Solving Equations Graphically (Edexcel IGCSE Further Pure Maths)

Revision Note

Dan Finlay

Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Solving Equations Graphically

How can I solve equations graphically?

  • A graph can be used to to help solve an equation like space straight f left parenthesis x right parenthesis equals straight g left parenthesis x right parenthesis

    • Draw the graphs of space y equals straight f left parenthesis x right parenthesis and space y equals straight g left parenthesis x right parenthesis

    • The solutions are the x-coordinates of the points of intersection

  • This can be used when an equation is difficult or impossible to solve algebraically

    • The solutions found will usually be approximations rather than exact answers

    • The more accurate the graph, the more accurate the approximation

How can I estimate a solution by drawing a line on a graph?

  • An exam question my ask you to estimate a solution by drawing a 'suitable' (or 'appropriate') straight line on a graph

  • Often this will be a horizontal line

    • For example solving straight f open parentheses x close parentheses equals k for some constant k

      • On a graph of y equals straight f open parentheses x close parentheses, draw the line y equals k

      • The solutions are the x-coordinates of any points of intersection

    • Finding roots by seeing where a graph crosses the x-axis is a special case of this

      • The x-axis is the horizontal line with equation  bold italic y bold equals bold 0

  • Sometimes it will be the line  bold italic y bold equals bold italic x

    • Draw this on the graph of y equals straight f open parentheses x close parentheses to find the solution(s) of straight f open parentheses x close parentheses equals x

  • But sometimes determining the line to draw will be more challenging

    • For example, 'By drawing an appropriate straight line on the graph of y equals 3 plus 2 straight e to the power of negative 2 x end exponent, estimate the root of the equation  ln open parentheses x minus 3 close parentheses cubed equals negative 6 x'

    • We need to rewrite the equation in the form  straight g open parentheses x close parentheses equals 3 plus 2 straight e to the power of negative 2 x end exponent, where y equals straight g open parentheses x close parentheses is the equation of a straight line

    • Take the exponential of both sides ('exp cancels log')

      • open parentheses x minus 3 close parentheses cubed equals straight e to the power of negative 6 x end exponent

    • Take the cube root of both sides

      • x minus 3 equals straight e to the power of negative 2 x end exponent

    • Multiply both sides by 2

      • 2 x minus 6 equals 2 straight e to the power of negative 2 x end exponent

    • Add 3 to both sides

      • 2 x minus 3 equals 3 plus 2 straight e to the power of negative 2 x end exponent

    • That equation is equivalent to  ln open parentheses x minus 3 close parentheses cubed equals negative 6 x

      • it will have the same solutions

    • So we need to draw the line  y equals 2 x minus 3  on the graph of  y equals 3 plus 2 straight e to the power of negative 2 x end exponent

      • the x-coordinates of the points of intersection will give the solution(s) for  2 x minus 3 equals 3 plus 2 straight e to the power of negative 2 x end exponent

      • But those are the same as the solution(s) for  ln open parentheses x minus 3 close parentheses cubed equals negative 6 x

Examiner Tips and Tricks

  • Be extra careful when drawing graphs on 'estimate solutions by using a graph' questions

    • The accuracy of your answer will depend on the accuracy of your drawing

    • Use a ruler for straight lines

Worked Example

A graph of  y equals 2 to the power of open parentheses x over 3 plus 1 close parentheses end exponent minus 1  in the interval  0 less or equal than x less or equal than 6  is shown in the following diagram

Graph of exponential function

By drawing a suitable straight line on the grid, show that the equation  log subscript 2 open parentheses 3 x minus 1 close parentheses squared minus 2 over 3 x equals 2  has a root in the interval  0 less or equal than x less or equal than 6, and obtain an estimate for the value of that root.

Be careful here – we cannot just draw the horizontal line y equals 2
That would only work if we had the graph of  y equals log subscript 2 open parentheses 3 x minus 1 close parentheses squared minus 2 over 3 x

Instead we must work on rearranging the equation
Start by getting the logarithm alone on the left-hand side

log subscript 2 open parentheses 3 x minus 1 close parentheses squared equals 2 over 3 x plus 2

Use laws of logarithms to bring the power down in front of the logarithm
Then divide both sides of the equation by 2

table row cell 2 log subscript 2 open parentheses 3 x minus 1 close parentheses end cell equals cell 2 over 3 x plus 2 end cell row cell log subscript 2 open parentheses 3 x minus 1 close parentheses end cell equals cell x over 3 plus 1 end cell end table

Now take both sides to the power of 2
This will cancel the logarithm on the left-hand side ('exp cancels log')

3 x minus 1 equals 2 to the power of open parentheses x over 3 plus 1 close parentheses end exponent

Finally subtract 1 from both sides

3 x minus 2 equals 2 to the power of open parentheses x over 3 plus 1 close parentheses end exponent minus 1

Now the right-hand side is the function that is graphed on the diagram
So we need to draw the straight line  y equals 3 x minus 2
We can estimate the root by considering the x-coordinate of the point of intersection

Solution graph for question


root:  bold italic x bold almost equal to bold 1 bold. bold 2

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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.

Lucy Kirkham

Author: Lucy Kirkham

Expertise: Head of STEM

Lucy has been a passionate Maths teacher for over 12 years, teaching maths across the UK and abroad helping to engage, interest and develop confidence in the subject at all levels.Working as a Head of Department and then Director of Maths, Lucy has advised schools and academy trusts in both Scotland and the East Midlands, where her role was to support and coach teachers to improve Maths teaching for all.