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Quadratic Simultaneous Equations (Cambridge O Level Additional Maths)

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Amber

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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
  • 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 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

Solution of quadratic simultaneous equations as the points of intersection of their graphs

Examiner Tip

  • If the resulting quadratic has a repeated root then the line is a tangent to the curve (as there is only 1 solution)
  • If the resulting quadratic has no roots then the line does not intersect with the curve (as there are 0 solutions) – 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

Worked example

Solve the equations

x2 + y2 = 36
x = 2y + 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 space open parentheses 1 close parentheses 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 space 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

[1]

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

[1]

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

[1]

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

The question does not give a specified degree of accuracy, so this can be factorised.
Take out the common factor of table 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

[1]

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 [1]

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    Amber

    Author: Amber

    Expertise: Maths

    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.