National 5MathsSQARevision NotesAlgebraic SkillsLinear Equations & InequationsLinear Equations

Linear Equations (SQA National 5 Maths): Revision Note

Exam code: X847 75

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

Solving linear equations with integer coefficients

What are linear equations?

  • A linear equation is one that can be written (or rewritten) in the form a x plus b equals c

    • a comma space b comma and c are numbers and xis the variable

      • 2x + 3 = 5

      • 3x + 4 = 1

      • x - 5 = -3

  • The greatest power of x is 1

    • There are no terms like x2

Examiner Tips and Tricks

You should be familiar with solving simple linear equations from your National 4 Maths course.

This is normally done by the 'balancing method', i.e. 'doing the same thing to both sides of the equation'. For example:

3 x minus 7 equals 35 open parentheses plus 7 close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses plus 7 close parentheses 3 x equals 42 open parentheses divided by 3 close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses divided by 3 close parentheses x equals 42 over 3 equals 14

How do I solve linear equations with brackets?

  • If a linear equation involves brackets, expand the brackets first

  • For example, solve 2 open parentheses x minus 3 close parentheses equals 10

    • Expand the brackets 

2 x minus 6 equals 10 space

  • Then solve as usual

    • Add 6 then divide by 2

table row cell 2 x end cell equals 16 row x equals 8 end table

How do I solve linear equations with x terms on both sides?

  • Collect the x terms (or whichever variable is involved) together on one side 

    • To do this, remove all the x terms from one side

      • It is easiest to remove the smallest x term to avoid negatives

      • This will be the most negative x term, if one or both of the x terms have a minus sign in front of them

  • For example, 4 x minus 7 equals 11 plus x

  • Remove the x term on the right-hand side, by subtracting x from both sides

table row cell 4 x minus 7 end cell equals cell 11 plus x end cell end table

table row blank blank cell open parentheses negative x close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses negative x close parentheses end cell end table

table row cell 3 x minus 7 end cell equals 11 end table

  • There are no longer any x terms on the right

  • This can now be solved as usual

    • Add 7 then divide by 3         

table row cell 3 x minus 7 end cell equals 11 row cell 3 x end cell equals 18 row x equals 6 end table

Worked Example

Solve the equation space 3 minus 5 x equals 6 open parentheses x minus 5 close parentheses.

Answer:

Start by expanding the brackets

3 minus 5 x equals 6 x minus 30

Get all the x terms onto one side

  • Here it is easiest to remove the -5x from the left hand side

  • -5x is 'the most negative x term' of the two x terms

Add 5x to both sides

table row 3 equals cell 11 x minus 30 end cell end table

Get 11x on its own by adding 30 to both sides

33 equals 11 x 

Divide both sides by 11 to find x

3 equals x 

x equals 3

Solving linear equations with fractional coefficients

How do I solve linear equations with fractions?

  • If a linear equation contains fractions (including algebraic fractions), multiply both sides by the lowest common denominator

  • For example, x over 5 plus 4 equals 9 over 2

    • The lowest common denominator of 5 and 2 is 10

    • Multiply all terms on both sides by 10

table row cell open parentheses 10 cross times x over 5 close parentheses plus open parentheses 10 cross times 4 close parentheses end cell equals cell 10 cross times 9 over 2 end cell row cell fraction numerator 10 x over denominator 5 end fraction plus 40 end cell equals cell 90 over 2 end cell end table

  • Simplify the fractions

table row cell 2 x plus 40 end cell equals 45 end table

  • Now solve as before, by subtracting 40, then dividing by 2

table row cell 2 x end cell equals 5 row x equals cell 5 over 2 end cell end table

  • Unless the question specifies otherwise, you can leave the answer as a fraction

    • An equivalent decimal or mixed number would usually also be accepted

Examiner Tips and Tricks

Instead of multiplying by the lowest common denominator, you can multiply first by one denominator, and then by the other (see the Worked Example).

Either method will lead to the correct answer if applied correctly.

  • The lowest common denominator method can save some time

  • But it is better to use the method you are most confident with

What if the unknown is in the denominator?

  • For example fraction numerator 4 over denominator x minus 2 end fraction equals 3

    • Multiply both sides of the equation by the denominator

fraction numerator 4 over denominator x minus 2 end fraction cross times open parentheses x minus 2 close parentheses equals 3 open parentheses x minus 2 close parentheses

  • Simplify the fractions, and expand any brackets

4 equals 3 open parentheses x minus 2 close parentheses 4 equals 3 x minus 6

  • Now solve as usual

    • Add 6 then divide by 3

table row 10 equals cell 3 x end cell row cell 10 over 3 end cell equals x end table

Worked Example

Solve the equation fraction numerator 4 x plus 1 over denominator 2 end fraction minus 7 equals fraction numerator 5 x over denominator 3 end fraction.

Answer:

Method 1 (using Lowest Common Denominator)

The lowest common denominator of 2 and 3 is 6, so multiply both sides by 6

table row cell fraction numerator 3 up diagonal strike 6 cross times open parentheses 4 x plus 1 close parentheses over denominator up diagonal strike 2 end fraction minus 6 cross times 7 end cell equals cell fraction numerator 2 up diagonal strike 6 cross times 5 x over denominator up diagonal strike 3 end fraction end cell row cell 3 open parentheses 4 x plus 1 close parentheses minus 42 end cell equals cell 10 x end cell end table

Expand the brackets and simplify

table row cell 12 x plus 3 minus 42 end cell equals cell 10 x end cell row cell 12 x minus 39 end cell equals cell 10 x end cell end table

Subtract 10x from both sides

table row cell 2 x minus 39 end cell equals 0 end table

Add 39 to both sides

2 x equals 39

Divide both sides by 2

x equals 39 over 2 space space open parentheses or space space x equals 19 1 half space space or space space x equals 19.5 close parentheses
 

Method 2 (multiplying by denominators separately)

Multiply both sides by 2 to get rid of the fraction on the left, then simplify

table row cell fraction numerator up diagonal strike 2 cross times open parentheses 4 x plus 1 close parentheses over denominator up diagonal strike 2 end fraction minus 2 cross times 7 end cell equals cell fraction numerator 2 cross times 5 x over denominator 3 end fraction end cell row cell 4 x plus 1 minus 14 end cell equals cell fraction numerator 10 x over denominator 3 end fraction end cell row cell 4 x minus 13 end cell equals cell fraction numerator 10 x over denominator 3 end fraction end cell end table

Multiply both sides by 3 to get rid of the fraction on the right

table row cell 3 open parentheses 4 x minus 13 close parentheses end cell equals cell fraction numerator up diagonal strike 3 cross times 10 x over denominator up diagonal strike 3 end fraction end cell row cell 3 open parentheses 4 x minus 13 close parentheses end cell equals cell 10 x end cell end table

Expand the brackets

table row cell 12 x minus 39 end cell equals cell 10 x end cell end table

Subtract 10x from both sides

table row cell 2 x minus 39 end cell equals 0 end table

Add 39 to both sides

2 x equals 39

Divide both sides by 2

x equals 39 over 2 space space open parentheses or space space x equals 19 1 half space space or space space x equals 19.5 close parentheses

Unlock more, it's free!

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

the (exam) results speak for themselves:

    Previous:Function Notation f(x)Next:Linear Inequations
    Author
    Reviewer
    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.