Adding and Subtracting Algebraic Fractions

How to add and subtract algebraic fractions with number, single-term or linear denominators: find the lowest common multiple of the denominators, write each fraction over it, and combine the numerators with care over the signs.

Before you start

Fractions can be added only when they have the same denominator. With numbers, the common denominator is the lowest common multiple (LCM) of the denominators:

$\dfrac{2}{3} + \dfrac{5}{8} = \dfrac{16}{24} + \dfrac{15}{24} = \dfrac{31}{24}\text{.}$ It is not $\dfrac{2 + 5}{3 + 8} = \dfrac{7}{11}\text{:}$ thirds and eighths are pieces of different sizes, and adding the denominators does not count them.

Fractions with the Same Denominator

Same denominator

$\dfrac{A}{C} + \dfrac{B}{C} = \dfrac{A + B}{C}$ and $\dfrac{A}{C} - \dfrac{B}{C} = \dfrac{A - B}{C}\text{,}$ where $C \neq 0$ .

When you subtract a numerator with more than one term, put it in brackets first. The minus sign applies to every term in it.

Example 1: Combining over One Denominator

Simplify (a) $\dfrac{3a}{2b} + \dfrac{a}{2b}$ (b) $\dfrac{2x + 1}{x - 4} - \dfrac{x - 3}{x - 4}$ (c) $\dfrac{2}{x + 1} + \dfrac{2x}{x + 1}$

Solution

(a) $\dfrac{3a + a}{2b} = \dfrac{4a}{2b} = \dfrac{2a}{b}\text{.}$

(b) $\dfrac{(2x + 1) - (x - 3)}{x - 4} = \dfrac{2x + 1 - x + 3}{x - 4} = \dfrac{x + 4}{x - 4}\text{.}$ The minus sign turns $-3$ into $+3$ .

(c) $\dfrac{2 + 2x}{x + 1} = \dfrac{2(1 + x)}{x + 1} = 2\text{,}$ for $x \neq -1$ . Always see whether the result can be simplified.

Check (b) with $x = 5$ : $\dfrac{11}{1} - \dfrac{2}{1} = 9$ , and $\dfrac{9}{1} = 9$ .

(a) $\dfrac{2a}{b}$ (b) $\dfrac{x + 4}{x - 4}$ (c) $2$

Finding a Common Denominator

When the denominators are different, write each fraction as an equivalent fraction (lesson 4.1) with a common denominator. The simplest one to use is the lowest common multiple (LCM) of the denominators: the simplest expression that each denominator divides into.

Adding or subtracting with different denominators

1. Factorise each denominator.

2. Find the LCM: take every factor that appears, each the greatest number of times it appears in any one denominator (its highest power), and multiply them.

3. Write each fraction over the LCM: multiply its numerator by what its denominator was multiplied by.

4. Add or subtract the numerators (with brackets), then simplify.

Example 2: Single-Term Denominators

Simplify $\dfrac{2}{3a} + \dfrac{3}{5a}$ .

Solution

The denominators are $3 \times a$ and $5 \times a$ . The LCM takes $3$ , $5$ and $a$ once each: $15a$ .

$3a \times 5 = 15a$ and $5a \times 3 = 15a$ , so multiply the numerators by $5$ and by $3$ :

$\dfrac{2 \times 5}{15a} + \dfrac{3 \times 3}{15a} = \dfrac{10 + 9}{15a} = \dfrac{19}{15a}\text{.}$

Check with $a = 1$ : $\dfrac{2}{3} + \dfrac{3}{5} = \dfrac{19}{15}\text{.}$

$\dfrac{19}{15a}$

Example 3: Different Letters, and Powers

Simplify (a) $\dfrac{3}{4a} - \dfrac{5}{6b}$ (b) $\dfrac{1}{x} + \dfrac{2}{x^2}$

Solution

(a) The LCM of $4$ and $6$ is $12$ , and both $a$ and $b$ are needed: the LCM is $12ab$ .

