Simultaneous Equations with 3 Variables (AQA GCSE Further Maths): Revision Note

Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on

Simultaneous Equations with 3 Variables

To be able to solve simultaneous equations with 3 unknowns; we will need 3 equations.

We can use similar methods to when we solve with 2 unknowns; elimination or substitution.

How do I solve simultaneous equations with 3 unknowns using elimination?

  • The general approach is to eliminate one of the variables from two equations; so that we are left with 2 equations in terms of 2 unknowns

  • Once we have 2 equations with 2 unknowns, we can use the methods we already know to solve them, and then substitute the values back in to find the 3rd unknown

  • It is very important to label the equations as there are lots to keep track of

  • Take for example the following simultaneous equations

    • circle enclose 1 space space space 2 x plus 3 y minus z equals 17

    • circle enclose 2 space space space x minus 3 y plus 2 z equals negative 12

    • circle enclose 3 space space space 3 x plus y plus z equals 9

  • We can eliminate z by adding equations circle enclose 1 and circle enclose 3

    • circle enclose 1 plus circle enclose 3 colon space space space 5 x plus 4 y equals 26

    • This is equation circle enclose 4

  • We can form another equation containing only x and y, by eliminating z, by taking equation circle enclose 2 and adding on 2 times equation circle enclose 1

    • circle enclose 2 plus space 2 cross times circle enclose 1 space colon space space space 5 x plus 3 y equals 22

    • This is equation circle enclose 5

  • Equations circle enclose 4 and circle enclose 5 can then be solved using the previously covered methods to find x equals 2 and y equals 4

  • Substitute these values into any of the original three equations to find z equals negative 1

  • As a check, we should substitute the found values for x comma space y comma space z into each of the original equations to make sure they all work

How do I solve simultaneous equations with 3 unknowns using substitution?

  • The general approach is the same; eliminate one of the variables from two equations; so that we are left with 2 equations in terms of 2 unknowns

    • With elimination we did this by adding or subtracting the equations (or multiples of the equations)

    • With substitution we do it by rearranging one of the equations to make x comma space y comma or z the subject, and then substituting this into the other two equations

  • Once we have 2 equations with 2 unknowns, we can use the methods we already know to solve them, and then substitute the values back in to find the 3rd unknown

  • It is very important to label the equations as there are lots to keep track of

  • Take for example the following simultaneous equations

    • circle enclose 1 space space space 2 x plus 3 y minus z equals 17

    • circle enclose 2 space space space x minus 3 y plus 2 z equals negative 12

    • circle enclose 3 space space space 3 x plus y plus z equals 9

  • We can rearrange equation circle enclose 1 to make z the subject

    • z equals 2 x plus 3 y minus 17

  • Substitute z equals 2 x plus 3 y minus 17 into equation circle enclose 2

    • x minus 3 y plus 2 open parentheses 2 x plus 3 y minus 17 close parentheses equals negative 12

    • Simplify into the form a x plus b y equals c

    • 5 x plus 3 y equals 22, call this equation circle enclose 4

  • Substitute z equals 2 x plus 3 y minus 17 into equation circle enclose 3

    • 3 x plus y plus open parentheses 2 x plus 3 y minus 17 close parentheses equals 9

    • Simplify into the form a x plus b y equals c

    • 5 x plus 4 y equals 26, call this equation circle enclose 5

  • Equations circle enclose 4 and circle enclose 5 can then be solved using the previously covered methods to find x equals 2 and y equals 4

  • Substitute these values into any of the original three equations to find z equals negative 1

  • As a check, we should substitute the found values for x comma space y comma space z into each of the original equations to make sure they all work

Examiner Tips and Tricks

  • Label each equation you use and write down exactly what you are doing in each step

    • e.g. circle enclose 1 minus circle enclose 2 or circle enclose 2 minus space 3 cross times circle enclose 2

  • Always check your final solutions work for all three original equations

    • You can leave the exam knowing that you got the answer correct!

Worked Example

Solve the simultaneous equations.

table row cell 2 a minus 3 b plus c end cell equals 19 row cell 3 a minus 4 b minus c end cell equals 15 row cell 5 a plus 3 b plus 2 c end cell equals 24 end table

Method 1: Elimination

Number the equations.

2 a minus 3 b plus c space equals space 19
3 a minus 4 b minus c space equals space 15
5 a plus 3 b plus 2 c equals 24space space space space space space space space space space open parentheses 1 close parentheses
space space space space space space space space space space open parentheses 2 close parentheses
space space space space space space space space space space open parentheses 3 close parentheses 

Eliminate c by adding equations (1) and (2). 

