Solving Equations with Fractions
Equations with fractions look harder than they are: the trick is to clear the denominators first by multiplying every term by the lowest common denominator, which turns the problem into an ordinary equation. This Edexcel IGCSE Maths guide covers both kinds you are examined on, fractions with numbers in the denominator and fractions with x in the denominator (including the case that becomes a quadratic), with worked examples and free auto-marked practice below.
How to solve an equation with fractions
The key move is to get rid of the fractions before you solve. Multiply every term on both sides by the lowest common denominator (LCD), and the fractions clear, leaving an ordinary equation.
- Find the LCD of all the denominators on both sides.
- Multiply every term by the LCD to clear the fractions.
- Expand any brackets, then collect like terms.
- Solve the resulting equation.
- If x was in a denominator, check your answer does not make any denominator zero.
Denominators \(2\) and \(4\): LCD \(=4\). Denominators \(3,5,6\): LCD \(=30\). With x in the bottom, multiply by \(x\) (or \(2x\), and so on).
Worked examples
💡 Example 1: numbers in the denominator
Solve \(\dfrac{x}{2} - \dfrac{x}{6} = 4\).
What is happening?
The LCD of \(2\) and \(6\) is \(6\). Multiply every term by \(6\): \(\dfrac{x}{2}\) becomes \(3x\), \(\dfrac{x}{6}\) becomes \(x\), and \(4\) becomes \(24\). Then solve as usual.
💡 Example 2: fractions with brackets
Solve \(\dfrac{3}{4}(x-2) = \dfrac{1}{2}(x+1)\).
What is happening?
The LCD of \(4\) and \(2\) is \(4\). Multiply both sides by \(4\): the left becomes \(3(x-2)\) and the right becomes \(2(x+1)\). Expand, then collect like terms.
💡 Example 3: a fraction on each side
Solve \(\dfrac{2x-1}{3} = \dfrac{x+4}{2}\).
What is happening?
The LCD of \(3\) and \(2\) is \(6\). Multiplying both sides by \(6\) leaves \(2(2x-1)\) on the left and \(3(x+4)\) on the right (the whole numerator is multiplied, so keep it in a bracket).
💡 Example 4: x in the denominator
Solve \(\dfrac{6}{x} = \dfrac{3}{4}\).
What is happening?
Multiply both sides by \(4x\) to clear both denominators (this is the same as cross-multiplying). Then solve the linear equation. Check: \(x=8\) does not make any denominator zero.
💡 Example 5: x in the denominator that becomes a quadratic
Solve \(\dfrac{9}{x} - x = 0\).
What is happening?
Multiply every term by \(x\): \(\dfrac{9}{x}\) becomes \(9\), and \(x \times x\) becomes \(x^2\). That leaves \(9 - x^2 = 0\), so \(x^2 = 9\). Square-rooting gives two answers, \(x = 3\) and \(x = -3\) (do not forget the negative root). Both check out, and neither makes the denominator zero.
🔑 Key Points
- Multiply every term on both sides by the LCD, including terms that are not fractions.
- When the whole numerator is multiplied, keep it in a bracket, then expand.
- With x in the denominator, multiply by x (or 2x, and so on) to clear it.
- If clearing x from the bottom gives an \(x^2\), there are usually two answers (\(\pm\)).
- Check your answer does not make an original denominator zero.
⚠️ Common Pitfalls
- Multiplying only the fractions and forgetting the whole-number terms.
- Forgetting to bracket a numerator like \(x+4\), then expanding wrongly.
- Giving only the positive root when \(x^2 = k\) has two solutions.
- Accepting a value that makes a denominator zero (x cannot be 0 when it is in the bottom).