$4a \times 3b = 12ab$ and $6b \times 2a = 12ab$ : $\dfrac{3 \times 3b}{12ab} - \dfrac{5 \times 2a}{12ab} = \dfrac{9b - 10a}{12ab}\text{.}$

(b) $x^2 = x \times x$ contains $x$ , so the LCM is $x^2$ , not $x^3$ . $x \times x = x^2$ :

$\dfrac{x}{x^2} + \dfrac{2}{x^2} = \dfrac{x + 2}{x^2}\text{.}$

Check (a) with $a = 1$ , $b = 1$ : $\dfrac{3}{4} - \dfrac{5}{6} = \dfrac{9 - 10}{12} = -\dfrac{1}{12}\text{,}$ and $\dfrac{9 - 10}{12} = -\dfrac{1}{12}\text{.}$

(a) $\dfrac{9b - 10a}{12ab}$ (b) $\dfrac{x + 2}{x^2}$

Example 4: Number Denominators, Expressions on Top

Simplify (a) $\dfrac{x + 1}{3} - \dfrac{x - 2}{4}$ (b) $\dfrac{2x - 1}{5} + \dfrac{x + 3}{2}$

Solution

(a) The LCM of $3$ and $4$ is $12$ . $\dfrac{4(x + 1)}{12} - \dfrac{3(x - 2)}{12} = \dfrac{4x + 4 - 3x + 6}{12} = \dfrac{x + 10}{12}\text{.}$

Notice $-3 \times {(-2)} = +6$ : the bracket keeps the sign right.

(b) The LCM of $5$ and $2$ is $10$ . $\dfrac{2(2x - 1) + 5(x + 3)}{10} = \dfrac{4x - 2 + 5x + 15}{10} = \dfrac{9x + 13}{10}\text{.}$

Check (a) with $x = 2$ : $\dfrac{3}{3} - \dfrac{0}{4} = 1$ , and $\dfrac{12}{12} = 1$ .

(a) $\dfrac{x + 10}{12}$ (b) $\dfrac{9x + 13}{10}$

Denominators with a Common Factor

Factorise the denominators first. If they share a bracket, the LCM needs that bracket only once.

Example 5: A Shared Bracket

Simplify $\dfrac{3}{2b - 4c} - \dfrac{2}{3b - 6c}\text{.}$

Solution

Factorise: $2b - 4c = 2{(b - 2c)}$ and $3b - 6c = 3{(b - 2c)}$ . The LCM is $2 \times 3 \times {(b - 2c)} = 6{(b - 2c)}\text{.}$

Multiply the first fraction top and bottom by $3$ , the second by $2$ :

$\dfrac{9}{6(b - 2c)} - \dfrac{4}{6(b - 2c)} = \dfrac{5}{6(b - 2c)}\text{.}$

Check with $b = 3$ , $c = 1$ : $\dfrac{3}{2} - \dfrac{2}{3} = \dfrac{5}{6}$ , and $\dfrac{5}{6 \times 1} = \dfrac{5}{6}$ .

$\dfrac{5}{6(b - 2c)}$ , for $b \neq 2c$

Example 6: Denominators That Are Negatives of Each Other

Simplify $\dfrac{2h}{h - 2k} - \dfrac{3k}{2k - h}\text{.}$

Solution

$2k - h = -{(h - 2k)}$ , so $\dfrac{3k}{2k - h} = \dfrac{3k}{-(h - 2k)} = -\dfrac{3k}{h - 2k}\text{.}$

Subtracting a negative is adding: $\dfrac{2h}{h - 2k} + \dfrac{3k}{h - 2k} = \dfrac{2h + 3k}{h - 2k}\text{.}$

Check with $h = 3$ , $k = 1$ : $\dfrac{6}{1} - \dfrac{3}{-1} = 6 + 3 = 9\text{,}$ and $\dfrac{9}{1} = 9$ .

$\dfrac{2h + 3k}{h - 2k}$

Different Linear Denominators

Two different brackets such as ${(x - 2)}$ and ${(x - 3)}$ have no common factor, so the LCM is their product ${(x - 2)}{(x - 3)}$ . Multiply each numerator by the bracket its denominator is missing.

Leave the denominator factorised: it shows at once whether anything cancels and where the fraction is undefined.

