Algebraic fractions are just fractions like regular fractions like \(\dfrac{3}{4}\), \(\dfrac{5}{7}\) etc. But they also have algebraic terms like \(3x\) or \((x-5)\). So they’re fractions with algebra instead of numbers.
Expressions like \(\dfrac{x+1}{x-3}\) or \(\dfrac{x^2+5x}{x^2+7x+10}\) are algebraic fractions. The rules for working with these fractions are the same as for fractions with numbers. They add, subtract, multiply, divide and cancel like regular fractions.
TLDR
- If there are two fractions being added or subtracted
- Find the LCM of the denominators
- Put both fractions over the new LCM denominator
- Add the numerators and simplify
- Cancel any factors present in the top and bottom
- If there are two fractions being multiplied or divided
- Factorise the numerator
This means turn it into expressions like \(2x(x-5)\) or \((x+3)(x+6)\)
The same thing you do when you are doing quadratic equations questions
- Factorise the denominator
- Flip the 2nd fraction if it’s a division
- Combine the fractions by multiplying the numerators together to form a new numerator
- Multiply the two denominators to form a new denominator
- Cancel any factors present in the new combined fraction’s top and bottom
So if there’s a \(2x+1\) on the top, and \(2x+1\) on the bottom, you can cross both of them out
- Factorise the numerator
- If it’s an equation with algebraic fractions that needs solving
- Find the LCM of the denominators
- Multiply each fraction and number in the equation by the LCM
- Expand and simplify
- Solve the equation
What You Need to Know (Taken From The Specification)
How to
- Simplify algebraic fractions by cancelling common factors
- Add and subtract algebraic fractions using a common denominator
- Multiply algebraic fractions (by multiplying the tops of each fractions, then the bottoms)
- Divide algebraic fractions (multiply the first fraction by the inverse of the 2nd fraction – i.e. the 2nd fraction flipped)
- Solve equations involving algebraic fractions
Summary Table – Algebraic Fractions Comparison
| Exam Board | Simplify | Add Subtract |
Multiply Divide |
Solve Equations | Identify Excluded Values | Quadratic Denominators | Tier |
|---|---|---|---|---|---|---|---|
| AQA | ✓ | ✓ | ✓ | ✓ | ✗ | ✓ | Higher |
| OCR | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ | Higher |
| Edexcel UK | ✓ | ✓ | ✓ | ✓ | ✗ | ✓ | Higher |
| Edexcel Int’l | ✓ | ✓ | ✓ | ✓ | ✗ | Possibly | ✗ |
| Cambridge IGCSE | ✓ | ✓ | ✓ | ✓ | ✗ | Some emphasis | ✗ |
Simplifying Algebraic Fractions
🔑 Key Fact
- To simplify an algebraic fraction, factorise the numerator and denominator.
This means turn it into expressions like \(2x(x-5)\) or \((x+3)(x+6)\).
The same thing you do when you are doing quadratic equations questions.
- Then cancel out any common factors.
So if there’s a \(2x+1\) on the top, and \(2x+1\) on the bottom, you can cross both of them out.
Example 1
Simplify: \(\dfrac{x^2 + 3x}{x + 3}\)
Step 1: Factorise the numerator.
\(x^2 + 3x = x(x + 3)\)
Step 2: Cancel out the common factor.
\(\dfrac{x^2 + 3x}{x + 3} = \dfrac{x(x + 3)}{x + 3} = x\)
Example 2
Simplify: \(\dfrac{x^2 – 4}{x – 2}\)
Step 1: Factorise the numerator. \(x^2 – 4 = (x + 2)(x – 2)\)
Step 2: Cancel out the common factor of \((x-2)\). It’s a common factor we can cancel out because it appears in the top half AND bottom half of the fraction.
\(\dfrac{x^2 – 4}{x – 2} = \dfrac{(x + 2)(x – 2)}{x – 2} = x + 2\)
Example 3
Simplify: \(\dfrac{3x^2 + 9x}{x^2 – 9}\)
Step 1: Factorise both the numerator (top) and denominator (bottom).
Top: \(3x^2 + 9x = 3x(x + 3)\)
Bottom: \(x^2 – 9 = (x + 3)(x – 3)\)
Step 2: Cancel the common factor \((x + 3)\) that’s in both the top and bottom.
