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

Revision Note

Did this video help you?

Linear Simultaneous Equations

What are linear simultaneous equations?

  • When there are two unknowns (x and y), we need two equations to find them both

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

      • The values that work are x = 3 and y = 1

  • These are called linear simultaneous equations

    • Linear because there are no terms like x2 or y2 

How do I solve linear simultaneous equations by elimination?

  • Elimination removes one of the variables, or y

  • To eliminate the x's from 3x + 2y = 11 and 2x - = 5, make the number in front of the x (the coefficient) in both equations the same (the sign may be different)

    • Multiply every term in the first equation by 2

      • 6x + 4y = 22

    • Multiply every term in the second equation by 3

      • 6- 3y = 15

    • Subtracting the second equation from the first eliminates x

      • When the sign in front of the term you want to eliminate is the same, subtract the equations

negative bottom enclose table row cell 6 x plus 4 y equals 22 end cell row cell 6 x minus 3 y equals 15 end cell end table end enclose
space space space space space space space space space space space space space 7 y equals 7

  • The y terms have become 4- (-3y) = 7(be careful with negatives

    • Solve the resulting equation to find y

    • = 1

  • Then substitute = 1 into one of the original equations to find x

    • 3x + 2 = 11, so 3= 9, giving x = 3

  • Write out both solutions together, = 3 and = 1

  • Alternatively, you could have eliminated the y's from 3+ 2= 11 and
    2- = 5 by making the coefficient of y in both equations the same 

    • Multiply every term in the second equation by 2

    • Adding this to the first equation eliminates y (and so on)

      • When the sign in front of the term you want to eliminate is different, add the equations

plus bottom enclose table row cell 3 x plus 2 y equals 11 end cell row cell 4 x minus 2 y equals 10 end cell end table end enclose space
space space space space 7 x space space space space space space space space space equals 21

How do I solve linear simultaneous equations by substitution?

  • Substitution means substituting one equation into the other

    • This is an alternative method to elimination

      • You can still use elimination if you prefer

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

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

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

    • Substitute this into the first equation

      • This means 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

      • = 1

How do I solve linear simultaneous equations graphically?

  • Plot both equations on the same set of axes

    • To do this, you can use a table of values

      • or rearrange into y = mx + if that helps

  • Find where the lines intersect (cross over)

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

  • For example, to solve 2- y = 3 and 3x + y = 4 simultaneously

    • First plot them both on the same axes (see graph)

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

    • The solution is x = 2 and = 1

The intersection of 2x-y=3 and 3x+y=7 is (2,1)

Examiner Tips and Tricks

  • Always check that your final solutions satisfy both original simultaneous equations!

  • Write out both solutions (x and y) together at the end to avoid examiners missing a solution in your working

Worked Example

Solve the simultaneous equations

table row cell 5 x plus 2 y end cell equals 11 row cell 4 x minus 3 y end cell equals 18 end table

It helps to number the equations

table row cell 5 x plus 2 y end cell equals cell 11 space space space space space space space space space space space space end cell row cell 4 x minus 3 y end cell equals 18 end tabletable row blank blank cell circle enclose 1 end cell end table
table row blank blank cell circle enclose 2 end cell end table 

We will choose to eliminate the terms
Make the y terms equal by multiplying all parts of equation 1 by 3 and all parts of equation 2 by 2

table row cell 15 x plus 6 y end cell equals cell 33 space space space space space space space space space end cell row cell 8 x minus 6 y end cell equals cell 36 space space space space space space space space end cell end tabletable row blank blank cell circle enclose 3 end cell end table
table row blank blank cell circle enclose 4 end cell end table

The 6terms 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 plus 6 y equals 33 space space space space space space space space space space space space
bottom enclose space plus space space space space space space space 8 x minus 6 y equals 36 space end enclose
space space space space space space space space space space 23 x space space space space space space space space space equals 69 space

Solve the equation to find x (divide both sides by 23)

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

Substitute x = 3 into either of the two original equations

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

Solve this equation to find y

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

Substitute x = 3  and = - 2 into the other equation to check that they are correct

table row blank blank cell circle enclose 2 space space space space space end cell end table table row cell 4 x minus 3 y end cell equals 18 end table
   table row cell 4 open parentheses 3 close parentheses minus 3 open parentheses negative 2 close parentheses end cell equals 18 row cell 12 minus open parentheses negative 6 close parentheses end cell equals 18 row 18 equals 18 end table

Write out both solutions together

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

This question can also be done by eliminating x first (multiplying equation 1 by 4 and equation 2 by 5 then subtracting)

How do I form simultaneous equations?

  • Introduce two letters, x and y, to represent two unknowns

    • Make sure you know exactly what they stand for (and any units)

  • Create two different equations from the words or contexts

    • 3 apples and 2 bananas cost $1.80, while 5 apples and 1 banana cost $2.30 

      • 3x + 2y = 180 and 5x + = 230
        x is the price of an apple, in cents
        y is the price of a banana, in cents

      • This question could also be done in dollars, $

  • Solve the equations simultaneously

  • Give answers in context (relate them to the story, with units)

    • x = 40, y = 30

    • In context: an apple costs 40 cents and a banana costs 30 cents

  • Some questions don't ask you to solve simultaneously, but you still need to

    • Two numbers have a sum of 19 and a difference of 5, what is their product?

      • x + y = 19 and - = 5

      • Solve simultaneously to get x = 12, = 7

      • The product is xy = 12 × 7 = 84

Examiner Tips and Tricks

  • Always check that you've answered the question! Sometimes finding and y isn't the end

    • E.g. you may have to state a conclusion

Worked Example

At a bakery a customer pays £9 in total for six bagels and twelve sausage rolls.

Another customer buys nine bagels and ten sausage rolls, which costs £12.30 in total.

Find the cost of 5 bagels and 15 sausage rolls.

The two variables are the price of bagels, b, and the price of sausage rolls, s

Write an equation for the first customer's purchases, and label it equation 1

circle enclose 1 space space space 6 b plus 12 s equals 9

Write an equation for the second customer's purchases, and label it equation 2

circle enclose 2 space space space 9 b plus 10 s equals 12.3

We will choose to eliminate the b terms
Make the b terms equal by multiplying all parts of equation 1 by 3 and all parts of equation 2 by 2
Label these as equations 3 and 4

table row cell circle enclose 1 cross times 3 space space space 18 b plus 36 s end cell equals cell 27 space space space space space space space space space space circle enclose 3 end cell row cell circle enclose 2 cross times 2 space space space 18 b plus 20 s end cell equals cell 24.6 space space space space space space space circle enclose 4 end cell end table

To eliminate b, subtract equation 4 from equation 3

circle enclose 3 minus circle enclose 4 space space space space space 16 s equals 2.4

Solve for s

s equals fraction numerator 2.4 over denominator 16 end fraction equals 0.15

Substitute this into either equation to find b, we will use equation 1

table row cell circle enclose 1 space space space 6 b plus 12 open parentheses 0.15 close parentheses end cell equals 9 row cell 6 b plus 1.8 end cell equals 9 row cell 6 b end cell equals cell 7.2 end cell row b equals cell 1.2 end cell end table

So sausage rolls cost £0.15 each and bagels cost £1.20 each

Use these values to find the price of 5 bagels and 15 sausage rolls

open parentheses 5 cross times 1.2 close parentheses plus open parentheses 15 cross times 0.15 close parentheses equals 8.25

£8.25

Last updated:

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

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.