Algebra
posted by Gemma .
A longdistance call using DEC costs 20 cents for the first minute and 16 cents for each additional minute. The same call, placed on LCS, costs 19 cents for the first minute and 18 cents for each additional minute. For what length of phone calls is DEC less expensive?
Here's my inequality and how I solved it:
.16c + .20 < .18c +.19
I subtracted .19 from each side
.16c + .01 < .18c
I subtracted .16c from each side
.01 < .02c
I divided each side by .02.
My final answer does not make sense to me: c > .5. Half a phone call, or half a minute? I think I must have messed up somewhere, so if you see a mistake, please tell me.
Thank you!

You actually have a rather silly question given to you.
You are dealing with two functions:
Cost(DEC) = 16t + 20 where t is in minutes and Cost is in cents
Cost(LCD) = 18t + 19
let's see where they intersect (are equal)
18t + 19 = 16t + 20
2t = 1
t = 1/2
so at 1/2 minute the cost would be the same, but apparently the cost would be calculated in whole minutes.
so for less than a minute (t=0)
for DEC it would cost 20 cents and for
LCS it would cost 19 cents.
for any vale of t ≥ 1, the cost for DEC would be less than the other agent.
What you did was mathematically correct, you just got yourself confused what your c stood for.
You got c > .5, I got t = .5, you solved an inequation, I solved an equation.
since our solution fell within the fixed cost rate of anything less than 1 minute , we cannot really use that result.
Hope that clears up your problem.