(1) + (2):

stack attributes charalign center stackalign right end attributes row none 2 a none minus 3 b up diagonal strike plus c end strike equals 19 none end row row none plus none left parenthesis 3 a none minus 4 b up diagonal strike space minus c end strike equals 15 right parenthesis end row horizontal line row 5 a none minus 7 b none equals 34 none end row end stack space space space space

5 a space minus space 7 b space equals space 34space space space space space space space space space space space space space open parentheses 4 close parentheses

Find a second equation in a and b by subtracting 2 times equation (1) from equation (3). 

(3) - 2(1):

stack attributes charalign center stackalign right end attributes row none 5 a none plus 3 b up diagonal strike plus space 2 c end strike equals 24 none end row row none minus none left parenthesis 4 a none minus 6 b up diagonal strike plus space 2 c end strike equals 38 right parenthesis end row horizontal line row a none plus 9 b none equals negative 14 end row end stack space space space space

a space plus space 9 b space equals space minus 14space space space space space space space space space space space space space open parentheses 5 close parentheses

Solve equations (4) and (5) using your preferred method of solving a pair of linear simultaneous equations.

table attributes columnalign right center left columnspacing 0px end attributes row cell 5 a space minus space 7 b space end cell equals cell space 34 end cell row cell a space plus space 9 b space end cell equals cell negative 14 end cell end table           table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 4 close parentheses end cell end table
table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 5 close parentheses end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank end table 

(4) -5(5):

space space space space space space space 5 a space minus space 7 b space equals space 34 space space space space space space
bottom enclose space minus open parentheses 5 a space plus space 45 b space equals negative 70 close parentheses space end enclose
space space space space space space space space space space space space space space space minus 52 b space equals space 104 space space space
b space equals space minus 2 space space space

Substitute into equation (4) or (5) and solve to find a.

table row blank blank cell open parentheses 4 close parentheses space space space space space end cell end table table row cell 5 a space minus space 7 open parentheses negative 2 close parentheses space end cell equals cell space 34 end cell end table
                table row cell 5 a space plus space 14 space end cell equals cell space 34 end cell row cell 5 a space end cell equals cell space 20 end cell row cell a space end cell equals cell space 4 end cell end table

Substitute a space equals space 4 and b space equals space minus 2 back into one of equation (1), (2) or (3) and solve to find c.

table row blank blank cell open parentheses 1 close parentheses space space space space space end cell end table table row cell 2 open parentheses 4 close parentheses space minus 3 open parentheses negative 2 close parentheses space plus space c space end cell equals cell space 19 end cell end table
                       table row cell 8 space plus space 6 space plus space c space end cell equals cell space 19 end cell row cell 14 space plus c space end cell equals cell space 19 end cell row cell c space end cell equals cell space 5 end cell end table

Substitute the three solutions into the other two equations to check that they are correct.

table row blank blank cell open parentheses 2 close parentheses space space space space space end cell end table table attributes columnalign right center left columnspacing 0px end attributes row cell 3 open parentheses 4 close parentheses space minus 4 open parentheses negative 2 close parentheses space minus 5 space end cell equals cell space 15 end cell end table
                    table row cell 12 space plus space 8 space minus space 5 space end cell equals cell space 15 end cell row cell 15 space end cell equals cell space 15 end cell end table

table row blank blank cell open parentheses 3 close parentheses space space space space space end cell end table table row cell 5 open parentheses 4 close parentheses space plus 3 open parentheses negative 2 close parentheses space plus 2 open parentheses 5 close parentheses space end cell equals cell space 24 end cell end table
                       table row cell 20 space minus space 6 space plus space 10 space end cell equals cell space 24 end cell row cell 24 space end cell equals cell space 24 end cell end table

bold italic a bold space bold equals bold space bold 4 bold comma bold space bold space bold italic b bold space bold equals bold minus bold 2 bold comma bold space bold space bold italic c bold equals bold space bold 5

Method 2: Substitution

Number the equations.

2 a minus 3 b plus c space equals space 19 space space space space space space space space space space space space
3 a minus 4 b minus c space equals space 15 space space space space space space space space space space space space
5 a plus 3 b plus 2 c equals 24 space space space space space space space space space space space space open parentheses 1 close parentheses
open parentheses 2 close parentheses
open parentheses 3 close parentheses

Rearrange equation (1) to make c the subject.  

(1):

c space equals space 19 space minus space 2 a space plus space 3 b

Substitute c space equals space 19 space minus space 2 a space plus space 3 b spaceinto equation (2) and simplify.

