Algebraic Fractions
Algebraic fractions obey exactly the rules you learned for numerical ones. The difficulty is not the fractions, it is remembering that a numerator with two terms is a single quantity and must be bracketed.
Work out $\dfrac{5}{6} - \dfrac{1}{4}$ and write down every step. Now write $\dfrac{5}{6} - \dfrac{x-1}{4}$ and try the same steps. At which step does the second one need something the first did not, and what is it?
There are no new rules. To add or subtract, find a common denominator; to simplify, factorise and cancel common factors. What changes is that a numerator is now an expression, and an expression with more than one term must be kept in brackets or the sign will only reach its first term.
$$\frac{A}{m} \pm \frac{B}{n} = \frac{n(A) \pm m(B)}{mn}$$
The brackets in that formula are the whole lesson. When $B$ is something like $x - 1$ and the operation is a subtraction, $-m(B)$ means $-m(x-1) = -mx + m$, not $-mx - m$. Writing the brackets and removing them deliberately is the difference between a correct answer and the most common error in the topic.
Know
- That adding or subtracting algebraic fractions requires a common denominator
- That only common FACTORS cancel, never individual terms
- That a numerator of more than one term must be bracketed
Understand
- Why a subtraction sign must be distributed across every term of the numerator that follows it
- Why an expression must be factorised before anything can be cancelled
Can Do
- Add and subtract algebraic fractions with binomial numerators and numerical denominators
- Simplify algebraic fractions by factorising numerator and denominator
- Multiply and divide algebraic fractions, cancelling correctly
Find the lowest common denominator, rewrite both fractions over it, then combine the numerators.
$$\frac{x+3}{4} - \frac{x-1}{6}$$
The lowest common denominator of $4$ and $6$ is $12$. Multiply the first numerator by $3$ and the second by $2$, keeping each in brackets:
$$= \frac{3(x+3) - 2(x-1)}{12}$$
Now remove the brackets, and notice what the minus does to BOTH terms of the second one:
$$= \frac{3x + 9 - 2x + 2}{12} = \frac{x + 11}{12}$$
Cancelling means dividing the top and bottom by the same factor. A factor multiplies the WHOLE expression, so nothing can be cancelled until both parts are written as products.
$$\frac{x^2 - 9}{x^2 + 7x + 12}$$
Neither part is a product yet, so factorise both:
$$= \frac{(x+3)(x-3)}{(x+3)(x+4)}$$
Now $(x+3)$ is a factor of both, so it cancels:
$$= \frac{x-3}{x+4}$$
Compare that with the tempting but wrong move on the unfactorised form: cancelling the $x^2$ terms. Those are terms, not factors, and cancelling them is not a legal operation.
The reason is worth seeing once with numbers, because it settles the question permanently.
$$\frac{2 + 6}{2} = \frac{8}{2} = 4$$
Cancelling the twos would give $\dfrac{6}{1} = 6$, which is wrong. The $2$ on the bottom divides the WHOLE numerator $2 + 6$, not just the first term of it.
The same happens in algebra. In $\dfrac{x + 5}{x}$ nothing cancels, because $x$ is not a factor of $x + 5$. But in $\dfrac{x^2 + 5x}{x}$ it does, because the numerator factorises as $x(x+5)$ and now $x$ IS a factor:
$$\frac{x^2 + 5x}{x} = \frac{x(x+5)}{x} = x + 5$$
Multiplication is the easy case: multiply the numerators, multiply the denominators, and cancel any factor appearing top and bottom. Factorise first so the factors are visible.
$$\frac{x^2 - 4}{3x} \times \frac{6x}{x + 2} = \frac{(x+2)(x-2)}{3x} \times \frac{6x}{x+2}$$
The $(x+2)$ cancels, and $\dfrac{6x}{3x} = 2$:
$$= 2(x - 2)$$
Division is the same after one move: multiply by the reciprocal of the second fraction.
$$\frac{x + 1}{5} \div \frac{x + 1}{10} = \frac{x+1}{5} \times \frac{10}{x+1} = 2$$
Watch Me Solve It · 3 examples
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1Find the lowest common denominator$\text{LCD of } 3 \text{ and } 5 \text{ is } 15$
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2Rewrite both, keeping the brackets$\frac{5(2x+1) - 3(x-4)}{15}$The brackets are what protect the second numerator from the minus sign.