Example 7: Adding over Two Brackets

Express $\dfrac{1}{x - 2} + \dfrac{2}{x - 3}$ as a single fraction in its simplest form.

Solution

LCM: ${(x - 2)}{(x - 3)}$ . The first fraction is missing ${(x - 3)}$ ; the second is missing ${(x - 2)}$ .

$\dfrac{1 \times (x - 3)}{(x - 2)(x - 3)} + \dfrac{2(x - 2)}{(x - 2)(x - 3)} = \dfrac{x - 3 + 2x - 4}{(x - 2)(x - 3)}$

$= \dfrac{3x - 7}{(x - 2)(x - 3)}$

The numerator $3x - 7$ has no factor in common with the denominator, so this is the simplest form. It is undefined when $x = 2$ or $x = 3$ .

Check with $x = 4$ : $\dfrac{1}{2} + \dfrac{2}{1} = \dfrac{5}{2}$ , and $\dfrac{12 - 7}{2 \times 1} = \dfrac{5}{2}\text{.}$

$\dfrac{3x - 7}{(x - 2)(x - 3)}$

Example 8: Subtracting over Two Brackets

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

Solution

LCM: ${(x + 1)}{(x - 4)}$ .

$\dfrac{3(x - 4) - 2(x + 1)}{(x + 1)(x - 4)} = \dfrac{3x - 12 - 2x - 2}{(x + 1)(x - 4)} = \dfrac{x - 14}{(x + 1)(x - 4)}\text{.}$

The minus sign multiplies the whole of $2{(x + 1)}$ : it gives $-2x - 2$ , not $-2x + 2$ .

Check with $x = 0$ : $\dfrac{3}{1} - \dfrac{2}{-4} = 3 + \dfrac{1}{2} = \dfrac{7}{2}\text{,}$ and $\dfrac{-14}{1 \times (-4)} = \dfrac{7}{2}\text{.}$

$\dfrac{x - 14}{(x + 1)(x - 4)}$

Example 9: Expressions on Top and Below

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

Solution

LCM: ${(x - 2)}{(x + 3)}$ . Numerator: ${(x + 1)}{(x + 3)} - {(x - 1)}{(x - 2)}\text{.}$

Expand each product first, then subtract the whole of the second: ${(x^2 + 4x + 3)} - {(x^2 - 3x + 2)} = x^2 + 4x + 3 - x^2 + 3x - 2 = 7x + 1\text{.}$

So the answer is $\dfrac{7x + 1}{(x - 2)(x + 3)}\text{.}$

Check with $x = 3$ : $\dfrac{4}{1} - \dfrac{2}{6} = \dfrac{11}{3}\text{,}$ and $\dfrac{22}{1 \times 6} = \dfrac{11}{3}$ .

$\dfrac{7x + 1}{(x - 2)(x + 3)}$

A Whole Number or Expression and a Fraction

A whole number or an expression is a fraction with denominator $1$ : $2 = \dfrac{2}{1}$ and $x = \dfrac{x}{1}$ . So the common denominator is the denominator of the fraction.

Example 10: Writing as One Fraction

Write as a single fraction (a) $2 + \dfrac{3}{x - 1}$ (b) $x - \dfrac{1}{x + 1}$

Solution

(a) $\dfrac{2(x - 1)}{x - 1} + \dfrac{3}{x - 1} = \dfrac{2x - 2 + 3}{x - 1} = \dfrac{2x + 1}{x - 1}\text{.}$

(b) $\dfrac{x(x + 1)}{x + 1} - \dfrac{1}{x + 1} = \dfrac{x^2 + x - 1}{x + 1}\text{.}$

Check (b) with $x = 1$ : $1 - \dfrac{1}{2} = \dfrac{1}{2}$ , and $\dfrac{1 + 1 - 1}{2} = \dfrac{1}{2}$ .

(a) $\dfrac{2x + 1}{x - 1}$ (b) $\dfrac{x^2 + x - 1}{x + 1}$

Where is the answer undefined?

