Quadratic Simultaneous Equations (Cambridge (CIE) IGCSE International Maths)

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Quadratic Simultaneous Equations

What are quadratic simultaneous equations?

  • When there are two unknowns (e.g. 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 substitution

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

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

  • E.g. To solve x squared plus y squared equals 25 and y minus 2 x equals 5

    • Rearrange the linear equation into y equals 2 x plus 5

    • Substitute this into the quadratic equation, replacing all y's with open parentheses 2 x plus 5 close parentheses

      •  x squared plus open parentheses 2 x plus 5 close parentheses squared equals 25

  • Expand and solve this quadratic equation

    • table row cell x squared plus 4 x squared plus 20 x plus 25 end cell equals 25 end table

    • table row cell 5 x squared plus 20 x end cell equals 0 end table

    • table row cell 5 x open parentheses x plus 4 close parentheses end cell equals 0 end table

    • x equals 0 and x equals negative 4

  • Substitute each value of x into the linear equation, y equals 2 x plus 5, to find the corresponding y values

    • y equals 2 open parentheses 0 close parentheses plus 5 equals 5

    • y equals 2 open parentheses negative 4 close parentheses plus 5 equals negative 3

  • 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 your final solutions satisfy both equations

What if the quadratic has repeated roots or no roots?

  • If the resulting quadratic after substituting has a repeated root,

    • then the line is a tangent to the curve

      • i.e. the curve and the line intersect in one place only

    • There is only one solution for x and y

  • If the resulting quadratic to be solved has no roots,

    • then the line does not intersect with the curve

    • There are no solutions to the simultaneous equations

    • If this happens it may be an indicator that your working is wrong!

What if I can't substitute one equation into the other straight away?

  • If the linear equation is not in the form y = ... or x = ...

    • You will need to rearrange it first, so that it can be substituted into the quadratic equation

  • Consider solving x y equals 3 and x plus y equals 4

  • Either:

    • Rearrange the second equation to y equals 4 minus x and substitute into x y equals 3

      • x open parentheses 4 minus x close parentheses equals 3

      • Expanding produces a quadratic that can be solved for x

      • 4 x minus x squared equals 3

    • Or rearrange the first equation to y equals 3 over x and substitute into x plus y equals 4

      • x plus 3 over x equals 4

      • Multiplying both sides by x produces a quadratic that can be solved for x

      • x squared plus 3 equals 4 x

How do I use a graph 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 it can help to rearrange into y = mx + c

  • Find the point where the lines intersect

    • 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 below)

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

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

Graphs of x^2 + 3x + 1 and y=2x+1 on the same graph

Examiner Tips and Tricks

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

Worked Example

Solve the equations

table row cell x squared plus y squared end cell equals 36 row cell x minus 2 y end cell equals 6 end table

Number the equations

table row cell x squared plus y squared end cell equals cell 36 space space space space space space space space space space space space end cell row cell x minus 2 y end cell equals cell 6 space space space space space space space space space space space space space space space end cell end tablecircle enclose 1
circle enclose 2 

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

Equation 2 can be rearranged to make x the subject, which can then be substituted into equation 1

You could rearrange to make y the subject instead, but this results in a fraction which can be more tricky to deal with

Rearranging equation 2

x equals 2 y plus 6

Substituting into equation 1

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

Expand the brackets
Remember that a bracket squared should be treated the same as double brackets

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

Simplify

table row cell 4 y squared plus 12 y plus 12 y plus 36 plus y squared end cell equals 36 row cell 5 y squared plus 24 y plus 36 end cell equals 36 end table

Rearrange to form a quadratic equation that is equal to zero
Do this by subtracting 36 from both sides

table row cell 5 y squared plus 24 y end cell equals 0 end table

Take out the 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 plus 24 close parentheses end cell equals 0 end table

Solve to find the values of y by equating each factor to zero

y equals 0 or 5 y plus 24 equals 0

Solve the linear equation above

y equals negative 24 over 5

So the two y values are

table row cell y subscript 1 end cell equals cell 0 space end cell row cell y subscript 2 end cell equals cell negative 24 over 5 end cell end table

Substitute the values of y into one of the equations (the linear equation is easiest) to find the values of x

x equals 2 y plus 6

x subscript 1 equals 2 left parenthesis 0 right parenthesis plus 6 equals 6 space space space space space space space space space space space space x subscript 2 equals 2 open parentheses negative 24 over 5 close parentheses plus 6 equals negative 18 over 5

Write the final solutions in clear pairs

bold italic x subscript bold 1 equals bold 6 comma space space bold italic y subscript 1 equals 0
bold italic x subscript bold 2 bold equals bold minus bold 18 over bold 5 bold comma bold space bold space bold italic y subscript bold 2 bold equals bold minus bold 24 over bold 5

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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.