Solving Linear Equations (Edexcel IGCSE Maths A (Modular))

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Flashcards

Solving Linear Equations

What are linear equations?

  • A linear equation is one that can be written 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

How do I solve linear equations?

  • You need to use operations like adding, subtracting, multiplying and dividing to get x on its own

  • Any operation you do to one side of the equation must also be done to the other side

  • For example, to solve  2 x plus 1 equals 9 look at the +1 on the left

    • Undo this by subtracting 1 from both sides and simplifying

table row cell 2 x plus 1 end cell equals 9 end table

table row blank blank cell left parenthesis negative 1 right parenthesis 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 end cell end table space space table row blank blank cell space space space space space space space space space space space space space space space space space space space left parenthesis negative 1 right parenthesis end cell end table

table row cell 2 x end cell equals 8 end table

  • This equation is now easier to solve

  • 2x is 2 × so undo this by dividing both sides by 2 and simplifying

table row cell 2 x end cell equals 8 end table

table row blank blank cell left parenthesis divided by 2 right parenthesis space space space space space space space space space space end cell end table space space table row blank blank cell space space space space space space space space space space space space space left parenthesis divided by 2 right parenthesis end cell end table

table row x equals 4 end table

  • The solution to the equation is x = 4

  • Adding 1 was undone by subtracting 1

  • Multiplying by 2 was undone by dividing by 2

    • Addition and subtraction are said to be inverse (opposite) operations

    • Multiplication and division are also inverse operations

Does the order of steps matter?

  • As long as each step is applied correctly, the order in which inverse operations are applied does not matter

    • Applying the operations in one order may be easier than another

  • Consider 4 x plus 8 equals 12

    • It is easier to first subtract 8 from both sides

4 x equals 4

  • Then divide both sides by 4

table row x equals 1 end table

  • If you want to first divide by 4, a common mistake is to write x plus 8 equals 3

    • This is incorrect as 8 has not been divided by 4

    • You must divide every term by 4

table row cell fraction numerator 4 x over denominator 4 end fraction plus 8 over 4 end cell equals cell 12 over 4 end cell row cell x plus 2 end cell equals 3 end table

  • Then subtract 2 from both sides

table row x equals 1 end table

How do I solve linear equations with negative numbers?

  • For example, 2 minus 3 x equals 10

    • Subtract 2 from both sides and simplify

table row cell 2 minus 3 x end cell equals 10 end table

table row blank blank cell left parenthesis negative 2 right parenthesis space space space space space space space space space space space space space space space space space space space space space end cell end table space space table row blank blank cell space space space space space space space space space space space space space space space space space space space space space left parenthesis negative 2 right parenthesis end cell end table

table row cell negative 3 x end cell equals 8 end table

  • Then divide both sides by -3 and simplify

table row cell negative 3 x end cell equals 8 end table

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell space left parenthesis divided by negative 3 right parenthesis space space space space space space space space space space space space space space space space space space space end cell end table space space table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell space space space space space space space space space space space space space space space space space space space space left parenthesis divided by negative 3 right parenthesis end cell end table

table row x equals cell negative 8 over 3 end cell end table

  • Some people prefer to write 2 minus 3 x equals 10 as negative 3 x plus 2 equals 10, which is equivalent

    • You then subtract 2 and divide by -3 as before

Examiner Tips and Tricks

Substitute your answer back into the original equation to check you got it right!

Worked Example

Solve the equation

9 minus 7 x equals 5 

Subtract 9 from both sides of the equation

table row cell negative 7 x end cell equals cell negative 4 end cell end table  

Divide both sides by -7
Remember that a negative divided by a negative will result in a positive number

x equals fraction numerator negative 4 over denominator negative 7 end fraction 

bold italic x bold equals bold 4 over bold 7

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 shown previously

    • Add 6 then divide by 2

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

  • Expanding brackets first will always work, but you can also divide first

    • Dividing both sides of 2 open parentheses x minus 3 close parentheses equals 10 by 2 gives open parentheses x minus 3 close parentheses equals 5

      • which gives x equals 8

    • This method works but can lead to harder fractions

How do I solve linear equations with fractions?

  • If a linear equation contains 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 like this

    • A decimal or mixed number would also be accepted

What if the unknown is on 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 before, by adding 6 to both sides, then dividing by 3

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

Worked Example

(a) Solve the equation

5 open parentheses 3 minus 4 x close parentheses plus 1 equals 26

Expand the bracket

table row cell 15 minus 20 x plus 1 end cell equals 26 end table

Simplify

table row cell 16 minus 20 x end cell equals 26 end table

It can be helpful to consider 16 - 20x as -20x + 16
Subtract 16 from both sides

table row cell negative 20 x end cell equals 10 end table

Divide both sides by -20 and simplify

table row x equals cell fraction numerator 10 over denominator negative 20 end fraction end cell row x equals cell negative 1 half end cell end table

bold italic x bold equals bold minus bold 1 over bold 2

-0.5 is also accepted

(b) Solve the equation

fraction numerator 5 x over denominator 4 end fraction equals 1 half

The lowest common denominator of 4 and 2 is 4
Multiply both sides by 4

table row cell 4 cross times fraction numerator 5 x over denominator 4 end fraction end cell equals cell 4 cross times 1 half end cell end table

Simplify (cancel) the fractions

table row cell 5 x end cell equals 2 end table 

To solve this equation, divide both sides by 5

bold italic x bold space bold equals bold 2 over bold 5

0.4 is also accepted

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

  • 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 now has the same form as previously seen

    • Solve by adding 7 then dividing 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

4 minus 5 x equals 6 x minus 29 

Remove the x terms from either side
We will remove them from the left as -5x is smaller than 6x

Add 5x to both sides

table row 4 equals cell 11 x minus 29 end cell end table

Get 11on its own by adding 29 to both sides

33 equals 11 x 

Divide both sides by 11 to find x

3 equals x 

bold italic x bold equals bold 3

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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.