A sum or difference has no value where any of its fractions is undefined. So $\dfrac{3}{x + 1} - \dfrac{2}{x - 4}$ is undefined when $x = -1$ or $x = 4$ : the factors of the LCM give these values.

Common Mistakes

Adding the numerators and adding the denominators: $\dfrac{2}{x} + \dfrac{3}{y} \neq \dfrac{5}{x + y}\text{.}$
Subtracting only the first term of the second numerator: $\dfrac{4x + 1}{x - 2} - \dfrac{x + 3}{x - 2}$ has numerator $4x + 1 - x - 3$ , not $4x + 1 - x + 3$ .
Changing the denominator but not the numerator: when $\dfrac{2}{x - 3}$ becomes a fraction over ${(x - 2)}{(x - 3)}$ , its numerator must become $2{(x - 2)}$ .
Taking more than needed for the LCM: for $\dfrac{1}{x}$ and $\dfrac{2}{x^2}$ it is $x^2$ , not $x^3$ .
Missing that $2k - h = -{(h - 2k)}$ : write it so, and the denominators become the same.

Practice

Write each answer as a single fraction in its simplest form. Check by putting a number in for each letter.

Question 1

Simplify (a) $\dfrac{5}{2x} - \dfrac{1}{2x}$ (b) $\dfrac{3x + 2}{x + 5} - \dfrac{x - 4}{x + 5}$ (c) $\dfrac{x}{x - 3} - \dfrac{3}{x - 3}$

ShowHide solution

(a) $\dfrac{4}{2x} = \dfrac{2}{x}$

(b) $\dfrac{3x + 2 - x + 4}{x + 5} = \dfrac{2x + 6}{x + 5} = \dfrac{2(x + 3)}{x + 5}$

(a) $\dfrac{2}{x}$ (b) $\dfrac{2(x + 3)}{x + 5}$ (c) $1$

Question 2

Simplify (a) $\dfrac{6}{5a} + \dfrac{3}{8a}$ (b) $\dfrac{5}{6x} - \dfrac{1}{4y}$ (c) $\dfrac{3}{a^2} + \dfrac{1}{2a}$

ShowHide solution

(a) $5$ and $8$ share no prime factor, so the LCM is $40a$ : $\dfrac{48}{40a} + \dfrac{15}{40a} = \dfrac{63}{40a}\text{.}$

(b) LCM $12xy$ : $\dfrac{10y}{12xy} - \dfrac{3x}{12xy} = \dfrac{10y - 3x}{12xy}\text{.}$

(a) $\dfrac{63}{40a}$ (b) $\dfrac{10y - 3x}{12xy}$ (c) $\dfrac{a + 6}{2a^2}$

Question 3

Simplify (a) $\dfrac{x + 2}{4} + \dfrac{x - 1}{6}$ (b) $\dfrac{2x + 1}{3} - \dfrac{x - 5}{2}$

ShowHide solution

(a) LCM $12$ : $\dfrac{3(x + 2) + 2(x - 1)}{12} = \dfrac{3x + 6 + 2x - 2}{12} = \dfrac{5x + 4}{12}\text{.}$

(b) LCM $6$ : $\dfrac{2(2x + 1) - 3(x - 5)}{6} = \dfrac{4x + 2 - 3x + 15}{6} = \dfrac{x + 17}{6}\text{.}$

(a) $\dfrac{5x + 4}{12}$ (b) $\dfrac{x + 17}{6}$

Question 4

Simplify (a) $\dfrac{4}{2b + 3c} + \dfrac{7}{6b + 9c}$ (b) $\dfrac{5}{3x - 6} - \dfrac{1}{x - 2}$

ShowHide solution

(a) $6b + 9c = 3{(2b + 3c)}\text{,}$ so the LCM is $3{(2b + 3c)}$ : $\dfrac{12}{3(2b + 3c)} + \dfrac{7}{3(2b + 3c)} = \dfrac{19}{3(2b + 3c)}\text{.}$

(b) $3x - 6 = 3{(x - 2)}$ : $\dfrac{5}{3(x - 2)} - \dfrac{3}{3(x - 2)} = \dfrac{2}{3(x - 2)}\text{.}$

