National 5MathsSQARevision NotesAlgebraic SkillsAlgebraic FractionsSolving Equations with Algebraic Fractions

Solving Equations with Algebraic Fractions (SQA National 5 Maths): Revision Note

Exam code: X847 75

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

Solving equations with 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 both sides of the equation

Worked Example

Solve the equation fraction numerator 5 over denominator x minus 4 end fraction plus fraction numerator 7 over denominator x plus 2 end fraction equals 6.

Answer:

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

Start by multiplying all terms in the equation by the denominator left parenthesis x minus 4 right parenthesis

  • The left parenthesis x minus 4 right parenthesis on top and bottom will cancel in the first term

table row cell fraction numerator 5 up diagonal strike open parentheses x minus 4 close parentheses end strike over denominator up diagonal strike x minus 4 end strike end fraction plus fraction numerator 7 open parentheses x minus 4 close parentheses over denominator x plus 2 end fraction end cell equals cell 6 open parentheses x minus 4 close parentheses end cell row cell 5 plus fraction numerator 7 open parentheses x minus 4 close parentheses over denominator x plus 2 end fraction end cell equals cell 6 open parentheses x minus 4 close parentheses end cell end table

Now multiply all terms on both sides by the next denominator, open parentheses x plus 2 close parentheses

  • The open parentheses x plus 2 close parentheses on top and bottom will cancel in the second term

table row cell 5 open parentheses x plus 2 close parentheses plus fraction numerator 7 open parentheses x minus 4 close parentheses up diagonal strike open parentheses x plus 2 close parentheses end strike over denominator up diagonal strike x plus 2 end strike end fraction end cell equals cell 6 open parentheses x minus 4 close parentheses open parentheses x plus 2 close parentheses end cell row cell 5 open parentheses x plus 2 close parentheses plus 7 open parentheses x minus 4 close parentheses end cell equals cell 6 open parentheses x minus 4 close parentheses open parentheses x plus 2 close parentheses end cell end table

Expand brackets on both sides

  • Use FOIL or another method to expand the double brackets on the right

table row cell 5 x plus 10 plus 7 x minus 28 end cell equals cell 6 open parentheses x squared minus 2 x minus 8 close parentheses end cell row cell 5 x plus 10 plus 7 x minus 28 end cell equals cell 6 x squared minus 12 x minus 48 end cell end table

Collect like terms on the left

table row cell 12 x minus 18 end cell equals cell 6 x squared minus 12 x minus 48 end cell end table

Subtract 12 x and add 18 to both sides of the equation to get zero on one side

0 equals 6 x squared minus 24 x minus 30

Divide both sides of the equation by 6 to simplify the coefficients

0 equals x squared minus 4 x minus 5

That is a quadratic equation that can be solved by factorising

0 equals open parentheses x plus 1 close parentheses open parentheses x minus 5 close parentheses

x equals negative 1 comma space space x equals 5

Those are the solutions you are looking for

  • You can substitute them back into the original equation to check that they are right

x equals negative 1 comma space space x equals 5

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    Author
    Reviewer
    Roger B

    Author: Roger B

    Expertise: Maths Content Creator

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

    Dan Finlay

    Reviewer: Dan Finlay

    Expertise: Maths Subject 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.