=0$ , the original is $3-(-1)=4$ , matching $(-8)/(-2)=4$ .

Here is a neat consequence: one unit fraction can split into two different ones. For a positive integer $n$ ,

\frac1{n+1}+\frac1{n(n+1)} =\frac{n+1}{n(n+1)} =\frac1n

Thus $\frac12=\frac13+\frac16$ , and splitting $\frac16$ again gives $\frac12=\frac13+\frac17+\frac1{42}$ . A small change in denominators can hide an unchanged total.

Watch out

Wrong: $3(x-2)-2(x+1)=3x-6-2x+2$ . The last sign is wrong. Right: $3x-6-2x-2=x-8$ .

The matched question changes the constants in the linear denominators, so choose the missing factors afresh.

Using shared factors in quadratic denominators

Factorise quadratic denominators before choosing a common denominator. Shared brackets need only appear once unless a higher power is actually required.

For denominators $(2x+3)(x-1)$ and $x-1$ , the quadratic is already a common denominator. Multiplying both denominators together would repeat $x-1$ , making more work for you.

Worked example 5: a denominator already contains the other

Question. Simplify fully $\dfrac{6x}{2x^2+x-3}-\dfrac3{x-1}$ .

Plan. Factorise the quadratic, then give the second fraction the missing factor $2x+3$ .

\begin{aligned} 2x^2+x-3&=(2x+3)(x-1) &&\text{factorise first}\\ \frac{6x}{2x^2+x-3}-\frac3{x-1}&=\frac{6x-3(2x+3)}{(2x+3)(x-1)} &&\text{common denominator}\\ &=\frac{6x-6x-9}{(2x+3)(x-1)} &&\text{expand every term}\\ &=\frac{-9}{(2x+3)(x-1)} &&\text{collect like terms} \end{aligned}

Why this step? The second denominator already contains $x-1$ ; only $2x+3$ is missing.

Answer. $\dfrac{-9}{2x^2+x-3}$ , for $x\ne-\frac32,1$ .

Check. At $x=0$ , the original is $0-(-3)=3$ , and the answer is $(-9)/(-3)=3$ .

The next example adds just one feature: a whole-number term $1$ . It needs the common denominator too.

Worked example 6: including a whole-number term

Question. Write $1+\dfrac{6x}{2x^2+x-3}-\dfrac3{x-1}$ as a single fraction in its simplest form.

Plan. Write $1$ with the quadratic denominator, then combine all three numerators.

\begin{aligned} 1+\frac{6x}{2x^2+x-3}-\frac3{x-1} &=\frac{(2x+3)(x-1)+6x-3(2x+3)}{(2x+3)(x-1)} &&\text{include all three terms}\\ &=\frac{2x^2+x-3+6x-6x-9}{(2x+3)(x-1)} &&\text{expand the numerator}\\ &=\frac{2x^2+x-12}{(2x+3)(x-1)} &&\text{collect like terms} \end{aligned}

Why this step? The numerator representing $1$ must equal the whole denominator, not just one of its factors.

Answer. $\dfrac{2x^2+x-12}{2x^2+x-3}$ , for $x\ne-\frac32,1$ .

Check. At $x=0$ , the original is $1+0+3=4$ , matching $(-12)/(-3)=4$ . The answer also equals $1$ plus Worked example 5.

The opening puzzle now uses the same process:

\frac1x-\frac1{x+1} =\frac{x+1-x}{x(x+1)} =\frac1{x(x+1)}\qquad x\ne0,-1

The numerator becomes $1$ because the two copies of $x$ cancel as terms after subtraction. No cancellation across a sum was needed.

Exam tip

If the question asks for a single fraction, finish with one fraction. An equivalent expression such as $1-\frac9{2x^2+x-3}$ has not yet answered that instruction.

The final question keeps the whole-number term and changes the quadratic's factors. Factorise before choosing the denominator.

Quick Recap

Match denominators before adding or subtracting; change the whole numerator with each denominator.
Keep brackets round a subtracted numerator until you distribute its minus sign.
Factorise quadratic denominators and use each required factor, avoiding unnecessary repetition.
A whole-number term also needs the common denominator. Finish in the form the question requests.

The opening difference has numerator $x+1-x=1$ , so it is $1/[x(x+1)]$ for $x\ne0,-1$ .

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2.9 - Adding and subtracting algebraic fractions

Making denominators match

Combining different linear denominators

Using shared factors in quadratic denominators

Quick Recap

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