GCSEFurther MathsAQARevision Notes2. Algebra & FunctionsSimultaneous EquationsSimultaneous Equations with 3 Variables

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

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Author
Reviewer
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 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.

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