Linear Inequations (SQA National 5 Maths): Revision Note

Exam code: X847 75

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

Solving linear inequations

What is an inequation?

  • An inequation (also known as an inequality) tells you that something is greater than (>) or less than (<) something else

    • x > 5 means x is greater than 5 

      • x could be 6, 7, 8, 9, ...

      • or any number greater than 5, like 5.25 or 97.8 or square root of 26 equals5.0990195...

        • x doesn't need to be an integer

  • Inequations may also include being equal (=) 

    • ⩾ means greater than or equal to

    • ⩽ means less than or equal to

      • x ⩽ 10 means x is less than or equal to 10

        • x could be 10, 9, 8, 7, 6,....

        • or 9.5 or -3

  • When inequations cannot be equal, they are called strict inequations

    • > and < are strict inequations

      • x > 5 does not include 5 (strict)

      • x ⩾ 5 does include 5 (not strict)

How do I solve linear inequations?

  • Solving linear inequations is just like Solving Linear Equations

    • Follow the same rules, but keep the inequality sign throughout

    • If you change the inequality sign to an equals sign you are changing the meaning of the problem

  • When you multiply or divide both sides by a negative number, you must flip the sign of the inequation 

    • E.g. 

space 1 less than 2
open parentheses cross times negative 1 close parentheses space space space space space space space space space space space space space space space space space space space open parentheses cross times negative 1 close parentheses
space minus 1 greater than negative 2

  • Never multiply or divide by a variable (x) as this could be positive or negative

  • The safest way to rearrange is simply to add and subtract to move

    • all the variable terms onto one side

    • and all the number terms onto the other side

Examiner Tips and Tricks

When solving inequations, remember:

  • Do not change the inequality sign to an equals sign

    • In an exam you can lose marks for doing this 

  • Reverse the direction of the inequality sign when multiplying or dividing both sides by a negative number!

Worked Example

Solve, algebraically, the inequation 4 open parentheses x minus 3 close parentheses plus 7 less than 9 x plus 10.

Answer:

Expand the brackets and simplify

table row cell 4 x minus 12 plus 7 end cell less than cell 9 x plus 10 end cell row cell 4 x minus 5 end cell less than cell 9 x plus 10 end cell end table

Subtract 4x from both sides

table row cell negative 5 end cell less than cell 9 x plus 10 minus 4 x end cell row cell negative 5 end cell less than cell 5 x plus 10 end cell end table

Subtract 10 from both sides

table row cell negative 5 minus 10 end cell less than cell 5 x end cell row cell negative 15 end cell less than cell 5 x end cell end table

Divide both sides by 5

  • 5 is not a negative number, so there is no need to flip the inequality sign

table row cell fraction numerator negative 15 over denominator 5 end fraction end cell less than x row cell negative 3 end cell less than x end table

That is the answer you are looking for

  • Although you could also 'flip the inequation over' to get x on the left

negative 3 less than x space space or space space x greater than negative 3 

Examiner Tips and Tricks

That Worked Example was solved in a way to avoid having to divide by a negative number. Instead the solution could have proceeded like this:

table row cell 4 x minus 5 end cell less than cell 9 x plus 10 end cell row cell 4 x minus 5 minus 9 x end cell less than 10 row cell negative 5 x minus 5 end cell less than 10 row cell negative 5 x end cell less than cell 10 plus 5 end cell row cell negative 5 x end cell less than 15 end table

That can be solved by dividing by -5, but then you need to remember to flip the inequality sign:

table row x bold greater than cell fraction numerator 15 over denominator negative 5 end fraction end cell row x greater than cell negative 3 end cell end table

Worked Example

Solve, algebraically, the inequation fraction numerator x minus 1 over denominator 4 end fraction plus 3 greater than fraction numerator 6 x over denominator 5 end fraction.

Answer:

Multiply both sides by 4 to get rid of the fraction on the left

  • 4 is not a negative number, so there is no need to flip the inequality sign

table row cell fraction numerator up diagonal strike 4 cross times open parentheses x minus 1 close parentheses over denominator up diagonal strike 4 end fraction plus 4 cross times 3 end cell greater than cell fraction numerator 4 cross times 6 x over denominator 5 end fraction end cell row cell open parentheses x minus 1 close parentheses plus 12 end cell greater than cell fraction numerator 24 x over denominator 5 end fraction end cell row cell x plus 11 end cell greater than cell fraction numerator 24 x over denominator 5 end fraction end cell end table

Multiply both sides by 5 to get rid of the fraction on the right

  • 5 is not a negative number, so there is no need to flip the inequality sign

table row cell 5 open parentheses x plus 11 close parentheses end cell greater than cell fraction numerator up diagonal strike 5 cross times 24 x over denominator up diagonal strike 5 end fraction end cell row cell 5 x plus 55 end cell greater than cell 24 x end cell end table

Subtract 5x from both sides

table row 55 greater than cell 24 x minus 5 x end cell row 55 greater than cell 19 x end cell end table

Divide both sides by 19

  • 19 is not a negative number, so there is no need to flip the inequality sign

55 over 19 greater than x

That is the answer you are looking for

  • Although you could also 'flip the inequation over' to get x on the left

55 over 19 greater than x space space or space space x less than 55 over 19 

How do I find integers that satisfy inequations?

  • Sometimes you may be interested in particular integers (whole numbers) that satisfy an inequation

  • If you are given two end points then look at whether each end point is included or not 

    • 3 ⩽ x ⩽ 6

      • x = 3, 4, 5, 6

    • 3 ⩽ x < 6

      • x = 3, 4, 5

    • 3 < x ⩽ 6

      • x = 4, 5, 6

    • 3 < x < 6

      • x = 4, 5

  • If only one end point is given, there are an infinite number of integers

    • x > 2

      • x = 3, 4, 5, 6, ...

    • x ⩽ 2

      • x = 2, 1, 0, -1, -2, ...

      • Remember zero and negative whole numbers are integers

      • If you only wanted positive integers then just list x = 2, 1

  • You can also find integers that satisfy two inequations

    • 0 < x < 5 and x ⩾ 3

      • List separately: x = 1, 2, 3, 4 and x = 3, 4, 5, 6,  ...

      • Find the values that appear in both lists: x = 3, 4 

  • Or you can find the smallest or largest integer

    • The smallest integer that satisfies x > 6.5 is 7

Examiner Tips and Tricks

If the question does not say x is an integer, do not assume x is an integer!

  • x > 3 actually means any value greater than 3

    • 3.1 is possible

    • pi = 3.14159... is possible

Worked Example

List all the integer values of x that satisfy 

negative 4 less or equal than x less than 2

Answer:

Integer values are whole numbers 

  • -4 ≤ x shows that x includes -4, so this is the first integer

x = -4

x < 2 shows that x does not include 2

  • Therefore the last integer is x = 1

x = 1

For the answer, list all the integers from -4 to 1

  • Remember integers can be zero and negative

x equals negative 4 comma space minus 3 comma space minus 2 comma space minus 1 comma space 0 comma space 1

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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.