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 onceExample 1
To solve this, we need to get the x on its own. Here is the working:
Always check your solution works in the original equation: 5 × 3 + 2 = 17 Example 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/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:
This time, three things have happened to the x: it has been multiplied by 2, it has had 7 added to it, and it Here is the working:
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 onceExample 4
To solve this, we need to get the all the “x“s on one side. To remove the “4x” on the right, we will subtract 4x on both sides. Here is the working:
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.
