Unit 16 Section 3 : Linear Equations
An equation is a statement such as 3x + 2 = 17 which contains an unknown number. In this case the unknown number is x.
The aim of this section is to show how to find the value of the unknown number x.
All equations contain an 'equals' sign. To solve the equation, you need to reorganise it so that the unknown value is by
itself on one side of the equation. This is done by performing operations on the equation. When you do this, in order to
keep the equality of the sides, you must remember that whatever you do to one side of an equation, you must also do
the same to the other side. This is often called the balance method.
Simple Examples
Some equations only require one operation to solve them. Look at how the following equations are solved:
(a) 11 = x + 6  11  =  x + 6  [subtract 6 from both sides to get x on its own] 
 11 – 6  =  x + 6 – 6  [note how the 'subtract 6' cancels out the 'add 6'] 
 5  =  x  [switch sides so the x is on the left] 
 x  =  5  

(b) y – 3 = 8  y – 3  =  8  [add 3 to both sides to get y on its own] 
 y – 3 + 3  =  8 + 3  [note how the 'add 3' cancels out the 'subtract 3'] 
 y  =  11  

(c) 15 = 5c  15  =  5c  [divide by 5 on both sides to get c on its own] 
 15 ÷ 5  =  5c ÷ 5  [note how the 'divide by 5' cancels out the 'multiply by 5'] 
 3  =  c  [switch sides so the c is on the left] 
 c  =  3  

(d)  w  = 10   2 
 w   2 
 =  10  [multiply by 2 on both sides to get w on its own] 
 w  × 2   2 
 =  10 × 2  [note how the 'multiply by 2' cancels out the 'divide by 2'] 
 w  =  20  
Harder Examples
Some equations require two operations to solve them. Look at these two examples:
Example 1
To solve this, we need to get the x on its own.
Think about what has happened to the x: it has been multiplied by 5 and has had 2 added to it.
To get x on its own, we need to reverse this process, by subtracting 2 and dividing by 5.
Here is the working:
5x + 2  =  17  [subtract 2 from both sides] 
5x + 2 – 2  =  17 – 2  [note how the subtract 2 cancels out the add 2] 
5x  =  15  [divide both sides by 5] 
5x ÷ 5  =  15 ÷ 5  [note how the divide by 5 cancels out the multiply by 5] 
x  =  3  
Always check your solution works in the original equation: 5 × 3 + 2 = 17
Example 2
p  + 3 = 7   2 

Think about what has happened to the p this time: it has been divided by 2 and has had 3 added to it.
To get p on its own, we again need to reverse this process, by subtracting 3 and multiplying by 2.
Here is the working:
p  + 3   2 
 =  7  [subtract 3 from both sides] 
p  + 3 – 3   2 
 =  7 – 3  [note how the subtract 3 cancels out the add 3] 
p    2 
 =  4  [multiply both sides by 2] 
p  × 2   2 
 =  4 × 2  [note how the multiply by 2 cancels out the divide by 2] 
p  =  8  
Remember to check your solution works in the original equation: 8 ÷ 2 + 3 = 7
Equations with the unknown letter on both sides
Look at this equation:
To solve this, we need to get the all the "x"s on one side.
There are more "x"s on the left, so we will aim to get all the "x"s on the left and all the "numbers" on the right.
To remove the "4x" on the right, we will subtract 4x on both sides.
To remove the "– 2" on the left, we will add 2 to both sides.
Here is the working:
6x – 2  =  4x + 8  [subtract 4x from both sides] [6x – 4x = 2x] 
2x – 2  =  4x + 8  [add 2 to both sides] [8 + 2 = 10] 
2x  =  4x + 10  [divide both sides by 2] 
2x  =  4x + 5  
As always, check your solution works in the original equation: 6 × 5 – 2 = 4 × 5 + 8
Note that many equations can be solved in more than one way, but all the methods will give the same solution.
Using an equation to solve a number puzzle
Look at this number puzzle:
James thinks of a number. He multiplies the number by 5 and then subtracts 7. The result is 43.
Use an equation to find his original number.

We can use x to represent the unknown number. If we multiply x by 5 and subtract 7 we will get 43.
This can be written as an equation:
Now we can solve the equation to find x:
5x – 7  =  43  [add 7 to both sides] 
5x  =  50  [divide by 5 both sides] 
x  =  10  
This gives us the answer to the original number puzzle:
James originally thought of the number 10. 
Practice Questions
Work out the answer to each of these questions then click on the button marked
to see whether you are correct.
Practice Question 1
Solve these equations:
(a) p + 10 = 17   

(b) q – 13 = 2   

(c) 4r = 14   

(d)  s  = 4   3 
  
Practice Question 2
Solve these equations:
(a) 
4x – 7 = 17 

(b) 
p + 3 
= 7 


2 
Practice Question 3
Solve these equations:
(a) 
3x + 2 = 4x – 3 

(b) 
2x + 7 = 8x – 11 

Practice Question 4
Tim thinks of a number, multiplies it by 3, and then adds 10. His final result is 28.
(a)  Write down an equation to represent this number puzzle. 
 
(b)  Solve your equation to find Tim's original number. 
 
Exercises
Work out the answers to the questions below and fill in the boxes. Click on the
button to find out whether you have answered correctly. If you are right
then will appear and you should move on to the next
question. If appears then your answer is wrong. Click
on to clear your original answer and have another go.
If you can't work out the right answer then click on
to see
the answer.
You will need a pen and paper with you so you can solve the equations on paper using the balance method.

You have now completed Unit 16 Section 3
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Produced by A.J. Reynolds April 2011
