Well this is definitely not the favourite part of algebra for most of my GCSE Maths students, and as you are here, rearranging formulae has probably got you muddled too.

Did you know you have actually been rearranging formulae and equations since early primary school? Yep, it just looked very different.

Here’s an example:

3 + ? = 7.

What is the ?

Yep its 4.

You did that so fast, you need to stop and think about *how* you actually worked that out!

Let’s take a closer look:

To work out the ?, you did 7 – 3 = 4.

So you moved the 3 from one side to the other and you made it the opposite of what it is – change the side, change the sign.

I.e. Here: 3 + ? = 7 The 3 is a positive number. We could write it as

? + 3 = 7 and its the same thing.

We need the ? on its own on one side of the equals, and we do that by moving the 3 over to the same side as the 7. However it cannot be a +3 when you move it to the other side, because that won’t work.

That would give ? = 7 + 3, which is 10 – that’s incorrect.

So when you move it over you need to make it the opposite of what it is. It was a + 3 on the left of the equals, so it becomes a subtract 3 on the right hand side of the equals.

? = 7 – 3.

Another way to look at it is you need to subtract 3 from both sides.

? + 3 – 3 = 7 – 3

The +3 and the -3 on the left hand side, cancel out and this leaves

? = 7 – 3

so ? = 4.

It doesn’t really matter which way you think about it, choose whatever makes sense to you.

The important thing is to move things across and make them the opposite of what they were orginally.

Here are some more examples:

7 x ? = 21

We are multiplying by 7 on the left hand side, so move that ‘multiply by 7’ to the other side and change it into a ‘divide by 7’.

? = 21/7

? = 3

Here are some questions to try:

**Question 1:** Rearrange the equation `3x = 9`

to solve for `x`

.

**Answer 1:**

`x = 9 / 3`

`x`

`= 3`

**Question 2:** If `2y = 8`

, find the value of `y`

.

**Answer 2:**

`y = 8 / 2`

`y`

`= 4`

**Question 3:** Solve for `a`

in the equation `4a = 12`

.

**Answer 3:**

`a = 12 / 4`

` `

a `= 3`

**Question 4:** Rearrange the equation `5z = 15`

to solve for `z`

.

**Answer 4:**

`z = 15 / 5`

`z`

`= 3`

**Question 5:** If `6m = 12`

, what is the value of `m`

?

**Answer 5:**

`m = 12 / 6`

` `

m `= 2`

**Question 6:** Solve for `x`

in the equation `2x = 8`

.

**Answer 6:**

`x = 8 / 2`

`x = 4`

Slightly harder:

**Question 1:** Rearrange the equation `2x + 3 = 11`

to solve for `x`

.

**Answer 1:**

`2x = 11 - 3`

`2x = 8`

`x = 8 / 2`

`x = 4`

**Question 2:** If `4y - 7 = 5`

, find the value of `y`

.

**Answer 2:**

`4y = 5 + 7`

`4y = 12`

`y = 12 / 4`

`y = 3`

**Question 3:** Solve for `a`

in the equation `3a + 5 = 14`

.

**Answer 3:**

`3a = 14 - 5`

`3a = 9`

`a = 9 / 3`

`a = 3`

**Question 4:** Rearrange the equation `2(x + 3) = 10`

to solve for `x`

.

**Answer 4:** `2(x + 3) = 10`

`2x + 6 = 10`

`2x = 10 - 6`

`2x = 4`

`x = 4 / 2`

`x = 2`

OR

(x+3) = 10/2

x + 3 = 5

x = 5 – 2

x = 2

**Question 5:** If `2(2p - 1) = 14`

, what is the value of `p`

?

**Answer 5:**

`2(2p - 1) = 14`

`4p - 2 = 14`

`4p = 14 + 2`

`4p = 16`

`p = 16 / 4`

`p = 4`

OR

(2p – 1) = 14/2

2p – 1 = 7

2p = 7 + 1

2p = 8

p = 8/4

p = 2

**Question 6:** Solve for `x`

in the equation `5(x - 2) = 15`

.

**Answer 6:** `5(x - 2) = 15`

`5x - 10 = 15`

`5x = 15 + 10`

`5x = 25`

`x = 25 / 5`

`x = 5`

OR

(x – 2) = 15/5

x – 2 = 3

x = 3 + 2

x = 5

**Question 7:** If `3(2q + 1) = 21`

, find the value of `q`

.

**Answer 7:**

`3(2q + 1) = 21`

`6q + 3 = 21`

`6q = 21 - 3`

`6q = 18`

`q = 18 / 6`

`q = 3`

OR

(2q + 1) = 21/3

2q + 1 = 7

2q = 7 – 1

2q = 6

q = 6/2

q = 3

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