Syllabus Edition

First teaching 2021

Last exams 2024

|

Linear Equations (CIE IGCSE Maths: Core)

Revision Note

Test Yourself
Amber

Author

Amber

Expertise

Maths

Solving Linear Equations

What are linear equations?

  • A linear equation is an equation that will produce a straight line when plotted on a graph
  • The greatest power of x in a linear equation is 1
    • This means there are no terms of x squared or a higher order
  • A linear equation is normally in form a x space plus space b space equals space c
    • where a comma space b comma and c are constants and xis a variable

How do I solve a linear equation?

  • To solve a linear equation you need to isolate the variable, usually x, by carrying out inverse operations to both sides of the equation
    • Inverse operations are just the opposite operations to what has already happened to the variable
  • The order in which the inverse operations are carried out is important
    • Most of the time, this will be BIDMAS in reverse
    • However it depends on the order in which the operations were applied to the variable to form the equation

How do I solve a linear equation of the form ax + b = c?

  • The operations that have been applied to x here are:
    • STEP 1
      Multiply by a
    • STEP 2
      Add b
  • To solve this, you must carry out the inverse operations in reverse order
    • STEP 1
      Subtract b
    • STEP 2
      Divide by a
  • For example, to solve the equation 2 x space plus space 1 space equals space 9
    • STEP 1
      Subtract 1

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x space plus space 1 space end cell equals cell space 9 end cell end table

table row blank blank cell left parenthesis negative 1 right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space end cell end table space space table row blank blank cell space space space space space space space space space space space space space space space space space space space left parenthesis negative 1 right parenthesis end cell end table

table row cell 2 x space end cell equals cell space 8 end cell end table

    • STEP 2
      Divide by 2

table row cell 2 x space end cell equals cell space 8 end cell end table

table row blank blank cell left parenthesis divided by 2 right parenthesis space space space space space space space space space space end cell end table space space table row blank blank cell space space space space space space space space space space space space space left parenthesis divided by 2 right parenthesis end cell end table

table row cell x space end cell equals cell space 4 end cell end table

  • Be extra careful if any of the terms have negatives
  • For example, to solve the equation 2 minus 3 x equals 10
    • STEP 1
      Subtract 2

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 space minus space 3 x space end cell equals cell space 10 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell left parenthesis negative 2 right parenthesis space space space space space space space space space space space space space space space space space space space space space end cell end table space space table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell space space space space space space space space space space space space space space space space space space space space space left parenthesis negative 2 right parenthesis end cell end table

table row cell negative 3 x space end cell equals cell space 8 end cell end table

      • Be careful not to drop the negative sign 
    • STEP 2
      Divide by -3

table row cell negative 3 x space end cell equals cell space 8 end cell end table

table row blank blank cell space left parenthesis divided by negative 3 right parenthesis space space space space space space space space space space space space space space space space space space space end cell end table space space table row blank blank cell space space space space space space space space space space space space space space space space space space space space left parenthesis divided by negative 3 right parenthesis end cell end table

table row cell x space end cell equals cell space minus 8 over 3 end cell end table

Exam Tip

  • If you have time in the exam, you should substitute your answer back into the equation to check you got it right

Worked example

Solve the equation
 

5 x space minus space 8 space equals space 22
 

 Add 8 to both sides of the equation
 

table attributes columnalign right center left columnspacing 0px end attributes row cell 5 x space end cell equals cell space 22 space plus space 8 end cell end table
  

Work out 22 + 8
 

5 x space equals space 30

Divide both sides by 5
 

x space equals space 30 over 5
 

Work out 30 ÷ 5
Keep the x on the left-hand side

bold italic x bold space bold equals bold space bold 6

Equations with Brackets & Fractions

How do I solve a linear equation that contains brackets?

  • If a linear equation contains brackets on one, or both, sides start by expanding the brackets
  • For example, to solve the equation 2 open parentheses x space minus space 1 close parentheses space equals space 10 space minus space open parentheses 3 space plus space x close parentheses
    • STEP 1
      Expand the brackets on both sides 

space space space space space space space space space space space space space space space space space space space 2 x space minus space 2 space equals space 10 space minus space 3 space minus space x

    • STEP 2
      Collect the xterms to one side by subtracting the term with the smaller coefficient of x

table row cell 2 x space minus space 2 space end cell equals cell space 7 space minus space x end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell negative space open parentheses negative x close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus space open parentheses negative x close parentheses end cell end table

table row cell 3 x space minus space 2 space end cell equals cell space 7 end cell end table

      • Be extra careful if any of the terms have negatives
    • STEP 3
      Solve the linear equation using the method above 

        table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x space minus space 2 space end cell equals cell space 7 end cell row cell 3 x space end cell equals cell space 9 end cell row cell x space end cell equals cell space 3 end cell end table

How do I solve a linear equation that contains fractions?