\(\dfrac{3x^2 + 9x}{x^2 – 9} = \dfrac{3x(x + 3)}{(x + 3)(x – 3)} = \dfrac{3x}{x – 3}\)
Remember to check if your answer can be simplified further. In this case, it can’t because nothing cancels on the top and bottom.
When you see a practice question, you’ll learn much more if you attempt it before you look at the solution. Use the solution to check your answer, or if you get stuck.
Practice Question 1
Simplify: \(\dfrac{4x^2 – 16}{2x – 4}\)
Solution to Question 1
Step 1: Factorise the numerator and denominator.
Numerator: \(4x^2 – 16 = 4(x^2 – 4) = 4(x + 2)(x – 2)\)
Denominator: \(2x – 4 = 2(x – 2)\)
Step 2: Cancel the common factor \((x – 2)\).
\(\dfrac{4x^2 – 16}{2x – 4} = \dfrac{4(x + 2)(x – 2)}{2(x – 2)} = \dfrac{4(x + 2)}{2} = 2(x + 2) = 2x + 4\)
Therefore, the simplified expression is \(2x + 4\).
Multiplying Algebraic Fractions
When multiplying algebraic fractions, we:
- Multiply the numerators together
- Multiply the denominators. These two steps are just like what we would do when multiplying two normal fractions with just numbers.
- Then we simplify if possible.
🔑 Key Rule \(\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}\)
Example 4
Multiply: \(\dfrac{x + 3}{x – 4} \times \dfrac{x^2 – 16}{x^2 + 3x}\)
Step 1: Multiply the numerators and denominators.
\(\dfrac{x + 3}{x – 4} \times \dfrac{x^2 – 16}{x^2 + 3x} = \dfrac{(x + 3)(x^2 – 16)}{(x – 4)(x^2 + 3x)}\)
Step 2: Factorise anything we can from the top and bottom.
Top: \(x^2 – 16 = (x + 4)(x – 4)\)
Bottom: \(x^2 + 3x = x(x + 3)\)
Step 3: Put these back in, then cancel any common factors.
\(\dfrac{(x + 3)(x + 4)(x – 4)}{(x – 4)(x)(x + 3)} = \dfrac{x + 4}{x}\)
Simplifying further:
\(\dfrac{x + 4}{x} = \dfrac{x}{x}+\dfrac{4}{x}=1 + \dfrac{4}{x}\)
Practice Question 2
Multiply: \(\dfrac{x^2 – 9}{x + 1} \times \dfrac{x – 2}{x – 3}\)
Solution to Question 2
Step 1: Multiply the numerators and denominators.
\(\dfrac{x^2 – 9}{x + 1} \times \dfrac{x – 2}{x – 3} = \dfrac{(x^2 – 9)(x – 2)}{(x + 1)(x – 3)}\)
Step 2: Factorise where possible.
\(x^2 – 9 = (x + 3)(x – 3)\)
Step 3: Rewrite the expression with the factorised terms.
\(\dfrac{(x^2 – 9)(x – 2)}{(x + 1)(x – 3)} = \dfrac{(x + 3)(x – 3)(x – 2)}{(x + 1)(x – 3)}\)
Step 4: Cancel the common factor \((x – 3)\).
\(\dfrac{(x + 3)(x – 3)(x – 2)}{(x + 1)(x – 3)} = \dfrac{(x + 3)(x – 2)}{(x + 1)}\)
Therefore, the result is \(\dfrac{(x + 3)(x – 2)}{(x + 1)}\).
Dividing Algebraic Fractions
To divide by a fraction, we multiply by its reciprocal (flip the second fraction upside down).
🔑 Key Rule \(\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}\)
Example 5
Divide: \(\dfrac{3x + 6}{x – 4} \div \dfrac{2x^2 + 9x + 10}{x^2 – 4x}\)
Step 1: Change division to multiplication by the reciprocal. Just flip the fraction after the \(\div\) sign.
\(\dfrac{3x + 6}{x – 4} \div \dfrac{2x^2 + 9x + 10}{x^2 – 4x} = \dfrac{3x + 6}{x – 4} \times \dfrac{x^2 – 4x}{2x^2 + 9x + 10}\)
Step 2: Factorise.
