Adding and Subtracting Algebraic Fractions
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Theory
An algebraic fraction is a fraction with a pronumeral in it, such as \(\dfrac{x + 3}{4}\) or \(\dfrac{5}{2x}\). This Year 10 maths page shows how to add and subtract them using a lowest common denominator, including numerators with two terms, pronumerals in the denominator and a whole number in the expression, and how to leave the answer in simplest form.
An algebraic fraction has a pronumeral in the numerator, the denominator or both. The rules are the same as for number fractions.
To add or subtract, the fractions need the same denominator. Use the lowest common denominator (LCD): the smallest number that every denominator divides into. The LCD of \(4\) and \(6\) is \(12\), not \(24\).
When a numerator has two terms, such as \(x - 1\), treat it as one quantity: keep it in brackets while you multiply it.
When a pronumeral is in a denominator, the LCD must contain it. Use the lowest common multiple of the numbers and the highest power of each pronumeral: for \(2x\) and \(3x\) the LCD is \(6x\), and for \(5a\) and \(2a^2\) it is \(10a^2\).
A fraction is in simplest form when the numerator and the denominator have no common factor. You can only cancel a factor of the whole numerator and the whole denominator.
Adding or subtracting over a common denominator:
With \(bd\) replaced by the LCD when \(b\) and \(d\) share a factor.
How to add or subtract algebraic fractions
- Find the LCD of the denominators.
- Rename each fraction: multiply its numerator, in brackets, by the factor that turns its denominator into the LCD. Write a whole number as a fraction over the LCD.
- Combine over one denominator. Expand the brackets, taking care with a minus sign in front of a numerator.
- Collect like terms in the numerator.
- Simplify: if every term of the numerator shares a factor with the denominator, factorise and cancel.
Finding the LCD
- Numbers only: the lowest common multiple, such as \(18\) for \(6\) and \(9\).
- Pronumerals: include each pronumeral at its highest power, such as \(12x^2\) for \(4x\) and \(3x^2\).
- Three fractions: the LCD must work for all three denominators. A whole number such as \(1\) sits over the LCD too.
(i) \(\dfrac{2a}{3} + \dfrac{a}{4}\)
(ii) \(\dfrac{5m}{6} - \dfrac{2n}{9}\)
(i) The LCD of \(3\) and \(4\) is \(12\). Rename each fraction, then add the numerators.
| \(\dfrac{2a}{3} + \dfrac{a}{4}\) | \(=\) | \(\dfrac{8a}{12} + \dfrac{3a}{12}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{11a}{12}}\) |
(ii) The LCD of \(6\) and \(9\) is \(18\). \(m\) and \(n\) are not like terms, so the numerator stays as two terms.
| \(\dfrac{5m}{6} - \dfrac{2n}{9}\) | \(=\) | \(\dfrac{15m}{18} - \dfrac{4n}{18}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{15m - 4n}{18}}\) |
The LCD is \(12\). Multiply each numerator, in brackets, by the factor that makes its denominator \(12\).
| \(\) | \(\) | \(\dfrac{2x + 5}{3} - \dfrac{x - 2}{4}\) |
| \(\) | \(=\) | \(\dfrac{4(2x + 5) - 3(x - 2)}{12}\) |
| \(\) | \(=\) | \(\dfrac{8x + 20 - 3x + 6}{12}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{5x + 26}{12}}\) |
\(-3 \times (-2) = +6\). Writing \(-6\) here is the most common mistake.
(i) \(\dfrac{5}{2x} + \dfrac{1}{3x}\)
(ii) \(\dfrac{4}{5a} - \dfrac{1}{2a^2}\)
(i) The LCD of \(2x\) and \(3x\) is \(6x\): the LCM of \(2\) and \(3\), times \(x\).
| \(\dfrac{5}{2x} + \dfrac{1}{3x}\) | \(=\) | \(\dfrac{15}{6x} + \dfrac{2}{6x}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{17}{6x}}\) |
(ii) The LCD needs the LCM of \(5\) and \(2\), which is \(10\), and the highest power of \(a\), which is \(a^2\). So it is \(10a^2\).
| \(\dfrac{4}{5a} - \dfrac{1}{2a^2}\) | \(=\) | \(\dfrac{4 \times 2a}{10a^2} - \dfrac{1 \times 5}{10a^2}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{8a - 5}{10a^2}}\) |
\(8a\) and \(5\) are not like terms, so the numerator stays as two terms.
The LCD of \(2\) and \(5\) is \(10\). Write the whole number \(2\) as \(\dfrac{20}{10}\).
| \(\) | \(\) | \(\dfrac{x}{2} - \dfrac{x + 3}{5} + 2\) |
| \(\) | \(=\) | \(\dfrac{5x - 2(x + 3) + 20}{10}\) |
| \(\) | \(=\) | \(\dfrac{5x - 2x - 6 + 20}{10}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{3x + 14}{10}}\) |
Check with \(x = 2\): the question gives \(1 - 1 + 2 = 2\) and the answer gives \(\dfrac{20}{10} = 2\).
Common pitfalls
Frequently asked questions
How do you add algebraic fractions?
Find the lowest common denominator, rename each fraction so it has that denominator, then add the numerators and collect like terms. Keep each numerator in brackets while you rename it.
Why do you put brackets around the numerator?
Because the whole numerator is multiplied by the same number. For \(\dfrac{x - 2}{4}\) renamed over \(12\), the numerator becomes \(3(x - 2)\), and a minus sign in front changes both signs.
How do you find the LCD when there is a pronumeral in the denominator?
Take the lowest common multiple of the numbers and the highest power of each pronumeral. For \(6a\) and \(4a^2\) the LCD is \(12a^2\).
Can you cancel the x in (x + 3) over x?
No. The \(x\) on top is only one term of the numerator, not a factor of it. You can only cancel a factor of the whole numerator and the whole denominator.
What is the lowest common denominator?
The smallest number that every denominator divides into exactly. For \(6\) and \(9\) it is \(18\). Using a bigger common denominator still works, but you have more simplifying to do at the end.