  • If a linear equation contains a fraction on one or both sides, remove the fractions by multiplying both sides by everything on the denominator
    • Remember to put brackets around the expression first  
  • For example, to solve the equation fraction numerator 2 over denominator x space plus space 3 end fraction space equals space fraction numerator space 3 over denominator 4 space minus space 2 x end fraction you will need to multiply both sides of the equation by both open parentheses x space plus space 3 close parentheses and open parentheses 4 space minus space 2 x close parentheses 
    • STEP 1
      Multiply both sides by open parentheses x space plus space 3 close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 over denominator x space plus space 3 end fraction open parentheses x space plus space 3 close parentheses space end cell equals cell space fraction numerator space 3 over denominator 4 space minus space 2 x end fraction open parentheses x space plus space 3 close parentheses end cell row cell 2 space end cell equals cell space fraction numerator 3 open parentheses x plus 3 close parentheses over denominator 4 space minus space 2 x end fraction end cell end table

    • STEP 2
      Multiply both sides by open parentheses 4 space minus space 2 x close parentheses

table row cell 2 open parentheses 4 space minus space 2 x close parentheses space end cell equals cell space fraction numerator 3 open parentheses x space plus space 3 close parentheses over denominator 4 minus 2 x end fraction open parentheses 4 space minus space 2 x close parentheses end cell row cell 2 open parentheses 4 space minus space 2 x close parentheses space end cell equals cell space 3 open parentheses x space plus space 3 close parentheses end cell end table

    • STEP 3
      Expand the brackets on both sides 

space space space space space space space space space space space space space space space space space space space 8 space minus space 4 x space equals space 3 x space plus space 9

    • STEP 4
      Collect the xterms to one side by subtracting the term with the smaller coefficient of x

table row cell 8 space minus space 4 x space end cell equals cell space 3 x space plus space 9 end cell end table

table row blank blank cell negative space open parentheses negative 4 x close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus space open parentheses negative 4 x close parentheses end cell end table

table row cell 8 space end cell equals cell space 7 x space plus space 9 end cell end table

    • STEP 5
      Solve the equation
      • You can swap the sides if it makes solving the equation easier    

    table row cell 7 x space plus space 9 space end cell equals cell space 8 end cell row cell 7 x space end cell equals cell space minus 1 end cell row cell x space end cell equals cell space minus 1 over 7 end cell end table

Exam Tip

  • If you have time in the exam, you should substitute your answer back into the equation to check you got it right

Worked example

a)
Solve the equation.

fraction numerator 3 x space minus space 2 over denominator 4 space minus space x end fraction space equals space minus 1

Get rid of the fraction by multiplying both sides by the denominator open parentheses 4 space minus space x close parentheses

table row cell 3 x space minus space 2 space space end cell equals cell space minus open parentheses 4 space minus space x close parentheses end cell end table

Expand the brackets

table row cell 3 x space minus space 2 space space end cell equals cell space minus 4 space plus space x end cell end table
 

Bring the terms to one side of the equation by subtracting from both sides
 

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x space minus space 2 space space end cell equals cell space minus 4 space end cell end table

Get the x  term by itself by adding 2 to both sides
 

2 x space equals space minus 2

Solve by dividing both sides by 2

bold italic x bold space bold equals bold space bold minus bold 1

b)
Solve the equation.

2 space plus space fraction numerator x space plus space 1 over denominator 3 end fraction space equals space 4

Isolate the fraction by subtracting 2 from both sides

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x space plus space 1 over denominator 3 end fraction space end cell equals cell space 2 end cell end table

Get rid of the fraction by multiplying both sides by the denominator (3)

table row cell x space plus space 1 space end cell equals cell space 6 end cell end table
 

Get the x  term by itself by subtracting 1 from both sides
 

bold italic x bold space bold equals bold space bold 5

Unknown on Both Sides

How do I solve a linear equation with the unknown variable, x, on both sides?