Top, first fraction: \(3x + 6 = 3(x + 2)\)
Top, 2nd fraction: \(x^2 – 4x = x(x – 4)\)
Bottom, 2nd fraction: \(2x^2 + 9x + 10 = (2x + 5)(x + 2)\)
Step 3: Cancel out the common factors \((x+2)\) and \((x-4)\).
\(\dfrac{3(x + 2)}{x – 4} \times \dfrac{x(x – 4)}{(2x + 5)(x + 2)} = \dfrac{3 \times x}{2x + 5} = \dfrac{3x}{2x + 5}\)
Practice Question 3
Divide: \(\dfrac{2x – 6}{x^2 – 4} \div \dfrac{x – 3}{x^2 – 4x + 4}\)
Solution to Question 3
Step 1: Change division to multiplication by the reciprocal of the second fraction.
\(\dfrac{2x – 6}{x^2 – 4} \div \dfrac{x – 3}{x^2 – 4x + 4} = \dfrac{2x – 6}{x^2 – 4} \times \dfrac{x^2 – 4x + 4}{x – 3}\)
Step 2: Factorise all expressions.
\(2x – 6 = 2(x – 3)\)
\(x^2 – 4 = (x + 2)(x – 2)\)
\(x^2 – 4x + 4 = (x – 2)^2\)
Step 3: Rewrite with factorised terms.
\(\dfrac{2x – 6}{x^2 – 4} \times \dfrac{x^2 – 4x + 4}{x – 3} = \dfrac{2(x – 3)}{(x + 2)(x – 2)} \times \dfrac{(x – 2)^2}{x – 3}\)
Step 4: Cancel common factors \((x – 3)\) and \((x – 2)\).
\(\dfrac{2(x – 3)}{(x + 2)(x – 2)} \times \dfrac{(x – 2)^2}{x – 3} = \dfrac{2 \times (x – 2)}{(x + 2)}\)
Therefore, the result is \(\dfrac{2(x – 2)}{x + 2}\) or simplified as \(\dfrac{2x – 4}{x + 2}\).
Adding and Subtracting Algebraic Fractions
To add or subtract algebraic fractions, we need to find a common denominator, just like with numerical fractions.
🔑 Key Rule \(\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad + bc}{bd}\)
Example 6
Add: \(\dfrac{2x}{x + 3} + \dfrac{4}{x – 2}\)
Step 1: Find the common denominator by multiplying the denominators. If they share factors you don’t have to (see Example 8).
Here they don’t share any, so the common denominator is \((x + 3)(x – 2)\)
Step 2: Rewrite each fraction with the common denominator.
In the first fraction, \((x-2)\) is missing from the denominator, so we multiply BOTH the top and bottom by \((x-2)\).
\(\dfrac{2x}{x + 3} = \dfrac{2x(x – 2)}{(x + 3)(x – 2)}\)
In the second fraction, \((x+3)\) is missing from the denominator, so we multiply the top and bottom by \((x+3)\).
\(\dfrac{4}{x – 2} = \dfrac{4(x + 3)}{(x – 2)(x + 3)}\)
Step 3: Add them together. As the denominators are the same this is just adding the numerators.
\(\dfrac{2x(x – 2)}{(x + 3)(x – 2)} + \dfrac{4(x + 3)}{(x – 2)(x + 3)} = \dfrac{2x(x – 2) + 4(x + 3)}{(x + 3)(x – 2)}\)
Note that we didn’t need to worry about order of the factors in the denominator. As they are being multiplied it doesn’t matter.
Think about it. Is \(3\times4\) different to \(4\times3\) ?
No.
In the same way, \((x+3)(x-2)\) is the same as \((x-2)(x+3)\).
Step 4: Expand and simplify the numerator.
\(2x(x – 2) + 4(x + 3) = 2x^2 – 4x + 4x + 12 = 2x^2 + 12\)
Step 5: Write the final answer.
\(\dfrac{2x^2 + 12}{(x + 3)(x – 2)} = \dfrac{2(x^2 + 6)}{(x + 3)(x – 2)}\)
Example 7
Subtract: \(\dfrac{5}{x – 3} – \dfrac{2}{x + 4}\)
Step 1: Find the common denominator. Common denominator = \((x – 3)(x + 4)\)
Step 2: Rewrite each fraction with the common denominator.