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3Expand carefully$\frac{10x + 5 - 3x + 12}{15}$$-3 \times (-4) = +12$, not $-12$.
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4Collect like terms$\frac{7x + 17}{15}$Nothing factorises, so this is the final form.
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1Factorise the numerator, common factor first$2x^2 - 8 = 2(x^2 - 4) = 2(x+2)(x-2)$The HCF of $2$ exposes a difference of two squares.
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2Factorise the denominator$x^2 + 5x + 6 = (x+2)(x+3)$Sum $5$, product $6$: the numbers are $2$ and $3$.
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3Write both as products$\frac{2(x+2)(x-2)}{(x+2)(x+3)}$Only now is anything cancellable.
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4Cancel the common factor$\frac{2(x-2)}{x+3}$$(x+2)$ divides both entirely. The $2$ does not cancel with anything.
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1Multiply by the reciprocal$\frac{x^2 - 1}{4x} \times \frac{8x^2}{x - 1}$Dividing by a fraction is multiplying by its reciprocal.
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2Factorise what can be factorised$\frac{(x+1)(x-1)}{4x} \times \frac{8x^2}{x-1}$$x^2 - 1$ is a difference of two squares.
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3Cancel common factors$(x-1) \text{ cancels}, \quad \frac{8x^2}{4x} = 2x$Both are genuine factors of the whole numerator and denominator.
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4Write the result$2x(x+1)$
Brain Trainer · 4 problems
Four quick problems. Work each one, then reveal the answer.
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1 Simplify $\dfrac{x+2}{3} + \dfrac{x+1}{4}$.
LCD $12$: $\dfrac{4(x+2) + 3(x+1)}{12}$.$\dfrac{7x + 11}{12}$ -
2 Simplify $\dfrac{3x}{5} - \dfrac{x - 2}{5}$.
Bracket first: $\dfrac{3x - (x-2)}{5}$.$\dfrac{2x + 2}{5}$ -
3 Simplify $\dfrac{x^2 - 16}{x + 4}$.
Factorise the numerator as a difference of two squares, then cancel.$x - 4$ -
4 Simplify $\dfrac{6x^2}{x + 5} \times \dfrac{x + 5}{3x}$.
$(x+5)$ cancels and $\dfrac{6x^2}{3x} = 2x$.$2x$
Multiple Choice · 5 questions
$\dfrac{5x}{7} - \dfrac{x - 3}{7}$ simplifies to:
$\dfrac{x + 6}{6}$ simplifies to:
$\dfrac{x^2 - 25}{x - 5}$ simplifies to:
$\dfrac{2}{x} + \dfrac{3}{5}$ equals:
$\dfrac{x + 4}{9} \div \dfrac{x + 4}{3}$ equals:
Short Answer · 3 questions
(a) $\dfrac{x + 5}{2} + \dfrac{x - 3}{4}$
(b) $\dfrac{3x - 1}{4} - \dfrac{x + 2}{6}$
(c) $\dfrac{2x}{3} - \dfrac{x - 6}{9} + 1$
(a) $\dfrac{x^2 + 6x + 8}{x^2 - 4}$
(b) $\dfrac{3x^2 - 27}{6x + 18}$
(c) $\dfrac{x^2 - 5x + 6}{2x - 6} \times \dfrac{4x}{x - 2}$
$\dfrac{4x}{7} - \dfrac{x - 5}{7} = \dfrac{4x - x - 5}{7} = \dfrac{3x - 5}{7}$
and separately
$\dfrac{x^2 + 4x}{x} = x^2 + 4$.
(a) Identify the error in the first line and give the correct answer.
(b) Identify the error in the second and give the correct answer.
(c) For the second, show the student a numerical test that exposes the error.
(b) Use your result to write $\dfrac{1}{n(n+1)}$ as a difference of two fractions.
(c) Hence evaluate $\dfrac{1}{1 \times 2} + \dfrac{1}{2 \times 3} + \dfrac{1}{3 \times 4} + \cdots + \dfrac{1}{9 \times 10}$ without adding nine fractions.
Add or subtract
Common denominator, then brackets on every numerator
The trap
A minus reaches every term of the numerator after it
Simplify
Factorise both parts, then cancel factors
Never
Cancel a term, or add denominators
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