How to calculate breakeven days or months

[rfbrenner]

My daughter has two problems that ask her to find the number of days or months for two different series to equal. I don't remember how to figure it out algebraically - only manually. In one case we're told that Verizon gives 4 free months and then $25 per month thereafter while Sprint gives 3 free months and charges $20 thereafter. It wants to know how many months it will take before two subscribers - one on Verizon and one on Sprint would have paid exactly the same total. The answer is that they would both pay $100 at the end of the 8th month.

In the second problem we're told that Blockbuster charges $2.25 for the first day and $.15 per day thereafter. Hollywood charges $5 for the first two days and then $ .05 per day thereafter. It wants to know the same thing - after how many days would a renter under each system spend the same amount.

As I said, I used a spreadsheet to set up a table and calculate a running balance under each scenario. But since my daughter is in pre-algebra I assume there must be some algebraic expression that would calculate this easier than doing the brute force method.

Any ideas are appreciated!

Thank you
rfbrenner

 

[wjm11]

My daughter has two problems that ask her to find the number of days or months for two different series to equal. I don't remember how to figure it out algebraically - only manually. In one case we're told that Verizon gives 4 free months and then $25 per month thereafter while Sprint gives 3 free months and charges $20 thereafter. It wants to know how many months it will take before two subscribers - one on Verizon and one on Sprint would have paid exactly the same total. The answer is that they would both pay $100 at the end of the 8th month.

Please do not double-post.

Assign variables:
Let x represent time (in this case the number of months)
Let y represent the cost in dollars.

“Verizon gives 4 free months and then $25 per month thereafter”: if we graph this, it means we have a point on the x-axis of (4,0) – representing the 4 free months. On the 5th month, we’d have a point at (5,25). Make sense? Our graph would be a line running through these two points, and the slope of the line would be 25.

Putting this into an equation in point-slope form, using the point (4,0) and a slope of 25:

y – y1 = m(x – x1)
y – 0 = 25(x – 4)

If you want to put this into slope-intercept form, y = mx + b, just rearrange it algebraically:

y = 25x – 100

Following the same logic for “Sprint gives 3 free months and charges $20 thereafter”, we’d get

y – 0 = 20(x – 3)

And rearranging get

y = 20x – 60

Solving for x we get

25x – 100 = 20x – 60
5x = 40
x = 8

Plugging 10 back into the original equations, we find that

y = 25(8) – 100 = 100
Or
y = 20(8) –60 = 100

So, at 8 months, we pay $100 for either package.

If you graph these two line equations, they will intersect at (8,100).

Hope that helps.

 

[rfbrenner]

Amazing! Thank you very much. (Sorry for the double post - I wasn't sure the pre-algebra topic would work).

I can see that I need to pull out a book and bone up on my algebra (slope of line, etc!). She's in seventh grade. They would definitely not know anything about this level of complexity in that grade, so I wonder how the teacher expected them to do it.

thanks again,

Rick

 

[Denis]

Rick, there's really no need for stuff like "slopes" here, IF the problems remain similar to the two you posted.
A problem has 2 choices, each with an "initial cost", and with a repeating cost; like:
choice 1: 20 initially, then 5 per day ; make that A initially, then X per day
choice 2: 10 initially, then 7 per day ; make that B initially, then Y per day

We need the point where both are equal; let that be D days:
A + XD = B + YD
YD - XD = A - B
D(Y - X) = A - B
D = (A - B) / (Y - X)

So, using my example above:
D = (20 - 10) / (7 - 5) = 5

So in 5 days, both plans have cost same; then choice 1 becomes cheapest, being 5 per day.

So it is really quite simple:
(difference in initial costs) / (difference in repeating costs)

Hope that helps...