In the first fraction, \((x+4)\) is missing from the denominator, so we multiply the top and bottom by \((x+4)\).
\(\dfrac{5}{x – 3} = \dfrac{5(x + 4)}{(x – 3)(x + 4)}\)
In the second fraction, \((x-3)\) is missing from the denominator, so we multiply the top and bottom by \((x-3)\).
\(\dfrac{2}{x + 4} = \dfrac{2(x – 3)}{(x + 4)(x – 3)}\)
Step 3: Subtract the numerators.
\(\dfrac{5(x + 4)}{(x – 3)(x + 4)} – \dfrac{2(x – 3)}{(x + 4)(x – 3)} = \dfrac{5(x + 4) – 2(x – 3)}{(x – 3)(x + 4)}\)
Step 4: Expand and simplify the numerator.
\(5(x + 4) – 2(x – 3) = 5x + 20 – 2x + 6 = 3x + 26\)
Step 5: Write the final answer.
\(\dfrac{3x + 26}{(x – 3)(x + 4)}\)
Example 8
Add: \(\dfrac{2}{x + 1} + \dfrac{3}{(x – 2)(x+1)}\)
Step 1: Find the common denominator. Watch out! Here there is a factor in both fractions, \((x+1)\)
We only need to use it once in the common denominator, not twice.
So the common denominator is \((x – 2)(x + 1)\)
Step 2: Rewrite each fraction with the common denominator.
The first fraction is missing an \((x-2)\) so multiply top and bottom by that.
\(\dfrac{2}{x + 1} = \dfrac{2(x – 2)}{(x+1)(x-2)}\)
The second fraction isn’t missing anything, so it stays as it is.
\(\dfrac{3}{(x – 2)(x+1)}\)
Step 3: Add the numerators.
\(\dfrac{2(x – 2)}{(x+1)(x-2)} + \dfrac{3}{(x – 2)(x+1)} = \dfrac{2(x-2) + 3}{(x + 1)(x – 2)}\)
Step 4: Expand and simplify the numerator.
\(2(x – 2) + 3 = 2x – 4 + 3 = 2x – 1\)
Step 5: Write the final answer.
\(\dfrac{2x – 1}{(x+1)(x-2)}\)
Practice Question 4
Add: \(\dfrac{3}{x^2 – 4} + \dfrac{2}{x – 2}\)
Solution to Question 4
Step 1: Factorise the denominator of the first fraction.
\(x^2 – 4 = (x + 2)(x – 2)\)
Step 2: Find the common denominator.
The common denominator is \((x + 2)(x – 2)\)
Step 3: Rewrite each fraction with the common denominator.
\(\dfrac{3}{x^2 – 4} = \dfrac{3}{(x + 2)(x – 2)}\)
For the second fraction, we need to multiply both numerator and denominator by \((x + 2)\):
\(\dfrac{2}{x – 2} = \dfrac{2(x + 2)}{(x – 2)(x + 2)} = \dfrac{2(x + 2)}{(x + 2)(x – 2)}\)
Step 4: Add the fractions by adding the numerators.
\(\dfrac{3}{(x + 2)(x – 2)} + \dfrac{2(x + 2)}{(x + 2)(x – 2)} = \dfrac{3 + 2(x + 2)}{(x + 2)(x – 2)}\)
Step 5: Simplify the numerator.
\(3 + 2(x + 2) = 3 + 2x + 4 = 2x + 7\)
Therefore, the result is \(\dfrac{2x + 7}{(x + 2)(x – 2)}\).
Solving Equations with Algebraic Fractions
When solving equations with algebraic fractions, we typically multiply all terms by the least common multiple (LCM) of the denominators to eliminate the fractions.
🔑 Key Step Find the LCM of all denominators, then multiply every term in the equation by this LCM.