  • If a linear equation contains the unknown variable, usually xon both sides start by collecting these terms together on one side of the equation
    • Moving the x term with the smallest coefficient (number in front of x) is easiest 
  • For example, to solve the equation  4 x space minus space 7 space equals space 11 space plus space x
    • STEP 1
      Move the x term with the smallest coefficient of x
      • The coefficients are 4 and 1 so move the x-term on the right hand side

table row cell 4 x space minus space 7 space end cell equals cell space 11 space plus space x end cell end table

table row blank blank cell open parentheses negative x close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses negative x close parentheses end cell end table

table row cell 3 x space minus space 7 space end cell equals cell space 11 end cell end table

    • STEP 2
      Solve the linear equation using the method above         

table row cell 3 x space minus space 7 space end cell equals cell space 11 end cell row cell 3 x space end cell equals cell space 18 end cell row cell x space end cell equals cell space 6 end cell end table

Exam Tip

  • If you have time in the exam, you should substitute your answer back into the equation to check you got it right

Worked example

Solve the equation
 

4 minus 5 x equals 6 x minus 29
 

5 is smaller than 6x  so it's easier to move this term first
Add 5 to both sides
 

4 equals 6 x minus 29 plus 5 x
 

Simplify the right-hand side by adding the 6x  and the 5x
 

4 space equals space 11 x space minus space 29
 

This could also have been written as 4 = -29 + 11x
Leave the x's on the right-hand side (avoid 'reflecting' the equation)
Get 11 on its own (by adding 29 to both sides)
 

4 space plus space 29 space equals space 11 x
 

Work out 4 + 29
 

33 space equals space 11 x
 

Get on its own (by dividing both sides by 11)
 

33 over 11 space equals space x
 

Work out 33 ÷ 11
 

3 space equals space x
 

The answer currently has on the right-hand side
At this point, you may reflect the equation to give your answer as x = ...
 

bold italic x bold space bold equals bold space bold 3

Forming Linear Equations

Before solving an equation you may need to form it from the information given in the question.

How do I form an expression?

  • An expression is an algebraic statement without an equals sign e.g. 3 x plus 7
  • Sometimes we need to form expressions to help us express unknown values
  • If a value is unknown you can represent it by a letter such as x
  • You can turn common phrases into expressions
    • Here you can represent the "something" by any letter

      2 less than "something" x minus 2
      Double the amount of "something" 2 x
      5 lots of "something" 5 x
      3 more than "something" x plus 3
      Half the amount of "something" 1 half x space or space x over 2
  • You might need to use brackets to show the correct order
    • "something" add 1 then multiplied by 3
      • left parenthesis x plus 1 right parenthesis cross times 3 which simplifies to 3 left parenthesis x plus 1 right parenthesis
    • "something" multiplied by 3 then add 1
      • left parenthesis x cross times 3 right parenthesis plus 1 which simplifies to 3 x plus 1
  • To make the expression as easy as possible choose the smallest value to be represented by a letter
    • If Adam is 10 years younger than Barry, then Barry is 10 years older than Adam
      • Represent Adam's age as x then Barry's age is x plus 10
      • This makes the algebra easier, rather than calling Barry's age x and Adam's age x minus 10
    • If Adam's age is half of Barry's age, then Barry's age is double Adam's age
      • So if Adam's age is x then Barry's age is 2 x
      • This makes the algebra easier, rather than using x for Barry's age and 1 half x for Adams's age

How do I form an equation?

  • An equation is simply an expression with an equals sign that can then be solved
  • You will first need to form an expression and make it equal to a value or another expression
  • It is useful to know alternative words for basic operations:
    • For addition: sum, total, more than, increase, etc
    • For subtraction: difference, less than, decrease, etc
    • For multiplication: product, lots of, times as many, double, triple etc
    • For division: shared, split, grouped, halved, quartered etc
  • Using the first example above 
    • If Adam is 10 years younger than Barry and the sum of their ages is 25, you can find out how old each one is
      • Represent Adam's age as x then Barry's age is x plus 10
      • We can solve the equation x plus x plus 10 space equals space 25  or  2 x plus 10 space equals space 25
  • Sometimes you might have two unrelated unknown values ( and y) and have to use the given information to form two simultaneous equations

Worked example

At a theatre the price of a child's ticket is £ x and the price of an adult's ticket is £ y.

Write equations to represent the following statements:

(a)
An adult's ticket is double the price of a child's ticket.
(b)
A child's ticket is £7 cheaper than an adult's ticket.
(c)
The total cost of 3 children's tickets and 2 adults' tickets is £45.

(a)
Adult = 2 × Child
bold italic y bold equals bold 2 bold italic x
Equivalently you could put x equals 1 half y
(b)
Rewrite as:
 
Adult = Child + £7
 
bold italic y bold equals bold italic x bold plus bold 7
Equivalently you could put x equals y minus 7 or y minus x equals 7


(c)
Total means add
 
3 × Child + 2 × Adult = £45
 
bold 3 bold italic x bold plus bold 2 bold italic y bold equals bold 45

You've read 0 of your 0 free revision notes

Get unlimited access

to absolutely everything:

  • Downloadable PDFs
  • Unlimited Revision Notes
  • Topic Questions
  • Past Papers
  • Model Answers
  • Videos (Maths and Science)

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Amber

Author: Amber

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.