In this blog, I will show you a method for solving linear equations.
Solving equations is the most useful aspect of algebra you will learn at school. Take the example below…
Solve 3x + 1 = 13
The x is just a number that we don’t know yet. Solving an equation basically just means find x. What this means is there is some number that when you multiply it by 3 and then add 1, you get 13. This may be something you can do in your head right away, but here I will show you a tried-and-tested method for solving equations that will work for even harder questions.
First of all, what are our goals? We want to change this equation to get x on its own on one side, and a number on the other side. That will mean we have solved the equation because whatever that number is, is x!
The way we will get to this is by eliminating numbers from the left-hand side.
Solve 3x + 1 = 13
The x currently lives on the left-hand side of the equation, so let’s leave it there. We now want to remove all the numbers of the left-hand side of the equation, hence isolating the x. There are two numbers we need to remove – the 3 and the 1.
First of all, we should remove the 1. To remove the 1, we have to do the opposite of adding 1 – taking away 1 (because 1 – 1 = 0). Now this part is very important – we are going to take away 1, but we need to remember to do this to both sides of the equation.
The reason for that is simple. It is an equation. The whole point of an equation is that it is equal! If you take away 1 from one side, you must do it to the other – otherwise it won’t be equal any more.
Take this set of scales for example. It has 13 kilograms on the left-hand side and 13 kilograms on the right. It is therefore in equilibrium.
If I take 1 kilogram off the left-hand side but don’t take it off the right-hand side too, then the scales will no longer be in equilibrium.
But if I take 1 kilogram off both the left and right, the scales will remain in equilibrium, as both sides have 12 kilograms. When we are solving equations, the scales must always be in equilibrium.
So, going back to our example, we are going to take away 1 from the left-hand side and the right-hand side.
This removes the 1 from the left-hand side and importantly also reduced the right-hand side from 13 to 12.
Now we have removed the 1, we need to remove the 3 next. 3x means 3 multiplied by x – so we need to do the opposite of that. To remove the 3, we need to divide by 3 – making sure that we do the same on both sides.
We have done it! We have isolated the x on the left-hand side and have 4 on the right. So we have solved the equation and found x. The answer is 4.
Solve 4x – 9 = 11
First we need to remove the 9 from the left-hand side. To do this, we add 9 (because -9 + 9 = 0).
Now we need to remove the 4. To do this, we divide by 4.
The answer is 5.
Solve 6x + 13 = 37
First we need to remove the 13. To do this, we take away 13.
Now we need to remove the 6. To do this, we divide by 6.
The answer is 6.
Solve 7x – 18 = 10
The answer is 4.
Hopefully after those examples, it’s starting to make sense.
Solve 5x + 4 = 3x + 10
Now this example is more difficult, because there are x’s on both sides of the equation.
What we need to do is move all the x’s onto the same side. The left-hand side has the most x’s already (5x is more than 3x) so it makes sense for the x’s to live on the left-hand side.
We need to remove the 3x from the right-hand side. To do this, we take away 3x.
Note that when we take away 3x from 5x + 4, the 4 does not change. This is very important.
Think of it this way – if I have 3 oranges and 4 apples and I take away 2 of the apples, I now have only 2 apples but the number of oranges I have is still 3. It hasn’t changed. It’s the same with x’s and numbers. If I have 5x and 4, and I take away 3x – I now only have 2x but I still have 4.
Now you will notice that the equation we are left with is just like the previous examples. So we can follow the same steps as before.
The answer is 3.
So, whenever there are x’s on both sides, immediately move them over to the same side.
Solve 7x + 11 = 4x + 23
Solve 9x – 20 = 4x + 5
So that’s a great method for solving linear equations. It may take a bit of time to get used to but once you’ve done a few examples you’ll realise just how easy it is.
Here are 10 further examples. Have a go at these, and put your solutions in the comments or email them to email@example.com and I’ll let you know if you got them right. If you need someone to explain this method in-person and work through some examples with you, book a free taster session.
1. 4x + 13 = 33
2. 5x – 19 = 11
3. 9x – 10 = 53
4. 11x + 12 = 8x + 27
5. 10x – 31 = x + 5
6. 7x – 22 = 27
7. 4x – 11 = 2x + 9
8. 12x + 41 = 161
9. 8x + 21 = 13
10. 12x – 19 = 7x – 4
If you feel confident on this and want to work through some more algebra problems, you can try our algebra worksheet.