Example 9
Solve: \(\dfrac{8}{x + 3} + \dfrac{3}{x + 8} = 1\)
Step 1: Find the LCM of the denominators. LCM = \((x + 3)(x + 8)\)
Step 2: Multiply all terms by the LCM. \((x + 3)(x + 8) \times \dfrac{8}{x + 3} + (x + 3)(x + 8) \times \dfrac{3}{x + 8} = (x + 3)(x + 8) \times 1\)
Step 3: Simplify.
\(8(x + 8) + 3(x + 3) = (x + 3)(x + 8)\)
\(8x + 64 + 3x + 9 = x^2 + 11x + 24\)
\(11x + 73 = x^2 + 11x + 24\)
\(0 = x^2 – 49\)
\(0 = (x + 7)(x – 7)\)
In the last step \(0=x^2 – 49\) is a difference of two squares, if that helps you understand how to factorise it into \(0 = (x + 7)(x – 7)\).
Step 4: Solve, giving us \(x= -7\) or \(x = 7\)
(OCR only) Step 5: Check for excluded values (values that make any denominator zero). \(x \neq -3\) and \(x \neq -8\)
Since neither of our solutions is excluded, the solutions are \(x = -7\) or \(x = 7\).
Practice Question 5
Solve: \(\dfrac{4}{x-2} – \dfrac{3}{x+1} = \dfrac{7}{(x-2)(x+1)}\)
Solution to Question 5
Step 1: Find the LCM of the denominators.
The LCM is \((x-2)(x+1)\)
Step 2: Multiply all terms by the LCM.
\((x-2)(x+1) \times \dfrac{4}{x-2} – (x-2)(x+1) \times \dfrac{3}{x+1} = (x-2)(x+1) \times \dfrac{7}{(x-2)(x+1)}\)
Step 3: Simplify.
\(4(x+1) – 3(x-2) = 7\)
\(4x + 4 – 3x + 6 = 7\)
\(x + 10 = 7\)
\(x = -3\)
(OCR only) Step 4: Check for excluded values.
\(x \neq 2\) and \(x \neq -1\)
So \(x = -3\) is not an excluded value
Step 5: Verify the answer.
\(x = -3\) is the solution to the equation.
Left side: \(\dfrac{4}{-3-2} – \dfrac{3}{-3+1} = \dfrac{4}{-5} – \dfrac{3}{-2} = -\dfrac{4}{5} + \dfrac{3}{2} = -\dfrac{8}{10} + \dfrac{15}{10} = \dfrac{7}{10}\)
Right side: \(\dfrac{7}{(-3-2)(-3+1)} = \dfrac{7}{(-5)(-2)} = \dfrac{7}{10}\)
Since \(\dfrac{7}{10} = \dfrac{7}{10}\), we confirm that \(x = -3\) is the solution.
Common Mistakes to Avoid
- Forgetting to factorise: Always try to factorise expressions before cancelling – you can only cancel factors, not terms.
- Cancelling terms instead of factors: You cannot cancel terms that are added or subtracted. For example:
- Correct: \(\dfrac{x(x+1)}{(x+1)} = x\)
- Correct: \(\dfrac{x+xy}{2x}=\dfrac{x(1+y)}{2\times x}=\dfrac{1+y}{2}\)
- Incorrect: \(\dfrac{x+y}{x} \neq \dfrac{1+y}{x}\)
- (OCR only) Not checking for excluded values: Always check which values of \(x\) make the denominator zero, as these are excluded from the solution.
- Forgetting to find a common denominator: When adding or subtracting fractions, you must have a common denominator.
Questions
Try these questions to practice your understanding of algebraic fractions:
- Simplify: \(\dfrac{x^2 – 9}{x^2 – 6x + 9}\)
- Multiply: \(\dfrac{x^2 – 4}{x^2 – 9} \times \dfrac{x^2 – 6x + 9}{x^2 – 16}\)
- Divide: \(\dfrac{x^2 + 2x – 8}{2x^2 + 5x – 3} \div \dfrac{x + 4}{2x – 1}\)
- Add: \(\dfrac{2x}{x^2 – 1} + \dfrac{3}{x – 1}\)
Solutions
Question 1
Simplify: \(\dfrac{x^2 – 9}{x^2 – 6x + 9}\)
Step 1: Factorise the numerator and denominator.
