IGCSEFurther MathsEdexcelRevision NotesEquations, Identities & InequalitiesEquations & IdentitiesSimultaneous Equations

Simultaneous Equations (Edexcel IGCSE Further Pure Maths)

Revision Note

Mark Curtis

Written by: Mark Curtis

Reviewed by: Dan Finlay

Linear Simultaneous Equations

What are linear simultaneous equations?

  • When there are two unknowns (say x and y) in a problem, we need two equations to be able to find them both: these are called simultaneous equations

    • you solve two equations to find two unknowns, x and y

      • for example, 3x + 2y = 11 and 2x - y = 5

      • the solutions are x = 3 and y = 1

  • If they just have x and y in them (no x2 or y2 or xy etc) then they are linear simultaneous equations

How do I solve linear simultaneous equations by elimination?

  • "Elimination" completely removes one of the variables, x or y

  • To eliminate the x's from 3x + 2y = 11 and 2x - y = 5 

    • Multiply every term in the first equation by 2

      • 6x + 4y = 22

    • Multiply every term in the second equation by 3

      • 6x - 3y = 15

    • Subtract the second result from the first to eliminate the 6x's

      • leaving 4y - (-3y) = 22 - 15, i.e. 7y = 7

    • Solve to find y (y = 1) then substitute y = 1 back into either original equation to find x (x = 3)

  • Alternatively, to eliminate the y's from 3x + 2y = 11 and 2x - y = 5 

    • Multiply every term in the second equation by 2

      • 4x - 2y = 10

    • Add this result to the first equation to eliminate the 2y's (as 2y + (-2y) = 0)

      • The process then continues as above

  • Check that your final solutions satisfy both equations

How do I solve linear simultaneous equations by substitution?

  • "Substitution" means substituting one equation into the other

  • Solve 3x + 2y = 11 and 2x - y = 5 by substitution

    • Rearrange one of the equation into y = ... (or x = ...)

      • For example, the second equation becomes y = 2x - 5 

    • Substitute this into the first equation (replace all y's with 2x - 5 in brackets)

      • 3x + 2(2x - 5) = 11

    • Solve this equation to find x (x = 3), then substitute x = 3 into y = 2x - 5 to find y (y = 1)

  • Check that your final solutions satisfy both equations

How do I use graphs to solve linear simultaneous equations?

  • Plot both equations on the same set of axes

    • to do this, you can use a table of values or rearrange into y = mx + c if that helps

  • Find where the lines intersect (cross over)

    • The x and y  solutions to the simultaneous equations are the x- and y-coordinates of the point of intersection

  • e.g. to solve 2x - y = 3 and 3x + y = 7 simultaneously, first plot them both (see graph)

    • find the point of intersection, (2, 1)

    • the solution is x = 2 and y = 1

Solving Equations Graphically Notes Diagram 1, A Level & AS Level Pure Maths Revision Notes

How do I solve linear simultaneous equations from worded contexts?

EPS Notes fig2 (1), downloadable IGCSE & GCSE Maths revision notes
EPS Notes fig2 (2), downloadable IGCSE & GCSE Maths revision notes

Examiner Tips and Tricks

  • Always check that your final solutions satisfy the original simultaneous equations - you will know immediately if you've got the right solutions or not.

Worked Example

Solve the simultaneous equations

5 x plus 2 y equals 11

4 x minus 3 y equals 18

Number the equations.

table row cell 5 x space plus space 2 y space end cell equals cell space 11 space space space space space space space space space space space space end cell row cell 4 x space minus space 3 y space end cell equals cell space 18 space space space space space space space space space space space space end cell end tabletable row blank blank cell open parentheses 1 close parentheses end cell row blank blank cell open parentheses 2 close parentheses end cell end table 

Make the y  terms equal by multiplying all parts of equation (1) by 3 and all parts of equation (2) by 2.
This will give two 6y  terms with different signs. The question could also be done by making the x  terms equal by multiplying all parts of equation (1) by 4 and all parts of equation (2) by 5, and subtracting the equations.

table row cell 15 x space plus space 6 y space end cell equals cell space 33 space space space space space space space space space space end cell row cell 8 x space minus space 6 y space end cell equals cell space 36 space space space space space space space space space space end cell end tabletable row blank blank cell open parentheses 3 close parentheses end cell row blank blank cell open parentheses 4 close parentheses end cell end table

The 6y  terms have different signs, so they can be eliminated by adding equation (4) to equation (3). 

space space space space space space space space space space 15 x space plus space 6 y space equals space 33 space space space space space space space space space space space space bottom enclose space plus open parentheses space space space space space space 8 x space minus space 6 y space equals space 36 close parentheses space end enclose space space space space space space space space space space 23 x space space space space space space space space space space space space space equals space 69 space

Solve the equation to find x  by dividing both sides by 23.

table row cell x space end cell equals cell space 69 over 23 equals space 3 end cell end table

Substitute x = 3 into either of the two original equations.

open parentheses 1 close parentheses space space space space space 5 open parentheses 3 close parentheses space plus space 2 y space equals space 11

Solve this equation to find y.

table row cell 15 space plus space 2 y space end cell equals cell space 11 end cell row cell 2 y space end cell equals cell space 11 space minus space 15 end cell row cell 2 y space end cell equals cell space minus 4 space end cell row cell y space end cell equals cell fraction numerator negative 4 over denominator 2 end fraction space equals space minus 2 end cell end table

Substitute x = 3  and y = - 2 into the other equation 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 row cell 4 x space minus space 3 y space end cell equals cell space 18 end cell end table
   table row cell 4 open parentheses 3 close parentheses space minus space 3 open parentheses negative 2 close parentheses space end cell equals cell space 18 end cell row cell 12 space minus open parentheses negative 6 close parentheses space end cell equals cell space 18 end cell row cell 18 space end cell equals cell space 18 end cell end table

bold italic x bold space bold equals bold space bold 3 bold comma bold space bold space bold italic y bold equals bold minus bold 2

Quadratic Simultaneous Equations

What are quadratic simultaneous equations?