(a) $\dfrac{19}{3(2b + 3c)}$ , or $\dfrac{19}{6b + 9c}$ (b) $\dfrac{2}{3(x - 2)}$

Question 5

Simplify (a) $\dfrac{3}{x - 5} + \dfrac{2}{5 - x}$ (b) $\dfrac{5x}{4x - 3y} - \dfrac{7y}{6y - 8x}$

ShowHide solution

(a) $\dfrac{2}{5 - x} = -\dfrac{2}{x - 5}\text{:}$ $\dfrac{3 - 2}{x - 5} = \dfrac{1}{x - 5}\text{.}$

(b) $6y - 8x = -2{(4x - 3y)}\text{,}$ so $-\dfrac{7y}{6y - 8x} = +\dfrac{7y}{2(4x - 3y)}\text{.}$

LCM $2{(4x - 3y)}$ : $\dfrac{10x}{2(4x - 3y)} + \dfrac{7y}{2(4x - 3y)} = \dfrac{10x + 7y}{2(4x - 3y)}\text{.}$

Check (b) with $x = 1$ , $y = 1$ : $\dfrac{5}{1} - \dfrac{7}{-2} = 5 + \dfrac{7}{2} = \dfrac{17}{2}\text{,}$ and $\dfrac{17}{2 \times 1} = \dfrac{17}{2}$ .

(a) $\dfrac{1}{x - 5}$ (b) $\dfrac{10x + 7y}{2(4x - 3y)}$

Question 6

Express each as a single fraction. (a) $\dfrac{1}{x + 1} + \dfrac{3}{x + 2}$ (b) $\dfrac{2}{x} + \dfrac{5}{x - 3}$ (c) $\dfrac{4}{x - 3} - \dfrac{1}{x + 2}$

ShowHide solution

(a) $\dfrac{(x + 2) + 3(x + 1)}{(x + 1)(x + 2)} = \dfrac{4x + 5}{(x + 1)(x + 2)}$

(b) $\dfrac{2(x - 3) + 5x}{x(x - 3)} = \dfrac{7x - 6}{x(x - 3)}$

(c) $\dfrac{4(x + 2) - (x - 3)}{(x - 3)(x + 2)} = \dfrac{4x + 8 - x + 3}{(x - 3)(x + 2)} = \dfrac{3x + 11}{(x - 3)(x + 2)}$

(a) $\dfrac{4x + 5}{(x + 1)(x + 2)}$ (b) $\dfrac{7x - 6}{x(x - 3)}$ (c) $\dfrac{3x + 11}{(x - 3)(x + 2)}$

Question 7

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

ShowHide solution

Numerator: ${(x + 2)}{(x + 1)} - {(x - 3)}{(x - 1)} = {(x^2 + 3x + 2)} - {(x^2 - 4x + 3)} = 7x - 1\text{.}$

So the answer is $\dfrac{7x - 1}{(x - 1)(x + 1)}\text{.}$

Check with $x = 2$ : $\dfrac{4}{1} - \dfrac{-1}{3} = \dfrac{13}{3}\text{,}$ and $\dfrac{13}{1 \times 3} = \dfrac{13}{3}$ .

$\dfrac{7x - 1}{(x - 1)(x + 1)}$

Question 8

Write as a single fraction (a) $3 - \dfrac{2}{x + 4}$ (b) $x + \dfrac{2}{x - 1}$

ShowHide solution

(a) $\dfrac{3(x + 4) - 2}{x + 4} = \dfrac{3x + 10}{x + 4}$

(b) $\dfrac{x(x - 1) + 2}{x - 1} = \dfrac{x^2 - x + 2}{x - 1}$

(a) $\dfrac{3x + 10}{x + 4}$ (b) $\dfrac{x^2 - x + 2}{x - 1}$

Question 9

Simplify $\dfrac{1}{x + 4} - \dfrac{2}{3x - 1}\text{,}$ and state the values of $x$ for which it is undefined.

