Here are my thoughts on how to tackle simultaneous equations.
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.