\(x^2 – 9 = (x + 3)(x – 3)\)
\(x^2 – 6x + 9 = (x – 3)^2 = (x – 3)(x – 3)\)
Step 2: Cancel the common factor \((x – 3)\).
\(\dfrac{x^2 – 9}{x^2 – 6x + 9} = \dfrac{(x + 3)(x – 3)}{(x – 3)(x – 3)} = \dfrac{x + 3}{x – 3}\)
Therefore, the simplified expression is \(\dfrac{x + 3}{x – 3}\).
Question 2
Multiply: \(\dfrac{x^2 – 4}{x^2 – 9} \times \dfrac{x^2 – 6x + 9}{x^2 – 16}\)
Step 1: Factorise all expressions.
\(x^2 – 4 = (x + 2)(x – 2)\)
\(x^2 – 9 = (x + 3)(x – 3)\)
\(x^2 – 6x + 9 = (x – 3)^2\)
\(x^2 – 16 = (x + 4)(x – 4)\)
Step 2: Multiply the fractions and cancel common factors.
\(\dfrac{x^2 – 4}{x^2 – 9} \times \dfrac{x^2 – 6x + 9}{x^2 – 16} = \dfrac{(x + 2)(x – 2)(x – 3)^2}{(x + 3)(x – 3)(x + 4)(x – 4)}\)
Step 3: Cancel the common factor \((x – 3)\).
\(\dfrac{(x + 2)(x – 2)(x – 3)^2}{(x + 3)(x – 3)(x + 4)(x – 4)} = \dfrac{(x + 2)(x – 2)(x – 3)}{(x + 3)(x + 4)(x – 4)}\)
Therefore, the result is \(\dfrac{(x + 2)(x – 2)(x – 3)}{(x + 3)(x + 4)(x – 4)}\).
Question 3
Divide: \(\dfrac{x^2 + 2x – 8}{2x^2 + 5x – 3} \div \dfrac{x + 4}{2x – 1}\)
Step 1: Change division to multiplication by the reciprocal.
\(\dfrac{x^2 + 2x – 8}{2x^2 + 5x – 3} \div \dfrac{x + 4}{2x – 1} = \dfrac{x^2 + 2x – 8}{2x^2 + 5x – 3} \times \dfrac{2x – 1}{x + 4}\)
Step 2: Factorise expressions where possible.
\(x^2 + 2x – 8 = (x + 4)(x – 2)\)
\(2x^2 + 5x – 3 = (2x – 1)(x + 3)\)
Step 3: Multiply the fractions and cancel common factors.
\(\dfrac{x^2 + 2x – 8}{2x^2 + 5x – 3} \times \dfrac{2x – 1}{x + 4} = \dfrac{(x + 4)(x – 2)(2x – 1)}{(2x – 1)(x + 3)(x + 4)}\)
Step 4: Cancel the common factors \((x + 4)\) and \((2x – 1)\).
\(\dfrac{(x + 4)(x – 2)(2x – 1)}{(2x – 1)(x + 3)(x + 4)} = \dfrac{x – 2}{x + 3}\)
Therefore, the result is \(\dfrac{x – 2}{x + 3}\).
Question 4
Add: \(\dfrac{2x}{x^2 – 1} + \dfrac{3}{x – 1}\)
Step 1: Factorise the denominator of the first fraction.
\(x^2 – 1 = (x + 1)(x – 1)\)
Step 2: Find the common denominator.
The common denominator is \((x + 1)(x – 1)\)
Step 3: Rewrite each fraction with the common denominator.
\(\dfrac{2x}{(x + 1)(x – 1)}\) stays the same.
For the second fraction:
\(\dfrac{3}{x – 1} = \dfrac{3(x + 1)}{(x – 1)(x + 1)} = \dfrac{3(x + 1)}{(x + 1)(x – 1)}\)
Step 4: Add the fractions.
\(\dfrac{2x}{(x + 1)(x – 1)} + \dfrac{3(x + 1)}{(x + 1)(x – 1)} = \dfrac{2x + 3(x + 1)}{(x + 1)(x – 1)}\)
Step 5: Simplify the numerator.
\(2x + 3(x + 1) = 2x + 3x + 3 = 5x + 3\)
Therefore, the result is \(\dfrac{5x + 3}{(x + 1)(x – 1)}\).
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