Adding and Subtracting Algebraic Fractions
To add or subtract algebraic fractions, rewrite them over a common denominator and then combine the numerators. This NSW Year 9 Advanced maths subtopic starts with numerical denominators (NESA outcome MA5‑ALG‑C‑01) and moves on to pronumerals in the denominator.
You will simplify sums and differences such as \(\dfrac{7a}{12} - \dfrac{5a}{18}\) and \(\dfrac{2x}{3} + \dfrac{y}{2}\), handle binomial numerators like \(\dfrac{3x + 2}{5} - \dfrac{x - 4}{2}\), and use denominators such as \(5x\), \(4a^2\) and \(x + 1\). You will find and fix a student's sign error when subtracting a numerator, and combine three fractions in one expression. It is part of Year 9 Advanced Algebra: Expanding and Factorising.
Theory
To add or subtract algebraic fractions, rewrite every fraction over the lowest common denominator, then add or subtract the numerators. This Year 9 Advanced maths page covers numerical denominators, binomial numerators with a minus sign in front, and denominators that contain pronumerals or binomials.
Fractions can only be added or subtracted when they have the same denominator. The lowest common denominator (LCD) is the lowest common multiple of the denominators.
To change a fraction to the LCD, multiply its numerator and denominator by the same thing: \(\dfrac{3a}{4} = \dfrac{3a \times 5}{4 \times 5} = \dfrac{15a}{20}\). This does not change its value.
Once the denominators match, combine the numerators and keep the denominator. Then collect like terms. Unlike terms, such as \(7x\) and \(4\), stay separate.
A whole number or term is a fraction over \(1\), so \(3 = \dfrac{3}{1}\) can be brought onto the LCD like any other fraction.
With the same denominator:
With different denominators, cross-multiply over the product:
How to add or subtract algebraic fractions
- Find the LCD of all the denominators (whole terms sit over \(1\)).
- Rewrite each fraction over the LCD, multiplying the whole numerator: use brackets for binomials.
- Combine the numerators over the single denominator. Put a bracket around every numerator after a minus sign.
- Expand and collect like terms. Watch \(-2 \times (-4) = +8\).
- Simplify: if the numerator has a common factor, factorise it and cancel with the denominator. For example \(\dfrac{x}{3} - \dfrac{x - 2}{4} + \dfrac{x + 6}{12} = \dfrac{2x + 12}{12}\) \(= \dfrac{2(x + 6)}{12} = \dfrac{x + 6}{6}\).
- Check by substituting a value, such as \(x = 1\), into the question and your answer. They must agree. Finding someone else's error works the same way.
(i) \(\dfrac{5k}{6} - \dfrac{3k}{10}\)
(ii) \(\dfrac{n}{4} + \dfrac{2n}{5} - 2\)
Write every term over the lowest common denominator, then combine the numerators.
(i) The LCD of \(6\) and \(10\) is \(30\).
| \(\dfrac{5k}{6} - \dfrac{3k}{10}\) | \(=\) | \(\dfrac{25k}{30} - \dfrac{9k}{30}\) |
| \(\) | \(=\) | \(\dfrac{16k}{30}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{8k}{15}}\) |
(ii) The \(2\) is \(\dfrac{2}{1}\). The LCD of \(4\), \(5\) and \(1\) is \(20\).
| \(\dfrac{n}{4} + \dfrac{2n}{5} - 2\) | \(=\) | \(\dfrac{5n}{20} + \dfrac{8n}{20} - \dfrac{40}{20}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{13n - 40}{20}}\) |
\(13n\) and \(40\) are not like terms, so the numerator stops there.
The LCD is \(12\). Multiply each whole numerator, so keep the brackets, and remember the minus applies to all of \(x - 3\).
| \(\) | \(\) | \(\dfrac{2(2x + 5)}{12} - \dfrac{3(x - 3)}{12}\) |
| \(\) | \(=\) | \(\dfrac{2(2x + 5) - 3(x - 3)}{12}\) |
| \(\) | \(=\) | \(\dfrac{4x + 10 - 3x + 9}{12}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{x + 19}{12}}\) |
\(-3 \times (-3) = +9\). Check with \(x = 1\): \(\dfrac{7}{6} - \dfrac{-2}{4} = \dfrac{14}{12} + \dfrac{6}{12} = \dfrac{20}{12}\), and \(\dfrac{1 + 19}{12} = \dfrac{20}{12}\).
The LCD uses the LCM of the numbers (\(12\)) and the highest power of each pronumeral (\(p^2\)), so it is \(12p^2\).
| \(\dfrac{5}{6p}\) | \(=\) | \(\dfrac{5 \times 2p}{6p \times 2p} = \dfrac{10p}{12p^2}\) |
| \(\dfrac{3}{4p^2}\) | \(=\) | \(\dfrac{3 \times 3}{4p^2 \times 3} = \dfrac{9}{12p^2}\) |
| \(\dfrac{5}{6p} - \dfrac{3}{4p^2}\) | \(=\) | \(\boldsymbol{\dfrac{10p - 9}{12p^2}}\) |
Using \(24p^3\) (the product of the denominators) also works, but then the answer must be cancelled back down.
The denominators have no common factor, so the common denominator is their product \((x + 3)(x - 2)\).
| \(\) | \(\) | \(\dfrac{4(x - 2)}{(x + 3)(x - 2)} - \dfrac{1(x + 3)}{(x + 3)(x - 2)}\) |
| \(\) | \(=\) | \(\dfrac{4x - 8 - x - 3}{(x + 3)(x - 2)}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{3x - 11}{(x + 3)(x - 2)}}\) |
Leave the denominator factorised. Nothing cancels, because \(3x - 11\) has no factor \((x + 3)\) or \((x - 2)\).
Common pitfalls
Frequently asked questions
How do you add algebraic fractions?
Find the lowest common denominator, rewrite each fraction over it by multiplying top and bottom by the same number or term, then add the numerators and keep the denominator. Collect like terms and simplify.
How do you subtract algebraic fractions with brackets?
Put the second numerator in brackets after the minus sign, then expand. The minus changes the sign of every term in the bracket, so minus (x minus 5) becomes minus x plus 5.
What is the lowest common denominator of 2x and 3x squared?
It is \(6x^2\). Take the lowest common multiple of the numbers, \(6\), and the highest power of \(x\), which is \(x^2\).
How do you add fractions with x in the denominator?
Use the denominators as factors of the common denominator. For \(\dfrac{1}{x} + \dfrac{1}{x + 2}\), the common denominator is \(x(x + 2)\). Multiply each numerator by the factor its denominator is missing, then add.
Can you add 2 over x and 3 over y?
Yes. Use the common denominator \(xy\): \(\dfrac{2}{x} = \dfrac{2y}{xy}\) and \(\dfrac{3}{y} = \dfrac{3x}{xy}\), so the sum is \(\dfrac{2y + 3x}{xy}\).