When a GCSE equation contains fractions, multiply every term by the lowest common multiple of the denominators. This clears all fractions in one step, leaving a straightforward equation you can solve in the normal way. The method works whether fractions appear on one side, on both sides, or in an algebraic expression.
Why do fractions make equations harder?
Fractions in equations are not a fundamentally different problem — they are ordinary linear or quadratic equations wearing an extra layer. The difficulty is that combining fractional terms before solving them invites errors with common denominators and sign changes.
The most reliable approach is to remove the fractions first by multiplying every single term in the equation — on both sides of the equals sign — by the LCM of all the denominators present. This converts a fractional equation into a whole-number equation, and only then do you solve it.
How do you clear a fraction on one side of the equation?
When fractions appear only on one side, find the LCM of the denominators on that side and multiply every term throughout.
Worked example: Solve x/3 + 2 = 7.
- The only denominator is 3. Multiply every term by 3: 3 × (x/3) + 3 × 2 = 3 × 7
- Simplify: x + 6 = 21.
- Subtract 6 from both sides: x = 15.
- Check: 15/3 + 2 = 5 + 2 = 7. ✓
How do you clear fractions on both sides?
When fractions appear on both sides, find the LCM of all the denominators across the whole equation, then multiply every term by it.
Worked example: Solve (x + 1)/2 = (x − 3)/4.
- The denominators are 2 and 4. LCM = 4. Multiply every term by 4: 4 × (x + 1)/2 = 4 × (x − 3)/4
- Simplify each term: 2(x + 1) = (x − 3).
- Expand: 2x + 2 = x − 3.
- Subtract x from both sides: x + 2 = −3.
- Subtract 2: x = −5.
- Check: (−5 + 1)/2 = −4/2 = −2; (−5 − 3)/4 = −8/4 = −2. ✓
What if an expression involving x is in the denominator?
When x appears in the denominator (e.g. 3/(x − 1)), you cannot simply cancel — instead, multiply both sides by that denominator and then solve the resulting equation.
Worked example: Solve 3/(x − 1) = 6.
- Multiply both sides by (x − 1): 3 = 6(x − 1).
- Expand the right side: 3 = 6x − 6.
- Add 6 to both sides: 9 = 6x.
- Divide by 6: x = 9/6 = 3/2.
- Check: 3/(3/2 − 1) = 3/(1/2) = 3 × 2 = 6. ✓
A warning: whenever x is in a denominator, check that your solution does not make that denominator equal to zero. In the example above, x = 3/2 gives a denominator of 1/2 — fine. Had the answer been x = 1, the original equation would have been undefined.
How do you handle fractions with algebraic numerators on both sides?
This is the type most likely to appear on a GCSE Higher paper. The approach is the same — find the LCM and clear — but expanding correctly after multiplying is where errors tend to creep in.
Worked example: Solve (2x + 3)/5 − (x − 1)/3 = 2.
| Step | Working |
|---|---|
| LCM of 5 and 3 | LCM = 15 |
| Multiply every term by 15 | 15 × (2x + 3)/5 − 15 × (x − 1)/3 = 15 × 2 |
| Simplify | 3(2x + 3) − 5(x − 1) = 30 |
| Expand | 6x + 9 − 5x + 5 = 30 |
| Collect like terms | x + 14 = 30 |
| Solve | x = 16 |
Check: (32 + 3)/5 − (16 − 1)/3 = 35/5 − 15/3 = 7 − 5 = 2. ✓
What are the most common errors when solving equations with fractions?
- Forgetting to multiply every term. If you multiply the fraction terms but not the whole-number terms, the equation changes and gives a wrong answer.
- Incorrect expansion after multiplying. After multiplying by 15, for example, 15 × (2x + 3)/5 becomes 3(2x + 3), which then expands to 6x + 9 — not 6x + 3. Always expand the bracket in full.
- Sign errors when a fraction is subtracted. Subtracting (x − 1)/3 means the whole bracket (x − 1) is subtracted; when you expand −5(x − 1) you get −5x + 5, not −5x − 5.
- Not checking the solution. Substituting your answer back into both sides takes less than a minute and immediately flags an error.
Frequently asked questions
Do I need to find the LCM, or can I use any common multiple?
You can multiply by any common multiple of the denominators and the fractions will clear. Using the LCM just keeps the numbers as small as possible, which reduces the risk of arithmetic errors later. For example, with denominators 4 and 6, multiplying by 24 works perfectly — you just end up with larger numbers than if you had used the LCM of 12.
What if the equation has three or more different denominators?
The method is identical. Find the LCM of all the denominators, then multiply every term (including whole-number terms) by that LCM. For denominators 2, 3, and 5, the LCM is 30. Multiply through by 30 and all three fractions disappear in one step.
Can solving equations with fractions lead to a quadratic?
Yes, particularly when x appears in a denominator. For example, 2/x + x = 3 multiplied through by x gives 2 + x² = 3x, which rearranges to x² − 3x + 2 = 0. You then factorise or use the quadratic formula to find x = 1 or x = 2. Always check whether either solution makes a denominator zero — if so, discard it.
How do I know my LCM is correct?
List the multiples of each denominator and find the smallest number that appears in both lists. For 4 and 6: multiples of 4 are 4, 8, 12, 16…; multiples of 6 are 6, 12, 18…. The LCM is 12. Alternatively, use prime factorisation: 4 = 2², 6 = 2 × 3, so LCM = 2² × 3 = 12.
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