I need help with this algebra problem.

Question

The telephone company offers two types of service. With plan A, you can make an unlimited number of

local calls per month for $18.50. With Plan B, you pay $6.50 monthly, plus 10 cents for each min of calls after

the first 40 minutes. At least how many minutes would you have to use the telephone each month to make Plan A the better option?

Answer 1

The main goal is to make A be the better option and since we are dealing with money the better option will be that A cost less than B.

Will follow y=mx+b where m is the slope and b is y-intercept

A = 18.50, where A does not have a slope

B = .10(x-40)+6.50

The reason why we do x-40 is because it states after 40 minutes you will be charge 10 cents per min so we take our x amount of minutes and subtract it from 40 and this gives us or minutes we are being charged for.

We want to know how many minutes will we need to make B > A

.10(x-40)+6.50 > 18.50 (we want to solve for x)

.10x-4+6.50 > 18.50 (combine like terms)

.10x+2.5 >18.50 (to get x alone must subtract 2.5 from both sides)

.10x > 16 (to get x to be 1x or x we must divide .10)

x > 160

This says plan A will be better if plan B goes over 160 minutes

To check our work we can plug 161 for x since x > 160 and we will see it be more than plan A thats 18.50

B = .10(161-40)+6.50

B= 18.60

Answer 2

Asked and answered!