Algebraic Solution to Simultaneous Equations (SQA National 5 Maths): Revision Note

Exam code: X847 75

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

Solving simultaneous equations algebraically

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 

    • Simultaneous because they have to be solved at the same time, to find answers that work for both

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 signs in front of the terms you want to eliminate is the same, subtract the equations

bottom enclose table row cell space space space 6 x plus 4 y equals 22 end cell row cell negative open parentheses 6 x minus 3 y equals 15 close parentheses 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 signs in front of the terms you want to eliminate are different, add the equations

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

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.

Examiner Tips and Tricks

An exam question will often tell you to solve a system of simultaneous equations algebraically.

This means that if you find the correct solution by trial and error, or by any other non-algebraic method, you will receive 0 marks for the question.

Worked Example

Solve, algebraically, the system of equations

table row cell 5 x plus 4 y end cell equals 13 row cell 4 x minus 6 y end cell equals 15 end table

Answer:

It helps to number the equations

table row cell 5 x plus 4 y end cell equals cell 13 space space space space space space space space space space space space end cell row cell 4 x minus 6 y end cell equals 15 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 

'Rescale' the equations to eliminate the terms

  • You could also choose to eliminate the x terms first

  • In that case you would multiply equation 1 by 4 and equation 2 by 5 and then subtract the equations

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 12 y end cell equals cell 39 space space space space space space space space space end cell row cell 8 x minus 12 y end cell equals cell 30 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 12terms have different signs, so they can be eliminated by adding equation 4 to equation 3 

space space space space space 15 x plus 12 y equals 39 space space space space space space space space space space space space
bottom enclose space plus open parentheses 8 x minus 12 y equals 30 close parentheses space end enclose
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 4 y equals 13

Solve this equation to find y

table row cell 15 plus 4 y end cell equals 13 row cell 4 y end cell equals cell 13 minus 15 end cell row cell 4 y end cell equals cell negative 2 space end cell row y equals cell fraction numerator negative 2 over denominator 4 end fraction equals negative 1 half end cell end table

Substitute x = 3  and = negative 1 half 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 6 y end cell equals 15 end table

   table row blank blank cell 4 open parentheses 3 close parentheses minus 6 open parentheses negative 1 half close parentheses end cell row blank equals cell 12 minus open parentheses negative 3 close parentheses end cell row blank equals cell 12 plus 3 end cell row blank equals 15 end table

Write out both solutions together

x equals 3 comma space space y equals negative 1 half

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Roger B

Author: Roger B

Expertise: Maths Content Creator

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Maths Subject 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.