ShowHide solution

$\dfrac{(3x - 1) - 2(x + 4)}{(x + 4)(3x - 1)} = \dfrac{3x - 1 - 2x - 8}{(x + 4)(3x - 1)} = \dfrac{x - 9}{(x + 4)(3x - 1)}\text{.}$

Undefined when $x + 4 = 0$ or $3x - 1 = 0$ : $x = -4$ or $x = \dfrac{1}{3}$ .

$\dfrac{x - 9}{(x + 4)(3x - 1)}\text{;}$ undefined when $x = -4$ or $x = \dfrac{1}{3}$

Question 10

Find and correct the error. (a) Daniel: $\dfrac{3}{a} + \dfrac{4}{b} = \dfrac{7}{a + b}$ (b) Priya: $\dfrac{5x + 2}{x + 3} - \dfrac{2x + 1}{x + 3} = \dfrac{3x + 3}{x + 3}$

ShowHide solution

(a) He added the tops and added the bottoms. Test $a = b = 1$ : $3 + 4 = 7$ , but his answer gives $\dfrac{7}{2}$ . Correct: the LCM is $ab$ , so $\dfrac{3b}{ab} + \dfrac{4a}{ab} = \dfrac{3b + 4a}{ab}\text{;}$ with $a = b = 1$ this is $7$ .

(b) She subtracted only the $2x$ : $-{(2x + 1)} = -2x - 1\text{,}$ not $-2x + 1$ . Correct: $\dfrac{5x + 2 - 2x - 1}{x + 3} = \dfrac{3x + 1}{x + 3}\text{.}$ Test $x = 1$ : $\dfrac{7}{4} - \dfrac{3}{4} = 1$ , and $\dfrac{4}{4} = 1$ ; her answer gives $\dfrac{6}{4}$ .

(a) $\dfrac{4a + 3b}{ab}$ (b) $\dfrac{3x + 1}{x + 3}$

Question 11

Wei Jie walks $12$ km to a reservoir at $x$ km/h and walks back at ${(x + 2)}$ km/h. (a) Write his total walking time as a single fraction in terms of $x$ . (b) Find the total time when $x = 4$ .

ShowHide solution

(a) Time $= \dfrac{\text{distance}}{\text{speed}}\text{:}$ $\dfrac{12}{x} + \dfrac{12}{x + 2} = \dfrac{12(x + 2) + 12x}{x(x + 2)} = \dfrac{24x + 24}{x(x + 2)} = \dfrac{24(x + 1)}{x(x + 2)}$ hours.

(b) $\dfrac{24 \times 5}{4 \times 6} = 5$ . Check: $12 \div 4 = 3$ hours there and $12 \div 6 = 2$ hours back.

(a) $\dfrac{24(x + 1)}{x(x + 2)}$ hours (b) $5$ hours

Question 12

Tap A fills a tank in $x$ hours and tap B fills it in ${(x + 3)}$ hours. (a) What fraction of the tank do the two taps fill together in one hour? Give a single fraction. (b) Find this fraction when $x = 6$ .

ShowHide solution

(a) In one hour, A fills $\dfrac{1}{x}$ of the tank and B fills $\dfrac{1}{x + 3}$ . Together: $\dfrac{(x + 3) + x}{x(x + 3)} = \dfrac{2x + 3}{x(x + 3)}\text{.}$

(b) $\dfrac{15}{6 \times 9} = \dfrac{15}{54} = \dfrac{5}{18}\text{.}$ Check: $\dfrac{1}{6} + \dfrac{1}{9} = \dfrac{3}{18} + \dfrac{2}{18} = \dfrac{5}{18}\text{.}$

(a) $\dfrac{2x + 3}{x(x + 3)}$ (b) $\dfrac{5}{18}$

Key Points

Fractions are added or subtracted over a common denominator: $\dfrac{A}{C} \pm \dfrac{B}{C} = \dfrac{A \pm B}{C}\text{.}$
Use the LCM of the denominators: factorise them, and take each factor the greatest number of times it appears in any one denominator.
Multiply each numerator by what its denominator was multiplied by.
Put brackets round a numerator that is subtracted: the minus sign changes every term.
Two different linear denominators have their product as the LCM; leave it factorised.
A whole number or expression has denominator $1$ .
Simplify the result if you can, and note where any denominator is zero.