# Rearranging Equations

In this section we’re going to be looking at rearranging equations. We shall explore how to arrange linear equations, brackets, fractions and quadratics. This entry explores the basic ideas of rearranging equations, more harder examples of rearranging equations can be found here

## What is the subject?

Consider the following SUVAT formula. The following formula is used to find the velocity of an object over time. In the formula V (velocity) is known as the subject of the formula. That is; VELOCITY is the subject we’re trying to solve or find.

The above formula is used to find velocity over time. Suppose we wanted to find time (t) in the formula, we would have to know what the value of v is.
Let’s rearrange the formula below to make t the subject.

First move u to the other side to leave the time (t) with the other letter a, we’re trying to make t the subject to do this we’ll have to make sure that t is on its own.

Now to leave t on its own we must divide both sides by a.

Now we have;

Lastly we rewrite the equation in a more easy readable form.

## Rearranging equations with brackets

Here we’re going to be looking at how to rearrange equations which involve brackets. Suppose we have an equation shown below and we wanted to make y the subject.

Here we can see a multiplication on one side of the equation. Since y is on the other side we have to find a way to leave it on its own. To get rid of k on the other side we must divide both sides by k.

That leaves;

We have y one the other side but with another term or letter which we have to get rid of. To get rid of this we must subtract z from both sides.

That leaves;

We rewrite it in an easy to approach form;

We have managed to make y the subject

## Rearranging equations with fractions

In the following example we’re going to look at rearranging equations with fractions. Suppose we have the following equation and are asked to make x the subject.

Here we have to get rid of the fraction; we do this by dividing both sides by x or cross multiplying.

…we multiply 5b with x and a+4 with 1 which results in;

Since we’re trying to make x the subject we have to get rid of 5b, we do this by dividing both sides by 5b.

…this leaves;

Rewrite it as;

We have managed to make x the subject.

The following shows an equation which involves quadratics;

Make x the subject…
In the above example we have a square. First we have to get rid of the fraction as we did before. We do this by dividing both sides by e or cross multipliying.

We multiply e with b and ax2 + f with 1 to get the following result.

Now we must leave x2 on its own since we’re trying to make x the subject. First we subtract f from both sides.

Now we divide both sides by a as shown below;

That leaves;

To make x2 x we must square root it, this means we must square root both sides if they’re to remain equal.

This results in;

That concludes this chapter on rearranging equations. If you would like to explore further with much harder examples click here.

## 20 thoughts on “Rearranging Equations”

1. Leon dunn

Thankyou for all the help on this but im still slightly confused on what you do if you have a more basic one like this x/2=a and you have to rearrange to make x the subject. please help my teacher didnt explain it very well.

This is a good method to use when rearranging fractions. I assume the question is arranged as below;

a is the same as a/1, So you rewrite it like this. I hope you realise why.

You then “cross multiply”. You multiply x with 1 and 2 with a.

This leaves;

It does not matter if 2a comes first or x comes first. It is the same.

1. Sam Dickinson

Is it always possible to rearrange an equation to get one term? In physics we have the equation of motion s=ut+1/2at^2. I can’t see how to rearrange that to describe t in terms of u, a and s.

1. corano

to type squared, di it like this x^2
multiply both sides by (x+b)^2
y*(x+b)^2=a
then divide both sides by y
(x+b)^2=a/y
square root both sides to get rid of the squared
x+b= root(a/y)
subtract b from both sides
x=root(a/y)-b

1. Author Post author

Hello Lucy,

I will answer that for you. First you will need to expand the expressions so you get;
$y = 12 + 4x$
Move 12 to the left hand side to get;
$y - 12 = 4x$
The divide both sides by 4 to get;
$\frac{y - 12}{4} = x$
It does not matter if x comes first because it is the same;
$x = \frac{y - 12}{4}$

1. Ed

a bit late to the party, but it’s very simple – what you do to one side, you have to do to the other. In this case, 4x = 8x-2/3

1. multiply both sides by 3 to remove the fraction

12x = 8x-2

2. you can now minus 8x from both sides so it will become:

4x = -2

I’ll let you figure out the rest

1. Helen

Hi Rebecca,
I am not sure if you have received the answer to your problem, but this is how I would have done it. 0.17=x+2.4(0.1-x)
1) take the last part 2.4(0.1-x) and multiply everything in the bracket with 2.4 you have;
$2.4 × 0.1- 2.4x = 0.24 -2.4x$
2) Then you put it back in your formula;
So you have;
$0.17 = x + 0.24 - 2.4x$
3) Then what I did, you collect all the terms with X;
so you have
$x - 2.4x =- 1.4x$
4) Then I had $0.17 = 0.24 - 1.4x$
5) Then you move those nu without X from right to left but remember to change the sign to a negative as 0.24 is on the right positive, you will have ;
$0.17 - 0.24 = - 1.4x$
6) Then you divide both sides by 1.4 and you are left with the fraction as 1.4 ÷ 1.4 X will be cancelled and you are left with X. So you are left with X = 0.17 -0.25 ÷ 1.4. Remember this is a fraction after the equal. I am having trouble with the keyboard, doesn’t have fraction on my tab. Hopefully it is correct but that was my logic in solving that . What is yours? Maybe I am wrong, maybe you get a better answer,let me know. Bye,good luck.

2. joel

Rearrange the following equation to ﬁnd x in terms of a and b, in
simplest form (assume that a ‘= 4b):
a(a − x) + 4b(x − 4b) = 0.

1. Author Post author

You have to try otherwise you may not get any reply. I also think you have written the question wrong. I will start you off.
If we have to write the new expression in terms of a and b then we won’t need this “Assume that a = 4b” because we have to include a and b in the final expression when x is the subject. So I will ignore that. We’re starting with;
$a(a - x) + 4b(x - 4b) = 0$
First will need to get rid of the factors ( expand the expression ).
$a^2 - ax + 4bx - 16b^2 = 0$
We collect like terms and make sure that all terms that contain x are on one side.
$-ax + 4bx = 16b^2 - a^2$
You should be able to finish off from here.