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Key Stage 4: Linear equations containing algebraic fractions

Author: John Pilkington

Solving equations containing fractions with a simple denominator
In the Lesson 'Algebraic fractions' we learned that algebraic fractions are fractions that contain variables, in contrast to numerical fractions, which contain numbers. So,

  • are algebraic fractions
  • are numerical fractions.

We also learned that the methods used when dealing with algebraic fractions are similar to those used for numerical fractions. So, whenever you are confronted with a problem involving algebraic fractions, ask yourself 'How would I deal with this if these fractions were numerical fractions?' The answer will serve as a useful guide as to how to proceed.

You will see throughout the Lesson that a key part of solving equations containing fractions is to eliminate those fractions. This is usually done by finding the least common multiple (lcm), although later in the Lesson we will look at another way. Once the fractions have all been eliminated, solving the equation becomes reasonably straightforward.

All the algebraic fractions shown above have a simple denominator (usually a number), and so we'll look first at how to solve equations containing fractions such as these.

Worked examples

  1. Solve

    The least common multiple (lcm) of 2 and 3 is 6. So multiply each term by 6, giving

    Cancelling gives

    3x + 2 = 24

    Now collect like terms, giving
    3x = 22

    Finally, divide both sides by 3, giving

  2. Solve (Tip: rewrite this as )

    The lcm of 2, 5 and 4 is 20, so multiply both sides by 20

    Cancel

    10y - 4y = 15

    Collect like terms
    6y = 15

    Divide both sides by 6, giving

  3. Solve

    The lcm of 3 and 4 is 12, so multiply both sides by 12

    Cancel

    8x = 24 + 3x

    Collect like terms
    8x - 3x = 24
    5x = 24

    Divide both sides by 5, giving

  4. Solve

    This time, there is a variable in the denominator

    The lcm of 2 and x is 2x, so multiply both sides by 2x

    Cancel

    x + 6 = 16x

    Collect like terms
    6 = 16x - x
    6 = 15x

    Divide both sides by 15, giving

    Note that the fraction is not in its simplest form, and so would not be acceptable as an answer.

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