# equations

http://www.cimt.org.uk/projects/mepres/book8/bk8i12/bk8_12i4.htm

In this section we solve linear equations where more than one step is needed to reach the solution.
Methods of solution vary from one equation to another, but there are two main groups:
(i) equations where the unknown letter only appears once (like 2x + 7 = 13), and
(ii) equations where the unknown letter appears more than once (like 5x – 3 = 2x+3).

### Equations where the unknown letter appears once

Example 1
Look at this equation:

 5x + 2 = 17

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
Look at this equation: p/2+ 3 = 7

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/2+ 3 = 7  [subtract 3 from both sides]

p/2  = 4 [note how the subtract 3 cancels out the add 3

p/2   x 2 = 4 x 2  [multiply both sides by 2]

p= 8  [note how the multiply by 2 cancels out the divide by 2]

Remember to check your solution works in the original equation: 8 ÷ 2 + 3 = 7

Example 3

Look at this equation:

 5(7 + 2x) = 65

This time, three things have happened to the x: it has been multiplied by 2, it has had 7 added to it, and it
has then been multiplied by 5. To reverse this process, we divide by 5, subtract 7 and then divide by 2.

Here is the working:

 5(7 + 2x) = 65 [divide by 5 on both sides] 5(7 + 2x = 13 [subtract 7 on both sides] 5(7 + 2x = 6 [divide by 2 on both sides] 5(7 + 2x = 3

Always check your solution works in the original equation: 5 × (7 + 2 × 3) = 65 Note that the last example could also have been solved by multiplying out the brackets first.

### Equations where the unknown letter appears more than once

Example 4
Look at this equation:

 6x – 2 = 4x + 8

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.