(2):

3 a space minus space 4 b space minus space open parentheses 19 space minus space 2 a space plus space 3 b close parentheses space equals space 15
3 a space minus space 4 b space minus space 19 space plus space 2 a space minus space 3 b space equals space 15
5 a space minus space 7 b space equals space 34

5 a space minus space 7 b space equals space 34 space space space space space space space space space space space space space blankopen parentheses 4 close parentheses

Substitute c space equals space 19 space minus space 2 a space plus space 3 b spaceinto equation (3) and simplify.

(3):

5 a space plus space 3 b space plus space 2 open parentheses 19 space minus space 2 a space plus space 3 b close parentheses space equals space 24
5 a space plus space 3 b space plus space 38 space minus space 4 a space plus space 6 b space equals space 24
a space plus space 9 b space equals negative 14

a space plus space 9 b space equals space minus 14 space space space space space space space space space space space space spaceopen parentheses 5 close parentheses space

Solve equations (4) and (5) using your preferred method of solving a pair of linear simultaneous equations.

table row cell 5 a space minus space 7 b space end cell equals cell space 34 space space space space space space space space space space space space space space space end cell row cell a space plus space 9 b space end cell equals cell negative 14 space space space space space space space space space space space space end cell end tabletable row blank blank cell open parentheses 4 close parentheses end cell end table
table row blank blank cell open parentheses 5 close parentheses end cell end table 

(4) -5(5):

space space space space space space space 5 a space minus space 7 b space equals space 34 space space space space space space
bottom enclose space minus open parentheses 5 a space plus space 45 b space equals negative 70 close parentheses space end enclose
space space space space space space space space space space space space space space space minus 52 b space equals space 104 space space space
b space equals space minus 2 space space space

Substitute into equation (4) or (5) and solve to find a.

table row blank blank cell open parentheses 4 close parentheses space space space space space end cell end table table row cell 5 a space minus space 7 open parentheses negative 2 close parentheses space end cell equals cell space 34 end cell end table
                table row cell 5 a space plus space 14 space end cell equals cell space 34 end cell row cell 5 a space end cell equals cell space 20 end cell row cell a space end cell equals cell space 4 end cell end table

Substitute a space equals space 4 and b space equals space minus 2 back into one of equation (1), (2) or (3) and solve to find c.

table row blank blank cell open parentheses 1 close parentheses space space space space space end cell end table table row cell 2 open parentheses 4 close parentheses space minus 3 open parentheses negative 2 close parentheses space plus space c space end cell equals cell space 19 end cell end table
                       table row cell 8 space plus space 6 space plus space c space end cell equals cell space 19 end cell row cell 14 space plus c space end cell equals cell space 19 end cell row cell c space end cell equals cell space 5 end cell end table

Substitute the three solutions into the other two equations to check that they are correct.

table row blank blank cell open parentheses 2 close parentheses space space space space space end cell end table table attributes columnalign right center left columnspacing 0px end attributes row cell 3 open parentheses 4 close parentheses space minus 4 open parentheses negative 2 close parentheses space minus 5 space end cell equals cell space 15 end cell end table
                    table row cell 12 space plus space 8 space minus space 5 space end cell equals cell space 15 end cell row cell 15 space end cell equals cell space 15 end cell end table

table row blank blank cell open parentheses 3 close parentheses space space space space space end cell end table table row cell 5 open parentheses 4 close parentheses space plus 3 open parentheses negative 2 close parentheses space plus 2 open parentheses 5 close parentheses space end cell equals cell space 24 end cell end table
                       table row cell 20 space minus space 6 space plus space 10 space end cell equals cell space 24 end cell row cell 24 space end cell equals cell space 24 end cell end table

bold italic a bold space bold equals bold space bold 4 bold comma bold space bold space bold italic b bold space bold equals bold minus bold 2 bold comma bold space bold space bold italic c bold equals bold space bold 5

👀 You've read 1 of your 5 free revision notes this week
An illustration of students holding their exam resultsUnlock more revision notes. It's free!

By signing up you agree to our Terms and Privacy Policy.

Already have an account? Log in

What’s included
  • Unlock 5 free revision notes per week

  • Add and manage all your subjects

  • Track your revision progress

  • Personalised to your ability

  • Exam board specific expert tips

Excellent

Did this page help you?

Jamie Wood

Author: Jamie Wood

Expertise: Maths Content Creator

Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.

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.

Download notes on Simultaneous Equations with 3 Variables