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Solving Equations with Algebraic Fractions (Edexcel IGCSE Maths A (Modular))

Revision Note

Amber

Written by: Amber

Reviewed by: Dan Finlay

Edexcel Maths A (Modular): Higher Unit 1Cambridge (CIE) International Maths: ExtendedCambridge (CIE) Maths: ExtendedEdexcel Maths A: Higher

Solving Algebraic Fractions

How do I solve an equation that contains algebraic fractions?

  • There are two methods for solving equations that contain algebraic fractions

  • One method is to add or subtract the algebraic fractions first and then solve as usual

    • For example, to solve fraction numerator 8 over denominator x plus 1 end fraction minus fraction numerator 5 over denominator x plus 2 end fraction equals 1

    • First subtract the fractions and simplify, fraction numerator 3 x plus 11 over denominator open parentheses x plus 1 close parentheses open parentheses x plus 2 close parentheses end fraction equals 1

    • Then cross-multiply, expand and solve

      table row cell 3 x plus 11 end cell equals cell 1 open parentheses x plus 1 close parentheses open parentheses x plus 2 close parentheses end cell row cell 3 x plus 11 end cell equals cell x squared plus 3 x plus 2 end cell row 0 equals cell x squared minus 9 end cell row 0 equals cell open parentheses x minus 3 close parentheses open parentheses x plus 3 close parentheses end cell row x equals cell 3 space or space x equals negative 3 end cell end table

  • Alternatively, you can remove the fractions first by multiplying everything on both sides of the equation by each expression in the denominators and then solve

    • For example, to solve the equation fraction numerator 4 over denominator x minus 3 end fraction plus fraction numerator 5 over denominator x plus 1 end fraction equals 5

    • First multiply every term in the equation by both open parentheses x minus 3 close parentheses and open parentheses x plus 1 close parentheses and cancel common factors where possible

      • Multiply every term by open parentheses x minus 3 close parentheses (this bracket goes in the numerator of any fractions)
        table row cell fraction numerator 4 over denominator up diagonal strike open parentheses x minus 3 close parentheses end strike end fraction up diagonal strike open parentheses x minus 3 close parentheses end strike plus fraction numerator 5 open parentheses x minus 3 close parentheses space over denominator x plus 1 end fraction end cell equals cell 5 open parentheses x minus 3 close parentheses end cell row cell 4 plus fraction numerator 5 open parentheses x minus 3 close parentheses space over denominator x plus 1 end fraction end cell equals cell 5 open parentheses x minus 3 close parentheses end cell end table

      • Then multiply every term by table row blank blank cell open parentheses x plus 1 close parentheses end cell end table

        table row cell 4 open parentheses x plus 1 close parentheses plus fraction numerator 5 open parentheses x minus 3 close parentheses over denominator open parentheses up diagonal strike x plus 1 end strike close parentheses end fraction up diagonal strike open parentheses x plus 1 close parentheses end strike end cell equals cell 5 open parentheses x minus 3 close parentheses open parentheses x plus 1 close parentheses end cell row cell 4 open parentheses x plus 1 close parentheses plus 5 open parentheses x minus 3 close parentheses end cell equals cell 5 open parentheses x minus 3 close parentheses open parentheses x plus 1 close parentheses end cell end table

    • Then solve

      table row cell 4 x plus 4 plus 5 x minus 15 end cell equals cell 5 open parentheses x squared minus 2 x minus 3 close parentheses end cell row cell 9 x minus 11 end cell equals cell 5 x squared minus 10 x minus 15 end cell row 0 equals cell 5 x squared minus 19 x minus 4 end cell row 0 equals cell open parentheses 5 x plus 1 close parentheses open parentheses x minus 4 close parentheses end cell row x equals cell negative 1 fifth space or space x equals 4 end cell end table

Examiner Tips and Tricks

  • When multiplying by an algebraic expression, use brackets around the expression, e.g. open parentheses 2 x plus 3 close parentheses

  • Multiplying by both denominators at once can speed up the process, but take care if choosing this technique in the exam!

    • and remember to multiply all terms on either side of the equation

Worked Example

fraction numerator 2 over denominator p plus 3 end fraction minus 5 over p equals 6 p

Show that this equation can be written as  6 p cubed plus 18 p squared plus 3 p plus 15 equals 0.

To clear the fractions, we multiply both sides of the equation by each denominator

Start by multiplying all terms in the equation by the denominator left parenthesis p plus 3 right parenthesis
The left parenthesis p plus 3 right parenthesis on top and bottom will cancel in the first term

2 minus fraction numerator 5 left parenthesis p plus 3 right parenthesis over denominator p end fraction equals 6 p left parenthesis p plus 3 right parenthesis

Now multiply all terms on both sides by the next denominator, p
The p on top and bottom will cancel in the second term

2 open parentheses p close parentheses minus 5 open parentheses p plus 3 close parentheses equals 6 p open parentheses p plus 3 close parentheses open parentheses p close parentheses

Expand brackets
Be careful with negative signs

table row cell 2 p minus 5 open parentheses p plus 3 close parentheses end cell equals cell 6 p squared open parentheses p plus 3 close parentheses end cell row cell 2 p minus 5 p minus 15 end cell equals cell 6 p cubed plus 18 p squared end cell end table

Collect like terms

negative 3 p minus 15 equals 6 p cubed plus 18 p squared

Add 3 p and 15 to both sides of the equation

0 equals 6 p cubed plus 18 p squared plus 3 p plus 15

bold 6 bold italic p to the power of bold 3 bold plus bold 18 bold italic p to the power of bold 2 bold plus bold 3 bold italic p bold plus bold 15 bold equals bold 0

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    Author
    Reviewer
    Amber

    Author: Amber

    Expertise: Maths

    Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.

    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.