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Equations with fractions

We multiply both sides by the lowest common denominator. Then the fractions disappear and we solve the equation as usual. Every term must be multiplied. If x is in the denominator, we multiply both sides by x.

Years 8–9 and upper secondary · Updated

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Multiply the denominator away

x/4 means x divided by 4. We do the opposite and multiply both sides by 4. Then the sides are still equal.

x/4 = 5 gives 4 · x/4 = 4 · 5, so x = 20

Several denominators

We want to remove all the denominators and then get x on its own.

We multiply by the lowest common denominator: the smallest number that all the denominators divide into exactly.

DenominatorsLowest common denominatorWhy
2 and 366 ÷ 2 = 3 and 6 ÷ 3 = 2
4 and 61212 ÷ 4 = 3 and 12 ÷ 6 = 2
5 and 101010 ÷ 5 = 2 and 10 ÷ 10 = 1
3 and 41212 ÷ 3 = 4 and 12 ÷ 4 = 3
  • Find the lowest common denominator.
  • Multiply every term on both sides by it, including terms without fractions.
  • Put brackets round any numerator with more than one term.
  • Solve as usual and check the answer.

When x is in the denominator

In 15/x = 3 we multiply both sides by x.

x · 15/x = 3 · x, so 15 = 3x

We divide both sides by 3 and get x = 5. Check: 15/5 = 3. Here x can never be 0.

Example

Example 1: x/4 + 2 = 7

We want x on its own. First we remove the denominator 4.

  1. We multiply every term on both sides by 4, including the two.

    4 · x/4 + 4 · 2 = 4 · 7

  2. 4 · x/4 = x, 4 · 2 = 8 and 4 · 7 = 28.

    x + 8 = 28

  3. We subtract 8 from both sides.

    x = 20

  4. Another way: subtract 2 first and then multiply both sides by 4.

    x/4 = 5, so x = 20

20/4 + 2 = 5 + 2 = 7. The left-hand side is 7, the same as the right. The answer is correct.

Example 2: x/2 + x/3 = 10

We want x on its own. The lowest common denominator of 2 and 3 is 6.

  1. We multiply every term on both sides by 6.

    6 · x/2 + 6 · x/3 = 6 · 10

  2. 6 ÷ 2 = 3 and 6 ÷ 3 = 2, so we get 3x and 2x.

    3x + 2x = 60

  3. Three x and two x make five x.

    5x = 60

  4. We divide both sides by 5.

    x = 12

12/2 + 12/3 = 6 + 4 = 10. The left-hand side is 10, the same as the right. The answer is correct.

Example 3: (x + 3)/4 = (x − 1)/2

We want x on its own. The lowest common denominator is 4, and the numerators get brackets.

  1. We multiply both sides by 4.

    4 · (x + 3)/4 = 4 · (x − 1)/2

  2. On the left the four cancels the denominator. On the right 4 ÷ 2 = 2.

    x + 3 = 2(x − 1)

  3. We multiply each term in the bracket by two: 2 · (−1) = −2.

    x + 3 = 2x − 2

  4. We subtract x from both sides.

    3 = x − 2

  5. We add 2 to both sides.

    x = 5

Left: (5 + 3)/4 = 8/4 = 2. Right: (5 − 1)/2 = 4/2 = 2. The sides are equal, so x = 5 is correct.

Pitfalls

  • Multiplying only the fraction, so that x/4 + 2 = 7 becomes x + 2 = 28.

    The whole side must become four times as big, including the two: x + 8 = 28.

  • Forgetting the brackets round the numerator, so that 6 · (x − 1)/2 becomes 3x − 1.

    The whole numerator is divided. Write 3(x − 1) and multiply out the bracket: 3x − 3.

  • Dividing the wrong way round when x is in the denominator, so that 15/x = 3 gives x = 3/15.

    Multiply both sides by x: 15 = 3x. Then divide by 3: x = 5.

Practise on your own

0 of 4 correct

  1. Solve the equation x/5 = 4.

  2. Solve the equation 36/x = 9.

  3. Solve the equation x/4 + x/6 = 10.

  4. Solve the equation (2x − 1)/3 = (x + 3)/2.

More walkthroughs with Fox

Common questions

Do I have to use the lowest common denominator?

No, any common denominator works, for example the denominators multiplied together. With the lowest one the numbers are smaller.

What do I do if the answer is a fraction?

That is fine. Write it in its simplest form, like x = 7/2, or as a decimal, x = 3.5.

Why can x not be zero when x is in the denominator?

We cannot divide by zero. So x = 0 can never be a solution to an equation like that.

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