**Rearrange formulae to change the subject** *KS3 Revision*

## What you need to know

**Things to remember:**

- We rearrange an equation in the same way that we solve an equation.
- Imagine that the letters we are like numbers, and move them as you would normally.

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Before we get to “rearranging”, let’s do some solving, which we will do in two steps.

*Solve 5x-3=27 by finding x.*

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**Step 1: **Are there any additions or subtractions on the same side as the x? If yes, do the opposite on both sides.

We have a -3, so we need to +3 to both sides.

5x-3=27

5x-3+3=27+3

5x=30

**Step 2: **Are there any multiplications or divisions on the same side as the x? If yes, do the opposite on both sides.

We have a multiplication by 5, so we need to divide by 5.

5x=30

5x\div5=30\div5

x=6

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And there we have it, the value of x. Sometimes though, there will be different types of letters, and we have to rearrange to make one the “subject”. Making something the “subject” just means to make it so you have it by itself, and we do this in the same way as solving.

*Rearrange 2y=4x+5 to make x the subject.*

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**Step 1: **Are there any additions or subtractions on the same side as the x? If yes, do the opposite on both sides.

We have a +3, so we need to -5 to both sides.

2y=4x+5

2y-5=4x+5-5

2y-5=4x

*Hint: **We can really subtract 5 from 2y, so we leave it as 2y-5*

**Step 2: **Are there any multiplications or divisions on the same side as the x? If yes, do the opposite on both sides.

We have a multiplication by 4, so we need to divide by 4.

2y-5=4x

(2y-5)\div4=4x\div4

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*Note: **We are dividing the whole of the left-hand side by 4, so we ned to put it in brackets.*

*Note: **Remember, we can write divisions as fractions.*

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\frac{2y-5}{4}=x

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Although this is correct, we’d usually write the letter we are making the subject on the left.

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x=\frac{2y-5}{4}

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This was quite a nice example with lots of numbers, but sometimes it will just be lots of letters. Remember though, letters are used to represent numbers, so we can treat them like that!

*Rearrange 5y=xz-z to make x the subject.*

**Step 0: **Swap sides of the equation so the variable you are making the subject is already on the left.

*Note: **This step isn’t necessary.*

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5y=xz-z

xz-z=5y

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**Step 1: **Are there any additions or subtractions on the same side as the x? If yes, do the opposite on both sides.

We have a -z, so we need to +z to both sides.

xz-z=5y

xz-z+z=5y+z

xz=5y+z

*Hint: **We can really subtract z from 5y, so we leave it as 5y+z*

**Step 2: **Are there any multiplications or divisions on the same side as the x? If yes, do the opposite on both sides.

We are multiplying x by z, so we need to divide both sides by z.

xz=5y+z

xz\div z=(5y+z)\div z

*Note: **We are dividing the whole of the left-hand side by z, so we ned to put it in brackets.*

*Note: **Remember, we can write divisions as fractions.*

x=\frac{5y+z}{z}

## Example Questions

**Question 1:** *Rearrange m=\frac{n}{3}-2 to make n the subject.*

**Step 0: **Swap sides of the equation so the variable you are making the subject is already on the left.

*Note: **This step isn’t necessary.*

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m=\frac{n}{3}-2

\frac{n}{3}-2 =m

**Step 1: **Are there any additions or subtractions on the same side as the x? If yes, do the opposite on both sides.

We have a -2, so we need to +2 to both sides.

\frac{n}{3}-2 =m

\frac{n}{3}-2+2 =m+2

\frac{n}{3}=m+2

**Step 2: **Are there any multiplications or divisions on the same side as the x? If yes, do the opposite on both sides.

We have a division by 3 (fractions are divisions), so we need to multiply both sides by 3.

\frac{n}{3}=m+2

\frac{n}{3}\times3=(m+2)\times3

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*Note: **We are multiplying the whole of the right-hand side by 3, so we ned to put it in brackets.*

*Note: **When we multiply a bracket by something, we put that “something” on the left.*

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n=3(m+2)

**Question 2:** *Rearrange ab=cd+e to make d the subject.*

**Step 0: **Swap sides of the equation so the variable you are making the subject is already on the left.

*Note: **This step isn’t necessary.*

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ab=cd+e

cd+e=ab

**Step 1: **Are there any additions or subtractions on the same side as the x? If yes, do the opposite on both sides.

We have a +e, so we need to -e to both sides.

cd+e=ab

cd+e-e=ab-e

cd=ab-e

**Step 2: **Are there any multiplications or divisions on the same side as the x? If yes, do the opposite on both sides.

We have a multiplication by c, so we need to divide both sides by c.

cd=ab-e

cd\div c=(ab-e)\div c

d=\frac{ab-e}{c}