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Equations with Brackets & Fractions (Cambridge (CIE) IGCSE International Maths)

Revision Note

Mark Curtis

Written by: Mark Curtis

Reviewed by: Dan Finlay

Cambridge (CIE) International Maths: CoreCambridge (CIE) Maths: Core

Equations with Brackets & Fractions

How do I solve linear equations with brackets?

  • If a linear equation involves brackets, expand the brackets first

  • For example, solve 2 open parentheses x minus 3 close parentheses equals 10

    • Expand the brackets 

2 x minus 6 equals 10

  • This now has the same form as before

    • Solve by adding 6 then dividing by 2

table row cell 2 x end cell equals 16 row x equals cell 16 over 2 end cell row x equals 8 end table

  • Expanding brackets first will always work, but you can also divide first

    • Dividing both sides of 2 open parentheses x minus 3 close parentheses equals 10 by 2 gives open parentheses x minus 3 close parentheses equals 5

      • which solves to x equals 8

    • This method works but can lead to harder fractions

How do I solve linear equations with fractions?

  • If a linear equation contains fractions, multiply both sides by the lowest common denominator

  • For example, solve x over 5 plus 4 equals 9 over 2

    • The lowest common denominator of 5 and 2 is 10

    • Multiply all terms on both sides by 10

      • Don't forget to do it to the 4

table row cell 10 cross times x over 5 plus 10 cross times 4 end cell equals cell 10 cross times 9 over 2 end cell end table

  • Simplify the fractions

    • Imagine the 10 is in the numerator and cancel

table row cell 2 x plus 40 end cell equals 45 end table

  • This now has the same form as before

    • Solve by subtracting 40 then dividing by 2

table row cell 2 x end cell equals 5 row x equals cell 5 over 2 end cell end table

  • Unless the question asks, you can leave the answer like this

    • A decimal or mixed number would also be accepted

  • Sometimes equations have the unknown on the denominator

  • For example fraction numerator 4 over denominator x minus 2 end fraction equals 3

    • Multiply both sides of the equation by the denominator

fraction numerator 4 over denominator x minus 2 end fraction cross times open parentheses x minus 2 close parentheses equals 3 open parentheses x minus 2 close parentheses

  • Simplify the fractions, and expand any brackets

4 equals 3 open parentheses x minus 2 close parentheses

4 equals 3 x minus 6

  • This now looks like a standard linear equation

    • Solve by adding 6 to both sides, then dividing by 3

10 equals 3 x 10 over 3 equals x

Examiner Tips and Tricks

  • In the exam, always substitute your answer back into the equation to check you got it right!

Worked Example

(a) Solve the equation

5 open parentheses 3 minus 4 x close parentheses plus 1 equals 26

First expand the brackets on the left-hand side

table row cell 15 minus 20 x plus 1 end cell equals 26 end table

Simplify the left-hand side (by adding 15 and 1)

table row cell 16 minus 20 x end cell equals 26 end table

Remember 16 - 20x is the same as -20x + 16
Subtract 16 from both sides

table row cell negative 20 x end cell equals cell 26 minus 16 end cell row cell negative 20 x end cell equals 10 end table

Divide both sides by -20 and simplify

table row x equals cell fraction numerator 10 over denominator negative 20 end fraction end cell row x equals cell negative 1 half end cell end table

bold italic x bold equals bold minus bold 1 over bold 2

-0.5 is also accepted

(b) Solve the equation

fraction numerator 5 x over denominator 4 end fraction equals 1 half

The lowest common denominator of 4 and 2 is 4
Multiply both sides by 4

table row cell 4 cross times fraction numerator 5 x over denominator 4 end fraction end cell equals cell 4 cross times 1 half end cell end table

Simplify (cancel) the fractions

table row cell 5 x end cell equals 2 end table 

To solve this equation, divide both sides by 5

bold italic x bold space bold equals fraction numerator bold space bold 2 over denominator bold 5 end fraction

0.4 is also accepted

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    Author
    Reviewer
    Mark Curtis

    Author: Mark Curtis

    Expertise: Maths

    Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

    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.