1. Simultaneous Equations

Ok, I have two telephone companies, one is offering calls at £0.10p with a monthly fee of £10, the second company is offering calls at £0.20p with a monthly fee of £5.

I am asked to create a formula using simultaneous equations, the question does not ask me to solve the equations.

I was told to use the letter "p" for the cost of calls per month, and "n" for the number of phone calls per month.

What I have learned so far is that the number of phone calls per month is not given, but the price per call and a monthly fee are, so based on this information I came up with the following formula.

N = 10p + 0.10
N = 5p + 0.20

Do these equations seem realistic, my understanding is that N would tell me the number of calls per month, and P would tell the total cost of calls per month.

Thanks

David

or should the equations be;

N = 10 + 0.10p
N = 05 + 0.20p

2. Originally Posted by David Green
Ok, I have two telephone companies, one is offering calls at £0.10p with a monthly fee of £10, the second company is offering calls at £0.20p with a monthly fee of £5.

I am asked to create a formula using simultaneous equations, the question does not ask me to solve the equations.

I was told to use the letter "p" for the cost of calls per month, and "n" for the number of phone calls per month.

What I have learned so far is that the number of phone calls per month is not given, but the price per call and a monthly fee are, so based on this information I came up with the following formula.

N = 10p + 0.10
N = 5p + 0.20

Do these equations seem realistic, my understanding is that N would tell me the number of calls per month, and P would tell the total cost of calls per month.

Thanks

David
You've got slightly confused with your variables, I think.

The first company charges £10 per month. For every call, you are charged an extra £0.10

This means that for N calls, you will be charged $10 + 0.10\times N$

So $p=10 + \0.10N$ for the first company.

3. Originally Posted by Quacky
You've got slightly confused with your variables, I think.

The first company charges £10 per month. For every call, you are charged an extra £0.10

This means that for N calls, you will be charged $10 + 0.10\times N$

So $p=10 + \0.10N$ for the first company.
Thanks for that information, but are you sure you don't mean; P = 10N + 0.10

Thanks

David

4. Originally Posted by David Green
Thanks for that information, but are you sure you don't mean; P = 10N + 0.10

Thanks

David
I don't believe I do. You have to pay £10 once every month, regardless of the number of calls used.

It is the £0.10 which is stacking with every call.

Also, if it was P=10N+0.10, I would hate to be the billpayer!

5. Originally Posted by Quacky
I don't believe I do. You have to pay £10 once every month, regardless of the number of calls used.

It is the £0.10 which is stacking with every call.

Also, if it was P=10N+0.10, I would hate to be the billpayer!
OK bit of a slight problem then here?

The equations I am given are;

N = 10P + 0.10, N = 5P + 0.20

N = 0.10p + 10, N = 0.20p + 5

P = 10N + 0.10, P = 5N + 0.20

P = 0.10N + 10, P = 0.20N + 5

Are you sure?

6. Originally Posted by David Green
OK bit of a slight problem then here?

The equations I am given are;

N = 10P + 0.10, N = 5P + 0.20

N = 0.10p + 10, N = 0.20p + 5

P = 10N + 0.10, P = 5N + 0.20

P = 0.10N + 10, P = 0.20N + 5

Are you sure?
P = 0.10N + 10 is the same as what I have - the order has been changed, but addition is commutative, silly!