Answer:
The constant charge for each minute used is $50
Explanation:
In order to solve this problem we will need to set two variables up. In this case:
F = constant Fee
R = rate per minute used
So the cost for the month of January is calculated like this:
F+300R=68
and the cost for February is calculated like this:
F+275R=66.5
So no we have a system of equations we can solve simultaneously. This can be solved by using different methods, elimination, substitution, graphically or by using matrices. I will solve this by substitution.
So let's solve the first equation for R:

and let's substitute this first equation into the second equation:

and now we can solve this for F:

We can multiply both sides by 12 so we get:
12F+11(68-F)=798
12F+748-11F=798
F= $50