# AAT Level 4 Applied Management Accounting: Simultaneous Equations

In this article, AAT Distance Learning Tutor, Phil Toomer discusses simultaneous equations, an extra resource section for AAT Level 4 Applied Management Accounting (AMAC).

# AAT Level 4 Applied Management Accounting: Simultaneous Equations

In this article, AAT Distance Learning Tutor, Phil Toomer discusses simultaneous equations, an extra resource section for AAT Level 4 Applied Management Accounting (AMAC).

## Let’s look at a question as an example.

A factory is manufacturing green and red toy cars. Plastic and labour are both limited.

They have managed to formulate the following linear programme. (G = Green, R = Red).

Plastic = 80G + 100R = 2,400.

Labour = 4G + 8R = 156.

## To work out the values of R and G, I need to make the above equations a little bit simpler.

First, I need an identical number in both labour and plastic.

To do this I need to see if one of the smaller labour figures will divide equally into the larger plastic figures…. I’m looking for an exact number…. No decimal places.

Green… Does 4 go into 80. (80 / 4 = 20). Yes, it does… 20 times… an exact number.

Red… Does 8 go into 100. (100 / 8 = 12.5). Yes, it does but it’s not an exact number, it has a decimal.

Because Green is an exact number, I will go with that.

Now I have an exact number, I can multiply every number in the Labour equation by 20.

4 x 20 = 80.

8 x 20 = 160.

156 x 20 = 3,120.

## Let’s now place them in the equation and look at both equations together.

Plastic = 80G + 100R = 2,400.

Labour = 80G + 160R = 3,120.

Excellent, I now have an identical number in both plastic and labour. 80G.

Because this number is the same in both, it won’t affect the difference in the total, so I can remove it and then see what that does to both equations.

Plastic = 80G + 100R = 2,400.

Labour = 80G + 160R = 3,120.

Now I can see a difference in red figures and a difference in the total figures.

I need to work out the differences.

160R – 100R = 60R.

3120 – 2,400 = 720.

This means that I now know that the difference in the total must relate to the difference in red.

60R = 720.

If I divide 720 by 60, I will get the value of 1R.

720 / 60 = 12.

R = 12.

## Now let’s look at the original equations.

Plastic = 80G + 100R = 2,400.

Labour = 4G + 8R = 156.

I know R = 12.

## I will now apply that to this equation.

I am going to use the smaller numbers within Labour…. 8R.

8 x 12 = 96.

8R = 96.  I can now put that into the Labour equation instead of 8R.

Labour = 4G + 96 = 156.

I now have a figure in place of 8R.

If I now deduct 96 from 156, that must be the value of 4G.

156 – 96 = 60.

4G = 60.

If I divide 60 by 4, I will get the value of 1G.

60 / 4 = 15.

G = 15.

R = 12.

G = 15.

## Do you see what I have done?

I’ve simplified the equation by finding an identical number in both equations.

Then, by removing that identical number, I remove 1 product completely and it leaves me with only 1 product (Red car).

There will be a difference in the amount and a difference in the total.

That difference can then be used to find the value of 1 item.

Then, by applying that into the original equations, I can then work out the value of the other product.

## Let’s do a final check on both equations to make sure I am correct.

I will enter actual figures based on the equation amounts to see if I get the same totals.

## Original equation.

Plastic = 80G + 100R = 2,400.

Labour = 4G + 8R = 156.

R = 12.

G = 15.

## Workings.

Plastic.

80G x 15 = 1,200.

100R x 12 = 1,200.

Plastic = 1,200 + 1,200 = 2,400… this is correct.

Labour.

4G x 15 = 60.

8R x 12 = 96.

Labour = 60 + 96 = 156… this is correct.

R = 12.

G = 15.

I hope this helps and lets you see how I work through this type of question.