What you need to know

Linear equations are a type of equation that appear all over the place in maths. They can look quite simple, like x+2=5, or they can look a little more complicated, such as

5p-7=3p+3.

The golden rule when solving these equations (and indeed, equations of all kinds) is that whenever you apply some operation to one side of the equation (e.g you add 5 to it, or multiply it by 3, etc), you must also do the same thing to the other side. Think of an equation like a perfectly balanced pair of scales – if you want to keep it balanced, any time you remove/add some weights to the one side, you must also do the same thing to the other side.

The aim is to get “p= something”, because then we’ve found p. To do this, we will collect terms of the same type together on the same side of the equals sign, a process known as collecting like terms. In the example above, this would mean getting only terms involving a p on the left-hand side, and only constant terms on the right. Let’s see how this is done.

Example: Solve the equation 5p-7=3p+3.

Our first step will be to collect the p terms on one side. To do this, we are going to get rid of the positive 3p (since it’s the smaller p term) on the right-hand side by subtracting 3p from both sides of the equation:

5p-7-3p=3p+3-3p

So, on the left-hand side we now have 5 lots of p and -3 lots of p, meaning that in total we have 2 lots of p. On the right-hand side, the 3p and -3p cancel each other out, so there is no p term on the right – this was the point of subtracting 3p in the first place. So, the equation becomes

2p-7=3

Now, our second step is to collect the constant terms on the other side. To do this, we will get rid of the -7 on the left-hand side by adding 7 to both sides of the equation:

2p-7+7=3+7

Then, the -7 and 7 cancel out on the left-hand side – this was the point of adding the 7 – and on the right-hand side we will be left with 3+7=10:

2p=10

The final step to determine what p is equal to, is to get rid of the 2 before it. Up until this point, the reason that every step worked in “getting rid” of the appropriate terms was because we did the opposite operation of what was being done (to get rid of -7, we did +7). This idea is key, and we do it again for the final step. Since p is being multiplied by the 2, we will divide both sides by 2:

2p\div2=10\div2

2p\div2 just becomes p – this was the aim – and then 10\div2=5, so we have that

p=5.

Generally, this process will move a lot more quickly than this example (we’ll see with the next one). Also, a common way of structuring an answer to these questions is to write two lines either side of your equation, and then track the operations that you’re doing (red) on the outside of the lines. In this case, that would look like

This is a neat way to both remind yourself what you’re doing and show the marker that you know what you’re doing.

 

Example: Solve the equation 3a+1=5-7a.

 

Alright, no dilly-dallying, let’s get to it. To get the a terms on one side, we are going to get rid of -7a (since it’s the smaller a term) by doing the opposite and adding 7a to both sides of the equation. 3a+7a=10a, so the equation becomes

 

10a+1=5

 

Now, to get the constant terms on the right-hand side, we will do the opposite of +1 and subtract 1 from both sides of the equation. Doing so, we get

 

10a=5-1=4

 

Now the equation is 10a=4, all that remains is to get rid of the 10. As it is being multiplied by the a, we will do the opposite and divide by 10:

 

a=\dfrac{4}{10}

 

This is not the nicest answer, but they often aren’t. At this stage, we’ve found the solution regardless. You can write it as a decimal, 0.4, if you’d prefer, or if the question asks for a simplified fraction, then the answer would be \frac{2}{5}.

 

If we were to answer this using the lines method, it would look like:

Example Questions

1) Solve the equation 3x+4=13.

Answer

There is only one x term, so we don’t have to worry about moving any of those about. So, to get the constant terms on the other side, we will get rid of the +4 by subtracting 4 from both sides. Then, the equation becomes

 

3x=9

 

Then, all that remains to get x on its own is to get rid of the 3, and since it is being multiplied by this 3, we will divide both sides by 3 to get

 

x=9\div3=3.

Thus, x=3 is the solution. If you had done this using the lines method, it would look like

 

 

 

2) Solve the equation 5m-2=16-4m.

Answer

Firstly, we want to get all the m terms on one side. -4m is the smaller of the two, so to get rid of -4m on the right-hand side, we will add 4m to both sides of the equation. 5m+4m=9m, so we get

9m-2=16

To get the constant terms on the other side of the equation, we must get rid of the -2 on the left. We will do this by adding 2 to both sides of the equation, which leaves us with

 

9m=16+2=18

 

So, now the equation is 9m=18, all that remains is to get rid of the 9. As it is being multiplied by the x, we will divide both sides by 9, to get

 

m=18\div9=2.

 

Thus, the solution is m=2. If you had done this using the lines method, it would look like

 

3) Solve the equation 4(z-2)=-20.

Answer

This looks a little different, but there’s only one extra step required: we need to expand the bracket. So, multiplying 4 by everything inside the bracket, the equation becomes

 

4z-8=-20

 

Suddenly, this equation looks more familiar. Now, there is only 1 z term, so that’s dealt with. To get the all constant terms over to the other side of the equation, we must get rid of the -8 on the left, and we will do this by adding 8 to both sides. So, our equation becomes

 

4z=-20+8=-12

 

So, now the equation is 4z=-12, all that remains is to get rid of the 4. As it is being multiplied by the z, we will divide both sides by 4, to get

 

z=(-12)\div4=-3.

 

Thus, the solution is z=-3. If you had done this using the lines method (after the initial bracket expansion), it would look like

 

For those maths teachers out there looking for GCSE Maths revision materials, our solving equations resources would be a great addition to anyone’s repertoire of work. You can use the solving equations worksheets as starters for GCSE Maths lessons or set them as homework.