  • When there are two unknowns (say x and y ) in a problem, we need two equations to be able to find them both: these are called simultaneous equations

  • If there is an x2 or y2 or xy  in one of the equations then they are quadratic (or non-linear) simultaneous equations

How do I solve quadratic simultaneous equations?

  • Use the method of substitution

    • Substitute the linear equation, y = ... (or x = ...), into the quadratic equation

      • Do not try to substitute the quadratic equation into the linear equation

  • For example solve x2 + y2 = 25 and y - 2x = 5 

    • Rearrange the linear equation into y = 2x + 5

    • Substitute this into the quadratic equation, replacing all y's with (2x + 5) in brackets

      •  x2 + (2x + 5)2 = 25

    • Expand and solve this quadratic equation (x = 0 and x = -4)

    • Substitute each value of x into the linear equation, y = 2x + 5, to get their value of y

    • Present your solutions in a way that makes it obvious which x  belongs to which y

      • x = 0, y = 5 or x = -4, y = -3

  • Check that your final solutions satisfy both equations

How do you use graphs to solve quadratic simultaneous equations?

  • Plot both equations on the same set of axes

    • to do this, you can use a table of values (or, for straight lines, rearrange into y = mx + c  if it helps)

  • Find where the lines intersect (cross over)

    • The x  and y  solutions to the simultaneous equations are the x- and y-coordinates of the point of intersection

  • e.g. to solve y = x2 + 3x + 1 and y = 2x + 1 simultaneously, first plot them both (see graph)

    • find the points of intersection, (-1, -1) and (0, 1)

    • the solutions are x = -1 and y = -1, or x = 0 and y = 1

Solving Equations Graphically - Notes Diagram 4, A Level & AS Level Pure Maths Revision Notes

Examiner Tips and Tricks

  • If the quadratic resulting from substitution has

    • a repeated root, then the line is a tangent to the curve

    • no roots, then the line does not intersect with the curve – or you have made a mistake!

  • When giving your final answer, make sure you indicate which x  and y  values go together

    • If you don’t make this clear you can lose marks for an otherwise correct answer.

  • Don't make the common mistake of thinking each squared term in x2 + y2 = 25 can be square-rooted to give x + y = 5

    • They can't, the most you can do is square root of x squared plus y squared end root equals plus-or-minus 5

    • But you shouldn't be trying to make x or y  the subject of this anyway!)

Worked Example

Solve the equations

x squared plus y squared equals 36

x equals 2 y plus 6

Number the equations.

x squared space plus space y squared space equals space 36 space space space space space space space space space space space x space equals space 2 y space plus space 6 space space space space space space space space space space space space space space space spaceopen parentheses 1 close parentheses open parentheses 2 close parentheses 

There is one quadratic equation and one linear equation so this must be done by substitution.

Equation (2) is equal to x so this can be eliminated by substituting it into the x part for equation (1).
Substitute x space equals space 2 y space plus space 6 into equation (1).

open parentheses 2 y space plus space 6 close parentheses squared space plus space y squared space equals space 36

Expand the brackets, remember that a bracket squared should be treated the same as double brackets.

open parentheses 2 y space plus space 6 close parentheses open parentheses 2 y space plus space 6 close parentheses space space plus space y squared space equals space 36 4 y to the power of 2 space end exponent plus space 6 open parentheses 2 y close parentheses space plus space 6 open parentheses 2 y close parentheses space plus space 6 squared space plus space y to the power of 2 space end exponent equals space 36

Simplify.

table row cell 4 y to the power of 2 space end exponent plus space 12 y space plus space 12 y space plus space 36 space plus space y to the power of 2 space end exponent end cell equals cell space 36 end cell row cell 5 y to the power of 2 space end exponent plus space 24 y space plus space 36 space end cell equals cell space 36 end cell end table

Rearrange to form a quadratic equation that is equal to zero.

table row cell 5 y to the power of 2 space end exponent plus space 24 y space plus space 36 space minus space 36 space end cell equals cell space 0 end cell row cell 5 y squared space plus space 24 y space end cell equals cell space 0 end cell end table

Both terms contain a power of y
So this can be factorised by taking out a common factor of table attributes columnalign right center left columnspacing 0px end attributes row blank blank y end table.

table row cell y open parentheses 5 y space plus space 24 close parentheses space end cell equals cell space 0 end cell end table

Solve to find the values of y.
Let each factor be equal to 0 and solve.

table row cell y subscript 1 space end cell equals cell space 0 space space space space space space space space space space space space space space end cell row cell 5 y subscript 2 space plus space 24 space end cell equals cell space 0 space space rightwards double arrow space space y subscript 2 equals space minus 24 over 5 space equals space minus 4.8 end cell end table

Substitute the values of y into one of the equations (the linear equation is easier) to find the values of x.
              x subscript 1 space equals space 2 left parenthesis 0 right parenthesis space plus space 6 space equals space 6 space space space space space space space space space space space space x subscript 2 space equals space 2 open parentheses negative 24 over 5 close parentheses space plus space 6 space equals space minus 9.6 space plus space 6

bold italic x subscript bold 1 bold space bold equals bold space bold 6 bold comma bold space bold space bold italic y subscript bold 1 bold equals bold space bold 0
bold italic x subscript bold 2 bold space bold equals bold minus bold 3 bold. bold 6 bold comma bold space bold space bold space bold italic y subscript bold 2 bold space bold equals bold minus bold 4 bold. bold 8

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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.